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Solutions Manual
forA Course in Ordinary Di!erential
Equations
byRandall J. Swift
Stephen A. Wirkus
2
Preface
This solutions manual is a guide for instructor’s using A Course in Ordinary Di!erential Equations.Many problems have their solution presented in its entirety while some merely have an answer andfew are skipped. This should provide su!cient guidance through the problems posed in the text.
As with the book, code for Matlab, Maple, or Mathematica is not given. It is our experience thatthe syntax given in the book is su!cient to learn the relevant commands used to obtain solutions tothe various problems in the book. Please give Appendix A a chance, if you have not done so already.
This solutions manual was put together by many people and we note a few of them here. We owea big thanks to our former students David Monarres, for help in preparing portions of this book,and Walter Sosa and Moore Chung, for their help in preparing solutions. More recently Scott Wildehas helped tremendously in shaping this manual. Jenny Switkes and other colleagues and studentshave also given feedback on various drafts of this manual and all have been helpful.
This book has evolved over the last few years and we have tried to make this solution manualstay in step. However, we realize that there are probably many typos throughout and we encourageyou to contact us with your corrections. Hopefully future printings of this manual will have anexponentially decreasing number of such errors.
We would appreciate any comments that you might have regarding the book and the manual.
Randall J. Swift (e-mail: [email protected])Stephen A. Wirkus (e-mail: [email protected])
URL for typos and errata: http://www.csupomona.edu/!swirkus/ACourseInODEs
Contents
1 Traditional First-Order Di!erential Equations 11.1 Some Basic Terminology . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.2 Separable Di"erential Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51.3 Physical Problems with Separable Equations . . . . . . . . . . . . . . . . . . . . . . 241.4 Exact Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 271.5 Linear Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 331.6 Chapter 1: Additional Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 43
2 Geometrical and Numerical Methods for First-Order Equations 512.1 Direction Fields—the Geometry of Di"erential Equations . . . . . . . . . . . . . . . 512.2 Existence and Uniqueness for First-Order Equations . . . . . . . . . . . . . . . . . . 562.3 First-Order Autonomous Equations—Geometrical Insight . . . . . . . . . . . . . . . 592.4 Population Modeling: An Application of Autonomous Equations . . . . . . . . . . . 822.5 Numerical Approximation with the Euler Method . . . . . . . . . . . . . . . . . . . . 832.6 Numerical Approximation with the Runge-Kutta Method . . . . . . . . . . . . . . . 872.7 An Introduction to Autonomous Second-Order Equations . . . . . . . . . . . . . . . 912.8 Chapter 2: Additional Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 91
3 Elements of Higher-Order Linear Equations 933.1 Some Terminology . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 933.2 Essential Topics from Linear Algebra . . . . . . . . . . . . . . . . . . . . . . . . . . . 943.3 Reduction of Order—The Case of n = 2 . . . . . . . . . . . . . . . . . . . . . . . . . 1013.4 Operator Notation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1043.5 Numerical Consideration for nth Order Equations . . . . . . . . . . . . . . . . . . . 1063.6 Chapter 3: Additional Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 112
4 Techniques of Higher-Order Linear Equations 1154.1 Homogeneous Equations with Constant Coe!cients . . . . . . . . . . . . . . . . . . . 1154.2 A Mass on a Spring . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1234.3 Cauchy-Euler (Equidimensional) Equation . . . . . . . . . . . . . . . . . . . . . . . . 1294.4 Nonhomogeneous Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1334.5 Method of Undetermined Coe!cients via Tables . . . . . . . . . . . . . . . . . . . . 1384.6 Method of Undetermined Coe!cients via the Annihilator Method . . . . . . . . . . . 1464.7 Variation of Parameters . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1544.8 Chapter 4: Additional Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 165
5 Fundamentals of Systems of Di!erential Equations 1675.1 Systems of Two Equations—Motivational Examples . . . . . . . . . . . . . . . . . . 1675.2 Useful Terminology . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1715.3 Linear Transformations and the Fundamental Subspaces . . . . . . . . . . . . . . . . 1745.4 Eigenvalues and Eigenvectors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1775.5 Matrix Exponentials . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1805.6 Chapter 5: Additional Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 182
6 Techniques of Systems of Di!erential Equations 1856.1 A General Method, Part I: Solving Systems with Real, Distinct Eigenvalues . . . . . 1856.2 A General Method, Part II: Solving Systems with Repeated Real or Complex Eigen-
values . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1886.3 Solving Linear Homogeneous and Nonhomogeneous Systems of Equations . . . . . . 1936.4 Nonlinear Equations and Phase Plane Analysis . . . . . . . . . . . . . . . . . . . . . 1956.5 Epidemiological Models . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2046.6 Chapter 6: Additional Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 207
7 Laplace Transforms 2137.1 Fundamentals of the Laplace Transform . . . . . . . . . . . . . . . . . . . . . . . . . 2137.2 Properties of the Laplace Transform . . . . . . . . . . . . . . . . . . . . . . . . . . . 2147.3 Step Functions, Translated Functions, and Periodic Functions . . . . . . . . . . . . . 2147.4 The Inverse Laplace Transform . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2157.5 Laplace Transform Solution of Linear Di"erential Equations . . . . . . . . . . . . . . 2167.6 Solving Linear Systems using Laplace Transforms . . . . . . . . . . . . . . . . . . . . 2177.7 The Convolution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2177.8 Chapter 7: Additional Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 218
8 Series Methods 2218.1 Power Series Representations of Functions . . . . . . . . . . . . . . . . . . . . . . . . 2218.2 The Power Series Method . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2238.3 Ordinary and Singular Points . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2248.4 The Method of Frobenius . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2258.5 Bessel Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2268.6 Chapter 8: Additional Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 229
B Graphing Factored Polynomials 231
C Selected Topics from Linear Algebra 233C.1 A Primer on Matrix Algebra . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 233C.2 Gaussian Elimination, Matrix Inverses, and Cramer’s Rule . . . . . . . . . . . . . . . 235C.3 Coordinates and Change of Basis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 238
Chapter 1
Traditional First-Order Di!erentialEquations
1.1 Some Basic Terminology
1. With y(x) = 2x3, we have y!(x) = 6x2. Substituting into the ODE gives x(6x2) = 3(2x3),which is true for x ! ("#,#).
2. y = 2,dy
dx=
d
dx(2) = 0. Substituting y and
dy
dxinto the ODE, we get 0 = x3(2 " 2)2, which
is true for all x. Thus y = 2 is a solution tody
dx= x3(y " 2)2 on ("#,#).
3. y(x) ="1
5x + 4= "(5x + 4)"1,
dy
dx= (5x + 4)"2(5) =
5(5x + 4)2
. y anddy
dxdo not exist when
5x + 4 = 0 $ x = "45. Substituting y and
dy
dxinto the ODE, we get
5(5x + 4)2
= 5!
"15x + 4
"2
=5("1)2
(5x + 4)2
=5
(5x + 4)2
which is true for {x | x %= " 45}. Thus y =
"15x + 4
is a solution to the ODE on (" 45 ,#).
4. Since y(x) = ex " x, we calculate y!(x) = ex " 1. Substituting into the ODE gives
(ex " 1) + (ex " x)2 = e2x + (1 " 2x)ex + x2 " 1$ ex " 1 + e2x " 2xex + x2 = e2x + ex " 2xex + x2 " 1
which is true for all x ! ("#,#).
5. y(x) = x3,dy
dx= 3x2. Substituting y and
dy
dxinto the ODE, we get
3x2 = 3(x3)2/3
= 3x2
1
2 Section 1.1
which is true for all x. Thus y = x3 is a solution to the ODE on ("#,#).
6. y(x) ="1
x " 3,
dy
dx= (x " 3)"2 =
1(x " 3)2
. y anddy
dxdo not exist when x " 3 = 0 $ x = 3.
Substituting y anddy
dxinto the ODE, we get
1(x " 3)2
=!
"1x " 3
"2
=("1)2
(x " 3)2
=1
(x " 3)
which is true for {x | x %= 3}. Thus y ="1
x " 3is a solution to the ODE on ("#, 3).
7. Taking the derivative of y(x) = x2 " x"1 gives us y!(x) = 2x " x"2 and y!!(x) = 2 + x"3.Substitution gives
x2(2 + x"3) = 2(x2 " x"1)$ 2x2 + x"1 = 2x2 " x"1
which is true where both sides are defined, which occurs when x %= 0.
8. y(x) = sin x + 2 cosx,dy
dx= cosx " 2 sinx, and
d2y
dx2= " sinx " 2 cosx. Substituting y and
d2y
dx2into the ODE, we get
(" sinx " 2 cosx) + (sin x + 2 cosx) = 0
which is true for all x. Thus y = sinx + 2 cosx is a solution to the ODE on ("#,#).
9. y(x) = x,dy
dx= 1, and
d2y
dx2= 0. Substituting y and
d2y
dx2into the ODE, we get
(0) + (x) = x
which is true for all x. Thus y = x is a solution to the ODE on ("#,#).
10. y(x) = x + C sin x,dy
dx= 1 + C cosx, and
d2y
dx2= "C sin x. Substituting y and
d2y
dx2into the
ODE, we get
("C sin x) + (x + C sinx) = x
x = x
which is true for all x. Thus y = x + C sin x is a solution to the ODE on ("#,#) for anyconstant C.
11. (a) y = ex, y! = ex, y!! = ex $ y!! " 3y! + 2y = ex " 3ex + 2ex = 0(b) y = e2x, y! = 2e2x, y!! = 4e2x
$ y!! " 3y! + 2y = 4e2x " 6e2x + 2e2x = 0
Section 1.1 3
12. (a) y = ex, y! = ex, y!! = ex $ y!! " 2y! + y = ex " 2ex + ex = 0(b) y = xex, y! = xex + ex, y!! = xex + 2ex
$ y!! " 2y! + y = xex + 2ex " 2xex " 2ex + ex = 0
13. (a) y = sin 3x, y! = 3 cos 3x, y!! = "9 sin 3x$ y!! + 9y = "9 sin3x + 9 sin 3x = 0 &YES
(b) y = sinx, y! = cosx, y!! = " sinx$ y!! + 9y = " sinx + 9 sinx %= 0 &NO
(c) y = cos 3x, y! = "3 sin3x, y!! = "9 cos 3x$ y!! + 9y = "9 cos 3x + 9 cos 3x = 0 &YES
(d) y = e3x, y! = 3e3x, y!! = 9e3x
$ y!! + 9y = 9e3x + 9e3x %= 0 &NO(e) y = x3, y! = 3x2, y!! = 6x
$ y!! + 9y = 6x + 9x3 %= 0(unless x = 0)&NO
14. y!! + 6y! + 9y = 0
(a) y = ex, y! = ex, y!! = ex $ex + 6ex + 9ex %= 0 &NO
(b) y = e"3x, y! = "3e"3x y!! = 9e"3x
$ 9e"3x " 18e"3x + 9e"3x = 0 &YES(c) y = xe"3x, y! = "3xe"3x + e"3x y!! = 9e"3x " 6e"3x
$ 9xe"3x " 6e"3x " 18xe"3x + 6e"3x + 9xe"3x = 0 &YES(d) y = 4e3x, y! = 12e3x, y!! = 36e3x
$ 36e3x + 72e3x + 36e3x %= 0 &NO(e) y = e"3x(x + 2), y! = e"3x("3x " 5), y!!e"3x(9x + 12)
$ e"3x(81x + 108 " 18x " 30 + 18 + 9x) %= 0 &NO
15. y!! " 7y! + 12y = 0
(a) y = e2x, y! = 2e2x, y!! = 4e2x
$ 4e2x " 14e2x + 12e2x %= 0 &NO(b) y = e3x, y! = 3e3x, y!! = 9e3x
$ 9e3x " 21e3x + 12e3x = 0 &YES(c) y = e4x, y! = 4e4x, y!! = 16e4x
$ 16e4x " 28e4x + 12e4x = 0 &YES(d) y = e5x, y! = 5e5x, y!! = 25e5x
$ 25e5x " 35e5x + 12e5x %= 0 &NO(e) y = e3x + 2e4x, y! = 3e3x + 8e4x, y!! = 9e3x + 32e4x
$ 32e4x + 9e3x " 21e3x " 56e3x + 12e3x + 24e4x = 0 &YES
16. y!! + 4y! + 5y = 0
(a) y = e"2x, y! = "2e"2x, y!! = 4e"2x
$ 4e"2x " 8e"2x + 5e"2x %= 0 &NO(b) y = e"2x sin 2x, y! = 2e"2x cos 2x " 2e"2x sin 2x, y!! = "8e"2x cos 2x
$ "8e"2x cos 2x + 8e"2x cos 2x " 8e"2x sin 2x + 5e"2x sin 2x %= 0&NO
4 Section 1.1
(c) y = e"2x cos 2x, y! = "2e"2x cos 2x " 2e"2x sin 2x, y!! = 8e"2x sin 2x$ 8e"2x sin 2x " 8e"2x cos 2x " 8e"2x sin 2x + 5e"2x cos 2x %= 0&NO
(d) y = cos 2x, y! = "2 sin2x, y!! = "4 cos 2x$ "4 cos 2x " 8 sin 2x + 5 cos 2x %= 0 &NO
17. y = erx, y! = rerx, and y!! = r2erx. The ODE gives
r2erx + 3rerx + 2erx = 0$ erx(r2 + 3r + 2) = 0$ erx(r + 2)(r + 1) = 0
$ r = "2,"1
18. y!! + 4y! + 4y = 0y = xerx, y! = rxerx + erx, y!! = r2xerx + 2rerx $
erx#r2x + 2r + 4rx + 4 + 4x
$= 0
$ xr2 + (2 + 4x)r + (4 + 4x) = 0
$ r ="2 " 4x ±
'4
2x
="4x
2xor
"4 " 4x
2x
= "2 or"2 " 2x
x
We can disregard the “value” involving x, which leaves only r = "2.
19. (a) 3y!! + y = sinx (i) 2nd order (ii) linear (iii) N/A(b) y!! + sin y = 0 (i) 2nd order (ii) nonlinear (iii) N/A(c) y(3) + (sin x)y(2) + y = x, y(0) = 1, y!(0) = 0, y!!(0) = 2
(i) 3rd order (ii) linear (iii) IVP(d) y! + exy = y4, y(0) = 0
(i) 1st order (ii) nonlinear (iii) IVP(e) y!! + y! " y (i) 2nd order (ii) linear (iii) N/A(f) y!! + exy! + y2 = 0, y(0) = 1, y(!) = 0
(i) 2nd order (ii) nonlinear (iii) BVP
20. (a) y!! " 3yy! = x (i) 2nd order (ii) nonlinear (iii) N/A(b) y!! = sin x (i) 2nd order (ii) linear (iii) N/A(c) y!! + 3y! = 0, y(0) = 1, y!(1) = 0
(i) 2nd order (ii) linear (iii) BVP(d) y!! = 0, y(1) = 1, y!(1) = 2
(i) 2nd order (ii) linear (iii) IVP(e) y!! " 4y! + 4y = 0, y(0) = 1, y!(0) = 1
(i) 2nd order (ii) linear (iii) IVP(f) x2y!! + y! + (lnx)y = 0 (i) 2nd order (ii) linear (iii) N/A
1.2. SEPARABLE DIFFERENTIAL EQUATIONS 5
1.2 Separable Di!erential Equations
1.
4xy dx + (x2 + 1) dy = 04x
x2 + 1dx = "1
ydy
"%
4x
x2 + 1dx =
%dy
y
Letting u = x2 + 1,
"2%
du
u=
%dy
y
"2 ln |u| + C1 = ln |y|"2 ln(x2 + 1) + C1 = ln |y|
e"2 ln(x2+1)+C1 = |y||y| = eC1(x2 + 1)"2, since eC1 > 0
y =C
(x2 + 1)2, where C = ±eC1
2.
tan xdy + 2y dx = 0
"%
dy
2y=
%dx
tan x
" ln |y| = 2 ln |sin x| + C1
1|y| = |sinx|2eC1
y = C · csc2 x, c = ±eC1
3. (ex + 1) cos y dy + ex(sin y + 1) dx = 0, y(0) = 3. Then
$%
cos y
sin y + 1dy =
% "ex
ex + 1dx
Substituteu1 = sin y + 1 u2 = ex + 1
du1 = cos y dy du2 = ex dx
Then
ln |sin y + 1| = " ln |ex + 1| + C $ sin y + 1 = A
&1
(ex + 1)
'
Apply IC: (sin 3) + 1 =A
(e0 + 1)$ A = 2(sin 3) + 2
Therefore,
sin y + 1 =2 + 2 sin 3
ex + 1
6 Section 1.2
4. We have 2x(y2 + 1) dx + (x4 + 1) dy = 0, y(1) = 1. Then
(x4 + 1) ds = "2x(s2 + 1) dr%dy
y2 + 1=
% "2x
x4 + 1dx =
(u = x2
du = 2xdx
"%
1u2 + 1
du
Therefore,
arctan(y) = " arctan(u) + C
arctan(y) = " arctan(x2) + C
arctan(1) = " arctan(1) + C
+2 arctan(1)( )* +!/4
= C $ C =!
2
Finally,
arctan(y) = " arctan(x2) +!
2y = tan
,!2" arctan(x2)
-
5. xy dx + (x + 1) dy = 0 &%
dy
y=% "x
x + 1dx. Substituting u = x + 1, x = u" 1, du = dx, we
have
ln |y| =%
" (u " 1)u
du =% !
1u" 1"
du = ln |x + 1|" x + C
$ y = A(x + 1)e"x
6.
(y2 + 1)1/2 dx = xy dy
dx
x=
y.y2 + 1
dy
%dx
x=
%y.
y2 + 1dy
Let u = y2 + 1 and du = 2y dy, so du2 = y dy. Then we have%
dx
x=
12
%du'
u
ln |x| + C =(1/2)
'u
(1/2)
ln |x| + C =.
y2 + 1
Section 1.2 7
7. We have (x2 " 1)y! + 2xy2 = 0 and y('
2) = 1. Then
(x2 " 1)dy
dx= "2xy2
dy
y2=
2xdx
x2 " 1
"%
dy
y2=
%2x
x2 " 1dx
1y
= ln |x2 " 1| + C
y =1
ln |x2 " 1| + C
IC:
11
= ln |2 " 1| + C
$ C = 1
y =1
ln |x2 " 1| + 1
8. We have y! cotx + y = 2 and y(0) = "1. Then
dy
dxcotx = "y + 2
%dy
y " 2= "
%tan xdx
ln |y " 2| = ln |cosx| + C
|y " 2| = eC |cos x|y = ±eC(cosx) + 2 = A(cosx) + 2
Since y(0) = "1, "1 = A + 2 $ A = "3. Thus
y = "3(cosx) + 2
9.
y! = 10x+y $ dy
dx= 10x10y
dy
10y= 10x dx &
%10"y dy =
%10x dx
Use au = eu ln a. Then
"10"y = 10x + C1
10y ="1
10x + C1
$ y = log10 |"10x + C|
8 Section 1.2
10.
xdx
dt+ t = 1
xdx = (1 " t) dt%xdx =
%(1 " t) dt
12x2 = t " 1
2t2 + C1
x2 = 2t " t2 + 2C1
x = ±.
2t " t2 + C, C = 2C1
11. Let u = y " x
& du
dx=
dy
dx" 1 & du
dx+ 1 = cosu
du
dx= cosu " 1
du
cosu " 1= dx
x =1
tan(u/2)
=1
tan(y"x2 )
$ y = x + 2 tan"1(1/x)
12. Let u = 2x + y " 3 & du
dx= 2 +
dy
dx
u! " 2 = u
u! = u + 2du
u + 2= dx
ln(u + 2) = x
ex = u + 2 = 2x + y " 1y = ex " 2x + 1
13. We have (x + 2y)y! = 1, y(0) = "2. Let u = 2y + x. Then
du
dx= 2
dy
dx+ 1
Substituting into the ODE gives
u
!u! " 1
2
"= 1
This is separable:
u! " 1 =2u
$ u! =2u
+ 1 =2 + u
u
$%
u
2 + udu =
%dx
Section 1.2 9
Note thatu
2 + u=
u + 2 " 22 + u
= 1 " 22 + u
Then % !1 " 2
2 + u
"du =
%dx $ u " 2 ln |2 + u| = x + C
Substitute back u = x + 2y:
(x + 2y) " 2 ln |2 + x + 2y| = x + C
Apply IC: 0 + 2("2) " 2 ln |2 + 0 + 2("2)| = 0 + C
"4 " 2 ln |"2| = C
The solution is given in implicit form as
2y " 2 ln |2 + x + 2y| + 4 + 2 ln 2 = 0
14. y! =.
4x + 2y " 1. Let u = 4x + 2y " 1 & dudx = 4 + 2 dy
dx $
u! = 2'
u + 4
dx =du
2'
u + 4x = "2
#ln#'
u + 2$$
"'
u
= "2,ln,.
4x + 2y " 1 + 2--
".
4x + 2y " 1
15. We have (y + 2) dx + y(x + 4) dy = 0, y("3) = "1. Then
(x + 4)y dy = "(y + 2) dx%y
y + 2dy = "
%dx
x + 4% !
1 " 2y + 2
"dy = "
%dx
x + 4y " 2 ln |y + 2| = " ln |x + 4| + C
ey"2 ln |y+2| = e" ln |x+4|+C
1|y + 2|2 · ey = eC · 1
|x + 4|ey
(y + 2)2= ± eC
(x + 4)=
A
(x + 4)where A = ±eC
Apply IC: y("3) = "1 $ e"1
("1 + 2)2=
A
("3 + 4)
$ A = e"1 $ ey
(y + 2)2=
e"1
(x + 4)
16. We have 8 cos2 y dx + csc2 xdy = 0, y(!/12) = !/4. Then
csc2 xdy = "8 cos2 y dx%sec2 y dy = "8
%sin2 xdx
tan y = "8!
12x " 1
4sin 2x + C1
"
tan y = 2 sin 2x " 4x + C, where C = "8C1
10 Section 1.2
Using y(!/12) = !/4, we solve for C:
tan,!
4
-= 2 sin
,!6
-" !
3+ C
C =!
3
Sotan y = 2 sin 2x " 4x +
!
3
17. We havedy
dx=
y3 + 2y
x2 + 3x, y(1) = 1. Separating variables and integrating gives
%1
y3 + 2ydy =
%1
x2 + 3xdx
$% ! "y
2(y2 + 2)+
12y
"dy =
% ! "13(x + 3)
+13x
"
$ "14
ln |y2 + 2| + 12
ln |y| = "13
ln |x + 3| + 13
ln |x| + C
Now apply IC:
"14
ln(12 + 2) +12
ln(1) = "13
ln(1 + 3) +13
ln(1) + C
$ C =13
ln 4 " 14
ln 3
The solution is then
"14
ln(y2 + 2) +12
ln(y) = "13
ln |x + 3| + 13
ln |x| + 13
ln 4 " 14
ln 3
18. We have y! = ex2, y(0) = 0. Then
dy
dx= ex2
%dy =
%ex2
dx
y =%
ex2dx
Apply IC:% x
0y dy =
% x
0et2 dt
y(x) " y(0) =% x
0et2 dt
y(x) =% x
0et2 dt
Section 1.2 11
19. We have y! = xyex2, y(0) = 1. Then
%dy
y=
%xex2
dx
ln |y| =12ex2
+ C
C = ln |y|" 12ex2
, when x = 0, y = 1
C = "12
Thus
ln |y| =12x2 " 1
2|y| = e"1/2e("1/2)x2
y = ±e1/2ex2"1
y is even since we have only an x2 term.
20. (a) We have x2y! " cos 2y = 1, limx#+$ y = 9!/4. Then
x2 dy
dx= cos 2y + 1
First we use a trigonometric identity as follows:
cos 2y + 1 = (cos2 y " sin2 y) + (cos2 y + sin2 y)= 2 cos2 y
Thus we obtain
dy
2 cos2 y=
dx
x2
2%
sec2 y dy =%
dx
x2
2 tan y = " 2x
+ C1
tan y = " 1x
+ C2
C2 = tan y +1x
limx#+$
C2 = tan!
9!4
"+ 0 = 1
Thustan y = 1 " 1
xor y = arctan
!1 " 1
x
"
12 Section 1.2
(b) We have 3y2y! + 16x = 2xy3. Then
3y2 dy
dx= 2x(y3 " 8)
%3y2
y3 " 8dy =
%2xdx
ln |y3 " 8| = x2 + C1
|y3 " 8| = eC1ex2
y3 = ±eC1ex2+ 8
y = [Cex2+ 8]1/3 where C = ±eC1
Notice that no value for C bounds y since eC1 %= 0.
21.
dN
dt= rN
1N
dN = r dt%
1N
dN =%
r dt
ln N = rt + C1
Note N is number of population, i.e. > 0.
elnN = ert+C1
N = eC1ert
N = C2ert
Note C2 is > 0, dealing with possible population. Thus N = C2ert is a solution to the ODE.Case 1. For values r ) 0 we have exponential growth as t & #.Case 2. For values r < 0 we have exponential decay as t & #.
Section 1.2 13
22.
dN
dt= r
!1 " N
K
"N
dN
rN#1 " N
K
$ = dt
(Let u = 1 " N/K)du
dt= " 1
K
dN
dt
"Kdu
dt= ruK(1 " u)
du
dt= ru(u " 1)
dt =du
ru(u " 1)
$ t =ln |u " 1|" ln |u|
r
=1r
ln
/1 " 1#
1 " NK
$0
$ etr = 1 " 1#1 " N
K
$
1 " N
K=
11 " etr
N = K " K
1 " etr
limt#$
N = K
23. We have (x + y) dx " xdy = 0. Then
(x + y)x
=dy
dx$ 1 +
y
x=
dy
dx
Letu =
y
x$ y! = u = xu!
Thenu + xu! = 1 + u $ u! =
1x$ u = ln |x| + C
Substituting for u givesy = x(ln |x| + C)
24. We havedy
dx=
y2 + 2xy
x2=
y2
x2+
2xy
x2
Letu =
y
x$ y! = u + xu!
Thenu + xu! = u2 + 2u $ u! =
u2 + u
x
14 Section 1.2
This equation is separable:%
du
u2 + u=%
1x
dx = ln |x| + C1
Use partial fractions to rewrite1
u2 + u.
1
223
1u(u + 1)
=A
u+
B
u + 1$ 1 = A(u + 1) + Bu = u(A + B) + A
$ A + B = 0A = 1
4$ B = "1
5
667
%1
u(u + 1)du =
% !1u" 1
u + 1
"du = ln |u|" ln |u + 1| = ln
8888u
u + 1
8888
Thus
ln8888
u
u + 1
8888 = ln |x| + C $8888
u
u + 1
8888 = eln |x|+C1 = eC1 · |x|
$ u
u + 1= C · x $ u = Cx(u + 1)
$ u(1 " Cx) = Cx
$ u =Cx
1 " Cx
Plugging in u = y/x then gives
y =Cx2
1 " Cx
25. We have2x2 dy
dx= x2 + y2 $ 2
dy
dx= 1 +
y2
x2
Let u = yx $ y! = u + xu!. Substituting gives
2(u + xu!) = 1 + u2 $ 2xu! = 1 + u2 " 2u = (u " 1)2
This is separable: %2 du
(u " 1)2=%
1x
dx
Letting w = u " 1, dw = du,
$ "2u " 1
= ln |x| + C $ u " 1 ="2
ln |x| + C
$ u = 1 " 2ln |x| + C
Plugging in u = y/x gives
y = x " 2x
ln |x| + C
Section 1.2 15
26.
xy! " y =.
x2 + y2
xy! =.
x2 + y2 + y
y! =.
x2 + y2
x+
y
x
=9
1 +, y
x
-2+, y
x
-(homogeneous)
Let y = vx, y! = v!x + v. Then
dv
dxx + v =
.1 + v2 + v
xdv
dx=
.1 + v2
1'1 + v2
dv =1x
dx
arcsinh v = ln |x| + C
Substituting back in, we have
y
x= sinh(ln |x| + C) $ y = x sinh(ln |x| + C)
27. We have (x + 2y) dx " xdy = 0. Then
x + 2y " xdy
dx= 0
x + 2y = xdy
dxdy
dx= 1 + 2
,y
x
-(homogeneous)
Let y = vx, y! = v!x + v. Then
dv
dxx + v = 1 + 2v
dv
dx= 1 + v
1v + 1
dv = dx
ln |v + 1| = x + C
v + 1 = Cex
y
x= Cex " 1
y = Cxex " x
28. We have(y2 " 2xy) dx + x2 dy = 0 $ dy
dx=
2xy " y2
x2= 2,y
x
-", y
x
-2
16 Section 1.2
Let u = yx , y! = u!x + u. Then
xu! + u = 2u " u2 $ u!x = u " u2
$ " ln |u " 1| + ln |u| = ln x + C1
$ u
u " 1= eln x+C1 = x · C
$ y/x
(y/x) " 1= xC
$ yx
y " x= Cx2
$ y
y " x= Cx
$ y = xCy " Cx2
$ y =Cx2
1 + Cx
29. We have 2x3y! = y(2x2 " y2). Then
dy
dx=
y3(2x2 " y2)2x3
$ dy
dx=, y
x
-!1 " 1
2
,y
x
-2"
Let u =y
x,
dy
dx= x
du
dx+ u. Then
xdu
dx+ u = u
!1 " 1
2u2
"$ du
dx· x = "1
2u3
$%
"2u"3 du =%
1x
dx
$ u2 = ln x + C
$ y2 = x2 ln x + Cx2
$ y = ±x'
ln x + C
Section 1.2 17
30. We have (x2 + y2)y! = 2xy. Then
dy
dx=
2xy
x2 + y2$ dv
dxx + v =
2x2v
x2 + v2x2
$ dv
dxx + v =
2v
1 + v2
$ dv
dxx =
2v
1 + v2" v
$ dv
dxx =
2v " v " v3
1 + v2
$ dv
dxx =
v " v3
1 + v2
$ 1 + v2
v " v3dv =
1x
dv
$ 1v(1 " v2)
dv +v
(1 " v2)dv =
1x
dx
$ ln |v|" ln |v2 " 1| = ln x + Cv
v2 " 1= Cx $ y
# yx
$2 " 1= Cx2
$ y = Cx2,y
x
-2" Cx2
y = Cy2 " Cx2 $ y = C(y2 " x2)
31.
xy! " y = x tan(y/x)
y! =y
x+ tan
, y
x
-(Homogeneous)
,y = vx, y! = v + xv!, v =
y
x
-
v + xv! = v + tan vdv
tan v=
dx
xln |x| = ln | sin v| + C
|x| = eC | sin v| = eC | sin y
x|
y
x= sin"1 | x
eC|
y = x sin"1 | x
eC|
32.
(2x + y) dx " (4x + 2y) dy = 0dy
dx=
12
y =x
2+ C
18 Section 1.2
33. We have y2 + x2y! = xyy!. Then
y2 = y!(xy " x2) $ y! =y2
xy " x2
$ y! =4
y " x
$ dy
dx=
y
(y " x)$ (y " x) dy = y dx
$ (y " x) dy + ("y) dx = 0
Let y! = v!x + v. Then
dv
dxx + v =
vx
vx " y=
v
v " 1$ dv
dxx =
v
v " 1" v
$ dv
dxx =
v " v(v " 1)v " 1
$ dv
dxx =
v " v2 + v
v " 1
$ dv
dxx =
2v " v2
v " 1v " 1
2x " v2dv =
1x
dx $ "12
ln(v " 2) " 12
ln |v| = ln x + C
$ ln8888
1'v " 2
8888+ ln8888
1'v
8888 = ln x + C
1.v(v " 2)
= Cx $ 1:#yx
$ #yx " 2
$ = Cx
1:1x2 (y)(y " 2x)
= Cx
=1.
y(y " 2x)= C
34. (x " y) + (y " x)y! = 0 $ y! = 1( for y %= x) $ y = x + C. If x = y, the equation is truetrivally.
Section 1.2 19
35.
(x + 4y)y! = 2x + 3y
y! =2x + 3y
x + 4y=
2 + 3(y/x)1 + 4(y/x)
,y = vx, y! = v + xv!, v =
y
x
-
v + xv! =2 + 3v
1 + 4vdx
x=
(1 + 4v)dv
2 + 2v " 4v2
(Partial fractions yields:)
ln |x| = "16;ln(1 + 2v)(2 " 2v)5
<
ln |x| = "16;ln(1 + 2(y/x))(2 " 2(y/x))5
<
36. We have(x " y) dx + (x + y) dy = 0 $ dy
dx=
y " x
x + y=
yx " 11 + y
x
Let u = yx , y! = xu! + u. Then
xu! + u =u " 11 + u
$ du
dx=
u " 11 + u
" u
$ du
dx=
"1 " u2
1 + u
$%
1 + u
1 + u2du =
%"dx
% !1
1 + u2+
u
1 + u2
"du =
%"dx $ arctanu +
12
ln |1 + u2| = "x
$ arctan, y
x
-+
12
ln!
1 +,y
x
-2"
= "x
20 Section 1.2
37.
y dx = (2x + y) dy
dy
dx=
(y/x)2 + (y/x)
(y = vx, y! = v + xv!)
v + xv! =v
2 + v
xv! = "v2 + v
v + 2
" 2 + v
v2 + vdv =
dx
x
ln!|v + 1|
v2
"= ln |x| + C
|v + 1|v2
= eC |x|
|(y/x) + 1|(y/x)2
= eC |x|
38.
y! = 2!
y
x + y
"2
= 2!
(y/x)1 + (y/x)
"2
,y = vx, y! = v + xv!, v =
y
x
-
v + xv! =2v2
(1 + v)2
xv! =v2 " v
(1 + v)2
(1 + v)2
v2 " vdv =
dx
x
ln!
(v " 1)4
v
"+ v = ln |x| + C
ln!
((y/x) " 1)4
(y/x)
"+ (y/x) = ln |x| + C
41. (a) M(1,y
x)
t= 1x= M(tx, ty)
Homog. deg. n= tnM(x, y)
=!
1x
"n
M(x, y) $ M(x, y) = xnM(1,y
x)
(b) Repeat above exchanging M for N .
(c) M(x, y)dx + N(x, y)dy = 0 $ dy
dx=
"M(x, y)N(x, y)
="M(1, y
x)N(1, y
x )
Define g1(t) = "M(1, t), g2(t) = N(1, t), g(t) ="g1(t)g2(t)
$ dy
dx=
"M(1, yx )
N(1, yx )
="g1( y
x)g2( y
x )= g(
y
x)
Section 1.2 21
42. In (23.)F (x, y) = (x + y)dx " xdy = 0 $F (tx, ty) = (tx + ty)t dx " tx t dy = t2((x + y)dx " xdy) = t2F (x, y) $Homogeneous of degree 2. The others are similar.
43. Given that M(x, y) dx + N(x, y) dy = 0 is homogeneous, thendy
dx=
"M(x, y)N(x, y)
="M(1, (y/x))N(1, (y/x))
and the substitution
x = r cos t
dx = "r sin t dt + cos t dr
y = r sin t
dy = r cos t dt + sin t dr
dy
dx=
"M(1, tan t)N(1, tan t)
="M
N
dy
dx=
r cos t dt + sin t dr
r sin t dt + cos t dr
$ "M
N=
r cos t dt + sin t dr
r sin t dt + cos t dr$ (r cos t dt + sin t dr)N = "M("r sin t dt + cos t dr)
N sin t dr + M cos t dr = "Nr cos t dt + Mr sin t dtdr
dt= r
!"N cos t + M sin t
N sin t + M cos t
"
Since the function M and N are functions of t, the right hand side is a function of r times afunction of t. Hence, it is seperable.
44. (a) From 43.
dr
dt= r
!r cos2 t + (cos t + sin t) sin t
" cos t sin t + (cos t + sin t) cos t
"
= r
!cos2 t + sin t cos t + sin2 t
cos2 t
"
= r
!1 + sin t cos t
cos2 t
"
dr
r=
!1 + sin t cos t
cos2 t
"dt
ln |r| = ln
/91
cos2 t
0+ tan t
|r| = exp
=ln
/91
cos2 t
0+ tan t + C
>
with r =.
x2 + y2 and t = tan"1,y
x
-, we finally have
.x2 + y2 = exp
=ln
/?1
cos2(tan"1# y
x
$)
0+ tan(tan"1
,y
x
-) + C
>
(b)
22 Section 1.2
45. (a)dy
dx=
y " x
y + x=
(y/x) " 1(y/x) + 1
which is homogeneous. As before, y = vx and
v + v!x =v " 1v + 1
v!x ="2
v + 1
(v + 1) dv ="2 dx
xv2
2+ v + C = = "2 ln |x|
|x| = Ke"(v2/4)"(v/2) = K exp(" y2
4x2" y
2x)
(b) i. (x " y " 1) dx + (y + x + 5) dy = 0 & M(x, y) = x " y " 1& M(tx, ty) = tx " ty " 1 %= tM(x, y). Hence, M is not homogeneous. Similarly forN , and so the equation is not homogeneous.
ii. x = X " 2, y = Y " 3, dx = dX , dy = dY and dydx = dY
dX = Y "XY +X
iii. |X | = K exp(" Y 2
4X2" Y
2X) $ |x + 2| =
K exp(" (y + 3)2
4(x + 2)2" y + 3
2(x + 2))
46. Lines are parallel, so there is no point of intersection.
47. Lines are parallel, so there is no point of intersection.
48.dy
dx=
2x + 3y " 5x + 4y
which gives point of intersection to be (4,"1). Thus, we let x = X " 4 and
y = Y + 1 and get to
dY
dX=
2X + 3Y
X + 4Y
=2 + 3(Y/X)1 + 4(Y/X)
Y = vX Y ! = v!X + v
v!X =1 " v
1 + 4vln |X | = "5 ln |1 " v|" 4X
ln |x + 4| = "5 ln88881 " y " 1
x + 4
8888" 4(x + 4)
Section 1.2 23
49. Point of intersection is (3,"2). Thus we let x = X + 3 and y = Y " 2 and get
dY
dX=
Y
2X + YdX
dY=
2X + Y
Y= 2(x/y) + 1
X = uYdX
dY= u!Y + u
u!Y + u = 2u + 1du
u + 1=
dY
Yln |u + 1| = ln |Y | + C
|Y | = K|u + 1| = K|(X/Y ) + 1|
y + 2 = ±K
!x " 3y + 2
+ 1"
50. y! = 2!
y + 2x + y " 1
"2
and as in 49.,
dY
dX= 2
!Y
X + Y
"2
dX
dY=
12
!X
Y+ 1"2
X = uYdX
dY= u!Y + u
u + u!X =12u2 + u +
12
u!X =12(u2 + 1)
du
u2 + 1=
12
dX
X
tan"1(u) =12
ln |X | + C
tan"1
!x " 3y + 2
"=
12
ln |x " 3| + C
24 Section 1.3
1.3 Physical Problems with Separable Equations
1. r = kv2, ve = 60m/s, v0 = 0, m = 75kg, s0 = 1800m, se = 500m
mdv
dt= mg " kv2
75v! = 75(9.8)" kv2
= 735 " kv2
75735 " kv2
dv = dt
"1.38 ln888%
kv"27.11%kv+27.11
888'
k= t + C
88888
'kv " 27.11'kv + 27.11
88888 = exp
='kt + C
"1.38
>
From the given information limt#$
v(t) = 60 $88888
'k60 " 27.11'k60 + 27.11
88888 = 0 & k = .20417
v0 = 0 & C = 0
Thus, we have.45v " 27.11.45v + 27.11
= e"(.326t)
and after some simplification
v(t) = 60!
1 + e".326t
1 " e".326t
"
2. r = kv2, r = .48N when v = 1 & k = .48
.4dv
dt= (.4)(9.8) " .48v2
dv
dt= 9.8 " 1.2v2
dv
9.8 " 1.2v2= dt
If air resistance is neglected, thens(t) = "4.9t2 + 20t, v(t) = "9.8t + 20 = 0& t = 2.04, s(2.04) = 20.4ft
4. v(t) = "gt + v0, v(3) = "3g + v0 = 0 &v0 = 3g = 3s(32ft/s2) = 96ft/s
5. (a) v0 = 0, v(t) = gt = 32t, v(4) = 128ft/s
(b) Ave velocity=14
% 4
032t dt = 64ft/s
(c) v(t) = 32t, s(t)16t2 + s0, s(4) = 0 & s(4) = 162 + s0 = 0& s0 = "256 & cli" was 256 ft tall
6. a(t) = 32, v(t) = 32t + v0 = 32t, s(t) = 16t2 + s0,s(5) = 16(25) + s0 = 0 & s0 = "400 & cli" was 400 ft tall
Section 1.3 25
7. s0 = "350, v0 = 0, s(t) = 16t2 " 350 = 0 & t ='
3504
& average velocity =2km'350/4
=8km'350s
= .428km/s
8. T (t) = (T0 " Ts)ekt + Ts, Ts = "2, T0 = 15, T (2) = 5
& T (2) = 17e2k " 2 = 5 & k =12
ln!
717
". Then to find eating time (t = eat),
T (eat) = 17ek(eat) " 2 = 0 & eat = 2ln(2/17)ln(7/17)
= 4.82hrs * 5pm
9. Let t = 0 be 3pm. T (0) = 79, T (3) = 68, Ts = 60.T (t) = 19ekt + 60, T (3) = 19e3k + 60 = 68
19e3k = 8
k =13
ln .421 = ".288
T (t) = 19e".288t + 60(Solve for t when T = 98.6) = 98.6
t = "2.46
Thus the person died about 12:30pm
10. T (t) = 80ekt + 20, T (10) = 80e10k + 20 = 60 & k = ".069
T (t) = 80e".069t + 20 = 25 & t =10 ln 16
ln 2= 40 min
13. N(t) = amount of N in tank after t seconds. N(0) = 16. dNdt = rate in - rate out = .1 " N
20(.1) &
200dN
20 " N= dt & N(t) = 20 " e("t/200)+C ,
N(0) = 16 = 20 " eC & C = ln 4 * 1.39 N(t) = 20 " e("t/200)+1.39. We want to solve N(t) =20(.99) = 19.8 for t. t = "200(ln(.2) " 1.39) * 599 seconds * 10 min.
14. s(t)=amount of salt at time t. s(0) = 10.ds
dt= 0 "
!s(t)kg
100L
"(5L/m) =
"s(t)20
& s(t) = e("t/20)+C & C = ln 10s(t) = 10e("t/20)+ln10 & s(60) = 4.97 * 5kg
15. (a)ds
dt= (4lb/gal)(2gal/m)" (s(t)lb/50gal)(2gal/min) = 8 " s(t)
25& s(t) = 16 + 9e("t/25) & concentration= s(t)/25.
(b)ds
dt= 8 " 3s
50& s(t) =
4003
" 3253
e("3t/50) & conc= s(t)/25 " t.
16. s(t) = amount of salt in pond, s0 = 0, v(t) =volume of pond, v0 = 10000, dvdt = "50 & v(t) =
10000" 50t.ds
dt= (1250m3/d)
!5kg
1000m3
""!
s(t)kg
(10000" 5t)m3
"(1300m3/d)
= 6.25 " 1300s(t)10000" 50t
Yes, after 200 days, there is no water in the pond which will result in no salt in the pond.
26 Section 1.3
17. C(t) = amount of CO2, C0 = (.0015)(200) = .3m3, dCdt =
,20m3
min
-(.0004)"
,Cm3
200m3
-(20m3/min) =
.008 " .1C This gives us that the concentration will be half of what it started (i.e. there willbe .15m3 of CO2) in 11.45 minutes.
18.dA
dt= kA,
dA
A= k dt & A(t) = ekt+C = A0e
kt.
A(30) = A0/2 = A0e30k
12
= e30k
k =ln(1/2)
30= ".023
A(t) = .01A0 = A0e".023t
t =ln(.01)".023
* 200 days
19. A(1) = .56 = ek & k = ".5798 & t = 1.195 years
20. p = amount of decayed tin, M=original amount of tin.
dp
dt= k(M " p)(p) where M " p is the amount of tin left
dP
p(M " p)= k dt
1M
ln!
p
M " p
"= kt + C
$ p(t) = M " M
1 + eMkt+C
where C would be determined by the original amount of decayed tin in the organ.
22. Let (x, y) be the point of tangency and c be the x-intercept of the tangent line. Then 12 |c "
x||y| = a2 & |c " x| = 2a2
|y| .
dy
dx=
y
|c " x| =y
2a2
|y|
=y|y|2a2
dy
y2=
12a2
dx
y = ±2a2
x
Because of symmetry, either graph will give the desired result.
23. Let (x, y) be the point of tangency and c be the x-intercept of the tangent line. Then |c"x|+|y| = b.
dy
dx=
y
|c " x| =y
b " |y|b ln |y|" |y| = x
This is an implicitly defined function.
1.4. EXACT EQUATIONS 27
33. Let A(t) be the amount of snow on the ground. Then A(t) = kt since it is falling at aconstant rate and there was none when it started. Let P (t) be the distance plowed. ThendPdt = m
A(t) = mkt . Thus dP = m
ktdt & P = mk ln t + C, P (1) = 2 = C & P (t) = m
k ln t + 2.P (2) = m
k ln 2 +2 = 1 & mk = "1.443. P (t) = "1.443 ln t + 2. Set P (t) = 0 to get t * 4 which
says it started snowing at about 8am.
1.4 Exact Equations
1. We have 2xy3 + (1 + 3x2y2) dydx = 0. Using M = 2xy3 and N = 1 + 3x2y2, we have
"M
"y= 6xy2,
"N
"x= 6xy2
so the equation is exact. We need to solve "f"x = M and "f
"y = N :
f =%"f
"xdx =
%2xy3 dx = x2y3 + g(y)
This f = x2y3 + g(y) must also satisfy "f"y = N .
1 + 3x2y2 ="
"y(x2y3 + g(y)) = 3x2y2 + g!(y)
$ g!(y) = 1 $ g(y) = y
Thus the solution is given by f = C:
x2y3 + y = C
2. We have (2xy + 1) + (x2 + 4y) dy = 0, y(0) = 1. Use M = 2xy + 1 and N = x2 + 4y. Then
"M
"y= 2x =
"N
"x
so the equation is exact. We set "f"x = M and "f
"y = N and solve.
f =%"f
"ydy =
%(x2 + 4y) dy = x2y + 2y2 + h(x)
This f must also satisfy "f"x = M :
2xy + 1 ="
"x(x2y + 2y2 + h(x)) = 2xy + h!(x)
$ h!(x) = 1 $ h(x) = x
The solution is given byx2y + 2y2 + x = C
Initial conditions yield C = 2, sox2y + 2y2 + x = 2
28 Section 1.4
3. We have (y sec2 x + secx tan x) + (tanx + 2y) dy = 0 with y(0) = 1. Setting M = (y sec2 x +secx tan x) and N = tan x + 2y leads to
"M
"y= sec2 x =
"N
"x
so the ODE is exact. Then,
F =%
(y sec2 x + secx tan x) dx
$ F = y tan x + secx + g(y)
tan x + 2y ="f
"y= tanx + g!(y)
$ 2y = g!(y) $ g(y) = y2
The solution is y tan x + secx + y2 = C. With initial conditions, 1(0) + 1 + 12 = C $ C = 2,so
y tan x + secx + y2 = 2
4. We have (2y sinx cos x + y2 sin x) + (sin2 x " 2y cosx) dy = 0 with y(0) = 3. Let M =2y sinx cos x + y2 sin x and N = sin2 x " 2y cosx. Then
"M
"y= 2 sinx cos x + 2y sinx
"N
"x= 2 sinx cos x " 2y(" sinx)
so the equation is exact. We set "f"x = M and "f
"y = N and solve.
f =%"f
"ydy =
%(sin2 x " 2y cosx) dy = y sin2 x " y2 cosx + h(x)
Then
2y sinx cos x + y2 sin x ="
"x(f) = 2y sin x cos x + y2 sin x + h!(x)
$ h!(x) = 0 $ h(x) = C1
The solution is given byy2 sin2 x " y2 cosx = C
Initial conditions give us 32 sin2 0 " 32 cos 0 = 0, so C = "9 and
y2 sin2 x " y2 cosx = "9
5. We have 2xy + (x2 " y2) dy = 0. Setting M = 2xy and N = x2 " y2,
"M
"y= 2x =
"N
"x
Then"f
"x= 2xy $ f(x, y) = x2y + g(y)
"f
"y= x2 + g!(y) = x2 " y2 $ g!(y) = "y2
$ g(y) = "13y3
Section 1.4 29
Thus the solution satisfiesx2y " 1
3y3 = C
6. We have (2 " 9xy2)x + (4y2 " 6x3)y dy = 0. Using M = (2 " 9xy2)x and N = (4y2 " 6x3)ygives
"M
"y= 18x2y =
"N
"x
Then
"f
"x= 2x " 9x2y2 $ f(x, y) = x2 " 3x3y2 + g(y)
"f
"y= "6x3y + g!(y) = 4y3 " 6x3y
$ g(y) = y4
Thus the solution satisfiesx2 " 3x3y2 + y4 = C
With initial conditions y(1) = 1, 12 " 3(1)3(1)2 + 14 = C gives C = "1, so
x2 " 3x3y2 + y4 = "1
7. We have e"y " (2y + xe"4)dy = 0 with y(1) = 3. Using M = e"y and N = "2y " xe"y wehave
"M
"y= "e"y =
"N
"x
We need to solvef =
%M dx =
%e"y dx = xe"y + g(y)
Substituting into "f"y = N ,
"
"y(xe"y + g(y)) = "xe"y + g!(y) = "2y " xe"y $ g!(y) = "2y
$ g(y) = "y2
The solution is given by "xe"y + y2 = C. Now apply initial conditions: "1e"3 " (3)2 = C, soC = "9 " e"3. The solution is then
"xe"y " y2 = "e"3 " 9
8. We have (1 + y2 sin 2x) + ("2y cos 2x) dy = 0. Then
"
"y[1 + y2 sin 2x] = 2y sin 2x
"
"x["2y cos 2x] = 4y sin 2x
so the equation is not exact.
30 Section 1.4
9. We have 3x2(1 " ln y) +!
x3
y" 2y
"dy = 0. Then
"
"y[3x2 + 3x2 ln y] =
3x2
y
"
"x
&x3
y" 2y
'=
3x2
y
so the equation is exact.
"f
"x= 3x2 + 3x2 ln y $ f(x, y) = x3 + x3 ln y + #(y)
$ "f
"y=
x3
y+ #!(y) =
x3
y" 2y $ #(y) = "y2 + C
Thenf(x, y) = x3 + x3 ln y " y2 + C
10.,2 +
y
x2
-dx +
!y " 1
x
"dy = 0 & "M
"y=
1x2
,"N
"x=
1x2
$ the equation is exact. Thus"F"x =
2 + yx2 & F = 2x " y
x + g(y). "F"y = "1
x + g!(y) = y " 1x $ g!(y) = y, g(y) = y2/2 + C $
F (x, y) = 2x " yx + y2
2 + C
11. We have!
x
sin y+ 2"
dx +(x2 + 1) cos y
cos 2y " 1dy = 0. Then
"
"y
&x
sin y+ 2'
= "x cot[y] csc[y]
"
"x
&(x2 + 1) cos y
cos 2y " 1
'=
2x cos y
cos 2y " 1
We find
cos2 x =12
+12
cos(2y) $ 12
cos 2y = cos2 x " 12
$ cos 2y = 2 cos2 y " 1
So"N
"x=
2x cos y
2(cos2 x " 1)= "x cos y
sin2 y= "x cot y csc y
so it is exact. Then"f
"x=
x
sin y= 2
$ f(x, y) =12
x2
sin y+ 2x + #(y)
$ "f
"y= "1
2x2 cot y csc y + #!(y)
=x2 cos y + cos y
cos 2y " 1
=x2 cos y
2 sin2 y+
cos y
2 sin 2y
=12x2 cot y csc y " 1
2cot y csc y
Section 1.4 31
So#!(y) = "1
2cot y csc y $ #(y) =
csc(y)2
+ C
andf(x, y) =
12x2 csc y + 2x +
12
csc(y) + C
12. Solve axa"1y1"a dx + ((1 " a)xay"a) dy = 0.
(i) As a separable equation,
axa"1y1"a dx = (a " 1)xay"a dy
axa"1
xadx =
(a " 1)y"a
y1"ady
a
xdx =
(a " 1)y
dy
a ln |x| + C = (a " 1) ln(y)y(a"1) = Cxa
(ii) As an exact equation,
"
"y[axa"1y1"a] = a(1 " a)xa"1y"a
"
"y[(1 " a)xay"a] = a(1 " a)xa"1y"a
"f
"x= aa"1y1"a $ f(x, y) = xay(1"a) + #(y)
$ "f
"y= (1 " a)xay"a + #!(y) = (1 " a)xay"a
#!(y) = C
f(x, y) = xay(1"a) + C
13. Determine A ! R such that the equation is exact. Then solve the resulting equation.
(a) (x2 + 3xy) dx + (Ax2 + 4y) dy = 0. Then
"
"y[x2 + 3xy] = 3x
"
"x[Ax2 + 4y] = 2Ax
so 2A = 3 and A = 32 . Then
"f
"x= x2 + 3xy $ f(x, y) =
13x3 +
32x2y + #(y)
$ "f
"y=
32x2 + #!(y) =
32x2 + 4y
$ #(y) = 2y2 + C
Sof(x, y) =
13x3 +
32x2y + 2y2 + C
32 Section 1.4
(b)!
Ay
x3+
y
x2
"dx +
!1x2
" 1x
"dy = 0. We have
"
"y
&Ay
x3+
y
x2
'=
A
x3+
1x2
"
"x
&1x2
" 1x
'= " 2
x3+
1x2
so A = "2. Then
"f
"x=
y
x2" 2y
x3$ f(x, y) = " y
x+
y
x2+ #(y)
"f
"y= " 1
x+
1x2
+ #!(y) =1x2
" 1x$ #(y) = C
Thereforef(x, y) =
y
x2" y
x+ C
14. Assume"M
"y="N
"x. We want to find F (x, y) such that
"F
"x= M,
"F
"y= N . As in the proof
of Thm. 1.4.1, proceed replacing the argument of"F
"x= M(x, y) with one for
"F
"y= N(x, y) and continuing in the same manner.
15. Determine the most general function N(x, y) so that
(x3 + xy2) dx + N(x, y) dy = 0
is exact. For the equation to be exact,
"
"y[x3 + xy2] = 2xy =
"N
"x
soN(x, y) =
%2xy dx = x2y + #(y)
16. Determine the most general function M(x, y) such that
M(x, y) dx + (2yex + y2e3x) dy = 0
is exact. We have"M
"y="
"x[2yex + y2e3x] = 2yex + 3y2e3x
SoM(x, y) =
%2yex + 3y2e3x dy = y2ex + y3e3x + #(x)
17. Show that(Ax + By) dx + (Cx + Dy) dy = 0
is exact if B = C. To prove this, assume that B = C. Then
"
"y[Ax + By] = B
"
"x[Cx + Dy] = C
1.5. LINEAR EQUATIONS 33
Then"
"y[Ax + By] =
"
"x[Cx + Dy]
which by Theorem 2.8.1 implies that the equation is exact.Now assume that the equation is exact. Then by Theorem 2.8.1 we know that
B ="
"y[Ax + By] =
"
"x[Cx + Dy] = C
or that B = C.
18. Show that the homogeneous equation
(Ax2 + Bxy + Cy2) dx + (Dx2 + Exy + Fy2) dy = 0
is exact if and only if B = 2D and E = 2C. To prove this, assume that B = 2D and E = 2C.Then
"
"y[Ax2 + Bxy + Cy2] = Bx + 2Cy = 2Dx + Ey =
"
"x[Dx2 + Exy + Fy2]
Therefore, by Theorem 2.8.1, the equation is exact. Now assume that the equation is exact.Then by Theorem 2.8.1,
Bx + 2Cy ="
"y[Ax2 + Bxy + Cy2]
="
"x[Dx2 + Exy + Fy2] = 2Dx + Ey
which implies that B = 2D and 2C = E.
1.5 Linear Equations
1. y! + 1xy = x: Linear equation with a(x) = 1
x , b(x) = x. Let A(x) =@
1x dx $ ln |x|. Then
eA(x) = |x| and e"A(x) = 1|x| . The solution is given by
y(x) =1|x|
&%x · |x| dx
'+
C
|x| =
ABBBC
BBBD
1"x
!"x3
3
"+
C
"xif x < 0
1x
!x3
3
"+
C
xif x ) 0
Theny(x) =
x2
3+
C
xfor all x.
2. y! " 2x1+x2 y = x2. Linear with a(x) = "2x
1+x2 , b(x) = x2. Then
A(x) =% "2x
1 + x2dx = " ln |1 + x2| = " ln(1 + x2)
because 1 + x2 is always positive. Substituting u = 1 + x2, du = 2xdx leads to
eA(x) = e" ln(1+x2) =1
1 + x2; e"A(x) = eln(1+x2) = 1 + x2
34 Section 1.5
Then%
b(x)eA(x) dx =%
x2
!1
1 + x2
"dx =
% !1 " 1
1 + x2
"dx = x " arctanx
The solution is therefore
y = (1 + x2)(x " arctanx) + (1 + x2) + C
3. (2x + 1)y! = 4x + 2y leads to
y! " 2y
2x + 1=
4x
2x + 1Using P = "2
2x+1 and Q = 4x2x+1 gives% "2
2x + 1dx = " ln |2x + 1|
e@
p(x) = e" ln |2x+1| =1
|2x + 1|e"@
p(x) = eln |2x+1| = |2x + 1|%4x
2x + 1|2x + 1| dx = 2x2 · sgn(2x + 1)
where
sgn(2x + 1) =
E"1 if 2x + 1 < 01 if 2x + 1 ) 0
The solution is y = |2x + 1|(2x2 · sgn(2x + 1) + C) or
y = (2x + 1)(2x2 + C1)
4. We have y! + y tan x = secx with y(!) = 1. Then
N(x) = secx $ d
dx[y secx] = sec2 x
so y secx = tanx + C.
y = sinx + C cosx
$ 1 = sin! + C cos! $ 1 = "C $ C = "1y = sinx " cosx
5. We have dy =!
2x +xy
x2 " 1
"dx. Then
dy
dx" x
x2 " 1y = 2x $ Q(x) = " x
x2 " 1
SoN(x) = e"
@(x/(x2"1))dx
Let u = x2 " 1 = du/2x = dx. Then
N(x) = e"(1/2)@
(1/u) du = eln |(1/%
x2"1)| =1'
x2 " 1
Section 1.5 35
Sod
dx
&y'
x2 " 1
'=
2x'x2 " 1
$ y'x2 " 1
=%
2x'x2 " 1
Let u = x2 " 1. Then du/2x = dx and%
2x'x2 " 1
dx =%
1'u
du =%
u"1/2 du = 2.
x2 " 1 + C
so y'x2 " 1
= 2.
x2 " 1 + C
andy = 2(x2 " 1) + C
.x2 " 1
6. y! " y = 4ex and y(0) = 4. Let P (x) = "1, Q(x) = 4ex. Then%
"dx = "x $ e@
p(x) = e"x; e"@
p(x) dx = ex
%4ex · e"x dx =
%4 dx = 4x
The solution isy = ex(4x + C)
With initial conditions, 4 = e0(4(0) + C) $ C = 4, thus
y = 4ex(x + 1)
7. y! + xy = 2x, P (x) = x, Q(x) = 2x, µ(x) = ex2/2 $%
2xex2/2 dx
= 2ex2/2 + C.Thus the general solution is y = e"x2/2
,2ex2/2 + C
-
8. P (x) =1x
, Q(x) = ex, µ(x) = eln x = x $%
xex dx = xex " ex.
Thus the general solution is y = e" ln x (xex " ex + C) =1x
(xex " ex + C)
9. P (x) = 2, Q(x) = "3x, µ(x) = e2x $%
"3xe2x dx
="3xe2x
2+
34e2x + C. Thus the general solution is
y = e"2x
!"3xe2x
2+
34e2x + C
"
10. We havedy
dx+
1x
y =cosx
xwith y
,!2
-=
4!
and x > 0. This time Q(x) = 1x so N(x) = x,
which implies thatd
dx[xy] = cos(x) $ xy = sin(x) + C
So with y(!2 ) = 4! , we have that
!
2· 4!
= sin,!
2
-+ C $ 2 = 1 + C
so C = 1 andy(x) =
sin(x)x
+ x"1
36 Section 1.5
11.
P (x) = "3x2
Q(x) = x2
µ(x) = e"x3
%Q(x)µ(x) dx = "e"x3
3+ C
y = ex3
/"e"x3
3+ C
0
12.
P (x) = "2x
Q(x) = x3
µ(x) = e"x2
%Q(x)µ(x) dx = " (x2 + 1)e"x2
2+ C
y =
13.
P (x) = 1Q(x) = cosx
µ(x) = ex
%Q(x)µ(x) dx =
ex
2(cos x + sinx) + C
y = e"x
!ex
2(cosx + sin x) + C
"
14.
P (x) = 1Q(x) = ex
µ(x) = ex
%Q(x)µ(x) dx =
e2x
2+ C
y = e"x
!e2x
2+ C
"
Section 1.5 37
15. y! + 2xy = 1 with y(2) = 1. Then we have
A(x) =%
2xdx = x2
eA(x) = ex2, e"A(x) = e"4x2
%b(x)eA(x) =
%ex2
dx% x
2y(t)eA(t) =
% x
2b(t)eAt
16. We have (x + y2) dy = y dx gives
x + y2 = ydx
dyor
dx
dy" 1
yx = y
Thus a(y) = " 1y , b(y) = y, A(y) =
@" 1
y dy = " ln y. Then
eA(y) = e" ln y =1y, e"A(y) = y
%b(y)eA(y) dy =
%y0
!1y
"dy =
%1 dy = y
Then x = y[y] + Cy andx = y2 + Cy
17. We have (2ey "x)y! = 1 or (2ey "x) dydx = 1. So 2ey"x = dx
dy and dxdy +x = 2ey. With Q(y) = 1
we have N(y) = ey.d
dy[xey] = 2e2y and xey = e2y + C
so the solution isx = ey + Ce"y
18. We have (sin 2y + x cot y)y! = 1. Rewriting, we have
(sin 2y + x cot y) =dx
dydx
dy" x cot y = sin 2y
With Q(x) = " cot y we have N(y) = e" ln(sin x) " csc y, so
d
dy[x csc y] =
sin 2y
sin y
$ d
dy[x csc y] =
2 sin y cos y
sin y
$ x csc y =%
2 cos y dy
$ x csc y = 2 sin y + C
$ x(y) = 2 sin2 y + C sin y
19. (a) y = 0, y! = 0,& y! + P (x)y = 0 + 0 = 0
38 Section 1.5
(b) y = y1 is a solution. y = ky1, y! = ky!1.
Then ky!1 + P (x)ky1 = k(y!
1 + P (x)y1) = k(0) = 0(c) y = y1 + y2, y! = y!
1 + y!2
Then y!1 + y!
2 + P (x)(y1 + y2) = (y!1 + P (x)y1) + (y!
2 + P (x)y2)= 0 + 0(Since y1 and y2 are solutions) = 0
20. (a) y = y1 + y2, y! = y!1 + y!
2
Then y!1 + y!
2 + P (x)(y1 + y2) = (y!1 + P (x)y1) + (y!
2 + P (x)y2)= 0 + r(x) (Since y1 and y2 are solutions) = r(x)
(b) y = y1 + y2, y! = y!1 + y!
2
Then y!1 + y!
2 + P (x)(y1 + y2) = (y!1 + P (x)y1) + (y!
2 + P (x)y2)= r(x) + q(x) (Since y1 and y2 are solutions)
(c) Solution to y! + 2y = e"x is y = e"2x(ex + C) and y! + 2y = cosx is
y = e"2x
!e2x
5(2 cosx + sin x) + C
".
Thus y = e"2x
!ex +
e2x
5(2 cosx + sin x) + C
"is a solution to the original equation.
21. (a)
y! = "P (x)ydy
y= "P (x) dx
ln |y| =%
"P (x) dx
yc = ±e@"P (x) dx
(b) y = A(x)yc, y! = A!(x)yc + A(x)y!c.
Then Q(x) = y! + P (x)y = A!(x)yc + A(x)y!c + P (x)A(x)yc =
A!(x)yc + A(x)(y!c + P (x)yc) = A!(x)yc
(c) A!(x)e@"P (x)dx = Q(x) & A!(x) = Q(x)e
@P (x) dx &
A(x) =%
Q(x)e@
P (x) dx dx
(d) yp = ycA(x), y = yc + yp = yc + ycA(x) = yc(1 + A(x)) =
e@"P (x) dx
!%Q(x)e
@P (x) dx dx
"
22. Let y be the amount of pollutant at time t. Then
y! = 12000(2)" 10000!
y
500000 + 2000t
"
= 24000" 10y
500 + 2t
Then y! +10y
500 + 2t= 24000 is linear with solution
y(t) =1
(t + 250)5#4000(t + 250)6 + C
$. C = "4000(250)6.
y(t) = 4000(t + 250) " 4000(250)6
(t + 250)5, y(10) = 218072 &
Concentration = 218072520000 = .419g/gal
Section 1.5 39
23. (a)dN
dt= rN + rA & dN
dt" rN = rA & N(t) = "a + Cert. N(0) = 0 &
A = C & N(t) = Aert " A
(b) r = ln 2. So, with A = 3, N(t) = 3ert " 3 = 3 · 2t " 3.N(24) = 50, 331, 645N(36) = 206, 158, 430, 205
27.dx
dt+ a(t)x = f(t), P (t) = a(t), µ(t) = e
@a(t) dt
$ x(t) = e@"a(t) dt
!%e@
a(t) dtf(t) dt + C
".
limt#$
x(t) = limt#$
!e@"a(t) dt
!%e@
a(t) dtf(t) dt + C
""= 0 since a(t) > 0, e
@"a(t) dt & 0
29. (a) See section 1.5.1. µ(x) = exp
!%1
x2y " x(1 " (2xy " 1)) dx
"=
1x2
which gives an exact
DE and F (x, y) = 2x " y
x+
y2
2+ C
(b)1N
!"M
"y" "N"x
"=
"1x
(1 " ("1)) ="2x
. So µ(x) =1x2
which gives an exact De and
F (x, y) =x3
3" ln |x|" 1
xy+ C
31. M and N change roles, but the argument is exactly the same as (1.31)
32."M
"y= 2y,
"N
"x= 0 & 1
N
!"M
"y" "N"x
"=
1y(2y) = 2 which a function only in x so µ(x) = e2x
33."M
"y=.
1 + x2,"N
"x=
x2
'1 + x2
+.
1 + x2 & 1N
!"M
"y" "N"x
"=
"x
1 + x2which a function
only in x so µ(x) =1'
1 + x2
34."M
"y= x + 2y,
"N
"x= y & 1
M
!"N
"x" "M"y
"=
"1y
which a function only in y so µ(y) =1y
35."M
"y= 3y2 + 1,
"N
"x= y2 " 2x + 1 Which does not yield anything of a function in x or y
36."M
"y= 1,
"N
"x= "1 & 1
N
!"M
"y" "N"x
"=
"2x
which a function only in x so µ(x) =1x2
37."M
"y= 1,
"N
"x= ex & 1
N
!"M
"y" "N"x
"= 1 which a function only in x so µ(x) = ex
38."M
"y= "2y + 1,
"N
"x= 2y " 1 & 1
N
!"M
"y" "N"x
"=
"2x
which a function only in x so µ(x) =1x2
40."M
"y= "2x cos y,
"N
"x= 2x cos y & 1
N
!"M
"y" "N"x
"=
"4x
(x2 + 1)2which a function only in x
so µ(x) =1
(x2 + 1)2
41. n = 2, v = y"1 $ v! " v = "x which is linear and gives solution
v = x + 1 + Cex $ y =1
x + 1 + Cex
40 Section 1.5
43. y! = y4 cosx + y tan x, sody
dx+ (y tan(x))y = y4 cosx
Let v = y"3. Thendv
dx= "3y"4 dy
dx= 7
dv
dx
&"1
3y4
'=
dy
dx
So
dv
dx
!"1
3y4
"+ (" tan(x))y = y4 cosx
dv
dx+ (3 tan(x))v = "3 cosx
N = e3@
tan x dx = eln(1/(cos3 x)) = sec3 xd
dx[v sec3 x] = "3 cos2 x
d
dx[v sec3 x] = "3
2+
32
cos 2x
v sec3 x =34
sin 2x " 32x + C
y"3 sec3 x =34
sin 2x " 32x + C
44. n = 4, v = y"3 $ v! " 9v = "3 which is linear and gives
v = Ce9x +13& y3 =
33Ce9x + 1
& y = 3
93
3Ce9x + 1
45. We have xy2y! = x2 + y3. Then
y! " y3
xy2=
x2
xy2$ y! " 1
xy3 = xy"2
This is a Bernoulli equation with r = "2, so 1 " r = 1 " ("2) = 3. Thus we need to solveu! + 3
#" 1
x
$u = 3(x). Take a(x) = " 3
x , b(x) = 3x, and A(x) =@" 3
x dx = "3 ln |x|. Then
eA(x) = e"3 ln |x| =1
|x|3 , e"A(x) = e3 ln |x| = |x|3
%b(x)eA(x) dx =
%3x
!1
|x|3
"dx
$x ) 0
%3x
!1x3
"dx =
%3!
1x2
"dx = " 3
x
x < 0%
3x
!" 1
x3
"dx =
%" 3
x2dx =
3x
FBBBG
BBBH= " 3
|x|
u = |x|3!" 3|x|
"+ |x|3 · C = "3x2 + C · |x|3
$ y3 = "3x2 + C · |x|3
$ y = ("3x2 + C|x|3)1/3
46. y = 0 is a trivial solution, so if y %= 0y!
y= 4 " 2x & dy
y= (4 " 2x) dx
ln |y| = 4x " x2 + C $ y = ±e4x"x2+C
Section 1.5 41
47. We have (x + 1)(y! + y2) = "y. Then
dy
dx+ y2 =
"y
x + 1$ dy
dx+
y
x + 1= y2
Let v = y"1, sodv
dx= "y"2 dy
dx$ dv
dx("y2) =
dy
dx
and thatdv
dx("y2) +
!1
x + 1
"y = y2
which implies that
dv
dx"!
1x + 1
"y"1 = "1 $ dv
dx"!
1x + 1
"v = "1
Q(x) = " 1x+1 , N(x) = e"
@(1/(x+1)) dx = 1
1+x , so
d
dx
&v · 1
1 + x
'=
11 + x
$ v
1 + x= ln
88881
1 + x
8888
v = (ln |1 + x|)(1 + x) $ 1y
= (ln |1 + x|)(1 + x)
$ y =1
(x + 1) ln |x + 1|
48. We have xy! " 2x2'y = 4y. Then
dy
dx" 2x
'y =
4x
y
dy
dx+!" 4
x
"y = 2xy1/2 (Bernoulli equation)
Let v = y1"(1/2) = y1/2, so dvdx = 1
2y"1/2 dydx . Then
dv
dx[2y1/2] +
!" 4
x
"y = 2xy1/2
dv
dx+!" 2
x
"v = x
!N(x) =
1x2
"
d
dx
&'y
x2
'=
1x
$'
y
x2= ln |x| + C
$ 'y = x2 ln |x| + Cx2
42 Section 1.5
49. We have xy! + 2y + x5y3ex = 0. Then
xdy
dx+ 2y + x5y3ex = 0
dy
dx+
2x
y = ("x5ex)y3 (Bernoulli equation)
v = y"2 $ dv
dx= "2y"3 dy
dx$ dv
dx
!"1
2y3
"=
dy
dx
dv
dx
!"1
2y3
"+
2x
y = ("x5e"x)y3
dv
dx" 4
xy = (2x5e"x)
Now use N(x) = 1x4 , so d
dx
;v · 1
x4
<= [2xe"x]. Then
v · 1x4
= 2e"x(x " 1) + C
$ v = 2x4e"x(x " 1) + Cx4
$ y =1
x2.
2e"x(x " 1) + C
50. We have xy dy = (y2 + x) dx. Then
dy
dx(xy) = y2 + x $ dy
dx= x"1y + y"1
dy
dx+!" 1
x
"y = y"1 (Bernoulli equation)
Let v = y1+1 = y2, so dvdx
#12y"1
$= dy
dx . Then
dv
dx
!12y"1
"+!" 1
x
"y = y"1 $ dv
dx+!" 2
x
"v = 2
Using N(x) = 1x2 we have d
dx
;vx2
<= 2
x2 , so
v
x2= " 2
x+ C
$ y2 = "2x + Cx2
51. Use the Bernoulli method to solve the logistic equation
dN
dt= rN
!1 " N
K
"
We havedN
dt= rN " r
KN2 $ dN
dt+ ("r)N = " r
KN2
Let v = N"1, so dvdt = "N"2 dN
dt and dvdt ("N2) = dN
dt . Then
dv
dt("N2) + ("r)N = " r
KN2
dv
dt+ rv =
r
K
1.6. CHAPTER 1: ADDITIONAL PROBLEMS 43
Using N = ert gives ddt [vert] = r
K ert, so
vert =1K
ert + C $ v =1k
+ Ce"rt
1N
=1K
+ ce"rt $ N(t) =1
1K + Ce"rt
1.6 Chapter 1: Additional Problems
1. False - It is linear, but no initial condition is given
2. True - y2 makes it non-linear
3. False - y(x) = f(x) is an explicit solution
4. The point is to get linear DE into an exact DE by multiplying by an integrating factor. Soeven if this were true, there would be no reason to do so.
5. False - Solutions can be defined on a restricted portion of the real line
6. True
7. True -dy
f(y)= dx
8. y! = C4x3ex4& 4x3(y " 1) = 4x3(1 + Cex4
" 1) = 4x3(Cex4) = y!
9. y! = Cex " 2x " 2 & x2 + y = x2 + Cex " x2 " 2x " 2= Cex " 2x " 2 = y!
10. y2 =1
C " 2xy3 = ±(C " 2x)"3/2, y = ±
91
C " 2x,
y! = ±(C " 2x)"3/2 = y3. Valid when C %= 2x
11. y! =1
(x + C)2, (y " 3)2 =
!"1
(x + C)
"2
=1
(x + C)2= y!
12. We have 2x2yy! + y2 = 2. Then
2x2ydy
dx= 2 " y2
2y dy
2 " y2=
dx
x2
$%
2y dy
y2 " 2=%
"dx
x2
ln |y2 " 2| =1x
+ C1
|y2 " 2| = eC1e1/x
y2 " 2 = Ce1/x, where C = ±C1
y = ±.
Ce1/x + 2
44 Chapter 1 Review
13. We have y! = 3 3.
y2 with y(2) = 0. Rewrite the equation as y! = 3y2/3. Separate variablesand integrate:
%y"2/3 dy =
%3 dx
3y1/3 = 3x + C
It is easiest to apply the initial conditions now: 3(0)1/3 = 3(2) + C leads to C = "6, so
3y1/3 = 3x " 6y1/3 = x " 2
y = (x " 2)3
14. We have y! " xy2 = 2xy. Then
dy
dx= x(y2 + 2y)
dy
y2 + 2y= xdx (by partial fractions)
% !"1/2y
+1/2
y " 2
"dy =
%xdx
"12
ln |y| + 12
ln |y " 2| =12x2 + C1
ln8888y " 2
y
8888 = x2 + 2C1
88881 " 2y
8888 = e2C1ex2
2y
= 1 ± e2C1ex2
y =2
1 ± e2C1x2
y =2
1 + Cx2where C = ±e2C1
15. We have e"x#1 + dx
dt
$= 1. Then
dx
dte"x = 1 " e"x
e"x
1 " e"xdx = dt
Letting v = 1 " e"x, du = e"x,%
du
u=%
dt
ln |u| = t + C1
|u| = eC1et
1 " e"x = ±eC1et
e"x = ±eC1et " 1"x = ln |±eC1et " 1|
x = " ln |Cet " 1| where C = ±eC1
Chapter 1 Review 45
16. We have xy! + y = y2 with y(1) = .5. Then
xy! = y2 " y $%
dy
y2 " y=%
dx
x
$% !
1y " 1
" 1y
"dy =
%dx
x
$ ln |y " 1|" ln |y| = ln |x| + C $ y " 1y
= Ax
Apply initial conditions:
12 " 1
12
= A(1) $ A = "1 $ 1 " 1y
= "x
Then we have1 + x =
1y$ 1
1 + x= y
17. We have xy! = y " xey/x. Rewriting we have y! = yx " ey/x. Let u = y
x , dydx = xdu
dx + u. Then
xdu
dx+ u = u " e"u
xdu
dx= "e"u $
%eu du =
%" 1
xdx
eu = " lnx + C $ eye1/x = " lnx + C
ey =" lnx + C
e1/x
y = ln!" lnx + C
e1/x
"
18. y! " y
x= (1 +
y
x) ln(1 +
y
x) which is Homogeneous. Let v =
y
x&
v + v!x " v = (1 + v) ln(1 + v)dv
(1 + v) ln(1 + v)=
dx
x
[u = ln(1 + v)]du
u=
dx
x|u| = |x|eC
| ln(1 +y
x)| = |x|eC
ln(1 +y
x) = ±xeC
46 Chapter 1 Review
19. We have xy! " y cos#ln# y
x
$$= 0. Then
yx! = y cos,ln, y
x
--
y! =y
xcos,ln,y
x
--(homogeneous)
dv
dxx + v = v cos(ln(v))
dv
dxx = v(cos(ln(v))) " v
dv
dxx = v(cos(ln(v)) " 1)
1v(cos(ln(v)) " 1)
dv =1x
dx
cot!
12
ln(v)"
= lnx + C
cot!
12
ln,y
x
-"= lnx + C
20. We have (y + 'xy) dx = xdy. We use M(x, y) = y + '
xy and N(x, y) = "x, so
M(tx, ty) = ty +.
t2xy = t(y +'
xy) + tM(x, y)N(tx, ty) = "tx = t("x) = tN(x, y)
We need to solve
dy
dx=
y
x+
'xy
x$ dv
dxx + v = v +
'v
$ dv
dxx =
'v
$ 1'v
dv =1x
dx
$ 2'
v = ln |x| + C
Then'
y = 2'
x ln x + 2C'
x
y = 4x(ln y + C)2
21. We have x dydx =
.x2 " y2 + y. Then
dy
dx=
91 ",y
x
-2+,y
x
-(homogeneous)
$ dv
dxx + v =
.1 " v2 + v
1'1 " v2
dv =1x
dx
$ arcsin(v) = lnx + C
$ arcsin, y
x
-= ln x + C
Chapter 1 Review 47
22.
dy
dx=
4y " 2x
x + y
=4(y/x) " 21 + (y/x)
Homogeneous (y = vx)
v + v!x =4v " 21 + v
1 + v
"v2 + 3v " 2dv =
dx
x
ln8888(1 " v)2
(2 " v)3
8888 = ln |x| + C
&(1 " (y/x))2
(2 " (y/x))3
'= ±eCx
23. We have (y/x) + (y3 + ln x) dy = 0. We find
"
"y
,y
x
-=
1x
"
"x[y3 + ln x] =
1x
so the equation is exact. Then,
"f
"x=
4x
$ f(x, y) = y ln x + #(y)
$ "f
"y= ln x + #!(y) = y3 + ln x
#(y) =14y4 + C
Sof(x, y) = y ln x +
14y4 + C
24. We have!
3x2 + y2
y2
"+!"2x3 " 5y
y3
"dy
dx= 0. Then
"
"y
&3x2 + y2
y2
'=
"6x2
y3
"
"x
&"2x3 " 5y
y3
'=
"6x2
y3
48 Chapter 1 Review
so the equation is exact. Solving, we get
"f
"x=
3x2
y2+ 1
$ f(x, y) =x3
y2+ x + #(y)
$ "f
"y=
"2x3
y3+ #!(y) =
"2x3
y3" 5
y2
$ #!(y) = " 5y2
#(y) =5y
+ C
f(x, y) =x3
y2+ x +
5y
+ C
25. We have 2x(1 +.
x2 " y) dx ".
x2 " y dy = 0. Then
"
"y
I2x + 2x
.x2 " y
J=
"x.x2 " y
"
"x
I".
x2 " yJ
= " 2x
2.
x2 " y=
"x.x2 " y
so the equation is exact. Solving,
"f
"x= 2x + 2x
.x2 " y
$ f(x, y) = x2 +%
2x.
x2 " y dx
$ f(x, y) = x2 +23(x2 " y)3/2 + #(y)
$ "f
"y= "
.x2 " y + #!(y)
= ".
x2 " y
#!(y) = 0 leads to #(y) = C, so
f(x, y) = x2 +23(x2 " y)3/2 + C
26. y! +1x
y =1x2
y"2 which is Bernoulli n = "2, v = y3,dv
dx= 3y2 dy
dx. Then we have
dv
dx+
3x
v =3x2
which is linear with solution v = y3 =32x
+C
x3
$ y = 3
932x
+C
x3
27."M
"y= 2y,
"N
"x= "y " 3x2 & 1
N
!"M
"y" "N"x
"=
3y + 3x2
"x(y + x2)which is only a function of x
which gives µ(x) = x"3 and F (x, y) ="y2
2x2" y + C
Chapter 1 Review 49
28.
dy
dx=
y
x" y2
(Homog) (y = vx)v + v!x = v " v2x2
dv
"v2= xdx
1v
=x2
2+ C
x
y=
x2
2+ C
29. Not linear, exact, homogeneous, Bernoulli, seperable
30. Not linear, exact, homogeneous, Bernoulli, seperable
31. y! +y
x= x3 Linear $ y =
x4
5+
C
x
32. Linear $ y = ex
&"e"x
2(cosx + sin x) + C
'
33. Linear $ y = e"2x2&58e2x2
(2x2 " 1) + C
'
34. Bernoulli n = 4, v = y"3, v! " 3x
v = "3 (linear) v = y"3 =3x
2+ Cx3
35. Bernoulli n = 3, v = y"2, v! + 2v = "2x (linear)
v = y"2 = e"2x
&"e2x
2(2x " 1) + C
'
36. Bernoulli n = 3, v = y"2, v! " xv = "4x (linear)v = y"2 = ex2/2
,4e"x2/2 + C
-
37. T (t) = (T0 " Ts)ekt + Ts = 95ekt + 5, k = ".055 $ soup will be edible in about 34 minutes.
38. T (t) = 50ekt + 70, k = ".105 $ co"ee is drinkable in about 5 minutes.
39. Let y(t) = amount of salt at time t, y(0) = 6
dy
dt= 2 " 6 + 2t
15 + 2t
=24 " 2t
15 + 2t
y =392
ln |2t + 15|" t + C
y(0) = 6 & C = "46.8The tank will fill in 15 minutes
y(15) = 12.43lbs of salt
50 Chapter 1 Review
40. y(t) lbs of sugar at time t, y(0) = 25
dy
dt= p " 25 + pt
100 " t
=100p" 25 " 2pt
100 " t
dy =100p" 25 " 2pt
100 " tdt
y = (25 " 100p) ln |100 " t|"200p ln |t " 100|" 2pt + C
C = 25 " 25 ln 100We want y(95) = 2.5 & p = .074lbs/min
41. y(t) amount of nitric acid, y(0) = 1
dy
dx= 1.2 " 8
!100 + 1.2t
200 " 2t
"
="560" 12t
200 " 2ty(t) = 880 ln |t " 100|+ 6t + 1 " 880 ln100
The solution reaches 10% when y(t) = 20" .2t which is when t = 138 minutes. Since the tankempties in 100 minutes, this will never happen.
42. If (x, y) is the point of tangency, then (x2 , 0) is the point of intersection and
dy
dx=
yx2
=2y
x.
Thendy
y=
2 dx
x$ y = x2 is such a curve.
Chapter 2
Geometrical and NumericalMethods for First-Order Equations
2.1 Direction Fields—the Geometry of Di!erential Equations
1. Looking at the point (2, 1), y! = 92 , which matches graph b.
2. Looking at the point (2, 1), y! = 18, which matches graph c.
3. Looking at the point (2, 1), y! = 29 , which matches graph a.
4. Looking at the point (2, 1), y! = 118 , which matches graph d.
5. Graph B
6. Graph C
7. Graph D
8. Graph A
9. y! = y4 #9 #10
10. y! = cos y
51
52 Section 2.1
11. y! = sin y
12. y! = e"y
13. y! = x4
Section 2.1 53
14. y! = cos x
15. y! = sin x
16. y! = e"x
54 Section 2.1
17. y! = xy
18. y! = x + y
19. y! = (x2 + 1)(y + 1)
Section 2.1 55
20. y! = ex2
21. y! =x " 1
y " 1
22. y! =x2 " 1
y2 + 1
56 Section 2.2
23. y! = y(y2 " 2)
24. xy(x2 + 2)
25. y! =x3(y2 + 1)
y2 + x2
2.2 Existence and Uniqueness for First-Order Equations
1. (a) y! = y2 " x2, y(0) = 0, f = y2 " x2,"f
"y= 2y. The theorem guarantees that a solution
exists and is unique on some interval.
Section 2.2 57
(b) y! = y"2 "x2, y(0) = 0, f =1y2
"x2,"f
"y= "y"3. The theorem does not guarantee that
a solution exists or is unique on some interval.
(c) y! = y +1
1 " x, y(1) = 0, f = y +
11 " x
,"f
"y= 1. The theorem does not guarantee that
a solution exists or is unique on some interval.
2. (a) f(x, y) = 3x(y + 2)2/3,"f
"y=
2x
(y + 2)1/3, which is discontinuous when y = "2, so a
solution exists and is unique everywhere except possibly along y = "2.
(b) f(x, y) = (x " y)1/5,"f
"y=
"15(x " y)4/5
, which is discontinuous when y = x, so solutions
exist and are unique everywhere except possibly along y = x. Alternatively, f(x, y) =(x " y)!
5,"f
"y= "1
5, so solutions exist and are unique everywhere.
(c) f(x, y) = x2y"1,"f
"y= "x2y"2, which are both discontinuous when y = 0, so solutions
exist and are unique everywhere except possibly along y = 0.
(d) f(x, y) = (x + y)"2,"f
"y=
"2(x + y)3
, which are both discontinuous when y = "x, so
solutions exist and are unique everywhere except possibly along y = "x.
3. (a)"f
"y=
43xy("1/3) which is not continuous at y = 0. Hence a solution exists and is unique
everywhere except possibly along y = 0.
Actual solution: y =x6
27. This solution crosses y = 0 when x = 0.
(b)"f
"y= 2x(2/3) which, along with f(x, y), are continuous everywhere. Hence solution exists
and is unique.Actual solution: y = 0.
(c)"f
"y= "2
3xy("5/3) which is not continuous at y = 0. Hence a solution exists and is unique
everywhere except possibly along y = 0.
Actual solution: y =!
56
"(3/5)
x(6/5) This solution crosses y = 0 when x = 0.
(d)"f
"y=
23xy("1/3) which is not continuous at y = 0. Hence a solution exists and is unique
everywhere except possibly along y = 0.
Actual solution: y =x6
216which crosses y = 0 when x = 0.
4. f(x, y) = 2'y is continuous for y ) 0 and"f
"y=
1'
yis continuous for y > 0. Hence, by the
theorem, a solution exists and is unique for y(1) = 3.Actual solution is given by: y = (x +
'3 " 1)2.
5. f(x, y) = 5(y " 2)3/5 and y(0) = 2."y
"y=
3(y " 2)2/5
is discontinuous when y = 2, so solutions
exist and are unique everywhere except possibly along y = 2. Solving for the initial value,
y! = 5(y " 2)3/5 $%
(y " 2)"3/5 dy =%
5 dx
$ 52(y " 2)2/5 = 5x + C
58 Section 2.2
Applying initial conditions,
52(2 " 2)2/5 = 5(0) + C $ C = 0
$ (y " 2)2/5 = 2x $ y " 2 = ±(2x)5/2
$ y = 2 ± (2x)5/2
Solutions passing through (0, 2) are NOT unique.
6."f
"y= x(1/3)("2y) which is continuous everywhere (as is f), therefore, unique solution will
exist everywhere.
7."f
"y= (y " 1)("2/3) which is continuous everywhere except y = 1, so solutions exist and are
unique everywhere except possibly along y = 1.Actual solution: 1 ± 2
'2x(3/2)
8."f
"y= "2xy"3 which is continuous everywhere except y = 0, so solutions exist and are unique
everywhere except possibly along y = 0.
Actual solution:!"32
+x2
2
"1/3
9. f(x, y) and"f
"yare not continuous at y = 1, so solutions exist and are unique everywhere
except possibly along y = 1.Actual solution: y = 1 + 3
'3 " 3 cosx which IS unique.
10."f
"yDNE at y = 0; if y > 0, then "f
"y = 1 and if y < 0, then "f"y = "1. Hence,
"f
"y= lim
h#0
f(x, y + h) " f(x, y)h
DNE. Therefore, the solutions will exist and be unique except
possibly at y = 0.If y ) 0, y = eCex which will never be 0. If y < 0, y = "eCe"x which will never be 0. Thus,the only solution to this DE is y = 0 which IS unique.
2.3. FIRST-ORDER AUTONOMOUS EQUATIONS—GEOMETRICAL INSIGHT 59
2.3 First-Order Autonomous Equations—Geometrical Insight
1. y! = 2y + 3, 2y + 3 = 0
Root y = " 32
Multiplicity 1
End behavior: +y
(i) y " y!
(ii) By the phase line diagram, y = " 32 is an unstable equilibrium point.
(iii) y > " 32 , y & # as x & #
y < " 32 , y & "# as x & #
(iv) xy-plane
60 Section 2.3
2. y! = y2 + 4y + 4, y2 + 4y + 4 = 0, (y + 2)2 = 0
Root y = "2Multiplicity 2
End behavior: y2
(i) y " y!
(ii) By the phase line diagram, y = "2 is a half-stable equilibrium point.(iii) y > "2, y & # as x & #
y < "2, y & "2 as x & #(iv) xy-plane
Section 2.3 61
3. x! = x2 " x " 6 = (x " 3)(x + 2)
(i) y " y!
(ii) x = "2 is a stable equilibriumx = 3 is an unstable equilibrium
(iii) If x0 > 3, then x(t) & # as t & #.If "2 < x0 < 3, then x(t) & "3 as t & #.If x0 < "2, then x(t) & "3 as t & #.
(iv) xy-plane
62 Section 2.3
4. x! = x(x + 2)(x " 3)
Equilibrium Multiplicityx = 0 1
x = "2 1x = 3 1
Highest power and coe!cient: x(+x)(+x) = +x3
(i) y " y!
(ii) x = "2, 3 unstable; x = 0 stable(iii) For x0 ! ("#,"2), x & "# as t & #
For x0 ! ("2, 3), x & 0 as t & #For x0 ! (3,#), x & # as t & #
(iv) xy-plane
Section 2.3 63
5. x! = (x " 2)3(x2 " 9) = (x " 2)3(x + 3)(x " 3)
Equilibrium 2 "3 3Multiplicity 3 1 1
Highest power coe!cient: (x)3(x2) = +x5
(i) y " y!
(ii) x = "3, 3 unstable, x = 2 stable(iii) For x0 ! ("#,"3), x & "# as t & #
For x0 ! ("3, 3), x & 2 as t & #For x0 ! (3,#), x & # as t & #
(iv) xy-plane
64 Section 2.3
6. y! = sin y, "2! < y < 2!sin y = 0, "2! < y < 2!
Roots y = "! y = 0 y = !Multiplicity 1 1 1
End behavior: trigonometric sine wave
(i) y " y!
(ii) By the phase line diagram, y = "! is a stable equilibrium point.y = 0 is an unstable equilibrium point.y = ! is a stable equilibrium point.
(iii) For y > !, y & ! as x & #For 0 < y < !, y & ! as x & #For "! < y < 0, y & "! as x & #For y < "!, y & "! as x & #
(iv) xy-plane
Section 2.3 65
7. y! = e"y2/2 " e"2, e"y2/2 " e"2 = 0. When
"y2
2= "2
"y2 = "4y2 = 4
(i) y " y!
(ii) By the phase line diagram, y = 2 is stabley = "2 is unstable
(iii) For y > 2, y &For "2 < y < 2, y &For y < "2, y &
(iv) xy-plane
66 Section 2.3
8. y! = "y2, "y2 = 0
Root y = 0Multiplicity 2
End behavior: "y2
(i) y " y!
(ii) By the phase line diagram, y = 0 is a half-stable equilibrium point.(iii) For y > 0, y & 0 as x & #.
For y < 0, y & "# as x & #.(iv) xy-plane
Section 2.3 67
9. x! = x2(2 " x)
Equilibrium Multiplicityx = 0 2x = 2 1
Highest power and coe!cient: x2("x) = "x3
(i) y " y!
(ii) x = 0 is a half-stable point; x = 2 is a stable point.(iii) For x ! ("#, 0), x & 0 as t & #.
For x ! (0,#), x & 2 as t & #(iv) xy-plane
68 Section 2.3
10. x! = (x " 1)2(x " 2)3(1 + x)
Equilibrium 1 2 "1Multiplicity 2 3 1
Highest power and coe!cient: (x)2(x)3(+x) = +x6
(i) y " y!
(ii) x = "1 is stable; x = 1 is half-stable; x = 2 is unstable.(iii) If x0 > 2, then x & # as t & #.
If 1 < x0 < 2, then x & 1 as t & #.If "1 < x0 < 1, then x & "1 as t & #.If x0 < "1, then x & "1 as t & #.
(iv) xy-plane
Section 2.3 69
11. y! = cos y + 1, "2! < y < 2!. cos y + 1 = 0, cos y = "1.
Roots y = "! y = !Multiplicity 1 1
End behavior: cosine wave
(i) y " y!
(ii) By the phase line diagram, y = "! is a half-stable equilibrium point. y = ! is a half-stableequilibrium point.
(iii) For y > !, y & # as x & #.For "! < y < !, y & ! as x & #.For y < "!, y & "! as x & #.
(iv) xy-plane
70 Section 2.3
12. x! = x3(2 + x)(5 + x)7(3 " x)(x " 2)4
Equilibrium 0 "2 "5 3 2Multiplicity 3 1 7 1 4
Highest power and coe!cient: x3(+x)(+x)7("x)(x)4 = "x16
(i) y " y!
!!
!!
!!
!
(ii) x = "5, 0 are unstable points; x = "2, 3 are stable points; x = 2 is a half-stable point.(iii) If x0 < "5, then x & "# as t & #.
If "5 < x0 < "2, then x & "2 as t & #.If "2 < x0 < 0, then x & "2 as t & #.If 0 < x0 < 2, then x & 2 as t & #.If 2 < x0 < 3, then x & 3 as t & #.If x0 > 3, then x & 3 as t & #.
(iv) xy-plane
Section 2.3 71
13. v! = g " kmv. Equilibrium satisfies g " (k/m)v = 0, so v =
gm
k
(i) v " v!
!
(ii) v =gm
kis a stable equilibrium point.
(iii) For v ! [0,#), v & gmk as t & #
(iv) Figures will vary. With g = 32, m = .25, k = 2 as in example 2 in 1.3 then
14. v! = g" kmv2. Equilibrium satisfies g" (k/m)v2 = 0, so v2 = gm/k yields v = ±
.gm/k. Note
that this is a free-fall problem where v > 0 in the downward direction. We thus ignore v < 0.
(i) Graphs will vary
(ii)9
gm
kis a stable equilibrium point
(iii) For v ! [0,#), v & +.
gm/k as t & #(iv) Graphs will vary
15. x! = x2(2 " x)(x + 3)2, x2(2 " x)(x + 3)2 = 0
Roots x = "3 x = 0 x = 2Multiplicity 2 2 1
End behavior: x2("x)(x)2 = "x5
72 Section 2.3
(i) x " x!
(ii) By the phase line diagram, x = 2 is a stable equilibrium point; x = 0 is a half-stableequilibrium point; x = "3 is a half-stable equilibrium point.
(iii) For x > 2, & 2 as t & #.For 0 < x < 2, & 2 as t & #.For "3 < x < 0, & 0 as t & #.For x < "3, & "3 as t & #.
(iv) tx-plane
Section 2.3 73
16. x! = (2 " x)3(x2 + 4), (2 " x)3(x2 + 4) = 0
Roots 2 ±2iMultiplicity 3 ignore
End behavior: ("x)3(x2) = "x5
(i) x " x!
(ii) By the phase line diagram, x = 2 is a stable equilibrium point.(iii) For x > 2, & 2 as t & #.
For x < "2, & 2 as t & #.(iv) tx-plane
74 Section 2.3
17. y! = (2 " y)3(y2 + 4)2
Root 2Multiplicity 3
Highest power and coe!cient: ("y)3(y2)2 = "y7
(i) x " x!
(ii) By the phase line diagram, y = 2 is a stable equilibrium point.(iii) If y0 < 2, then y & 2 as x & #.
If y0 > 2, then y & 2 as x & #.(iv) tx-plane
Section 2.3 75
18. y! = "y2(4 " y)(9 " y2)
Roots 0 4 3 "3Multiplicity 2 1 1 1
Highest power and coe!cient: "y2("y)("y2) = "y5
(i) x " x!
(ii) y = "3, 4 are stable; y = 0 is half-stable; y = 3 is unstable.(iii) If y0 < "3, y & "3 as x & #.
If "3 < y0 < 0, y & "3 as x & #.If 0 < x < 3, y & 0 as x & #.If 3 < x < 4, y & 4 as x & #.If x > 4, y & 4 as x & #.
(iv) tx-plane
76 Section 2.3
19. x! = x5(1 " x)(1 " x3), x5(1 " x)(1 " x3) = 0
Roots 0 1 two normalMultiplicity 5 2 ignore
End behavior: x5("x)("x3) = x9
(i) x " x!
(ii) By the phase line diagram, = 1 is a half-stable equilibrium point; = 0 is an unstableequilibrium point.
(iii) For x > 1, & # as t & #.For 0 < x < 1, & 1 as t & #.For x < 0, & "# as t & #.
(iv) tx-plane
Section 2.3 77
20. x! = x(x " 3)(1 + x3)(1 " x2)2, x(x " 3)(1 + x3)(1 " x2)2 = 0
Roots 0 3 "1 two non-real ±1Multiplicity 1 1 1 ignore 2
End behavior: x(x)(x3)("x2)2 = x9
(i) x " x!
(ii) By the phase line diagram, x = 3 is an unstable equilibrium point; x = 1 is a half-stableequilibrium point; x = 0 is a stable equilibrium point; x = "1 is an unstable equilibriumpoint.
(iii) If x > 3, & # as t & #.If 1 < x < 3, & 1 as t & #.If 0 < x < 1, & 0 as t & #.If "1 < x < 0, & 0 as t & #.If x < "1, & "# as t & #.
(iv) tx-plane
78 Section 2.3
21. x! = x2(1 " 2x)3(x2 " 1), x2(1 " 2x)3(x2 " 1) = 0
Roots 0 12 ±1
Multiplicity 2 3 1
End behavior: x2("x)3(x2) = "x7
(i) x " x!
(ii) By the phase line diagram, x = 1 is a stable equilibrium point; x = 12 is an unstable
equilibrium point; x = "1 is a stable equilibrium point.(iii) If x > 1, & 1 as t & #.
If 12 < x < 1, & 1 as t & #.
If "1 < x < 12 , & "1 as t & #.
If x < "1, & "1 as t & #.(iv) tx-plane
Section 2.3 79
22. x! = x3(x2 + 5)(x " 4)2(x + 5)
Roots 0 4 "5Multiplicity 3 2 1
End behavior: x3x2x2x = x8
(i) x " x!
(ii) By the phase line diagram, x = 1 is a stable equilibrium point; x = 12 is an unstable
equilibrium point; x = "1 is a stable equilibrium point.(iii) If x > 1, & 1 as t & #.
If 12 < x < 1, & 1 as t & #.
If "1 < x < 12 , & "1 as t & #.
If x < "1, & "1 as t & #.(iv) tx-plane
80 Section 2.3
23. y& = ±1, f !(y) = "2y, f !("1) = 2 $ Unstablef !(1) = "2 $ Stable
24. y& = ±1, f !(y) = "2y, f !("1) = 2 $ Unstablef !(1) = "2 $ Stable
25. y& = 0, ±!, f !(y) = cos y, f !(0) = 1 $ Unstablef !(±!) = "1 $ Stable
26. y& = "1, f !(y) = 3y2, f !("1) = 3 $ Unstable
27. y& = 0, f !(y) = "3y2, f !(0) = 0 $ Inconclusive. However, from the phase line diagram, y& = 0is stable.
28. If we Taylor expand the function about y& and keep the lowest order non-zero term, we seethat we have y! = f (3)(y&)y3 as the approximate solution near the equilibrium point. Phaseline analysis then shows the equilibrium point is stable.
29. (a) y! = ry " y3
"
"
Figure 2.1: Phase line for y! = ry " y3 : r < 0, r = 0, r > 0
Figure 2.2: Pitchfork bifurcation for y! = ry " y3
Section 2.3 81
(b) y! = ry + y3
"
"
Figure 2.3: Phaseline for y! = ry + y3 : r < 0, r = ", r > 0
Figure 2.4: Pitchfork bifurcation for y! = ry + y3
82 Section 2.4
2.4 Population Modeling: An Application of AutonomousEquations
1. (a) x = 0 is half-stable, x = 1 is stable. For logistic equation, x = 0 is unstable and x = k isstable. Yes, the are di"erent.
(b) For small x, logistic model growth is larger.
2. x = 0 - Stable, x = 1 - Unstable, x = 6 - Stable
3. x = 0 - Stable, x = 2 - Unstable, x = 10 - Stable
4. (a) x = 0 - Half-stable, x = 1 - Unstable, x = 4 - Stable(b) For small x, Allee e"ect model has larger growth rate.
5. (a) Exponential growth - Unlimited growth rate. No limitations placed on organisms.(b) Logistic model - Growth rate dependent on factors such as population amount or food
availability.(c) Allee e"ect - Growth rate dependent on factors such as population amount or food avail-
ability as well as a su!cient population to sustain itself.
6. (a) x = 0 - Unstable, x = a - Stable, x = 5 - Unstable(b) For x0 > 5, bacteria grows uninhibited.(c) The parameter a could represent the strength of the immune system or ability of the
body to fight o" the given bacteria. A healthy person would likely have an a-value that iscloser to 0 than to 5 because the lower value of a represents a lower value of the bacteria(that is stable).
7. (a) For a > 0,x = 0 - Unstablex = 5 "
'a - Stable
x = 5 +'
a - Unstablex = 10 - Stable
(b) For a = 0,x = 0 - Unstablex = 5 - Half-stablex = 10 - Stable
(c) For a < 0,x = 0 - Unstablex = 10 - Stable
(d) Saddle-node(e) Bacteria grows unchecked to a level of 10. No, the bacteria population eventually reaches
and levels o" at 10, which above the fatal level.(f) The parameter a could again represent the strength of the immune system or ability of
the body to fight o" the given bacteria.
8. x! = x(1 " x)(x " 6)(x " 10)
9. x! = x(2x " 1)(x " 1)2(x " 2)(x " 8) or x! = x(2x " 1)2(x " 1)(x " 2)(x " 8)
10. (a) x! = x2(2 " x)2(x " 4)
2.5. NUMERICAL APPROXIMATION WITH THE EULER METHOD 83
(b) x = 0 - Half-stable, x = 2 - Half-stable, x = 4 - Unstable
11. (a) x! = rx(x " a)(x " 1) r, a > 0(b) i. 0 < a < 1, x = 0 - Unstable, x = a - Stable, x = 1 - Unstable
ii. a = 1, x = 0 - Unstable, x = 1 - Half-stableiii. a > 1, x = 0 - Unstable, x = 1 - Stable, x = a - Unstable
(c) When a < 1, the bacteria goes to the stable level a if it starts with a level less than 1and grows without bound otherwise. When a = 1, the bacteria goes to the half-stablelevel a = 1 if it starts with a level less than 1 and grows without bound otherwise. Whena > 1, the bacteria goes to the stable level of 1 if it starts with a level less than a andgrows without bound otherwise.
2.5 Numerical Approximation with the Euler Method
1. dy/dx = x3, y(1) = 1; explicit solution: y = 14 (x4 + 3)
Euler Explicitxi yi y(xi)1.0 1 11.1 1.1 1.11601.2 1.2331 1.26841.3 1.4059 1.46401.4 1.6256 1.7104
2. dy/dx = x4y, y(1) = 1; explicit solution: y = e(x5"1)/5
Euler Explicitxi yi y(xi)1.0 1 11.1 1.1 1.12991.2 1.2611 1.34671.3 1.5225 1.72051.4 1.9574 2.4004
3.dy
dx= "y2 cosx, h = 0.1, y(0) = 1; explicit solution y =
11 + sinx
.
x0 = 0x1 = 0 + (0.1)(1) = 0.1x2 = 0 + (0.1)(2) = 0.2x3 = 0 + (0.1)(3) = 0.3x4 = 0 + (0.1)(4) = 0.4y0 = 1y1 = 1 + (0.1)("(1)2 cos(0)) = .9y2 = .9 + (0.1)("(.9)2 cos(0.1)) = .8194y3 = .8194 + (0.1)("(.8194)2 cos(0.2)) = .7536y4 = .7536 + (0.1)("(.7536)2 cos(0.3)) = .6993
84 Section 2.5
Explicit:
y(0) =1
1 + sin 0= 1
y(0.1) =1
1 + sin(0.1)= .909228
y(0.2) =1
1 + sin(0.2)= .834258
y(0.3) =1
1 + sin(0.3)= .7718907
y(0.4) =1
1 + sin(0.4)= .719725
Euler Explicitxi yi y(xi)0.0 1 10.1 .9 .9092280.2 .8194 .8342580.3 .7536 .77189070.4 .6993 .719725
4. y! =sin x
y3, y(!) = 2, h = .1.
x0 = !
x1 = ! + .1x2 = ! + .2x3 = ! + .3x4 = ! + .4y0 = 2
y1 = 2 + (.1)!
sin(!)23
"= 2
y2 = 2 + (.1)!
sin(! + .1)23
"= 1.99875
y3 = 1.995 + (.1)!
sin(! + .2)(1.995)3
"= 1.99626
y4 = 1.985 + (.1)!
sin(! + .3)(1.985)3
"= 1.99255
Explicit:
y = (12 " 4 cosx)1/4
y(!) = 2y(! + .1) = (12 " 4 cos(! + .1))1/4 = 1.99937y(! + .2) = (12 " 4 cos(! + .2))1/4 = 1.99750y(! + .3) = (12 " 4 cos(! + .3))1/4 = 1.99439y(! + .4) = (12 " 4 cos(! + .4))1/4 = 1.99006
Section 2.5 85
Euler Explicitxi yi y(xi)! 2 2
! + .1 2 1.99937! + .2 1.99875 1.99750! + .3 1.99626 1.99439! + .4 1.99255 1.99006
5. dy/dx = ye"x, y(0) = 1; explicit solution: y = exp(1 " e"x)Euler Explicit
xi yi y(xi)1.0 1 11.1 1.1 1.09981.2 1.1995 1.19871.3 1.2977 1.29591.4 1.3939 1.3905
6. dy/dx = xy2, y(0) = 1; explicit solution: y =2
2 " x2
Euler Explicitxi yi y(xi)0.0 1 10.1 1.1 1.00500.2 1.01 1.02040.3 1.0304 1.04710.4 1.0623 1.08700.5 1.1074 1.14290.6 1.1687 1.21950.7 1.2507 1.32450.8 1.3602 1.4706
7. dy/dx = y + cosx, y(0) = 0; explicit solution: y = 12 (sin x " cosx + ex)
Euler Explicitxi yi y(xi)0.0 0 00.1 0.1 0.10500.2 0.2095 0.22000.3 0.3285 0.34500.4 0.4568 0.48010.5 0.5946 0.62530.6 0.7418 0.78070.7 0.8986 0.94660.8 1.0649 1.1231
8. dy/dx = y + sin x, y(0) = 2; explicit solution: y = "12 (cosx + sin x " 5ex)
86 Section 2.5
Euler Explicitxi yi y(xi)0.0 2 20.1 2.2 2.21550.2 2.4300 2.46410.3 2.6929 2.74920.4 2.9917 3.07430.5 3.3298 3.44330.6 3.7107 3.86030.7 4.1383 4.32990.8 4.6165 4.8568
9. dy/dx = e"y, y(0) = 2; explicit solution: y = ln(x + e2)Euler Explicit
xi yi y(xi)0.0 2 20.1 2.0135 2.01340.2 2.0269 2.02670.3 2.0401 2.03980.4 2.0531 2.05270.5 2.0659 2.06550.6 2.0786 2.07810.7 2.0911 2.09050.8 2.1034 2.1028
10. dy/dx = x + y, y(0) = 0; explicit solution: y = "x " 1 + ex
Euler Explicitxi yi y(xi)0.0 0 00.1 0 0.00520.2 .01 0.02140.3 .031 0.04990.4 .0641 0.09180.5 .1105 0.14870.6 .1716 0.22210.7 .2487 0.31380.8 .3436 0.4255
11. dy/dx = (x + 1)(y2 + 1), y(0) = 0; explicit solution: y = tan(12x2 + x)
Euler Explicitxi yi y(xi)0.0 0 00.1 0.1 0.10540.2 0.2111 0.22360.3 0.3364 0.35940.4 0.4812 0.52060.5 0.6536 0.72150.6 0.8677 0.98930.7 1.1481 1.38370.8 1.5422 2.0660
2.6. NUMERICAL APPROXIMATION WITH THE RUNGE-KUTTA METHOD 87
12. a. x! = 1.7x#1 " x
1000
$, b. x! = 2.7x
#1 " x
1000
$
h (a.) x(1) (b.) x(1)1 100 100
0.5 100.425 100.6750.25 100.637 101.0120.1 100.765 101.214
2.6 Numerical Approximation with the Runge-Kutta Method
1. dy/dx = x3, y(1) = 1; explicit solution: y = 14 (x4 + 3)
Runge-Kutta Explicitxi yi y(xi)1.0 1 11.1 1.1160 1.11601.2 1.2684 1.2684
2. dy/dx = x4y, y(1) = 1; explicit solution: y = e(x5"1)/5
Runge-Kutta Explicitxi yi y(xi)1.0 1 11.1 1.1299 1.12991.2 1.3467 1.3467
3. y! = "y2 cosx, h = .1, y(0) = 1. We have x0 = 0, x1 = .1, x2 = .2.y1:
k1 = f(0, 1) = "(1)2 cos 0 = "1
k2 = f
!0 + .05, 1 +
(.1)("1)2
"= "(.95)2 cos(.05) = ".901372
k3 = f
!0 + .05, 1 +
(.1)2
(".902499)"
= ".9107543
k4 = f(0 + .1, 1 + (.1)(".911786)) = ".8220166
y1 = 1 +(.1)6
("1 + 2(".901372) + 2(".910754) + (".8220166))
= .9092288
y2:
k1 = ".822567k2 = ".745136k3 = ".751797k4 = ".68177
$ y2 = .8342587
4. y! =sin x
y3, h = .1, y(!) = 2
88 Section 2.6
y1:
k1 = 0k2 = ".049979k3 = ".0503557k4 = ".101356889
$ y1 = .994966
y2:
k1 = ".101356k2 = ".15405988k3 = ".1552968k4 = ".211447
$ y2 = .9794409
5. dy/dx = ye"x, y(0) = 1; explicit solution: y = exp(1 " e"x)Runge-Kutta Explicit
xi yi y(xi)1.0 1 11.1 1.0998 1.09981.2 1.1987 1.1987
6. dy/dx = xy2, y(0) = 1; explicit solution: y =2
2 " x2
Runge-Kutta Explicitxi yi y(xi)0.0 1 10.1 1.0050 1.00500.2 1.0204 1.02040.3 1.0471 1.04710.4 1.0870 1.08700.5 1.1429 1.14290.6 1.2195 1.21950.7 1.3245 1.32450.8 1.4706 1.4706
7. y! = y + cosx, y(0) = 0, y = 12 (sin x " cosx + ex)
x Runge-Kutta Explicit0 0 0.1 .1049999 .1050000.2 .22000 .22000.3 .34502 .34502.6 .78071 .78071.7 .946563 .946564.8 1.123094 1.123095.9 1.310658 1.3106601.0 1.509723 1.509725
8. dy/dx = y + sin x, y(0) = 2; explicit solution: y = "12 (cosx + sin x " 5ex)
Section 2.6 89
Runge-Kutta Explicitxi yi y(xi)0.0 2 20.1 2.2155 2.21550.2 2.4641 2.46410.3 2.7492 2.74920.4 3.0743 3.07430.5 3.4433 3.44330.6 3.8603 3.86030.7 4.3299 4.32990.8 4.8568 4.8568
9. dy/dx = e"y, y(0) = 2; explicit solution: y = ln(x + e2)Runge-Kutta Explicit
xi yi y(xi)0.0 2 20.1 2.0134 2.01340.2 2.0267 2.02670.3 2.0398 2.03980.4 2.0527 2.05270.5 2.0655 2.06550.6 2.0781 2.07810.7 2.0905 2.09050.8 2.1028 2.1028
10. y! = x + y, y(0) = 0, y = "x " 1 + ex
x Runge-Kutta Explicit.1 .00517 .00517.2 .02140 .02140.5 .14872 .14872.6 .22211 .22211.7 .31375 .31375.8 .42554 .42554.9 .55960 .559601.0 .71828 .71828
11. dy/dx = (x + 1)(y2 + 1), y(0) = 0; explicit solution: y = tan(12x2 + x)
Runge-Kutta Explicitxi yi y(xi)0.0 0 00.1 0.1054 0.10540.2 0.2236 0.22360.3 0.3594 0.35940.4 0.5206 0.52060.5 0.7215 0.72150.6 0.9893 0.98930.7 1.3837 1.38370.8 2.0660 2.0660
90 Section 2.6
12. y! = e"x2
13. y! = e"x2
14. y! = x3ey + 3x2 sin y
15. y! = e"x2
16. y! = e"x2
2.7. AN INTRODUCTION TO AUTONOMOUS SECOND-ORDER EQUATIONS 91
17.
h (a.) x(1) (b.) x(1)1 100.75 101.249
0.5 100.75 101.2490.25 100.75 101.2490.1 100.75 101.249
2.7 An Introduction to Autonomous Second-Order Equa-tions
1. y = cekx, y! = kcekx, y!! = k2cekx $yy!! = cekx + k2cekx = c2k2e2kx = (y!)2
2. If v = $!, then v =C + g cos $
mand
tanh"1
!(C"g) tan(#/2)'
g2"C2
"
.g2 " C2
=t
m+ K
4. (a) If u(x) =dy
dx, then we have u! =
.1 + u2.
(b) This is separable, so%
du'1 + u2
= sinh"1(u) = x " x0
$ u =e(x"x0) " e"(x"x0)
2for some constants c, d.
(d) Then, if u =dy
dx=
e(x"x0) " e"(x"x0)
2= sinh(x " x0),
then y = cosh(x " x0) + C.
2.8 Chapter 2: Additional Problems
1. False. It may not be unique.
2. False. RK use four function evaluations to calculate the next step.
3. False. Euler’s method is not superior.
4. False. Neither f nor "f"y is continuous everywhere.
5. True.
6. True. Phase line analysis will work and gives information on long-term behavior.
8. (a) (i) Solutions exist everywhere; (ii) solutions are unique everywhere except possibly alongy = 0.
(b) (i) Solutions exist everywhere; (ii) solutions are unique everywhere.(c) (i) Solutions exist everywhere except possibly when y = !
2 ± n! for n = 0, 1, 2, · · · ; (ii)solutions are unique everywhere except possibly when y = !
2 ± n! for n = 0, 1, 2, · · · .(d) (i) Solutions exist everywhere except possibly along x = "1; (ii) solutions are unique
except possibly along x = "1.
92 Chapter 2 Review
17. y! = 1"y2
2x , y(1) = !.xi RK:yi Euler:yx
1.0000 3.1416 3.14161.1000 2.7742 2.69811.2000 2.5144 2.41271.3000 2.3210 2.21181.4000 2.1713 2.06211.5000 2.0522 1.94591.6000 1.9550 1.85311.7000 1.8743 1.77701.8000 1.8061 1.7135
20. y! = y"y2
x+1 , y(2) = 0.All answers are 0 because y0 = 0 is an equilibrium point.
23. y! = 1"y2
2x , y(1) = !.xi RK:yi Euler:yx
1.4142 1.0000 1.00001.5142 1.1251 1.12071.6142 1.2592 1.24951.7142 1.4030 1.38681.8142 1.5571 1.53321.9142 1.7222 1.68912.0142 1.8992 1.85522.1142 2.0889 2.03212.2142 2.2924 2.2207
26. y = Cx2e"3/x
27. y = 0, (x±C)2
4x2
30. y = ln,
"1ln(x"2)+C
-
32. y = ±1'1+Cex2
33. y = x"lnx+C(x"1)2
34.
36.
38. y = ±'
6 sin3 x+9C3 sin x
41. y2/3 " x3 " 1
6 " Ce2x
42. y = ±'
e!2x(x2+C)e!2x
Chapter 3
Elements of Higher-Order LinearEquations
3.1 Some Terminology
1. By Theorem 3.1.1, a unique solution exists (a0(x) = 1 %= 0). y = sin x is the solution.
2. a0(x) = x2 = 0 when x = 0, so Thm 3.1.1 does not apply.For the given initial conditions, y = c2x
2 is a solution. (Notice it is not unique).
3. a0(x) = x2 = 0 when x = 0, so Thm 3.1.1 does not apply.y(0), y!(0) are undefined, so there is no solution.
4. a0(x) = x2 = 0 when x = 0, so Thm 3.1.1 does not apply.Any constants c1 and c2 will work.
5. a0(x) = 1 %= 0, so Thm 3.1.1 guarantees a unique solution does exist.y = 2e2x + 2e"2x is the solution.
6. (a) x ! ("#,#)
(b) x %= k!
2, k ! Z
(c) x %= 0(d) x %= 0
7. (a) x ! ("#,#)(b) x > 0(c) x ! ("#,#)(d) x %= 1
93
94 Section 3.2
3.2 Essential Topics from Linear Algebra
1. Any solution of y! + p(x)y = 0 looks like y = Ce"@
p(x) dx. If y1 and y2 are solutions, they areof the form y1 = c1e"
@p(x) dx and y2 = c2e"
@p(x) dx. Then
W (x) = det&
c1e"@
p(x) dx c2e"@
p(x) dx
"p(x)c1e"@
p(x) dx "p(x)c2e"@
p(x) dx
'
= e@
p(x) dx · det&
c1 c2
"c1p(x) "c2p(x)
'
= 0 $ linearly dependent
2. {x, 2x}. c1x + c2(2x) = 0. Choose c1 = "2, c2 = 1. This will hold for all x on [0, 1].
3. {e2x, xe2x}. By definition,
c1e2x + c2xe2x = 0 $ e2x(c1 + c2x) = 0
$ c1 + c2x = 0$ c1 = c2 = 0$ linearly independent
Using the Wronskian,
det&
e2x xe2x
2e2x e2x(1 + x)
'
x=0
= det&
1 02 1
'= 1 $ linearly independent
4. c1x + c2 sinx = 0. Then
d
dx(c1x + c2 sin x = 0) = c1 + c2 cosx = 0
We can solve for c1 and c2 as long as
W (x) =8888
x sin x1 cosx
8888
is not zero. Choose x = !4 . Then
W,!
4
-=
88888
!4 sin !
4
1 cos !4
88888 ='
22
,!4" 1-%= 0
Thus {x, sin x} are linearly independent.
5. By definition, c1ex + c2(x + 1) = 0. Then
x = 0 $ c1 + c2 = 0x = "1 $ c1e
"1 = 0
We then havec1 = 0 $ c2 = "c1 $ c2 = 0
Thus c1 = c2 = 0. Taking the determinant,
W (ex, x + 1) =8888
ex x + 1ex 1
8888 = ex + ex(x + 1) = ex(x + 2) %= 0 if x %= "2.
Therefore {ex, x + 1} is linearly independent.
Section 3.2 95
6. Using the Wronskian,
W (x) =8888
ex ex+5
ex ex+5
8888$ W (x) = e2x+5 " e2x+5 = 0
Therefore the set is linearly dependent. (Take c2 = "c1e5 )
7. Using the Wronskian,
W (x) = det&
sin 2x cos 2x2 cos 2x "2 sin2x
'$ W (0) = det
&0 12 0
'= "2 %= 0.
Therefore {sin 2x, cos 2x} is linearly independent.
8. c1(x3 " 4) + c2x + c33x = 0. Grouping like terms and choosing c1 = 0, c2 = "3, and c3 = 1shows the set is linearly dependent.
9. {x3 " 4x, x, 2x3}. By definition,
c1(x3 " 4x) + c2x + c3(2x3) = 0x3(c1 + 2c3) + x("4c1 + c2) = 0
Choosing c1 = 2, c3 = "1, and c2 = 8 shows the set is linearly dependent. Using the Wronskianwill not work because the set is linear dependent.
10. Letting c1 = 2, c2 = 8, c3 = "1 shows the set is linearly dependent.
11. Letting c1 = 1, c2 = "2, c3 = 2 shows the set is linearly dependent.
12. Letting c1 = 4, c2 = "1.5, c3 = "1 shows the set is linearly dependent.
13. Letting c1 = 1, c2 = "1, c3 = "3 shows the set is linearly dependent.
14. W (x) = 2 $ linearly independent.
15. W (x) = 2x3 " 3x2 & W (1) = "1 %= 0 $ linearly independent.
16. W (x) = 2 $ linearly independent.
17. Letting c1 = 2, c2 = "1, c3 = "2 shows the set is linearly dependent.
18. Letting c1 = 0, c2 = e"4, c3 = "1 shows the set is linearly dependent.
19. Letting c1 = 0, c2 = 1, c3 = e7 shows the set is linearly dependent.
20. W (x) = ex ("5 cosx cos 2x + 3 cos 2x sinx " 4 sinx sin 2x) & W (0) %= 0 $ linearly independent.
21. Letting c1 = 1, c2 = "1, c3 = "1, c4 = "1 shows the set is linearly dependent.
22. (a) On [0, 1], |x| = x. Then c1x+ c2x = 0 when c1 = "c2 shows the set is linearly dependent.(b) On ["1, 0], |x| = "x. Then c1x+c2("x) when c1 = c2 shows the set is linearly dependent.(c) On ["1, 1], we need c1x + c2|x| = 0. This can only happen when c1 = "c2 and when
c1 = c2. These are both true only if c1 = c2 = 0. Thus the set is linearly independent on["1, 1].
(d) W (x) = det&
x "x1 "1
'= 0 on ["1, 0]
96 Section 3.2
W (x) = det&
x x1 1
'= 0 on [0, 1]
23. (a) On [0, 1], |x|3 = x3. Then c1x3 + c2x3 = 0 when c1 = "c2 shows the set is linearlydependent.
(b) On ["1, 0], |x|3 = "x3. Then c1x3 + c2("x)3 when c1 = c2 shows the set is linearlydependent.
(c) On ["1, 1], we need c1x3 + c2|x|3 = 0. This can only happen when c1 = "c2 and whenc1 = c2. These are both true only if c1 = c2 = 0. Thus the set is linearly independent on["1, 1].
(d) W (x) = det&
x3 "x3
1 "1
'= 0 on ["1, 0]
(e) W (x) = det&
x3 x3
1 1
'= 0 on [0, 1]
24. y = c1x + c2x ln x, y! = c1 + c2(1 + lnx), y!! =c2
xx2y!! " xy! + y = c2x " c1x " c2x " c2x ln x + c1x + c2x ln x = 0Letting c1 = 3, c2 = "4 satifies the IC. This does not violate theorem since a0(x) %= 0 on thespecified interval.
25. y = c1 + c2x2, y! = 2c2x, y!! = 2c2
xy!! " y! = x(2c2) " 2c2x = 0Theorem DOES guarantee existance of unique solution on (0,#) since a0(x) %= 0 on this
inteval. y =x2
2is the solution that satisfies IC.
26. y = c1x2 + c2x, y! = 2c1x + c2, y!! = 2c1
x2y!! " 2xy! + 2y = 2x2(c1) " 2x(2c1x2 + c2) + 2(c1x
2 + c2x) = 0Theorem does NOT guarantee existance of unique solution on (0,#) since a0(x) = 0 at x = 0.y = c1x
2 + x is the general solution that satisfies IC (not unique).
27. There are two functions and the equation is 2nd order. Both functions are solutions andW (x) = "17 %= 0 $ functions are linearly independent. Hence, the set is a fundamental solu-tion set.
28. There are two functions and the equation is 2nd order. Both functions are solutions andW (x) = 4 %= 0 $ functions are linearly independent. Hence, the set is a fundamental solutionset.
29. There are two functions and the equation is 2nd order. We have
(e"2x)!! " (e"2x)! " 6(e"2x) = 4e"2x + 2e"2x " 6e"2x = 0
and(e3x)!! " (e3x)! " 6(e3x) = 0
so both are solutions to the ODE. Taking the determinant,
det&
e"2x e3x
"2e"2x 3e3x
'
x=0
= det&
1 1"2 3
'= 3 " ("2) = 5 %= 0
Therefore they are linearly independent, and {e"2x, e3x} forms a fundamental set of solutions.
Section 3.2 97
30. There are two functions and the equation is 2nd order. Both functions are solutions andW (x) = 3e7x %= 0 $ functions are linearly independent. Hence, the set is a fundamental solu-tion set.
31. Let y1 = sin(2x), y2 = cos(2x). Then
y!1 = 2 cos(2x), y!!
1 = "4 sin(2x), y!2 = "2 sin(2x), y!!
2 = "4 cos(2x)
There are 2 functions and the ODE is 2nd order. Substitution shows that y!!1 + 4y1 = 0 and
y!!2 + 4y2 = 0. Thus both are solutions to the ODE. Taking the determinant,
det&
sin 2x cos 2x2 cos 2x "2 sin 2x
'
x=0
= det&
0 12 0
'= "2 %= 0
Therefore they are linearly independent, and the equations form a fundamental set of solutions.
32. We have
y1 = ex $ y!1 = ex, y!!
1 " ex
y2 = xex, y!2 = (1 + x)ex, y!!
2 = (2 + x)ex
Then
y!!1 " 2y!
1 + y1 = ex " 2ex + ex = 0y!!2 " 2y!
2 + y2 = ex(2 + x) " 2(1 + x)ex + xex = 0
so both ex and xex are in the solution space of the ODE. Using the Wronskian,
W (x) = det&
ex xex
ex (1 + x)ex
'$ W (0) = det
&1 01 1
'= 1 %= 0
so they are linearly independent. Because we have two functions in a two-dimensional spaceand they are linearly independent, it follows that they span. Thus {ex, xex} is a basis.
33. We have
y1 = e"3x, y!1 = "3e"3x, y!!
1 = 9e"3x
y2 = ex/2, y!2 =
12ex/2, y!
3 =14ex/2
Then
2y!!1 + 5y!
1 " 3y1 = 2(9e"3x) + 5("3e"3x) " 3(e"3x)= (18 " 15 " 3)e"3x
= 0
2y!!2 + 5y!
2 " 3y2 = 2!
14ex/2
"+ 5!
12ex/2
"" 3(ex/2)
=!
12
+52" 3"
ex/2 = 0
Thus both e"3x and ex/2 are in the solution space of the ODE. Using the Wronskian,
W (x) = det&
e"3x ex/2
"3e"3x 12ex/2
'$ W (0) = det
&1 1"3 1
2
'%= 0
Thus they are linearly independent. Because we have two functions in a two-dimensional spaceand they are linearly independent, it follows that they span. Thus {e"3x, ex/2} is a basis.
98 Section 3.2
34. There are three functions and the equation is 3rd order. All functions are solutions andW (x) = 1 %= 0 $ functions are linearly independent. Hence, the set is a fundamental solutionset.
35. There are three functions and the equation is 3rd order. All functions are solutions andW (x) = "2e"3x %= 0 $ functions are linearly independent. Hence, the set is a fundamentalsolution set.
36. There are three functions and the equation is 3rd order. All functions are solutions andW (x) = 8 cos4 x + 8 cos2 x sin2 x %= 0 $ functions are linearly independent. Hence, the set is afundamental solution set.
37. No. y = x2 + 4 is not a solution.
38. There are two functions and the equation is 2nd order. All functions are solutions andW (x) = 3e3x %= 0 $ functions are linearly independent. Hence, the set is a fundamental solu-tion set.
39. NO. y = x"1 is not a solution.
40. There are two functions and the equation is 2nd order. All functions are solutions andW (x) = "15 %= 0 $ functions are linearly independent. Hence, the set is a fundamental solu-tion set.
41. There are two functions and the equation is 2nd order. All functions are solutions andW (x) = "10 %= 0 $ functions are linearly independent. Hence, the set is a fundamental solu-tion set.
42. No. cos 3x is not a solution.
43. No, because sinx is not a solution to y!! " y = 0.
44. There are two functions and the equation is 2nd order. All functions are solutions andW (x) = 6x %= 0 on interval $ functions are linearly independent. Hence, the set is a fun-damental solution set.
45. We have {x2, x, e"x} and the ODE y(4)(x)"y!!!(x) = 0. The number of possible solutions doesnot match the order of the ODE (4th order), thus {x2, x, e"x} does not form a fundamentalset of solutions for y(4)(x) " y!!!(x) = 0.
46. No. e"3x is not a solution.
47. There are two functions and the equation is 2nd order. All functions are solutions andW (x) = e"2x %= 0 $ functions are linearly independent. Hence, the set is a fundamental solu-tion set.
48. There are two functions and the equation is 2nd order. All functions are solutions andW (x) = e4x %= 0 $ functions are linearly independent. Hence, the set is a fundamental so-lution set.
49. There are three functions and the equation is 3rd order. We have
(x)!!! " (x)!! = 0, (3)!!! " (3)!! = 0, (ex)!!! " (ex)!! = 0
so all are solutions. Taking the determinant,
det
K
Lx 3 ex
1 0 ex
0 0 ex
M
N
x=0
= det
K
L0 3 11 0 10 0 1
M
N = "3 det&
1 10 1
'= 3 %= 0
Section 3.2 99
so they are linearly independent. Thus {x, 3, ex} form a fundamental set of solutions.
50. No. It is a 2nd degree ODE with 3 functions.
51. There are two functions and the equation is 2nd order. All functions are solutions andW (x) = 2e6x %= 0 $ functions are linearly independent. Hence, the set is a fundamental solu-tion set.
52. There are two functions and the equation is 2nd order. All functions are solutions andW (x) = "e"4x %= 0 $ functions are linearly independent. Hence, the set is a fundamentalsolution set.
53. No. W (x) = 0.
54. There are four functions and the equation is 4th order. All functions are solutions andW (x) = "16 %= 0 $ functions are linearly independent. Hence, the set is a fundamental solu-tion set.
55.&
xy
'= c1
&0"1
'+ c2
&11
'$&
xy
'=&
0 1"1 1
' &c1
c2
'. Because the determinant is
nonzero, we can always solve this system for c1, c2 given any x, y. Thus, the setO!
0"1
",
!11
"4
spans R2. The functions are linearly independent because c1
&0"1
'+ c2
&11
'=&
00
'only
when c1 = c2 = 0. Thus they form a basis for R2.For the solutions to 56-73, the reader is referred to Thm 5.4.1, which says (in ourneed) that if detA %= 0 then the column vectors of A form a basis.
56. det&
2 "1"1 1
'= 1
57. det& '
2 00
'5
'=
'10
58. det&
10 31 2
'= 17
59. det
K
L1 1 11 1 01 0 0
M
N = "1
60. det
K
L2 1 01 0 "2"1 1 0
M
N = 6
61. det
K
PPL
1 0 3 10 1 2 00 0 0 22 "1 1 0
M
QQN = 6
62. det
K
PPL
1 "1 1 "10 1 "1 00 0 1 "20 0 0 2
M
QQN = 6
100 Section 3.2
63. Yes,detA = 3
64. No, detA = 0
65. No, 3 vectors for R2
66. Yes, detA = "3
67. No, 3 vectors for R2
68. No, the vectors are not 3 dimensional.
69. No, only 3 vectors for R4
70. No, only 3 vectors for R4
71. Yes, detA = "3
72. No, detA = 0
73. Yes, detA = "2
74. We need to show that R2 satisfies Definition 3.5. So, let %u =!
u1
u2
", %v =
!v1
v2
"and
%w =!
w1
w2
"! R2, u1, u2, v1, v2, w1, w2 ! R
(a) %u + %v =!
u1 + v1
u2 + v2
"! R2
(b) c%v =!
cv1
cv2
"! R2
(c) %u + %v =!
u1 + v1
u2 + v2
"=!
v1 + u1
v2 + u2
"= %v + %u
(d) %u + (%v + %w) =!
u1 + (v1 + w1)u2 + (v2 + w2)
"=!
(u1 + v1) + w1
(u2 + v2) + w2
"= (%u + %v) + %w
(e) Let %0 =!
00
". Then %u +%0 =
!u1 + 0u2 + 0
"=!
0 + u1
0 + u2
"= %0 + %u = %u
(f) Let "%u =!
"u1
"u2
". Then %u + ("%u) =
!u1
u2
"+!
"u1
"u2
"
=!
u1 " u1
u2 " u2
"=!
00
"= %0
(g) c(%u + %v) =!
c(u1 + v1)c(u2 + v2)
"=!
cu1 + cv1
cu2 + cv2
"= c%u + c%v
(h) (c1 + c2)%u =!
(c1 + c2)u1
(c1 + c2)u2
"=!
cu1 + c2u1
c1u2 + c2u2
"
=!
c1u1
c2u2
"+!
c2u1
c2u2
"= c1%u + c2%u
(i) c1(c2%u) =!
c1(c2u1)c1(c2u2)
"=!
(c1c2)u1
(c1c2)u2
"= (c1c2)%u
(j) 1%u =!
1u1
1u2
"=!
u1
u2
"= %u
3.3. REDUCTION OF ORDER—THE CASE OF N = 2 101
75. Same as 74., except with 3D vectors instead of 2D.
76. Same as 74, except with 2 , 2 matrices.
77. Label the matrices A1 · · ·A4. It is clear that
c1A1 + c2A2 + c3A3 + c4A4 =!
c1 c2
c3 c4
"spans R2'2 and it is also clear that
c1A1 + c2A2 + c3A3 + c4A4 =!
c1 c2
c3 c4
"=!
0 00 0
"only if
c1, c2, c3, c4 all are 0. Hence, they are linearly independent and so they form a basis.
78. All vectors in this subset are of the form
1
3xy0
5
7. Set x = y = 0. Thus
1
3000
5
7 is in this
subset W = set of all vectors of the form
1
3xy0
5
7. Let
1
3x1
y1
0
5
7 and
1
3x2
y2
0
5
7 be in W . Then
1
3x1
y1
0
5
7+
1
3x2
y2
0
5
7 =
1
3x1 + x2
y1 + y2
0
5
7
is also in W . For any c ! R, c
1
3xy0
5
7 =
1
3cxcy0
5
7 is still in W . Thus
W is a subspace of R2.
79. Repeat argument for 78, changing all the vectors to
1
30yz
5
7.
80. Let W be the set of planes passing through the origin. Since each plane passes through the
origin,
1
3000
5
7 is a member of each element of W . Also, if %u and %v pass through the origin,
then %u + %v must pass through the origin. Hence, W is closed under addition. Finally, if %upasses through the origin, then any scalar multiple of %u will also pass through the origin, soc%u ! W . So W is a subspace of R3.
3.3 Reduction of Order—The Case of n = 2
FOR EXERCISES 1-21, THE GENERAL SOLUTION IS GIVEN BYy = v + p(x) where p(x) is the particular solution given in the exercises.
1. y = vex
y!! " 2y! + y = v!! = 0 & v = Cx + K
2. y = ve"x/2
4y!! + 4y! + y = 4v!! = 0 & v = Cx + K
3. y = v cos 5xy!! + 25y = v!! cos 5x " 10 sin 5xv! = 0, w = v!, w! = v!!
102 Section 3.3
& dw
w= 10 tan 5x & ln w = ln(cos"2 5x) + C & w = v! =
C
cos2 5xand
thus v =C
5tan 5x + K.
4. y = ve3x
y!! " 9y = v!! + 6v! = 0 $ v!!
v!= "6 & ln v! = "6x + C & v! = e"6x+C
& v ="16
e"6x+C + K
5. y = ve3x
y!! " 6y! + 9y = v!! = 0 & v! = C & v = Cx + K
6. y = ve4x
y!! " 8y! + 16y = v!! = 0 & v! = C & v = Cx + K
7. y = vex/2
2y!! + 5y! " 3y = 2v!! + 7v! = 0 & v!!
v!= "7
2& v! = e"
72 x+C
& v = "27e"
72 x+C + K
8. y = vx2
xy!! " y! = x3v!! + 3x2v! = 0 & v!!
v!=
"3x
& ln v! = ln x"3 + C
& v! =C
x3& v =
C
x2+ K
9. y = v ln x
xy!! + y! = v!! + v!!
2 + lnx
x ln x
"= 0 & v!!
v!= "
!2 + lnx
x ln x
"
& v! =C
x ln2 x& v =
C
ln x+ K
10. y = vx3
x2y!! " xy! " 3y = x5v!! + 5x4v! = 0 & v!!
v!=
"5x
& ln v! = ln x"5 + C
& v! =C
x5& v =
C
x4+ K
11. y = vx ln x
x2y!! " xy! + y = x3 ln xv!! + v!(2x2 + x2 ln x) = 0 & v!!
v!=
"(2x2 + x2 ln x)x3 ln x
& ln v! = ln x"1 + ln!
1ln2 x
"+ C & v! =
C
x ln2 x
& v =C
ln x+ K
12. y =v'x
2x2y!! + 5xy! + y = 2x3/2v!! + 3x1/2v! = 0 & v!!
v!=
"32x
& v! = Cx"3/2
& v =C'x
+ K
Section 3.3 103
13. y = v(x + 1)
(x + 1)2y!! " 3(x + 1)y! + 3y = (x + 1)3v!! " (x + 1)2v! = 0 & v!!
v!=
1x + 1
& v! = C(x + 1) & v =Cx2
2+ Cx + K
14. y = ve2x
(2x + 1)y!! " 4(x + 1)y! + 4y = e2x (4xv! + (1 + 2x)v!!) = 0 & v!!
v!=
"4x
1 + 2x& ln v! = "2x + ln(1 + 2x) + C & v! = ce"2x(1 + 2x)& v = ce"2x(1 " x) + K
15. y = v(x2 + 1)
(x2 " 1)y!! " 2xy! + 2y = v!!(x4 " 1) + v!(2x"6x) = 0 & v!!
v!=
6x " 2x3
x4 " 1
& ln v! = ln!
x2 " 1(x2 + 1)2
"+ C
& v! =C(x2 " 1)(x2 + 1)2
& v ="Cx
x2 + 1+ K
17. y = ve2x
y!!! " 2y!! " y! + 2y = v!!! + 4v!! + 3v! = 0 which is a 2nd degree ODE of v!. Hence, from thecharacteristic equation, v! = e"3x and v! = e"x are two solutions. So a general solution wouldbe v! = c1e"3x + c2e"x & v = k1e"3x + k2e"x + k3
18. y = vex
y!!! " 4y!! + 5y! " 2y = v!!! " v!! = 0 & v!!!
v!!= 1 & ln v!! = x + C
& v!! = Cex & v! = Cex + K& v = Cex + Kx + D
19. If xe2x is a solution, then so is e2x. So y = ve2x
y!!! " 5y!! + 8y! " 4y = e2x(v!!! + v!!) = 0 & v!!!
v!!= "1 & v!! = Ce"x
& v! = Ce"x + K & v = Ce"x + Kx + D
20. y!!! + 5y!! = 0 & y!!!
y!! = "5 & y!! = Ce"5x & y = Ce"5x + Kx + D is another solution.
21. The substitution y = 2v doesn’t simplify the ODE any. So just going ahead straightforward
gives: y!!! " 3y!! = 0 & y!!!
y!! = 3 & y!! = Ce3x
& y! = Ce3x + K, y = Ce3x + Kx + D
22. For simplicity, all functions depend on x. Thus we have:
104 Section 3.4
a0fw!
w= "2a0f
! " a1f
w!
w=
"2a0f !
a0f" a1f
a0f
= "2f !
f" a1
a0
ln w = "2 ln f "%
a1
a0+ C
= ln f"2 "%
a1
a0+ C
w = Cexp,"@
a1a0
-
f2
24. (a)d
dxcoshx =
d
dx
!ex + e"x
2
"=!
ex " e"x
2
"= sinhx
(b)d
dxsinh x =
d
dx
!ex " e"x
2
"=!
ex + e"x
2
"= coshx
(c) cosh2 x " sinh2 x =!
e2x
4+
12
+e"2x
4
""!
e2x
4" 1
2+
e"2x
4
"= 1
(d) det8888
coshx sinh xsinh x coshx
8888 = 1 %= 0 so they are linearly independent.
(e) y = c1 coshx + c2 sinh x, y! = c1 sinh x + c2 coshx,y!! = c1 coshx + c2 sinh x, & y!! " y = 0
3.4 Operator Notation
1. P (D) = D " 5
(a) P (D)(3x + 7) = 3 " 5(3x + 7) = "15x" 32(b) P (D)(cos x) = "5 sinx " 5 cosx
(c) P (D)(e5x) = 5e5x " 5e5x = 0
2. P (D) = D + 1
(a) P (D)(e"x) = "e"x + e"x = 0(b) P (D)(4 + sin x) = cosx + 4 + sin x
(c) P (D)(e"2x) = "2e"2x + e"2x = "e"2x
3. P (D) = D2 " 1
(a) P (D)(4e"x + e2x) = 3e2x
(b) P (D)(x3 + 4) = 6x " x3 " 4(c) P (D)(ex " 5e"x) = 0
4. P (D) = D2 + 1
Section 3.4 105
(a) P (D)(x3 + 2x) = x3 + 8x
(b) P (D)(cos x + sin x) = 0(c) P (D)(sin 2x) = "3 sin2x
5. P (D) = D2 " 4D + 5
(a) P (D)(e2x cosx) = e2x ("4 sinx + 3 cosx " 8 cosx + 4 sinx + 5 cosx)= 0
(b) P (D)(4x2 + ex) = 2ex + 20x2 " 32x + 8(c)
(D2 " 4D + 5)(x3 " cos 2x) = D2(x3 " cos 2x)"4D(x3 " cos 2x) + 5(x3 " cos 2x)
= (6x + 4 cos 2x) " 4(3x2 + 2 sin 2x)+(5x3 " 5 cos 2x)
6. P (D) = D2 + 2D + 1
(a) P (D)(ex) = 4ex
(b) P (D)(xe"x) = 0(c) P (D)(sin x) = 2 cosx
7.
[D3 + 2D](D2 " D)(x6 " 2x2) = [D3 + 2D](D6x5 " Dx4 " 6x5 + 4x)= (D3 + 2D)(30x4 " 4 " 6x5 + 4x)= 720x" 0 " 360x2 " 0
+240x3 " 0 " 60x4 + 8= "60x4 + 240x3 " 360x2 + 720x + 8
(D2 " D)(D3 + 2D)(x6 " 2x2) = (D2 " D)(120x3 " 0 + 12x5 " 8x)= (720x + 240x3 " 360x2 " 60x4 + 8)
Both P (D)Q(D)(y) and Q(D)P (D)(y) yield the same answer.
8. PQ(y) = QP (y) = 3x2 " 8x + 3 sin x " 5 cosx
9. PQ(y) = 12x2 + 3QP (y) = 6x
10. PQ(y) = 3x cosx + 2 sinxQP (y) = 3x cosx " sinx
11. PQ(y) = ex(2x + 7)QP (y) = 2ex(x + 3)
12. PQ(y) = 5x3 + 15x2 + 21x + 21QP (y) = 5x3 + 12x2 + 21x + 16
13. PQ(y) = e"x2(2x2 " 1)
QP (y) = 2e"x2(x2 " 1)
106 Section 3.5
14. PQ(y) = 1 " x sin xQP (y) = " cosx " x sin x
15. D3(D " 2)(D + 3)(x2 " 2x + 7) = (D " 2)(D + 3)D3(x2 " 2x + 7) = 0
16. 8e2x (The only function involved that has 3 derivatives is e2x)
17. (a) (D " 3)(D + 2)
(b)
/D "
/11 +
'73
2
00/D "
/11 "
'73
2
00
(c) (D2 + 1)2
18. (a) (D2 " 1)2
(b) D(D2 " D + 4)(c) D(D " 4)(D " 1)
19. (a) (D " 1)(D2 + D + 1)(b) (D + 2)(D2 " 2D + 4)
(c) (D " 1)
/D "
/1 +
'5
2
00/D "
/1 "
'5
2
00
20. (a) (D2 + 10D + 16)(y) = (D + 8)(D + 2)(y) = 0(b) (D5 + 4D3 + 4D)(y) = D(D2 + 2)2(y) = 0
21. (a) (D3 + 27)(y) = (D + 3)(D2 " 3D + 9)(y) = 0(b)
y!! + 3y! " 4y = 0 $ (D2 + 3D " 4)(y) = 0$ (D + 4)(D " 1)(y) = 0
22. (a) (D3 " 8)(y) = (D " 2)(D + 2D + 4)(y) = 0(b)
y!!! + 6y!! + 9y! = 0 $ (D3 + 6D2 + 9D)(y) = 0$ D(D + 3)2(y) = 0
23. See discussion on page 187.
3.5 Numerical Consideration for nth Order Equations
1. (a)u!
1 = u2
u!2 = "u2 + x(1 " 2u1)
andu1("1) = 1
u2("1) = 0
(b) h 1 .2 .01y(5) .434067 .433995 .433995
Section 3.5 107
(c) All plots on same graph
2. (a)u!
1 = u2
u!2 = " 4
7u2 + 37u1
andu1(0) = 0
u2(0) = 1
(b) h 1 .2 .01y(5) 5.96191 5.96191 5.96191
(c) All plots on same graph
3. (a)u!
1 = u2
u!2 = x"3u1
x+2
andu1(0) = 0
u2(0) = 4
(b) h 1 .2 .01y(5) -1.99941 -1.99941 -1.99941
(c) All plots on same graph
4. (a)u!
1 = u2
u!2 = "x2u2 " 12u1
andu1(0) = 0
u2(0) = 7
(b) h 1 .2 .01y(5) 1.76 , 1019 .0111645 .0111255
(c) All plots on same graph
108 Section 3.5
5. (a)u!
1 = u2
u!2 = 1 " 4u2 " 3 sinu1
andu1(0) = "1
u2(0) = !
(b) h 1 .2 .01y(5) .335637 .335636 .335636
(c) All plots on same graph
6. (a)u!
1 = u2
u!2 = ex"3u1
x2+2
andu1(0) = 1
u2(0) = 2
(b) h 1 .2 .01y(5) 5.73168 5.7317 5.7317
(c) All plots on same graph
7. (a)u!
1 = u2
u!2 = 3x"u2 sin x"u1
x3
andu1(1) = 1
u2(1) = 32
(b) h 1 .2 .01y(5) 8.90804 8.90804 8.90804
(c) All plots on same graph
Section 3.5 109
8. (a)u!
1 = u2
u!2 = sin x " 2u2 " 10u1
andu1(!) = e
u2(!) = 1
(b) h 1 .2 .01y(5) .0825329 .0825356 .0825358
(c) All plots on same graph
9. (a)u!
1 = u2
u!2 =u3
u!3 = ex " xu3u2 " u1
and
u1(0) = 0
u2(0) = 1
u3(1) = 1
(b) h 1 .2 .01y(5) "1.09 , 1029732975 "1.99, 102135498 19.5423
(c) All plots on same graph
!
!
!
!
10. (a)u!
1 = u2
u!2 =u3
u!3 = e"x " xu3u2 " sin u1
and
u1(1) = 1
u2(1) = 0
u3(1) = 1
(b) h 1 .2 .01y(5) 4.91114 4.91114 4.91114
(c) All plots on same graph
110 Section 3.5
11. (a)u!
1 = u2
u!2 =u3
u!3 = "u3
8
and
u1(0) = 1
u2(0) = 0
u3(0) = 2
(b) h 1 .2 .01y(5) 21.5135 21.5135 21.5135
(c) All plots on same graph
12. (a)u!
1 = u2
u!2 =u3
u!3 = ln x " xu3 " xu2 " u3
1
and
u1(1) = 0
u2(1) = 0
u3(1) = 1
(b) h 1 .2 .01y(5) 1.42093 1.42093 1.42093
(c) All plots on same graph
13. (a)u!
1 = u2
u!2 =u3
u!3 = e"x2
+ u23 " u2 " xu1
and
u1(0) = 1
u2(0) = 0
u3(0) = 0
(b) h 1 .2 .01y(5) "4.67 , 10588 -1.68357 -1.68355
(c) All plots on same graph
Section 3.5 111
14. (a)u!
1 = u2
u!2 =u3
u!3 = 8u1
x3
and
u1(1) = 1
u2(1) = 1
u3(1) = 0
(b) h 1 .2 .01y(5) 45.6552 45.6554 45.6554
(c) All plots on same graph
15. (a)u!
1 = u2
u!2 =u3
u!3 = u4
u!4 = "7u4 " 6u3 + 32u2 " 32u1
and
u1(1) = "1
u2(1) = 0
u3(1) = 0
u4(1) = 1
(b) h 1 .2 .01y(5) 66.4945 66.4945 66.4945
(c) All plots on same graph
16. (a)u!
1 = u2
u!2 =u3
u!3 = u4
u!4 = 1
x+2 (u4 " 6u3 " 3u2 + 2u1)
and
u1("1) = 3
u2("1) = "1
u3("1) = 0
u4("1) = 1
(b) h 1 .2 .01y(5) 7.33357 7.33351 7.33351
(c) All plots on same graph
112 Chapter 3 Review
17. (a)u!
1 = u2
u!2 =u3
u!3 = u4
u!4 = u5
u!5 = 9u2
and
u1(1) = 2
u2(1) = 0
u3(1) = 0
u4(1) = 1
u5(1) = 1
(b) h 1 .2 .01y(5) 79.3336 79.3335 79.3335
(c) All plots on same graph
3.6 Chapter 3: Additional Problems
1. False. We only can conclude that this is a possibility.
2. False. This equation has no problems.
3. True.
4. False. Only constant-coe!cient di"erential operators commute.
5. False.
7. It is guaranteed to have a unique solution through the given IC. The constants are c1 =1 " e"1,c2 = e"1.
9. Linearly independent.
10. Linearly independent.
11. Linearly independent.
12. Linearly independent.
13. Linearly dependent.
14. Linearly dependent.
15. Linearly dependent.
16. Not a fundamental set because x is not a solution.
17. Yes, it forms a fundamental set of solutions.
18. Not a fundamental set because neither is a solution.
19. Not a fundamental set because e"x is not a solution.
Chapter 3 Review 113
20. Yes, it forms a fundamental set of solutions.
21. Yes, it forms a fundamental set of solutions.
22. Not a fundamental set because only 3 is a solution.
23. Yes, it forms a basis.
24. Yes, it forms a basis.
25. No, it is not a basis because vectors are linearly independent.
26. Yes, it forms a basis.
27. Yes, it forms a basis.
28. No, it is not a basis because vectors are linearly independent.
29. No, it is not a basis because there are more than 3 vectors.
30. No, it is not a basis because the vectors are not in the space.
31. No, it is not a basis because vectors are linearly independent.
35. y = e5x
37. y = e"5x
40. (a) (D " 4)(D " 3)(y) = 0(b) D(D " 4)(D + 4)
114 Chapter 3 Review
Chapter 4
Techniques of Higher-Order LinearEquations
4.1 Homogeneous Equations with Constant Coe"cients
1. Assuming y = y(x) or y = y(t),
y!! + 8y! + 12y = 0 $ r2 + 8r + 12 = 0$ (r + 6)(r + 2) = 0$ r = "6,"2
Linearly independent solutions are e"6t, e"2t and the general solution is
y = C1e"6t + C2e
"2t
2.
7r2 + 4r " 3 = 0(7r " 3)(r + 1) = 0
r = "1,37
General solution $ y = c1e"x + c2e
3x7
3.
8y!!! + y!! = 0 $ 8r3 + r2 = 0$ r2(8r + 1) = 0
$ r = 0, 0,"18
We have three linearly independent solutions: e0·x, xe0·x, e"x/8 and the general solution is
y = C1 + C2x + C3e"x/8
4.
y!! " 2y! = 0 $ r2 " 2r = 0$ r(r " 2) = 0$ r = 0, 2
115
116 Section 4.1
The two linearly independent solutions are {1, e2x} and the general solution is
y = C1 + C2e2x
5. y!! + y! " 12y = 0, y(0) = 0, y!(0) = 7.
r2 + r " 12 = 0 $ (r + 4)(r " 3) = 0 $ r = "4, 3
Two linearly independent solutions are e"4x and e3x, and the general solution is
y = C1e"4x + C2e
3x
y! = "4C1e"4x + 3C2e
3x
Applying initial conditions,
0 = C1 + C2 $ C1 = "C2
7 = "4C1 + 3C2
7 = "4("C2) + 3C2 $ C2 = 1, C1 = "1
Therefore,y = "e"4x + e3x
! !
!
!
6. r = 1, "2 $ y = c1ex + c2e
"2x and IC give y = ex.
! !
7. y!! + 4y! + 4y = 0, y(0) = 0, y!(0) = 1. The characteristic equation is
r2 + 4r + 4 = 0 $ (r + 2)2 = 0$ r = "2,"2$ e"2x, xe"2x
Section 4.1 117
are two linearly independent solutions. The general solution is
y = C1e"2x + C2xe"2x
y! = "2C1e"2x + C2(e"2x " 2xe"2x)
by the product rule. Applying initial conditions, 0 = y(0) = C1, 1 = y!(0) = "2(0)+C2 yieldsC2 = 1 and
y = xe"2x
! !
!
!
!
!
!
!
!
8. y!! + 4y! + 3y = 0, y(1) = 1, y!(1) = 2. The characteristic equation is
r2 + 4r + 3 = 0 $ (r + 3)(r + 1) = 0$ r = "3,"1$ e"3x, e"x
are two linearly independent solutions. The general solution is then
y = C1e"3x + C2e
"x
Applying initial conditions,
1 = y(1) = C1e"3 + C2e
"1
2 = y!(1) = "3C1e"3 " C2e
"1
Adding the equations we get 3 = "2C1e"3, so C1 = " 32e3 and then
C2 = e1
!1 "!"3
2e3
"e"3
"
= e1
!1 +
32
"=
52e1
Theny = "3
2e"3x+3 +
52e"x+1
118 Section 4.1
!
!
!
9. 4y!! + 4y! + y = 0, so the characteristic equation is 4r2 + 4r + 1 = 0 or (2r + 1)2 = 0.Thus r = " 1
2 ," 12 . Two linearly independent solutions are e"x/2 and xe"x/2, and the general
solution isy = C1e
"x/2 + C2xe"x/2
10. 2y!!"5y!+2y = 0, so the characteristic equation is 2m2"5m+2 = 0. We have a = 2, b = "5,and c = 2. Roots are
m =5 ±
'25 " 164
$ m = 2,12
Two linearly independent solutions are e2x and ex/2, and the general solution is
y(x) = C1e2x + C2e
x/2
2y!! + 5y! + 2y = 0 $ 2r2 + 5r + 2 = 0 & r = "2, "12
$ y = c1e"2x + c2e
"x/2.
With IC given, y = e"2x
!"2 + !
3
"+ e"x/2
!2 + 4!
3
"
! !
!
!
!
!
11. (D2 " 4D + 5)(y) = 0; the characteristic equation is r2 " 4r + 5 = 0, or
r =4 ±.
16 " 4(5)(1)2
= 2 ± i
Two linearly independent solutions are e2x cosx and e2x sin x, and the general solution is
y = C1e2x cosx + C2e
2x sin x
Section 4.1 119
12. (D2 + 2D + 10)(y) = 0; the characteristic equation is m2 + 2m + 10. We have a = 1, b = 2,and c = 10, so the roots are
m ="2 ±
'4 " 40
2$ m = "1 ± 3i
Two linearly independent solutions are e"x cos 3x and e"x sin 3x, and the general solution is
y(x) = C1e"x cos 3x + C2e
"x sin 3x
IC give y = "e"!"x
3sin 3x
! ! ! ! !
!
!
!
13. (D2 + 1)2(y) = 0; the characteristic equation is (m2 + 1)2 = 0, the roots are m = ±i,±i, thelinearly independent solutions are cosx, sin x, x cos x, and x sin x, and the general solution is
y(x) = C1 cosx + C2 sinx + C3x cosx + C4x sin x
14. (D2 + 4)y = 0, y(!) = 1, y!(!) = 1; the characteristic equation is r2 + 4 = 0 so r = ±2i andthe two linearly independent solutions are cos 2x, sin 2x. The general solution is then
y = C1 cos 2x + C2 sin 2x
y! = "2C1 sin 2x + 2C2 cos 2x
Applying initial conditions,
1 = y(!) = C1 cos(2!) + C2 sin(2!)1 = y!(!) = "2C1 sin 2! + 2C2 cos 2!
$ C1 = 1, C2 =12
The solution isy = cos 2x +
12
sin 2x
! ! !
!
!
120 Section 4.1
15. y!!! " 8y = 0; the characteristic equation is r3 " 8 = 0, or (r " 2)(r2 + 2r + 4) = 0, which
leads to r = 2,"2 ±
.4 " 4(4)2
or 2, "1 +'
3 i, and "1 "'
3 i. Three linearly independent
solutions are e2x, e"x cos('
3x), e"x sin('
3x), and the general solution is
y = C1e2x + C2e
"x cos('
3x) + C3e"x sin(
'3 x)
Initial conditions are y(0) = 0, y!(0) = 1, y!!(0) = 0, and the solution is
y =16e2x " 1
6e"x cos(
'3x) +
16'
3 e"x sin('
3x)
! ! !
16. y(4) " y = 0; the characteristic equation is r4 " 1 = 0, which yields
r2 = +1 $ r = ±1r2 = "1 $ r = ±i
Four linearly independent solutions are {cosx, sin x, ex, e"x} and the general solution is
y = C1 cosx + C2 sin x + C3ex + C4e
"x
17. r4 " 16 = 0 & r = ±2, ±2i
y = c1e2x + c2e
"2x + c3 sin 2x + c4 cos 2x
18. (D2 " 2D + 1)(y) = 0; the characteristic equation is r2 " 2r + 1 = 0 or (r " 1)2 = 0, r = 1, 1.Two linearly independent solutions are ex and xex, and the general solution is
y = C1ex + C2xex
19. r4 + 7r3 + 6r2 " 32r " 32 = 0 & r = "1, 2, "4, "4
y = c1e"4x + c2xe"4x + c3e
"x + c4e2x
20. D3(D2"6D+9)y = 0; the characteristic equation is r3(r2 "6r+9) = 0, so r = 0, 0, 0,"3,"3.Five linearly independent solutions are e0x, xe0x, x2e0x, e"3x, and xe"3x. This simplifies as1, x, x2, e"3x, xe"3x. The general solution is then
y = C1 + C2x + C3x2 + C4e
"3x + C5xe"3x
Section 4.1 121
21. D2(D " 1)2(D2 + 1)(y) = 0; the characteristic equation is r2(r " 1)2(r2 + 1) = 0, which leadsto r = 0, 0, 1, 1,±i. We have 6 linearly independent solutions, 1, x, ex, xex, sinx, and cosx.The general solution is
y = C1 + C2x + C3ex + C4xex + C5 sin x + C6 cosx
22. y(5) " 10y!!! + 9y! = 0; the characteristic equation is r5 " 10r3 + 9r = r(r4 " 10r2 + 9) = 0 orr(r2 " 9)(r2 " 1) = 0, which leads to r = 0,±1,±3. Five linearly independent solutions are 1,ex, e"x, e3x, e"3x. The general solution is
y = C1 + C2ex + C3e
"x + C4e"3x + C5e
"3x
23. y(4) + 2y!! + y = 0; the characteristic equation is m4 + 2m2 + 1 = 0, so (m2)2 + 2(m2) + 1 = 0or (m2 +1)2 = 0. The roots are m = ±i,±i, which leads to four linearly independent solutionscosx, sin x, x cosx, and x sin x. The general solution is then
y(x) = C1 cosx + C2 sinx + C3x cosx + C4x sin x
24. (D2 + 4)2(y) = 0; the characteristic equation is (m2 + 4)2 = 0. The roots are m = ±2i,±2i,so the four linearly independent solutions are cos 2x, sin 2x, x cos 2x, x sin 2x, and the generalsolution is
y(x) = C1 cos 2x + C2 sin 2x + C3x cos 2x + C4x sin 2x
25. r3 " 3r2 + 3r " 1 = 0 & (r " 1)3 = 0 & r = 1, 1, 1
y = c1ex + c2xex + c3x
2ex
26. r3 " r2 " r + 1 = 0 & r = "1, 1, 1
y = c1e"x + c2e
x + c3xex
27. r4 " 5r2 + 4 = 0 & r = ±2, ±1
y = c1e2x + c2e
"2x + c3ex + c4e
"x
28. D(D2 + 4)(D2 " 2D + D)(y) = 0; the characteristic equation is r(r2 + 4)(r2 " r) = 0, sor = 0, 0, 1,±2x. Five linearly independent solutions are 1, x, ex, cos 2x, sin 2x, and the generalsolution is
y = C1 + C2x + C3ex + C4 cos 2x + C5 sin 2x
29. y(5) + 8y!!! + 16y! = 0; the characteristic equation is r5 + 8r3 + 16r = 0 or r(r4 + 8r2 + 16) = 0or r(r2 +4)2 = 0. The roots are r = 0 (once), r = ±2i (twice). Linearly independent solutionsare 1, cos 2t, sin 2t, t cos 2t, t sin 2t. The general solution is thus
y = C1 + C2 cos 2t + C3 sin 2t + C4t cos 2t + C5t sin 2t
30. r4 + 16 = 0 & r ='
2 ± i'
2, "'
2 ± i'
2
y = e%
2x(c1 sin'
2x + c2 cos'
2x) + e"%
2x(c1 sin'
2x + c2 cos'
2x)
122 Section 4.1
31. r4 + 64 = 0 & r = 2 ± 2i,"2± 2i
y = e2x(c1 sin 2x + c2 cos 2x) + e"2x(c1 sin 2x + c2 cos 2x)
32. r2(r + 2) = 0y!!! + 2y!! = 0
33. Since r1 = 3i is a root, then r4 = "3i must also be a root with multiplicity 2. Hence,(r " 3i)2(r + 3i)2(r " (1 + i))(r " (1 " i)) = 0
y(6) " 2y(5) + 20y(4) " 36y!!! + 117y!! " 162y! + 162y = 0
34. r4(r " (2 + 3i))3(r " (2 " 3i))3 = 0
y(10) " 12y(9) + 87y(8) " 376y(7) + 1131y(6) " 2028y(5) + 2197y(4) = 0
35. (r " (2 + 3i))2(r " (2 " 3i))2(r + 5)(r " 2)3 = 0
y(8) " 9y(7) + 32y(6) + 50y(5) " 939y(4) + 4207y!!! " 10131y!! + 12948y! " 6760y = 0
36. ar2 + br + c = 0 &
r ="2b ±
'4b2 " 4ac
2a=
"b ±'
b2 " ac
a
y = e"bxa
!c1e
x%
b2!aca + c2e
x%
b2!aca
"= e"
bxa#c1e
( + c2e"($
Now,
y = c1 cosh- + c2 sinh-
= c1e( + e"(
2+ c2 e( " e"(
2
= e(!
c1 + c2
2
"+ e"(
!c1 " c2
2
"
= k1e( + k2e
"(
which is the desired form.
37.
ex = 1 + x +x2
2!+
x3
3!+
x4
4!+ · · ·
ei# = 1 + (i$) +(i$)2
2!+
(i$)3
3!+
(i$)4
4!+ · · ·
= 1 + (i$) " $2
2!" i$3
3!+$4
4!+ · · ·
=!
1 " $2
2!+$4
4!+ · · ·
"+ i
!$ " $
3
3!+$5
5!+ · · ·
"
= cos $ + i sin $
38. z1 = ei#, z!1 = iei# = iz, z(0) = 1z2 = cos $ + i sin $, z!2 = "sin$ + i cos $ = iz, z(0) = 1Since this is a first order ODE, there must only be one linearly independent solution. Hence,z1 = Kz2.
4.2. A MASS ON A SPRING 123
4.2 A Mass on a Spring
1.1.512
k = 12
k = 160
m =12lb
32ft/s2=
38
x!! +160
38
x = 0
r2 +1280
3= 0
r = ±i
91280
3
x(t) = A sin
/t
91280
3
0+ B cos
/t
91280
3
0
x(0) =16, x!(0) = 0
x(t) =16
cos(t9
12803
)
Amplitude =16ft = 2in
Period =2!:1280
3
Frequency =
:1280
3
2!
2.
.5k = 16k = 32
m =16lb
32ft/s2=
12
x!! + 64x = 0r2 + 64 = 0
r = ±8i
x(t) = A sin (8t) + B cos (8t)
x(0) =13, x!(0) = 2
x(t) =14
sin 8t +13
cos 8t
Amplitude =
?!14
"2
+!
13
"2
=15ft
Period =!
4
Frequency =4!
124 Section 4.2
3.
.5k = 4k = 8
m =4lb
32ft/s2=
18
x!! + 64x = 0r2 + 64 = 0
r = ±8i
x(t) = A sin (8t) + B cos (8t)x(0) = 0, x!(0) = 2
x(t) =14
sin 8t
Amplitude =14ft = 3in
Period =!
4
Frequency =4!
6. 12!
:K( 1
I1+ 1
I2)
7. A = B1"m
k $2
8. I = VR (1 " e
RtL )
11. I = q$CLe"
Rt2L sin&t, CR2 < 4L
14. x!! + x! + x = 0 leads to
r2 + r + 1 = 0
r ="1 ±
.1 " 4(1)(1)2(1)
= "12± i
'3
2
which is underdamped.
!
Section 4.2 125
15. 4x!! + 12x! + 3x = 0 leads to
b2 " 4mk =14" 4(4)(3) < 0
which is underdamped.
!
16. x!! + 4x! + 3x = 0 leads to
r2 + 4r + 3 = 0(r + 3)(r + 1) = 0
r = "3,"1
which is overdamped.
17. !x!! + 4x! + 2x = 0 leads tob2 " 4mk = 16 " 4(2)(!) < 0
which is underdamped.
126 Section 4.2
18. x!! + 2x! + x = 0 leads to
r2 + 2r + 1 = 0(r + 1)2 = 0
r = "1,"1
which is critically damped.
19. 3x!! + 7x! + x = 0 leads tob2 " 4mk = 49 " 4(3)(1) = 37
which is overdamped.
20. x!! + 6x! + 9x = 0 leads tob2 " 4mk = 36 " 9(4)(1) = 0
which is critically damped.
Section 4.2 127
21. Overdamped: x = c1er1t + c2er2t, x! = c1r1er1t + c2r2er2t. Substitute t = 0 and solve forconstants to get a solution
x = ""v0 + x0r2
r1 " r2er1t +
r1x0 " v0
r1 " r2er2t.
Set this equal to 0 (corresponding to crossing through rest position) and solve for t:
t =ln,
r1x0"v0"v0+x0r2
-
r1 " r2
In order for this crossing to be possible, the argument inside the natural log must be nonneg-ative. This gives the desired condition
|v0| > |x0r2|, where r2 = ("b ".
b2 " 4mk)/2m
is the smaller of the characteristic roots and v0 and x0 must have opposite sign.Critically damped: the argument is similar to the above one to obtain |v0| > |bx0/(2m)|where v0 and x0 must again have opposite sign.
23. mx!! + bx! + kx = 0 leads to m = 1, b = 2, and
r2 + 2r + k = 0
r ="2 ±
.22 " 4(1)(k)2(1)
which is underdamped for 4 " 4k < 0, 4 < 4k, 1 < k.
24. m = 1, k = 3 leads to r2 + br + 3 = 0, or
r ="b ±
.b2 " 4(1)(3)2(1)
which is critically damped with b2 " 12 = 0 or b = 2'
3.
25. We need b2 " 4mk > 0 to be overdamped. Then
22 " 4m(1) > 0 $ 4 " 4m > 0$ 4 > 4m
$ 1 > m
26. We need b2 " 4mk < 0 for underdamped. Then
22 " 4m(2) < 0 $ 4 " 8m < 0 $ 4 < 8m $ 12
< m
27 .
!
128 Section 4.2
28 .
!
!
29 .
!
30. (a) & = 10
!
!
(b) & = 1
!
!
4.3. CAUCHY-EULER (EQUIDIMENSIONAL) EQUATION 129
(c) It seems & = 1 gives the largest oscillations.
31. (a) & = 10
!
!
(b) & = 1
!
!
(c) It seems & = 1 gives the largest oscillations. Amplitude * 10.
4.3 Cauchy-Euler (Equidimensional) Equation
1.
x2y!! + 4xy! + 2y = 0.
r2 + (4 " 1)r + 2 = 0(r + 2)(r + 1) = 0
r = "2 r = "1y = Ax"2 + Bx"1
2.
x2y!! + 3xy! + y = 0.
r2 + (3 " 1)r + 1 = 0(r + 1)(r + 1) = 0
r = "1 r = "1y = Ax"1 + Bx"1 ln x
130 Section 4.3
3.
x2y!! + xy! + 4y = 0.
r2 + 4 = 0(r + 2i)(r " 2i) = 0
r = ±2i
y = A sin(2 lnx) + B cos(2 lnx)
4.
x2y!! + 3xy! + 2y = 0.
r2 + 2r + 2 = 0
r ="2 ±
'4 " 8
2= "1 ± i
y = Asin(ln x)
x+ B
cos(ln x)x
5.
x2y!! + 11xy! + 21y = 0.
r2 + 10r + 21 = 0(r + 7)(r + 3) = 0
r = "7 r = "3y = Ax"7 + Bx"3
6.
2x2y!! + xy! " y = 0.
2r2 " r " 1 = 0(2r + 1)(r " 1) = 0
r = "12
r = 1
y = Ax + Bx" 12
7.
2x2y!! " 3xy! + 3y = 0.
2r2 " 5r + 3 = 0(2r " 31)(r " 1) = 0
r = "12
r = 1
y = Ax32 + Bx
Section 4.3 131
8.
x2y!! " 3xy! + 3y = 0.
r2 " 4r + 3 = 0(r " 3)(r " 1) = 0
r = 3 r = 1y = Ax + Bx3
9.
x2y!! " 3xy! + 4y = 0.
r2 " 4r + 4 = 0(r " 2)2 = 0
r = 2 r = 2y = Ax2 + Bx2 ln x
10.
x2y!! + 3xy! " 2y = 0.
r2 + 2r " 2 = 0
r ="2 ±
'12
2r = "1 ±
'3
y = Ax"1+%
3 + Bx"1"%
3
11.
x2y!! + 5xy! " 3y = 0.
r2 + 4r " 3 = 0
r ="4 ±
'28
2= "2 ±
'7
y = Ax"2+%
7 + Bx"2"%
7
12.
3x2y!! + 3xy! + 9y = 0.
3r2 + 9 = 0r = ±i
'3
y = A sin('
3 ln x) + B cos('
3 ln x)
132 Section 4.3
13.
7x2y!! + 5xy! + y = 0.
r2 " 2r + 1 = 0
r =2 ±
'"24
14
=1 ± i
'6
7
y = Ax17 sin
/'6
7ln x
0+ Bx
17 cos
/'6
7ln x
0
14.
x2y!! + 5xy! + 8y = 0.
r2 + 4r + 8 = 0
r ="4 ±
'"16
2= "2 ± 2i
y = Ax"2 sin(2 ln x) + Bx"2 cos(2 ln x)
15.
3x2y!! + 13xy! + 11y = 0.
3r2 + 10r + 11 = 0
r ="10 ±
'"32
6
="5 ± i
'8
3
y = Ax" 53 sin(
'8
3ln x) + Bx" 5
3 cos('
83
ln x)
16.
3x2y!! + 5xy! + y = 0.
3r2 + 2r + 1 = 0
r ="2 ±
'"8
6
="1 ± i
'2
3
y = Ax" 13 sin(
'2
3ln x) + Bx" 1
3 cos('
23
ln x)
4.4. NONHOMOGENEOUS EQUATIONS 133
17. Examine the Wronskian, W (x)
W (x) =8888
y1 ln x y2 ln xy1x + y!
1 ln x y2x + y!
2 ln x
8888
=y1y2 ln x
x+ y1y
!2(ln x)2
"y1y2 ln x
x" y!
1y2(ln x)2
= (lnx)2(y1y!2 " y!
1y2)%= 0
since y1 and y2 are linearly independent for all x %= 1.
4.4 Nonhomogeneous Equations
1. Since 3ex("53 ) = "5ex, we need to multiply Yp = yp("5
3 ) = ex
2
4. Since 2x(A2 ) = Ax, we need to multiply Yp = yp(A
2 ) = Ax2
2 " Ax
5. We multiply y1p by 2, y2p by "12 and y3p by 6 to get
Yp = 2y1p " 12y2p + 6y3p = 3ex " 2x " 43
6. We multiply y1p by 8 and y2p by "3 to get
Yp = 8y1p " 3y2p = 2x " 2 " 3x2
2e"2x
134 Section 4.4
7. & = 1
!
& = 2
!
& = 3
!
!
From the graphs, resonant frequency happens when & = 2 and the amplitude is about .22.
Section 4.4 135
8. & = 1
!
& = 2
!
& = 3
!
!
From the graphs, resonant frequency happens when & = 1 and the amplitude is about .4.
136 Section 4.4
9. & =.
1/18
!
!
& =.
5/18
!
& =.
1/2
!
From the graphs, resonant frequency happens when & =.
1/2 and the amplitude is about .6.
Section 4.4 137
10. & =.
5/9
!
!
& =.
7/9
!
!
& =.
10/9
!
!
From the graphs, resonant frequency happens when & =.
10/9 and the amplitude is about.45.
138 Section 4.5
11. & = 1
!
& ='
2
!
& ='
3
!
From the graphs, resonant frequency happens when & ='
3 and the amplitude is about .25.
4.5 Method of Undetermined Coe"cients via Tables
1. y!!+y = e"x+x2; homogeneous solution r2+1 = 0 $ r = ±i; fundamental set is {cosx, sin x}.Particular UC sets: {e"x}, {x2, x, 1}, so no modification is necessary.
yp = Ae"x + Bx2 + Cx + E
Section 4.5 139
2. y!! + 3y! = x2e2x + cosx; homogeneous solution r2 + 3r = 0 $ r = 0, "3; fundamental set is{1, e"3x}. Particular UC sets: {e2x, xe2x, x2e2x},{cosx, sin x}, so no modification is necessary.
yp = Ae2x + Bxe2x + Cx2e2x + D cosx + E sin x
3. y!! + y = 4 sinx + ex cosx; homogeneous solution r2 + 1 = 0; linearly independent solutions{cosx, sin x}, yc = C1 cosx + C2 sin x. Particular solution:
4 sinx $ {sinx, cosx} $ modify as {x cosx, x sin x}ex cosx $ {ex cosx, ex sin x} $ no modification necessary
Henceyp = A1x cosx + A2x sin x + B1e
x cosx + B2ex sinx
4. y!!+4y!+y = sin 2x+cosx; homogeneous solution r2+4r+1 = 0 $ r = "2±'
3; fundamentalset is {e("2+
%3)x, e("2"
%3)x}. Particular UC sets: {sin 2x, cos 2x}, {cosx, sin x}, so no
modification is necessary.
yp = A sin 2x + B cos 2x + C sinx + D cosx
5. y!!! + y = xe2x cosx + sinx; UC set is {sinx, cos x, xe2x cosx, xe2x sin x, e2x cosx, e2x sin x}.From the homogeneous part, r3 + 1 = (r + 1)(r2 " r + 1). The linearly independent solutionsare e"x, ex/2 cos
%3
2 x, ex/2 sin%
32 x. Thus
yp = A sinx + B sin x + Cxe2x cosx + Exe2x sin x + Fe2x cosx + Ge2x sinx
6. y!! " y = ex + xe"x; homogeneous solution r2 " 1 = 0 $ r = ±1; fundamental set is{ex, e"x}. Particular solution: UC sets are {ex}, {xe"x, e"x}, both need modification, to{xex}, {x2e"x, xe"x}. Thus
yp = Axex + Bx2e"x + Cxe"x
7. y!!! + 8y = 4 sinx + cosx; homogeneous solution r3 + 8 = 0 or (r + 2)(r2 + 2r + 4) = 0, sor = "2, 1 ±
'3 i. Linearly independent solutions are {e"2x, ex cos
'3x, ex sin
'3x}, and
yc = C1e"2x + C2e
x cos'
3 x + C3ex sin
'3 x
Particular solution: 4 sinx + cosx $ {sinx, cos x} (UC set), no modification necessary. Thus
yp = A sinx + B cosx
8. y(5) + y!! = x3 + xex cosx; homogeneous solution
r5 + r2 = 0r2(r3 + 1) = 0
r2(r + 1)(r2 " r + 1) = 0
r = 0, 0,"1,1 ±
'3 i
2
140 Section 4.5
Fundamental set is {1, x, e"x, ex/2 cos%
32 x, ex/2 sin
%3
2 x}. Particular solution: UC sets {x3, x2, x, 1},{xex cosx, xex sin x, ex cosx, ex sin x}; the first one needs modification. Multiplying by x2 willgive linearly independent solutions, so {x5, x4, x3, x2}. Then
yp = Ax5 + Bx4 + Cx3 + Ex2 + Gxex cosx
+ Hxex sin x + Iex cosx + Jex sin x
9. y!! + 4y! + 13y = e"2x. We have
yp = Ae"2x $ y!p = "2Ae"2x, y!!
p = 4Ae"2x
y!!p + 4y!
p + 13yp = (4Ae"2x) + 4("2Ae"2x) + 13(Ae"2x) = e"2x
$ 4A " 8A + 13A = 1 $ 9A = 1 $ A =19
10. y!! + 4y! = x2 " 3, yp = Ax3 + Bx2 + Cx. Substitution gives
12Ax2 + (8B + 6A)x + 4C + 2B = x2 " 3
Equating coe!cients of like terms,
12A = 18B + 6A = 04C + 2B = "3
so A = 112 , B = " 1
16 , C = " 2332 .
11. y!! " 4y! = cosx + 3 sinx, yp = A cosx + B sin x. Substitution gives
("A " 4B) cosx + ("B + 4A) sin x = cosx + 3 sinx
Equating coe!cients of like terms,
"A " 4B = 1"B + 4A = 3
so A = 1117 , B = " 7
17 .
12. A = 1/8, B = 0, C = 0, E = "1/35
13. A = 0, B = 8/441, C = 1/21, E = 0
14. y!! + 2y! + 17y = ex + 2, yp = Aex + B. Substituting into the ODE and equating coe!cientsgives 20Aex + 17B = ex + 2, so A = 1
20 , B = 217 .
15. A = "5/36, B = "1/6, C = 1/5, E = 3/50
16. yc = Ae2x + Bex
UC sets:{x2, x, 1} No modificationyp = Cx2 + Dx + E and equating coe!cients givesyp = 2x2 + 6x + 7
Section 4.5 141
17. y!! " 2y! " 8y = 4e2x " 21e"3x; homogeneous solution r2 " 2r " 8 = 0 or (r " 4)(r + 2) = 0, sor = 4,"2. The fundamental set is {e4x, e"2x}, so yh = C1e4x + C2e"2x. Particular solution:UC set is {e2x, e"3x}, which does not need modification, so yp = Ae2x + Be"3x. Substitutionand simplification gives
"8Ae2x + 7Be"3x = 4e2x " 21e"3x
Equating
"8A = 47B = "21
gives A = " 12 , B = "3, so
yp = "12e2x " 3e"3x
The general solution is
y = C1e4x + C2e
"2x " 12e2x " 3e"3x
18. yc = Ae"x sin 2x + Be"x cos 2xUC sets:{cos 2x, sin 2x} No modificationyp = C cos 2x + D sin 2x and equating coe!cients givesyp = " cos 2x + 2 sin 2x
19. yc = Ae"4x + Be"3x cosx + Ce"3x sin xUC sets:{xe"4x, e"4x} modify & {x2e"4x, xe"4x}{e"3x cosx, e"3x sinx} modify & {xe"3x cosx, xe"3x sin x}yp = Dx2e"4x + Exe"4x + Fxe"3x cosx + Gxe"3x sin x and equating coe!cients gives
yp =14x2e"4x +
12xe"4x " 1
2xe"3x cosx +
12xe"3x sin x
20. y!! " 4y = 32x, y(0) = 0, y!(0) = 6. Homogeneous solution: r2 " 4 = 0 so r = ±2. Thefundamental set is {e2x, e"2x}, yh = C1e2x + C2e"2x. Particular solution: UC set is {x, 1},and does not need modification, so yp = Ax + B. Substitution gives
"4Ax " 4B = 32x $ A = "8, B = 0$ yp = "8x
General solution:
y = C1e2x + C2e
"2x " 8x
y! = 2C1e2x " 2C2e
"2x " 8
Applying initial conditions,
0 = y(0) = C1 + C2
6 = y!(0) = 2C1 " 2C2 " 8
so C1 = 72 and C2 = " 7
2 . The solution is
y =72e2x " 7
2e"2x " 8x
142 Section 4.5
21. y!! " 2y! + 2y = ex + x cosx. Homogeneous solution:
m2 " 2m + 2 = 0
$ m =2 ±.
4 " 4(2)2
=2 ±
'"4
2= 1 ± i
yh = C1ex cosx + C2e
x sinx
Particular solution: UC set is {ex, x cosx, x sin x, cosx, sin x}, so
yp = Aex + Bx cosx + Cx sin x + D cosx + E sin x
Substituting into the ODE and equating coe!cients:
ex: A = 1sin x: 2B + E " 2C + 2D = 0cosx: 2C + D " 2B " 2E = 1
x sin x: C + 2B = 0x cosx: B " 2C = 1
Hence A = 1, B = 15 , C = " 2
5 , D = " 125 , and E = " 28
25 , so
y = C1ex cosx + C2e
x sin x + ex +15x cosx " 2
5x sin x " 1
25cosx " 28
25sin x
22. y!! +6y! +10y = 3xe"3x "2e3x cosx, y(0) = 1, y!(0) = "2. Homogeneous solution: m2 +6m+10 = 0, so
m ="6 ±
.36 " 4(10)2
="6 ± 2i
2= "3 ± i
soyh = C1e
"3x cosx + C2e"3x sin x
Particular solution: UC set is {xe"3x, e"3x, e3x cosx, e3x sin x}, so
yp = Bxe"3x + Ce3x cosx + De3x sin x
Substituting into the ODE and equating coe!cients:
xe"3x: B = 3e3x cosx: 36C + 12D = "2e3x sin x: "12C + 36D = 0
so B = 3, C = " 120 , D = " 1
60 , giving
y = C1e"3x cosx + C2e
"3x sinx + 3xe"3x " 120
e3x cosx " 160
e3x sin x
Applying initial conditions,
1 = C1 "120
"2 = "3C1 + C2 +176
so C1 = 2120 , C2 = " 101
60 , leading to
y =2120
e"3x cosx " 10160
e"3x sinx + 3xe"3x " 120
e3x cosx " 160
e3x sin x
Section 4.5 143
23. yc = Ae4x cos 2x + Be4x sin xUC sets:{xe4x sin 2x, e4x sin 2x, xe4x cos 2x, e4x cos 2x} modify& {x2e4x sin 2x, xe4x sin 2x, x2e4x cos 2x, xe4x cos 2x}yp =
516
xe4x sin x " 58x2e4x cos 2x
24. yc = Ae"5x + Be"2x
UC sets:{xe"2x cos 5x, e"2x cos 5x, xe"2x sin 5x, e"2x sin 5x} No modificationyp = Cxe"2x cos 5x + De"2x cos 5x + Exe"2x sin 5x + Fe"2x sin 5xyp = "1
34 xe"2x cos 5x + 637225e"2x cos 5x + 3
170xe"2x sin 5x + 5578e"2x sin 5x
25. yc = Aex cos 2x + Bex sin 2xUC sets:{xex, ex} No modification{ex cos 2x, ex sin 2x} modify & {xex cosx, xex sin x}yp = Cxex + Dex + Exex cos 2x + Fxex sin 2x
yp =12xex " 1
4xex sin 2x
26. yc = Aex + Bxex
UC sets:{xex, ex} modify & {x3ex, x2ex}{ex sin 2x, ex cos 2x} No modificationyp = Cx3ex + Dx2ex + Eex cos 2x + Fex sin 2x
y =x3
3ex " ex
4sin 2x
27. yc = Ae2x + Bex
UC sets:{ex} modify & {xex}yp = Cxex
yp = "xex
y = Ae2x + Bex " xex, y(0) = 1, y!(0) = 0 & A = 0, B = 1y = ex " xex
28. yc = Aex + Be"x
UC sets:{ex} modify & {xex}{e"x} modify & {xe"x}yp = Cxex + Dxe"x
yp = xex + xe"x
29. yc = Ae"3x + Be"x
UC sets:{ex} No modification{e"x} modify & {xe"x}yp = Cex + Dxe"x
yp =ex
16+
xe"x
430. yc = A cos 2x + B sin 2x
UC sets:
144 Section 4.5
{ex sin 2x, ex cos 2x} No modification{e"x sin 2x, e"x cos 2x} No modificationyp = Cex sin 2x + Dex cos 2x + Ee"x sin 2x + Fe"x cos 2xyp = 1
34ex sin 2x " 217ex cos 2x " 1
34e"x sin 2x " 217e"x cos 2x
31. yc = Ae"x sin x + Be"x cosxUC sets:{ex sin x, ex cosx} No modification{e"x sin x, e"x cosx} Modify & {xe"x sin x, xe"x cosx}yp = Cex sin x + Dex cosx + Ee"x sin x + Fe"x cosxyp = 1
16ex sin x " 116ex cosx " 1
4xe"x cosx
32. y(4) " 18y!! + 81y = e3x; Homogeneous solution: r4 " 18r2 + 81 = 0, or (r2 " 9)2 = 0, sor = 3, 3,"3,"3. The four linearly independent solutions are {e3x, xe3x, e"3x, xe"3x}, giving
yc = C1e3x + C2xe3x + C3e
"3x + C4xe"3x
Particular solution: e3x leads to the UC set {e3x}, which needs to be modified as {x2e3x}, so
yp = Ax2e3x
Substitution into the ODE gives
72Ae3x = e3x $ 72A = 1 $ A =172
Thus yp = 172x2e3x and the general solution is
y = yc + yp = C1e3x + C2xe3x + C3e
"3x + C4xe"3x +172
x2e3x
33. (a) yc = Ae4x, UC = {x2, x, 1}
yp = Bx2 + Cx + D & yp = "x2
4" x
8" 1
32(b) yc = Ae"x, UC = {cos 2x, sin 2x}
yp = B cos 2x + C sin 2x & yp =15
cos 2x +25
sin 2x
(c) yc = Aex, UC = {e4x}
yp = Be4x & yp =e4x
336.
A(w) =F0m.
4'2w2 + (w2 " w20)2
A!(w) = "F0
m
!8'2w + 4w(w2 " w2
0)4'2w2 + (w2 " w2
0)3
"
A!(w) = 0.
0 = 8'2w + 4w3 " 4ww20
= 4w3 + 4w(2'2 " w20)
= 4w(w2 + 2'2 " w20)
w = 0 or w2 = w20 " 2'2
w =:
w20 " 2'2
Section 4.5 145
37. (b) If b = 0, the terms of yc must involve sin $ and cos $ $ limx#$(y2 " y1) may not exist.(c) yc has terms involving e"Ax, UC= {1} & cyp = k & yp = k
c . Therefore y = yp + yc =kc +terms of e"Ax and limx#$ y = k
c
(d) r(ar + b) = 0 & yc = A + Be"ba x
UC= {1} modify & {x} & yp = Cx, bC = k & C = k/b & yp = kb x $ y =
A + Be"ba x + k
b x
(e) ay!! = k & y!! = ka & y! = k
ax + C & y = k2ax2 + Cx + D
limx#$ y = #
38. (a) y!!+ay!+by =Rn
i=0 cnxn. Homogeneous solution from r2+ar+b = 0 & r = "a±%
a2"4b2 =
±-. Thus yc = Ae(x + Be"(x. UC= {xn, xn"1, . . . , 1} $ yp =Rn
i=0 dnxn
(b) yc = Ae"2x + Be"x
UC= {x2, x, 1}yp = "x2 + 4x " 1
2
39. (a)
0 / x / 1 | x > 1yc = A cosx + B sin x | yc = A cosx + B sin x
UC = {x, 1} | UC = {1}yp = Cx + D | yp = C
0 | 0y!!
p = 0 | yp = 1yp = x | yp = 1
y = A cosx + B sin x + x | y = A cosx + B sin x + 1
y =O
A cosx + B sin x + x, 0 / x / 1A cosx + B sin x + 1, x > 1
(b)
y(0) = A = 0y!(0) = B + 1 = 1 & B = 0
y =O
x, 0 / x / 11, x > 1
146 Section 4.6
4.6 Method of Undetermined Coe"cients via the Annihila-tor Method
1.
y!! + y = e"x + x2
(D2 + 1)y =(Annihilator or rhs = (D + 1)(D3)
$
(D2 + 1)(D + 1)(D3)y = 0r3(r + 1)(r2 + 1) = 0
$ y = A + Bx + Cx2 + De"x + E cosx + F sin x
yc = a cosx + b sin x
$ yp = A + Bx + Cx2 + De"x
2.
y!! + 3y! = x2e2x + cosx
(D2 + 3D)y =(Annihilator or rhs = (D " 2)3(D2 + 1)
$
(D2 + 3D)(D " 2)3(D2 + 1)y = 0r(r + 3)(r " 2)3(r2 + 1) = 0
$ y = A + Be"3x + Ce2x + Dxe2x
+Ex2e2x + F cosx + G sin x
yc = a + be"3x
$ yp = Ce2x + Dxe2x + Ex2e2x + F cosx + G sin x
3.
y!! + y = 4 sinx + ex cosx
(D2 + 1)y =(Annihilator or rhs = (D2 " 2D + 2)(D2 + 1)
$
(D2 + 1)2(D2 " 2D + 2)y = 0(r2 + 1)2(r2 " 2r + 2) = 0
$ y = A sinx + B cosx + Cx sin x + Dx cos x
+Eex sin x + Fex cosx
yc = a cosx + b sinx
$ yp = Cx sin x + Dx cosx
+Eex sin x + Fex cosx
Section 4.6 147
4.
y!! + 4y! + y = sin 2x + cosx
(D2 + 4D + 1)y =(Annihilator or rhs = (D2 + 1)(D2 + 4)
$
(D2 + 1)(D2 + 4)(D2 + 4D + 1)y = 0(r2 + 1)(r2 + 4)(r2 + 4r + 1) = 0
$ y = A cosx + B sin x
+C cos 2x + D sin 2x
+Ee("2+%
3)x + Fe("2"%
3)x
yc = ae("2+%
3)x + be("2"%
3)x
$ yp = A cosx + B sin x
+C cos 2x + D sin 2x
5.
y!!! + y = xe2x cosx + sinx
(D3 + 1)y =(Annihilator or rhs = (D2 + 1)(D2 " 4D + 5)2)
$
(D2 + 1)(D2 " 4D + 5)2(D3 + 1)y = 0(r2 + 1)(r2 " 4r + 5)2(r3 + 1) = 0
$ y = Ae"x + B cosx + C sin x
+De2x cos 2x + Ee2x sin 2x
+Fex/2 cos('
32
)x
+Gex/2 sin('
32
)x
yc = ae"x + bex/2 cos('
32
)x
+cex/2 sin('
32
)x
$ yp = B cosx + C sin x
+De2x cos 2x + Ee2x sin 2x
6.
y!! " y = ex + xe"x
(D2 " 1)y =(Annihilator or rhs = (D " 1)(D + 1)2
$
(D2 " 1)(D " 1)(D + 1)2y = 0(r " 1)2(r + 1)3 = 0
$ y = Aex + Bxex
+Ce"x + Dxe"x + Ex2e"x
yc = aex + be"x
$ yp = Bxex + Dxe"x + Ex2e"x
148 Section 4.6
7.
y!!! + 8y = 4 sinx + cosx
(D3 + 8)y =(Annihilator or rhs = (D2 + 1)
$
(D2 + 1)(D3 + 8)y = 0(r3 + 8)(r2 + 1) = 0
$ y = A cosx + B sinx + Ce"2x
+Dex cos'
3x + Eex sin'
3xyc = ae"2x + bex cos
'3x + cex sin
'3x
$ yp = A cosx + B sinx
8.
y(5) + y!! = xex cosx + x3
(D5 + D2)y =(Annihilator or rhs = D4(D2 " 2D + 2)2
$
(D5 + D2)(D4)(D2 " 2D + 2)2y = 0(r5 + r2)(r4)(r2 " 2r + 2)2 = 0
$ y = A + Bx + Cx2 + Dx3 + Ex4 + Fx5
+Gex cosx + Hxex cosx
+Iex sin x + Jxex sin x
+Kex2 cos(
'3
2)x + Le
x2 sin(
'3
2)x
+Me"x
yc = a + bx + cex2 cos(
'3
2)x
+dex2 sin(
'3
2)x + fe"x
$ yp = Cx2 + Dx3 + Ex4 + Fx5
+Gex cosx + Hxex cosx
+Iex sin x + Jxex sin x
9. y!! + 4y! + 13y = e"2x. We have
yp = Ae"2x $ y!p = "2Ae"2x, y!!
p = 4Ae"2x
y!!p + 4y!
p + 13yp = (4Ae"2x) + 4("2Ae"2x) + 13(Ae"2x) = e"2x
$ 4A " 8A + 13A = 1 $ 9A = 1 $ A =19
10. y!! + 4y! = x2 " 3, yp = Ax3 + Bx2 + Cx. Substitution gives
12Ax2 + (8B + 6A)x + 4C + 2B = x2 " 3
Section 4.6 149
Equating coe!cients of like terms,
12A = 18B + 6A = 04C + 2B = "3
so A = 112 , B = " 1
16 , C = " 2332 .
11. y!! " 4y! = cosx + 3 sinx, yp = A cosx + B sin x. Substitution gives
("A " 4B) cosx + ("B + 4A) sin x = cosx + 3 sinx
Equating coe!cients of like terms,
"A " 4B = 1"B + 4A = 3
so A = 1117 , B = " 7
17 .
12. A = 1/8, B = 0, C = 0, E = "1/35
13. A = 0, B = 8/441, C = 1/21, E = 0
14. y!! + 2y! + 17y = ex + 2, yp = Aex + B. Substituting into the ODE and equating coe!cientsgives 20Aex + 17B = ex + 2, so A = 1
20 , B = 217 .
15. A = "5/36, B = "1/6, C = 1/5, E = 3/50
16. y!! " 3y! + 2y = 4x2; Homogeneous solution: r2 " 3r + 2 = 0 or (r " 2)(r " 1) = 0, so r = 2, 1.Two linearly independent solutions are e2x, ex. We have P (D) = (D " 2)(D " 1). What givesrise to x2 on the right-hand side?
r = 0 (3 times) $ Q(D) = D3
Q(D)P (D)(y) = 0 $ D3(D " 2)(D " 1)(y) = 0$ r3(r " 2)(r " 1) = 0
Roots are then r = 0 (3 times), 2, 1, which gives the linearly independent solutions 1, x, x2,e2x, ex. Ignoring those that were part of the homogeneous solution,
yp = a1 + a2x + a3x2
Then
y!p = a2 + 2a3x
y!!p = 2a3
Substituting into the ODE we get
(2a3) " 3(a2 + 2a3x) + 2(a1 + a2x + a3x2) = 4x2
Equating coe!cients gives
2a3 " 3a2 + 2a1 = 0"6a3 + 2a2 = 0
2a3 = 4
Thus a3 = 2, a2 = 6, and a1 = 7. The general solution is then
yh + yp = C1e2x + C2e
x + 7 + 6x + 2x2
150 Section 4.6
17. y!! " 2y! " 8y = 4e2x " 21e"3x; homogeneous solution r2 " 2r " 8 = 0 or (r " 4)(r + 2) = 0, sor = 4,"2. The fundamental set is {e4x, e"2x}, so yh = C1e4x + C2e"2x. Particular solution:UC set is {e2x, e"3x}, which does not need modification, so yp = Ae2x + Be"3x. Substitutionand simplification gives
"8Ae2x + 7Be"3x = 4e2x " 21e"3x
Equating
"8A = 47B = "21
gives A = " 12 , B = "3, so
yp = "12e2x " 3e"3x
The general solution is
y = C1e4x + C2e
"2x " 12e2x " 3e"3x
18. yc = Ae"x sin 2x + Be"x cos 2xyp = " cos 2x + 2 sin 2x
19. yc = Ae"4x + Be"3x cosx + Ce"3x sin x
yp =14x2e"4x +
12xe"4x " 1
2xe"3x cosx +
12xe"3x sin x
20. yc = C1e2x + C2e"2x
yp = "8x
y =72e2x " 7
2e"2x " 8x
21. y!! " 2y! + 2y = ex + x cosx; Homogeneous solution: r2 " 2r + 2 = 0, so
r =2 ±.
4 " 4(2)2
= 1 ± i
Linearly independent solutions are ex cosx, ex sin x. For particular solution, ex comes fromr = 1 $ (D " 1). x cos x comes from r = ±i (twice), which leads to
(r2 + 1)2 = 0 $ (D2 + 1)2
Q(D) = (D " 1)(D2 + 1)2
Q(D)P (D)(y) = 0
Linearly independent solutions that come from the characteristic equation are ex, cosx, sinx,x cosx, x sin x. Ignoring those from the homogeneous solution, ex cosx, ex sinx. This gives
yp = a1ex + a2 cosx + a3 sin x + a4x cosx + a5x sin x
y!p = a1e
x " a2 sinx + a3 cosx + a4(cosx " x sin x) + a5(sin x + x cosx)y!!
p = a1ex " a2 cosx " a3 sin x + a4(" sin x " sinx " x cos x)
+ a5(cosx + cosx " x sin x)
Substituting into the ODE gives
ex(a1 " 2a1 + 2a1) + (cosx)(a2 " 2a5 " 2a3 + 2a4)+ (sin x)(a3 " 2a4 + 2a2 " 2a5) + (x sin x)(a4 + 2a5)+ (x cos x)(a5 " 2a4) = ex + x cosx
Section 4.6 151
Equating coe!cients,
a1 = 1a2 " 2a3 + 2a4 " 2a5 = 02a2 + a3 " 2a4 " 2a5 = 0
a4 + 2a5 = 0"2a4 + a5 = 1
which gives a1 = 1, a2 = 225 , a3 = " 14
25 , a4 = 15 , a5 = " 2
5 , so
yp = ex +225
cosx " 1425
sin x " 25x sin x +
15x cos x
The general solution isC1e
x cosx + C2ex sinx + yp
22. y!! + 6y! + 10y = 3xe"3x " 2e3x cosx; Homogeneous solution: r2 + 6r + 10 = 0,
r ="6 ±
.36 " 4(10)2
= "3 ± i
Linearly independent solutions e"3x cosx, e"3x sin x. The annihilator of xe"3x is (D +3)2 andof e3x cosx is (D2 " 6D + 10). We calculate the characteristic equation for Q(D)P (D), and,ignoring the linearly independent solutions that occurred in the homogeneous part we havee"3x, xe"3x, e3x cosx, e3x sin x. Thus
yp = a1e"3x + a2xe"3x + a3e
3x cosx + a4e3x sin x
y!p = "3a1e
"3x + a2(e"3x " 3xe"3x) + a3(3e3x cosx " e3x sin x)
+ a4(3e3x sin x + e3x cosx)y!!
p = 9a1e"3x + a2("3e"3x " 3e"3x + 9xe"3x)
+ a3(9e3x cosx " e3x sin x " 3e3x sin x " e3x cosx)+ a4(9e3x sin x + 3e3x cosx + 3e3x cosx " e3x sin x)
Substituting into the ODE and equating coe!cients gives
e"3x: 10a1 " 18a1 + 6a2 + 9a1 " 6a2 = 0xe"3x: 10a2 " 18a2 + 9a2 = 3
e3x cosx: 10a3 + 18a3 + 6a4 + 8a3 + 6a4 = "2e3x sin x: 10a4 " 6a3 + 18a4 " 6a3 + 8a4 = 0
which gives a1 = 0, a2 = 3, a3 = " 120 , a4 = " 1
60 , so
yp = 3xe"3x " 120
e3x cosx " 160
e3x sin x
The general solution isC1e
"3x cosx + C2e"3x sinx + yp
23. yc = Ae4x cos 2x + Be4x sin x
yp =516
xe4x sin x " 58x2e4x cos 2x
152 Section 4.6
24. yc = Ae"5x + Be"2x
yp = "134 xe"2x cos 5x + 63
7225e"2x cos 5x + 3170xe"2x sin 5x + 5
578e"2x sin 5x
25. yc = Aex cos 2x + Bex sin 2x
yp =12xex " 1
4xex sin 2x
26. yc = Aex + Bxex
y =x3
3ex " ex
4sin 2x
27. yc = Ae2x + Bex
y = ex " xex
28. yc = Aex + Be"x
yp = xex + xe"x
29. yc = Ae"3x + Be"x
yp =ex
16+
xe"x
430. yc = A cos 2x + B sin 2x
yp = 134ex sin 2x " 2
17ex cos 2x " 134e"x sin 2x " 2
17e"x cos 2x
31. yc = Ae"x sin x + Be"x cosxyp = 1
16ex sin x " 116ex cosx " 1
4xe"x cosx
32. yc = C1e3x + C2xe3x + C3e"3x + C4xe"3x
yp = 172x2e3x
33. (a) yc = Ae4x
yp = "x2
4" x
8" 1
32(b) yc = Ae"x
yp =15
cos 2x +25
sin 2x
(c) yc = Aex
yp =e4x
336.
A(w) =F0m.
4'2w2 + (w2 " w20)2
A!(w) = "F0
m
!8'2w + 4w(w2 " w2
0)4'2w2 + (w2 " w2
0)3
"
A!(w) = 0.
0 = 8'2w + 4w3 " 4ww20
= 4w3 + 4w(2'2 " w20)
= 4w(w2 + 2'2 " w20)
w = 0 or w2 = w20 " 2'2
w =:
w20 " 2'2
Section 4.6 153
37. (a)(b) If b = 0, the terms of yc must involve sin $ and cos $
$ limx#$(y2 " y1) may not exist.(c) yc has terms involving e"Ax, UC= {1} & cyp = k & yp = k
c . Therefore y = yp + yc =kc +terms of e"Ax and limx#$ y = k
c
(d) r(ar + b) = 0 & yc = A + Be"ba x
UC= {1} modify & {x} & yp = Cx, bC = k & C = k/b & yp = kb x $ y =
A + Be"ba x + k
b x
(e) ay!! = k & y!! = ka & y! = k
ax + C & y = k2ax2 + Cx + D
limx#$ y = #
38. (a) y!!+ay!+by =Rn
i=0 cnxn. Homogeneous solution from r2+ar+b = 0 & r = "a±%
a2"4b2 =
±-. Thus yc = Ae(x + Be"(x. yp =Rn
i=0 dnxn
(b) yc = Ae"2x + Be"x
yp = "x2 + 4x " 12
39. (a)
0 / x / 1 | x > 1yc = A cosx + B sin x | yc = A cosx + B sin x
UC = {x, 1} | UC = {1}yp = Cx + D | yp = C
0 | 0y!!
p = 0 | yp = 1yp = x | yp = 1
y = A cosx + B sin x + x | y = A cosx + B sin x + 1
y =O
A cosx + B sin x + x, 0 / x / 1A cosx + B sin x + 1, x > 1
(b)
y(0) = A = 0y!(0) = B + 1 = 1 & B = 0
y =O
x, 0 / x / 11, x > 1
154 Section 4.7
4.7 Variation of Parameters
1.
yc = cx + d
yp = u1x + u2Ou!
1x + u!2 = 0
u!1 = 1
1+x2
u!1 =
11 + x2
u!2 = " x
1 + x2
u1 = tan"1 x
u2 = ln!
11 + x2
"
yp = x tan"1 x + ln!
11 + x2
"
2.
yc = Aex + Bxex
yp = u1ex + u2xex
Ou!
1ex + u!
2xex = 0u!
1ex + u!
2(ex(x + 1)) = ex'x
u!1 = "x3/2
u!2 =
'x
u1 = "25x5/2
u2 =23x3/2
yp =415
x5/2ex
3.
yc = A sinx + B cosx
yp = u1 sin x + u2 cosxO
u!1 sin x + u!
2 cosx = 0u!
1 cosx " u!2 sin x = sec x
u!1 = 1
u!2 = " tanx
u1 = x
u2 = ln(cosx)yp = x sin x + cosx ln(cosx)y = A sinx + B cosx + x sin x + cosx ln(cosx)
y(0) = B = 1y!(0) = A = 2
y = 2 sinx + cosx + x sin x + cosx ln(cosx)
Section 4.7 155
4.
yc = A sinx + B cosx
yp = u1 sin x + u2 cosxO
u!1 sin x + u!
2 cosx = 0u!
1 cosx " u!2 sin x = tan x
u!1 = sinx
u!2 = " sinx tan x
u1 = " cosx
u2 = "2 tanh"1,tan,x
2
--+ sin x
yp = "2 tanh"1,tan,x
2
--
y = A sinx + B cosx " 2 tanh"1,tan,x
2
--
y(0) = B = "1y!(0) = A " 1 = 1 & A = 2
y = 2 sinx " cosx " 2 tanh"1,tan,x
2
--cosx
5.
yc = A sin x + B cosx
yp = u1 sin x + u2 cosxO
u!1 sin x + u!
2 cosx = 0u!
1 cosx " u!2 sin x = sec2 x
u!1 = sec x
u!2 = " secx tan x
u1 = " cosx
u2 = 2 tanh"1,tan,x
2
--
yp = 2 tanh"1,tan,x
2
--sin x " 1
y = A sin x + B cosx " 2 tanh"1,tan,x
2
--
y(0) = B " 1 = 0 & B = 1y!(0) = A = 1
y = sin x + cosx + 2 tanh"1,tan,x
2
--sinx " 1
156 Section 4.7
6.
yc = A sin 2x + B cos 2x
yp = u1 sin 2x + u2 cos 2xO
u!1 sin 2x + u!
2 cos 2x = 02u!
1 cos 2x " 2u!2 sin 2x = sec x
u!1 =
cos 2x
2 sinx
u!2 = " sin 2x
2 sinx= " cosx
u1 = cosx +12
ln,tan
x
2
-
u2 = " sinx
yp = sin 2x cosx + sin x cosx ln,tan
x
2
-" cos 2x sin x
7.
yc = Aex + Bxex
yp = u1ex + u2xex
Ou!
1ex + u!
2xex = 0u!
1ex + u!
2(xex + ex) = ex
x2
u!1 = " 1
x
u!2 =
1x2
u1 = " lnx
u2 = " 1x
yp = "ex ln x " ex
8.
yc = Ae"x + Bxe"x
yp = u1e"x + u2xe"x
Ou!
1e"x + u!
2xe"x = 0"u!
1e"x + u!
2("xe"x + e"x) = e!x
x4
u!1 = " 1
x3
u!2 =
1x4
u1 =1
2x2
u2 = " 13x3
yp =e"x
6x2
Section 4.7 157
9. y!! " 7y! + 10y = e3x; Homogeneous part: r2 " 7r + 10 = 0, or (r " 2)(r " 5) = 0 and r = 2, 5.Linearly independent solutions are e2x, e5x. Let #1 = e2x, #2 = e5x. Then
u!1 =
det&
0 e5x
e3x 5e5x
'
det&
e2x e5x
2e2x 5e5x
' ="e8x
5e7x " 2e7x=
"ex
3$ u1 = "1
3ex
u!2 =
det&
e2x 02e2x e3x
'
3e7x=
e5x
3e7x=
13e"2x $ u2 = "1
6e"2x
This gives
yp = u1#1 + u2#2 = "13exe2x +
!"1
6
"e"2xe5x = "1
3e3x " 1
6e3x = "1
2e3x
Then the general solution is
y = C1e2x + C2e
5x " 12e3x
y(0) = C1 + C2 " 12 = 1, y!(0) = 5C1 + 2C2 " 3
2 = 2 $ C1 = 16 , C2 = 4
3
$ y =e5x
6+
4e2x
3" e3x
210.
yc = Ae"3x + Be"2x
yp = u1e"3x + u2e
"2x
Ou!
1e"3x + u!
2e"2x = 0
"3u!1e
"3x " 2u!2e
"2x = e"x
u!1 = "e2x
u!2 = ex
u1 = "12e2x
u2 = ex
yp =e"x
2
y = Ae"3x + Be"2x +12e"x
y(0) = A + B +12
= 1
y!(0) = "3A " 2B " 12
= 2
A = "72
B = 4
y = "72e"3x + 4e"2x +
e"x
2
158 Section 4.7
11.
yc = A sin x + B cosx
yp = u1 sinx + u2 cosxO
u!1 sin x + u!
2 cosx = 0u!
1 cosx " u!2 sin x = 6x
u!1 = 6x cosx
u!2 = "6x sinx
u1 = 6 cosx + 6x sin x
u2 = 6x cosx " 6 sinx
yp = 6x
y = A sin x + B cosx + 6x
y(0) = B = 1y!(0) = A + 6 = 1
A = "5 B = 1y = "5 sinx + cosx + 6x
12.
yc = A sin x + B cosx
yp = u1 sinx + u2 cosxO
u!1 sin x + u!
2 cosx = 0u!
1 cosx " u!2 sin x = sin2 x
u!1 = cosx sin2 x
u!2 = " sin3 x
u1 =sin3 x
3
u2 =cos 3x
12" 3 cosx
4
yp =sin4 x
3" cosx cos 3x
12+
3 cos2 x
4y = A sin x + B cosx + yp
y(0) = B " 112
+34
= 1
y!(0) = A = 0
A = 0 B =13
y =cosx
3+ yp
Section 4.7 159
13.
yc = Aex + Bxex + Cx2ex
yp = u1ex + u2xex + u3x
2ex
AC
D
u!1e
x + u!2xex + u!
3x2ex = 0
u!1e
x + u!2(xex + ex) + u!
3(x2ex + 2xex) = 0u!
1ex + u!
2(xex + 2ex) + u!3(x2ex + 4xex + 2ex) = ex
x
u!1 =
x
2u!
2 = "1
u!3 =
12x
u1 =x2
4u2 = "x
u3 =12
ln x
yp = "34x2ex +
12x2 ln xex
14.
yc = A + Be"x + Ce2x
yp = u1 + u2e"x + u3e
2x
AC
D
u!1 + u!
2e"x + u!
3e2x = 0
"u!2e
"x + 2u!3e
2x = 0u!
2e"x + 4u!
3e2x = x3
u!1 = "x3
2
u!2 =
exx3
3
u!3 =
e"2xx3
6
u1 = "x4
8
u2 =x3ex
3" x2ex + 2xex " 2ex
u3 = "x3e"2x
12" x2e"2x
8" xe"2x
8" e"2x
16
yp = "x4
8+
x3
4" 9x2
8+
15x
8" 17
16
160 Section 4.7
15.
yc = A sin 3'
3x + B cos 3'
3xyp = u1 sin 3
'3x + u2 cos 3
'3x
Ou!
1 sin 3'
3x + u!2 cos 3
'3x = 0
3'
3u!1 cos 3
'3x " 3
'3u!
2 sin 3'
3x = e3x
u!1 =
e3x cos 3'
3x
3'
3
u!2 = "e3x sin 3
'3x
3'
3
u1 =e3x
36'
3
,cos 3
'3x +
'3 sin 3
'3x-
u2 =e3x
36'
3
,'3 cos 3
'3x " sin 3
'3x-
yp =e3x
36
16.
yc = A sinx + B cosx + C
yp = u1 sin x + u2 cosx + u3AC
D
u!1 sin x + u!
2 cosx + u!3 = 0
u!1 cosx " u!
2 sin x = 0"u!
1 sinx " u!2 cosx = tanx
u!1 = " sinx tan x
u!2 = " sinx
u!3 = sinx tan x + tan2 x
u1 = sinx " 2 tanh("1),tan
x
2
-
u2 = cosx
u3 = sinx " 2 tanh("1),tan
x
2
-+ tanx " x
yp = 1 " 4 tanh("1),tan
x
2
-+ sinx + tanx " x
Section 4.7 161
17.
y(4) " 16y = e4x =$ y = C1 sin(2x) + C2 cos(2x) + C3e2x + C4e
"2x +1
240e4x
18. y!! +4y! +3y = 65 cos(2x). The characteristic equation is r2 +4r +3 = 0 or (r +1)(r +3) = 0,so r = "1,"3. This gives
yh = C1e"x + C2e
"3x
Variation of parameters: Let #1 = e"x, #2 = e"3x. Then
u!1 =
det&
0 #2
65 cos(2x) #!2
'
det #2
#!1 #!2
' ="65e"3x cos(2x)
"2e"4x
u!2 =
det 0#!1 65 cos(2x)
'
det #2
#!1 #!2
' =65e"x cos(2x)
"2e"4x
u1 =%
u!1 dx =
132
ex (cos(2x) + 2 sin(2x))
u2 =%
u!2 dx = "15
2e3x cos(2x) " 5e3x sin(2x)
This givesyp = u1#1 + u2#2
The general solution is
y = C1e"x + C2e
"3x " cos(2x) + 8 sin(2x)
Undetermined coe!cients: Roots that gave cos(2x) are ±2i, which gives Q(D) = D2 + 4; so
Q(D)P (D)y = 0 $ (r2 + 4r + 3)(r2 + 4) = 0$ r = "1,"3,±2i
$ yp = a1 cos(2x) + a2 sin(2x)
Plug into the ODE and simplify:
("a1 + 8a2) cos(2x) + ("a2 " 8a1) sin(2x) = 65 cos(2x)
Equating coe!cients:
"a1 + 8a2 = 65"a2 " 8a1 = 0
so a2 = "8a1, "a1 = 8("8a1) = 65, so a1 = "1 and a2 = 8. Then
yp = " cos(2x) + 8 sin(2x)
162 Section 4.7
19. Plugging both functions into the left-hand side of the DE results in a solution to the homoge-neous DE.
yp = u1 + u2(xe"x + e"x)O
u!1 + u!
2(xe"x + e"x) = 0u!
2("xe"x) = ex
1+x
u!1 =
ex
x
u!2 = " e2x
x(1 + x)
u1 =%
ex
xdx
u2 =% "e2x
x(1 + x)yp = u1 + u2(xe"x + e"x)
20. Plugging both functions into the left-hand side of the DE results in a solution to the homoge-neous DE.
yp = u1x + u2x ln xO
u!1x + u!
2x ln x = 0u!
1 + u!2(1 + lnx) = 1
x
u!1 = " ln x
x
u!2 =
1x
u1 = "12
ln2 x
u2 = lnx
yp =x
2ln2 x
21. Plugging both functions into the left-hand side of the DE results in a solution to the homoge-neous DE.
Section 4.7 163
yp = u1 sin(ln x) + u2 cos(ln x)O
u!1 sin(ln x) + u!
2 cos(ln x) = 0u!
1cos(ln x)
x " u!2
sin(ln x)x = 1
x
u!1 = cos(ln x)
u!2 = " sin(ln x)
u1 =x
2cos(ln x) +
x
2sin(ln x)
u2 =x
2cos(ln x) " x
2sin(ln x)
yp =x
2
22. Plugging all three functions into the left-hand side of the DE results in a solution to the ho-mogeneous DE.
We need to rewrite the DE as:
y!!! +3x
y!! =1
2x2
yp = u1 + u2
!1x
"+ u3x
AC
D
u!1 + u!
2
#1x
$+ u!
3x = 0"u!
2
#1x2
$+ u!
3 = 02u!
2
#1x3
$= 1
2x2
u!1 =
12
u!2 =
x
4
u!3 =
14x
u1 =x
2
u2 =x2
8
u3 =14
ln x
yp =5x
8+
x
4ln x
23. As the exercise is written, u(x) = yc(x) and v(x) = yp(x)
24. Note that yc will come from Cauchy-Euler material earlier, but variation of parameters will
164 Section 4.7
require dividing everything by x2.
yc =A
x+
B
x2
yp =u1
x+
u2
x2E
u"1
x + u"2
x2 = 0"u"
1x2 " 2u"
2x3 = ex
x2
u!1 = ex
u!2 = "xex
u1 = ex
u2 = "xex + ex
yp =ex
x2
25. Note that yc will come from Cauchy-Euler material earlier, but variation of parameters willrequire dividing everything by x2.
yc =A
x+
B
xln x
yp =u1
x+
u2
xln x
Eu"1
x + u"2
x ln x = 0"u"
1x2 + u!
21"ln x
x2 = ln xx2
u!1 = " ln2 x
u!2 = lnx
u1 = "2x + 2x ln x " x ln2 x
u2 = x ln x " x
yp = lnx " 2
26. Note that yc will come from Cauchy-Euler material earlier, but variation of parameters willrequire dividing everything by x2.
yc = A sin(2 ln x) + B cos(2 ln x)yp = u1 sin(2 lnx) + u2 cos(2 ln x)
Ou!
1 sin(2 lnx) + u!2 cos(2 ln x) = 0
u!1 cos(2 lnx)( 2
x ) + u!2 sin(2 ln x)( 2
x ) = 2x2
u!1 =
cos(2 ln x)x
u!2 = " sin(2 lnx)
x
u1 =sin(2 ln x)
2
u2 =cos(2 ln x)
2
yp =12
4.8. CHAPTER 4: ADDITIONAL PROBLEMS 165
4.8 Chapter 4: Additional Problems
1. False, this is true with constant coe!cients.
2. True.
3. False, it corresponds to the particular solution.
4. False.
5. False.
6. y(x) = c1e2x + c2e3x
8. y(x) = (3x + 1)e"3x
9. y(x) =12e"4x(3e2x " 1)
11. y(x) =13e2(x"!/2)(5 cos 3x " 6 sin 3x)
13. y(x) = c1e"x + c2 + c3x + c4e4x
15. y(x) = c1e"x + c2xe"x + c3x2e"x + c4
19. y(x) = c1e"7x + c2e"2x + c3 + c4e2x
21. y(x) =c1
x+
c2'x
22. y =c1
x+ c2
'x
23. y = c1 sin#'
2 lnx$
+ c2 cos#'
2 lnx$)
25. y(x) = c1e4x + c2 "
117
(4 sinx + cosx)
27. y(x) = c1e2x + c2xe2x +
116
e"2x
29. y(x) = c1 sin(5x
2) + c2 cos(
5x
2) +
229
e"x
30. yh + yp = C1e"x cos 4x + C2e
"x sin 4x +18xe"x sin 4x
31. y(x) = c1e"5x + c2e
2x " 1100
(10 + 3e3x + 10xe3x)
33. Overdamped.
34. Underdamped.
35. Overdamped.
36. Overdamped.
37. Critically damped.
38. Critically damped.
166 Chapter 4 Review
39. b < 2'
3
40. b < 4
41. k <18
42. k >94
43. m =14
44. m <4912
45. (a) y(x) = (3/4)e"x + (1/2)xe"x + (1/4)ex
(b) y(4) = 13.70
47. (a) y = (7/6) sin(2x) + cos(2x) + (1/3) sin(x),(b) y(4) = .7565
50. (a) y(x) = c1x + c2%
x
52. y(x) = c1 sin#'
2 lnx$
+ c2 cos#'
2 lnx$
Chapter 5
Fundamentals of Systems ofDi!erential Equations
5.1 Systems of Two Equations—Motivational Examples
1. (a) ! = !3 " Saddle
(b) ! = 6, " = 5, "2 ! 4! = 1 " Unstable Spiral1.(a) xy-plane 1.(b) xy-plane
2. (a) ! = 5," = 0, " Center(b) ! = 1, " = 0 " Center
2.(a) xy-plane 3.(b) xy-plane
167
168 Section 5.1
3. (a) ! = !5 " Saddle
(b) ! = !3 " Saddle3.(a) xy-plane 3.(b) xy-plane
4. (a) ! = 5," = 6,"2 ! 4! = 16 " Unstable Node
(b) ! = 13," = 6, "2 ! 4! ! 16 " Unstable Spiral4.(a) xy-plane 4.(b) xy-plane
5. (a) ! = 11," = 4, "2 ! 4! = !28 " Unstable Spiral
(b) ! = 8," = 4,"2 ! 4! = !16 " Unstable Spiral5.(a) xy-plane 5.(b) xy-plane
Section 5.1 169
6. (a) ! = 8," = !4,"2 ! 4! = !16 " Stable Spiral
(b) ! = 26," = !10, "2 ! 4! = !4 " Stable Spiral6.(a) xy-plane 6.(b) xy-plane
7. (a) ! = 7," = 4,"2 ! 4! = !12 " Unstable Spiral
(b) ! = !15 " Saddle7.(a) xy-plane 7.(b) xy-plane
8. (a) ! = !40 " Saddle
(b) ! = 1," = 2,"2 ! 4! = 0 " Unstable Node8.(a) xy-plane 8.(b) xy-plane
170 Section 5.1
9. (a) ! = 1," = !2,"2 ! 4! = 0 " Stable Node
(b) ! = 10," = !2,"2 ! 4! = !44 " Stable Spiral9.(a) xy-plane 9.(b) xy-plane
10. (a) ! = 4," = !4,"2 ! 4! = 0 " Stable Node
(b) ! = !18 " Saddle10.(a) xy-plane 10.(b) xy-plane
11. (a) ! = 3," = !4,"2 ! 4! = 4 " Stable Node
(b) ! = 29, " = !10, "2 ! 4! = !16 " Stable Spiral11.(a) xy-plane 11.(b) xy-plane
5.2. USEFUL TERMINOLOGY 171
12. (a) ! = 19," = !8,"2 ! 4! = !12 " Stable Spiral
(b) ! = 7, " = 8, "2 ! 4! = 36 " Unstable Node12.(a) xy-plane 12.(b) xy-plane
13. u1 = x, u2 = x!, u3 = x!! " x!! = !bx! ! kx, u3 = !bu2 ! ku1 " u!1 = u2 = 0u1 + u2, u
!2 =
!ku1 ! bu2 " ! = k," = !b,"2 ! 4! = b2 ! 4k "k > 0, b < 0 " Stablek > 0, b > 0 " Unstable
14. x = 2u1 + u2, y = u1 + u2 " u!1 = 2u1, u
!2 = 3u2 " u1 = c1e
2t, u2 = c2e3t " x = 2c1e
2t + c2e3t, y =
c1e2t + c2e3t
15. x = u1 + u2, y = u1 + 2u2 " u!1 = !u1, u!
2 = !3u2 " u1 = c1e"t, u2 = c2e"3t " x = c1e"t +c2e
"3t, y = c1e"t + 2c2e
"3t
16. x = 4u1, y = u1 + u2 " u!1 = 5u1, u
!2 = !3u2 " u1 = c1e
5t, u2 = c2e"3t " x = 4c1e
"t, y =c1e
"t + c2e"3t
17. x = u1+3u2, y = u1+u2 " u!1 = !2u1, u
!2 = 2u2 " u1 = c1e
"2t, u2 = c2e3t " x = c1e
"2t+3c2e2t, y =
c1e"2t + c2e
2t
18. x = u1 + u2, y = u1 + 3u2 " u!1 = !u1, u
!2 = u2 " u1 = c1e
"t, u2 = c2et " x = c1e
"t + c2et, y =
c1e"t + 3c2et
19. See Section 5.1.2. This is done verbatim there.
5.2 Useful Terminology
1.dAdt
=
!cos t et
2t 3
"
dBdt
=
!2 sin t cos t !te"t + e"t
0 1
"
%A dt =
!! cos t et
t3
33t2
2
"+
!C11 C12
C21 C22
"
%B dt =
! t2 ! 1
4 sin(2t) e"t(!1 ! t)
0 3t + t2
2
"+
!C11 C12
C21 C22
"
2.dAdt
=
!cos t et
2t 3
"
dBdt
=
!2e2t cos t ! e2t sin t 2te"t ! t2e"t
0 1t
"
%A dt =
!! cos t et
t3
33t2
2
"+
!C11 C12
C21 C22
"
%B dt =
!2e2t cos t
5 + e2t sin t5 e"t(!2 ! 2t ! t2)
3t !t + t ln t
"+
!C11 C12
C21 C22
"
172 Section 5.2
3.dAdt
=
1
30 !e"t 3e3t
1 0 6e3t
et !e"t 9e3t
5
7
dBdt
=
1
3cos t !te"t + e"t !e"t
0 0 6e3t
et 2t ! e"t ! sin t
5
7
%A dt =
1
23t !e"t e3t
3t2
2 0 2e3t
3et !e"t e3t
5
67 +
1
3C11 C12 C13
C21 C22 C23
C31 C32 C33
5
7
%B dt =
1
23! cos t e"t(!1 ! t) !e"t
0 0 2e3t
3
et t3
3 ! e"t sin t
5
67 +
1
3C11 C12 C13
C21 C22 C23
C31 C32 C33
5
7
4.dAdt
=
1
3e"t(cos t ! sin t !3e3t !3e"3t
12#
t4t3 ! sin t
43 t1/3 0 3te3t + e3t
5
7
dBdt
=
1
3sec2 t !te"t + e"t !e"t
0 0 6e3t
et 2t ! e"t ! sin t
5
7
%A dt =
1
23t !e"t e3t
3t2
2 0 2e3t
3et !e"t e3t
5
67 +
1
3C11 C12 C13
C21 C22 C23
C31 C32 C33
5
7
%B dt =
1
23! e!t
2 (cos t + sin t) ! e3t
3 ! e!3t
323 t3/2 t5
5 sin t37 t7/3 0 e3t( t
3 ! 19 )
5
67 +
1
3C11 C12 C13
C21 C22 C23
C31 C32 C33
5
7
5.dx1
dt=
!!4e4t
!8e4t
"= Ax1
dx2
dt=
!et
3et
"= Ax2
6.dx1
dt=
!4e"t
!2e"t
"= Ax1
dx2
dt=
!6e2t
!2e2t
"= Ax2
7.dx1
dt=
!!4e"2t
!6e"2t
"= Ax1
dx2
dt=
!!3e"3t
!6e"3t
"= Ax2
8.dx1
dt=
!4e2t
10e2t
"= Ax1
dx2
dt=
!et
3et
"= Ax2
9.dx1
dt=
1
3et
00
5
7 = Ax1
dx2
dt=
1
30
!e"t
0
5
7 = Ax2
dx3
dt=
1
3!2e2t
02e2t
5
7 = Ax3
Section 5.2 173
10.dx1
dt=
1
32e2t
2e2t
2e2t
5
7 #= Ax1
dx2
dt=
1
3e"t
!2e"t
!e"t
5
7 #= Ax2
dx3
dt=
1
3!2e"2t
00
5
7 #= Ax3
11.dx1
dt=
1
3!2e"2t
6e"2t
!2e"2t
5
7 = Ax1
dx2
dt=
1
3e"t
!2e"t
0
5
7 = Ax2
dx3
dt=
1
3et
0!et
5
7 = Ax3
12. If c1x!1 = c1Ax1, c2x
!2 = c2Ax2 then c1x
!1 + c2x
!2 = c1Ax1 + c2Ax2 = A(c1x1 + c2x2) " c1x1 + c2x2 is
a solution.
13. (a) W (t) =
8888et et
0 et
8888 = e2t #= 0 " Linearly Independent
(b) W (t) =
88881 21 !1
8888 = !3 #= 0 " Linearly Independent
(c) W (t) =
88883 !6!1 2
8888 = 0 " Linearly Dependent
14. (a) W (t) =
88883et !6e"t
!et 2e"t
8888 = 0 " Linearly Dependent
(b) W (t) =
8888!1 21 !1
8888 = !1 #= 0 " Linearly Independent
(c) W (t) =
888888
3 1 0!1 1 00 0 1
888888= 4 #= 0 " Linearly Independent
15. (a) W (t) =
888888
3et !3e2t 0et !e2t 00 0 e"t
888888= 0 " Linearly Dependent
(b) W (t) =
888888
2 !2 0!1 1 00 1 1
888888= 0 " Linearly Dependent
16. x!1 =
!e3t cos t + 3e3t sin t!e3t sin t + 3e3t cos t
"#= Ax1 =
!3e3t sin t + 2e3t cos t!2e3t sin t + 3e3t cos t
"" NOT a fundamental solution
set.
17. x!1 =
!2e3t cos 2t + 3e3t sin 2t!2e3t sin 2t + 3e3t cos 2t
"= Ax1
x!2 =
!!2e3t sin 2t + 3e3t cos 2t!2e3t cos 2t ! 3e3t sin 2t
"= Ax2 "
W (t) = !e3t #= 0 " Linearly Independent " this set IS a fundamental solution set.
18. x!1 =
!!5e5t
5e5t
"= Ax1
174 Section 5.3
x!2 =
!et
et
"= Ax2 "
W (t) = !2e6t #= 0 " Linearly Independent " this set IS a fundamental solution set.
19. x!1 =
!!e"t
e"t
"= Ax1
x!2 =
!!2e"t
!2e"t
"#= Ax2 =
!6e"t
!2e"t
"" this set is NOT a fundamental solution set.
20. x!1 =
1
3et
00
5
7 = Ax1
x!2 =
1
3e"t
!e"t
0
5
7 = Ax2
x!3 =
1
300
!2e"2t
5
7 = Ax3
W (t) =
888888
et !e"t 00 e"t 00 0 e"2t
888888= e"2t #= 0 " Linearly Independent " this set IS a fundamental solution
set.
21. x!1 =
1
300
!e"t
5
7 #= Ax1 =
1
300
!2e"t
5
7 " this set is NOT a fundamental solution set.
5.3 Linear Transformations and the Fundamental Subspaces
1. (a)
!3/2 00 3/2
"!13
"=
!3/29/2
"
(b) A =
!cos #/2 ! sin #/2sin #/2 cos #/2
"!13
"=
!0 !11 0
"!13
"=
!!31
"
(c) A =
!cos2 #/2 cos#/2 sin #/2
cos #/2 sin #/2 sin2 #/2
"!13
"=
!0 00 1
"!13
"=
!03
"
(d) A =
!2 cos2 #/2 ! 1 2 cos #/2 sin #/2
2 cos #/2 sin #/2 2 sin2 #/2 ! 1
"!13
"=
!!1 00 1
"!13
"=
!!13
"
2. (a)
!!2 00 !2
"!!1!2
"=
!24
"
(b) A =
!cos # ! sin #sin # cos #
"!!1!2
"=
!!1 00 !1
"!!1!2
"=
!12
"
(c) A =
!cos2 # cos # sin #
cos # sin # sin2 #
"!!1!2
"=
!1 00 0
"!!1!2
"=
!!10
"
(d) A =
!2 cos2 # ! 1 2 cos# sin #2 cos# sin # 2 sin2 # ! 1
"!!1!2
"=
!1 00 !1
"!!1!2
"=
!!12
"
3. (a)
!1/2 00 1/2
"!1!3
"=
!1/2!3/2
"
(b) A =
!cos #/3 ! sin #/3sin #/3 cos #/3
"!1!3
"=
!1/2 !
$3/2$
3/2 1/2
"!1!3
"=
!1/2 + 3
$2/2$
3/2 ! 3/2
"
(c) A =
!cos2 #/3 cos#/3 sin #/3
cos #/3 sin #/3 sin2 #/3
"!1!3
"=
!1/4
$3/4$
3/4 3/4
"!1!3
"=
/1"3
#3
4#3"94
0
Section 5.3 175
(d) A =
!2 cos2 #/3 ! 1 2 cos #/3 sin #/3
2 cos #/3 sin #/3 2 sin2 #/3 ! 1
"!1!3
"=
!!1/2
$3/2$
3/2 1/2
"!1!3
"
=
/"1"3
#3
2#3"32
0
4. (a)
!3 00 3
"!!23
"=
!!69
"
(b) A =
!cos #/3 ! sin #/3sin #/3 cos #/3
"!!23
"=
!1/2 !
$3/2$
3/2 1/2
"!!23
"=
!!1 ! 3
$3/2
!2$
3/2 + 3/2
"
(c) A =
!cos2 #/3 cos#/3 sin #/3
cos #/3 sin #/3 sin2 #/3
"!!23
"=
!1/4
$3/4$
3/4 3/4
"!!23
"
=
!!1/2 + 3
$3/4
!2$
3/4 + 9/4
"
(d) A =
!2 cos2 #/3 ! 1 2 cos #/3 sin #/3
2 cos #/3 sin #/3 2 sin2 #/3 ! 1
"!!23
"=
!!1/2
$3/2$
3/2 1/2
"!!23
"
=
!1 + 3
$3/2
!$
3 + 3/2
"
5. (a)
!!1 00 !1
"!21
"=
!!2!1
"
(b) A =
!cos!#/4 ! sin!#/4sin!#/4 cos!#/4
"!21
"=
! $2/2
$2/2
!$
2/2$
2/2
"!21
"=
! $2 +
$2/2
!$
2 +$
2/2
"
(c) A =
!cos2 0 cos 0 sin 0
cos 0 sin 0 sin2 0
"!21
"=
!1 00 1
"!21
"=
!21
"
(d) A =
!2 cos2 0 ! 1 2 cos 0 sin 02 cos 0 sin 0 2 sin2 0 ! 1
"!21
"=
!1 00 !1
"!21
"=
!2!1
"
6. (a)
!2 00 2
"!!31
"=
!!62
"
(b) A =
!cos!#/2 ! sin!#/2sin!#/2 cos!#/2
"!13
"=
!0 1!1 0
"!!31
"=
!13
"
(c) A =
!cos2 #/3 cos#/3 sin #/3
cos #/3 sin #/3 sin2 #/3
"!!31
"=
!1/4
$3/4$
3/4 3/4
"!!31
"
=
!!3/4
$3/4
!3$
3/4 + 3/4
"
(d) A =
!2 cos2 #/6 ! 1 2 cos #/6 sin #/6
2 cos #/6 sin #/6 2 sin2 #/6 ! 1
"!!31
"=
!1/2
$3/2$
3/2 !1/2
"!!31
"
=
!!3/2 +
$3/2
!3$
3/2 ! 1/2
"
7. A =
!2 cos2 $ ! 1 2 cos $ sin $2 cos $ sin $ 2 sin2 $ ! 1
""
A2 =
!(2 cos2 $ ! 1)2 + (2 cos $ sin $)2 B
B 4 cos2 $ sin2 $ + (2 sin2 $ ! 1)2
"
where B = 4 cos3 $ sin $ ! 2 cos $ sin $ + 4 cos $ sin3 $ ! 2 cos $ sin $
=
!(cos2 2$)2 + (sin2 2$) 2 cos $ sin $(2 cos2 $ ! 1 + 2 sin2 $ ! 1)
2 cos $ sin $(2 cos2 $ ! 1 + 2 sin2 $ ! 1) (cos2 2$)2 + (sin2 2$)
"=
!1 00 1
"
8. Column space A =
O!10
",
!01
"4, Null space A = %
Column space B =
O!1!3
"4, Null space B =
O!31
"4
176 Section 5.3
9. Column space A =
O!10
",
!01
"4, Null space A =
AC
D
1
34!35
5
7
FG
H
Column space B =
O!10
",
!01
"4, Null space B =
AC
D
1
3!1!11
5
7
FG
H
10. Column space A =
O!10
",
!01
"4, Null space A =
ABBC
BBD
1
223
3!504
5
667 ,
1
223
!1120
5
667
FBBG
BBH
Column space B =
O!10
",
!01
"4, Null space B =
ABBC
BBD
1
223
7!202
5
667 ,
1
223
!2110
5
667
FBBG
BBH
11. Column space A =
AC
D
1
310
5/4
5
7 ,
1
301
1/8
5
7
FG
H, Null space A = %
Column space B =
AC
D
1
31!21
5
7
FG
H, Null space B =
O!31
"4
12. Answers will vary
13. Answers will vary
14. Answers will vary
15. Use code from Section to complete.
16. NullSpace(A) =
AC
D
1
31!21
5
7
FG
H
ColumnSpace(A) =
AC
D
1
310!1
5
7 ,
1
3012
5
7
FG
H
RowSpace(A) =
AC
D
1
310!1
5
7 ,
1
3012
5
7
FG
H
LeftNullSpace(A) =
AC
D
1
31!21
5
7
FG
H
17. NullSpace(A) =%
ColumnSpace(A) =
AC
D
1
3100
5
7 ,
1
3010
5
7 ,
1
3001
5
7
FG
H
RowSpace(A) =
AC
D
1
3100
5
7 ,
1
3010
5
7 ,
1
3001
5
7
FG
H
LeftNullSpace(A) =%18. See Linear Algebra Text
19. See Linear Algebra Text
5.4. EIGENVALUES AND EIGENVECTORS 177
20. m = tan%
1m2 + 1
!1 ! m2 2m
2m m2 ! 1
"=
1tan2 %+ 1
!1 ! tan2 % 2 tan%
2 tan% tan2 %! 1
"
=1
sec2 %
!1 ! tan2 % 2 tan%
2 tan% tan2 %! 1
"
=
!cos2 %! sin2 % 2 sin% cos%2 sin% cos% sin2 %! cos2 %
"
=
!2 cos2 %! 1 2 sin% cos%2 sin % cos% 2 sin2 %! 1
"
5.4 Eigenvalues and Eigenvectors
1. (a) det(A ! &I) = &2 ! &! 2 = 0 " &1 = 2, &2 = !1
&1 = 2; (A ! &I)v1 = 0 " v1 =
!52
"
&2 = !1; (A! &I)v2 = 0 " v2 =
!11
"
(b) det(A ! &I) = &2 + 3&! 18 = 0 " &1 = !6, &2 = 3
&1 = !6; (A! &I)v1 = 0 " v1 =
!29
"
&2 = 3; (A ! &I)v2 = 0 " v2 =
!00
"
(c) det(A ! &I) = &2 + 4&+ 3 = 0 " &1 = !3, &2 = !1
&1 = !3; (A! &I)v1 = 0 " v1 =
!12
"
&2 = !1; (A! &I)v2 = 0 " v2 =
!11
"
2. (a) det(A ! &I) = (!2 ! &)(!5! &) = 0 " &1 = !2, &2 = !5
&1 = !2; (A! &I)v1 = 0 " v1 =
!10
"
&2 = !5; (A! &I)v2 = 0 " v2 =
!1!1
"
(b) det(A ! &I) = (!4 ! &)(!1! &) = 0 " &1 = !4, &2 = !1
&1 = !4; (A! &I)v1 = 0 " v1 =
!10
"
&2 = 3; (A ! &I)v2 = 0 " v2 =
!73
"
(c) det(A ! &I) = (1 ! &)(7 ! &) = 0 " &1 = 1, &2 = 7
&1 = 1; (A ! &I)v1 = 0 " v1 =
!10
"
&2 = 7; (A ! &I)v2 = 0 " v2 =
!!12
"
3. (a) det(A ! &I) = &2 ! 4&+ 3 = 0 " &1 = 3, &2 = 1
&1 = 3; (A ! &I)v1 = 0 " v1 =
!11
"
&2 = 1; (A ! &I)v2 = 0 " v2 =
!1!1
"
(b) det(A ! &I) = &2 ! 4 = 0 " &1 = 2, &2 = !2
&1 = 2; (A ! &I)v1 = 0 " v1 =
!11
"
178 Section 5.4
&2 = !2; (A! &I)v2 = 0 " v2 =
!1!1
"
(c) det(A ! &I) = &2 ! 2&+ 8 = 0 " &1 = 4, &2 = !2
&1 = 4; (A ! &I)v1 = 0 " v1 =
!11
"
&2 = !2; (A! &I)v2 = 0 " v2 =
!1!1
"
4. (a) det(A ! &I) = &2 ! 2&! 3 = 0 " &1 = 3, &2 = !1
&1 = 3; (A ! &I)v1 = 0 " v1 =
!11
"
&2 = !1; (A! &I)v2 = 0 " v2 =
!31
"
(b) det(A ! &I) = &2 ! 5&+ 6 = 0 " &1 = 3, &2 = 2
&1 = 3; (A ! &I)v1 = 0 " v1 =
!11
"
&2 = 2; (A ! &I)v2 = 0 " v2 =
!21
"
(c) det(A ! &I) = &2 + 2&! 15 = 0 " &1 = !5, &2 = 3
&1 = !5; (A! &I)v1 = 0 " v1 =
!!31
"
&2 = 3; (A ! &I)v2 = 0 " v2 =
!11
"
5. (a) det(A ! &I) = &2 ! 2&! 3 = 0 " &1 = 3, &2 = !1
&1 = 3; (A ! &I)v1 = 0 " v1 =
!13
"
&2 = !1; (A! &I)v2 = 0 " v2 =
!1!1
"
(b) det(A ! &I) = &2 ! 6&+ 5 = 0 " &1 = 5, &2 = 1
&1 = 5; (A ! &I)v1 = 0 " v1 =
!11
"
&2 = 1; (A ! &I)v2 = 0 " v2 =
!!13
"
(c) det(A ! &I) = &2 + 2&! 5 = 0 " &1 = !1 +$
6, &2 = !1 !$
6
&1 = !1 +$
6; (A ! &I)v1 = 0 " v1 =
!1
2 +$
6
"
&2 = !1 !$
6; (A ! &I)v2 = 0 " v2 =
!!1
!2 +$
6
"
6. (a) det(A ! &I) = (5 ! &)(!3 ! &) = 0 " &1 = 5, &2 = !3
&1 = 5; (A ! &I)v1 = 0 " v1 =
!41
"
&2 = !3; (A! &I)v2 = 0 " v2 =
!01
"
(b) det(A ! &I) = &2 ! 3&! 40 = 0 " &1 = 8, &2 = !5
&1 = 8; (A ! &I)v1 = 0 " v1 =
!61
"
&2 = !5; (A! &I)v2 = 0 " v2 =
!1!2
"
(c) det(A ! &I) = &2 + 2&+ 1 = 0 " &1 = !1
&1 = !1; (A! &I)v2 = 0 " v2 =
!1!1
"
Section 5.4 179
7. (a) det(A ! &I) = &2 + 1 = 0 " &1 = i, &2 = !i
&1 = i; (A ! &I)v1 = 0 " v1 =
!1!i
"" v2 =
!1!i
"
(b) det(A ! &I) = &2 + 2&+ 5 = 0 " &1 = !1 + 2i, &2 = !1 ! 2i
&1 = !1 + 2i; (A ! &I)v1 = 0 " v1 =
!1i
"" v2 =
!1!i
"
8. (a) det(A ! &I) = &2 + 1 = 0 " &1 = i, &2 = !i
&1 = i; (A ! &I)v1 = 0 " v1 =
!2
3+i10
"v2 = 0 " v2 =
!2
3"i10
"
(b) det(A ! &I) = &2 ! 6&+ 13 = 0 " &1 = 3 + 2i, &2 = 3 ! 2i
&1 = 3 + 2i; (A! &I)v1 = 0 " v1 =
! "1+i21
"" v2 =
! "1"i21
"
9. (a) det(A ! &I) = &2 + 4&+ 8 = 0 " &1 = !2 + 2i, &2 = !2 ! 2i
&1 = !2 + 2i; (A ! &I)v1 = 0 " v1 =
!1
1"2i5
"" v2 =
!1
1+2i5
"
(b) det(A ! &I) = &2 + 4&+ 8 = 0 " &1 = !2 + 2i, &2 = !2 ! 2i
&1 = !2 + 2i; (A ! &I)v1 = 0 " v1 =
!1
! 5+2i29
"" v2 =
!1
! 5"2i29
"
10. (a) det(A ! &I) = &2 + 10& + 26 = 0 " &1 = !5 + i, &2 = !5 ! i
&1 = !5 + i; (A ! &I)v1 = 0 " v1 =
!1
2"i5
"" v2 =
!1
2+i5
"
(b) det(A ! &I) = &2 + 5 = 0 " &1 = i$
5, &2 = !i$
5
&1 = !8; (A! &I)v1 = 0 " v1 =
!1
1"i#
56
"" v2 =
!1
1+i#
56
"
11. (a) det(A ! &I) = &2 + 10& + 29 = 0 " &1 = !5 + 2i, &2 = !5 ! 2i
&1 = !5 + 2i; (A ! &I)v1 = 0 " v1 =
!1i
"" v2 =
!1!i
"
(b) det(A ! &I) = &2 ! 4&+ 11 = 0 " &1 = 2 + i$
7, &2 = 2 ! i$
7
&1 = 2 + i$
7; (A! &I)v1 = 0 " v1 =
!1
"1+i#
78
"" v2 =
!1
"1"i#
78
"
12. (a) &1 = 1; v1 =
1
3001
5
7
&2 = 2; v2 =
1
31!11
5
7
&3 = 5; v3 =
1
31/210
5
7
(b) &1 = !4; v1 =
1
31!11
5
7
&2 = 1; v2 =
1
3001
5
7
&3 = 2; v3 =
1
31/210
5
7
180 Section 5.5
13. (a) &1 = !1; v1 =
1
3!132
5
7
&2 = !1 ! 2i; v2 =
1
3!i10
5
7
&3 = !1 + 2i; v3 =
1
3i10
5
7
(b) &1 = 1; v1 =
1
3!3!15
5
7
&2 = 2 ! 2i; v2 =
1
3!1 ! 2i
10
5
7
&3 = 2 + 2i; v3 =
1
3!1 + 2i
10
5
7
14. (a) &1 = !1; v1 =
1
3001
5
7
&2 = 3; v2 =
1
33/201
5
7
&3 = 3; v3 =
1
31/210
5
7
(b) &1 = !2; v1 =
1
31/210
5
7
&2 = !1; v2 =
1
3001
5
7
&3 = 0; v3 =
1
31!11
5
7
5.5 Matrix Exponentials
1. (a) & = 3, !6; p(3) = 3t'1 + '2 = e3t; p(!6) = !6t'1 + '2 = e"6t "
'1 =e3t ! e"6t
9t, '2 =
e"6t + 2e3t
3"
eAt = '1At + '2I =
!e3t 2
9 (!e3t + e"6t)0 e"6t
"
(b) & = !3, !1; p(!3) = (!3t)'1 + '2 = e"3t; p(!1) = !t'1 + '2 = e"t "
'1 =e"t ! e"3t
2t, '2 =
3e"t ! e"3t
2"
eAt = '1At + '2I =
!2e"t ! e"3t e"3t ! e"t
2e"t ! 2e"3t !e"t + 2e"3t
"
2. (a) & = 3, 1; p(3) = 3t'1 + '2 = e3t; p(1) = t'1 + '2 = et "
'1 =e3t ! et
2t, '2 =
3et ! e3t
2"
Section 5.5 181
eAt = '1At + '2I =
/e3t+et
2e3t"et
2e3t"et
2e3t+et
2
0
(b) & = 1, !1; p(1) = (t)'1 + '2 = et; p(!1) = !t'1 + '2 = e"t "
'1 =et ! e"t
2t, '2 =
et + e"t
2"
eAt = '1At + '2I =
/et+e!t
2et"e!t
2et"e!t
2et+e!t
2
0
3. From Corollary 5.5.1,
(a) a = 1, b = 1 " eAt = et
!cos t ! sin tsin t cos t
"
(b) a = 3, b = !2 " eAt = e3t
!cos(2t) sin(2t)! sin(2t) cos(2t)
"
4. (a) & = 5, !3; p(5) = 5t'1 + '2 = e5t; p(3) = 3t'1 + '2 = e"3t "
'1 =e5t ! e"3t
8t, '2 =
3e5t + 5e"3t
8"
eAt = '1At + '2I =
!e5t 0
e5t"e!3t
4 e"3t
"
5. (a) eAt =
1
23
2e!2t+et
3"2e!2t+2et+3tet
9"2e!2t+2et
30 et 0
"e!2t+et
3e!2t"et+3tet
9e!2t+2et
3
5
67
(b) eAt =
1
23et 0 0
"5+4et+e4t
21+3e4t
43"3e4t
4"15+16et"e4t
61"e4t
43+e4t
4
5
67
6. eAt =
1
2223
e2t !e"t + e2t "e!3t"5e!t+6e2t
10 e"t ! e2t
e2t"e4t
22e!t+5e2t+3e4t
10"3e!3t+e!t+3e2t"e4t
10"2e!t"5e2t+7e4t
100 0 e"3t 0
e2t"e4t
2"8e!t+5e2t+3e4t
102e!3t"4e!t+3e2t"e4t
108e!t"5e2t+7e4t
10
5
6667
11. (a) i. det(A ! &I) = &2 + 3&! 18 " A2 + 3A ! 18I =
!9 60 36
"+
!9 !60 18
"!!
18 00 18
"
=
!0 00 0
"
ii. det(A ! &I) = &2 + 4&+ 3 " A2 + 4A + 3I =
!!7 8!16 17
"+
!4 !816 20
"+
!3 00 3
"
=
!0 00 0
"
(b) i. det(A ! &I) = &2 ! 4&+ 3 " A2 ! 4A + 3I =
!5 44 5
"!!
8 44 8
"+
!3 00 3
"
=
!0 00 0
"
ii. det(A ! &I) = &2 ! 1 " A2 ! I =
!1 00 1
"!!
1 00 1
"
=
!0 00 0
"
(c) i. det(A ! &I) = &2 ! 2&+ 2 " A2 ! 2A + 2I =
!0 !22 0
"!!
2 !22 2
"+
!2 00 2
"
=
!0 00 0
"
182 Chapter 5 Review
ii. det(A ! &I) = &2 ! 6&+ 13 " A2 ! 6A + 13I =
!5 12
!12 5
"!!
18 12!12 18
"+
!13 00 13
"
=
!0 00 0
"
(d) i. det(A ! &I) = &2 ! 2&! 15 " A2 ! 2A ! 15I =
!25 04 9
"!!
10 04 !6
"!!
15 00 15
"
=
!0 00 0
"
ii. det(A ! &I) = &2 ! 4&+ 5 " A2 ! 4A + 5I =
!!1 !84 7
"!!
4 !84 12
"+
!5 00 5
"
=
!0 00 0
"
5.6 Chapter 5: Additional Problems
1. False. Only those that have the proper form have those specific interpretations.
2. True.
3. True. See Definition 5.1.
4. True.
5. True. See Theorem 5.3.2.
6. False.
7. (a) unstable node, &1,2 = 4, 2
(b) unstable node, &1,2 = 3, 1
8. (a) unstable spiral, &1,2 = 5 ± i
(b) unstable spiral, &1,2 = 1 ± i
9. (a) unstable spiral, &1,2 = 1 ± i
(b) saddle, &1,2 = 1,!2
10. (a) saddle, &1,2 = 2,!1
(b) saddle, &1,2 = 2,!2
11. (a) unstable spiral, &1,2 = 1 ± i
(b) center, &1,2 = ±i
12. (a) unstable spiral, &1,2 = 2 ± i$
3
(b) unstable node, &1,2 = 2, 3
13. (a) unstable spiral, &1,2 = 2 ± i$
3
(b) stable spiral, &1,2 = !4 ± i$
3
17. &1,2,3 = !2, 2, 3, v1 =
1
32!12
5
7, v2 =
1
31!11
5
7, v3 =
1
32!11
5
7
18. &1,2,3 = 1, 1, 2, v1 =
1
3!210
5
7, v2 =
1
3001
5
7, v3 =
1
31!11
5
7
19. &1,2,3 = !1, 1, 5, v1 =
1
3111
5
7, v2 =
1
311!1
5
7, v3 =
1
31!11
5
7
20. &1,2,3 = !3, 5, 1, v1 =
1
3!1!11
5
7, v2 =
1
31!11
5
7, v3 =
1
3111
5
7
Chapter 5 Review 183
21. &1,2,3 = 5, 1, 3, v1 =
1
31!11
5
7, v2 =
1
3111
5
7, v3 =
1
311!1
5
7
22. &1,2,3 = !27,!18,!9, v1 =
1
31/3!4/3
1
5
7, v2 =
1
321!3
5
7, v3 =
1
3216
5
7
23. Yes, it’s a fund set of solutions
24. No, it’s NOT a fund set of solutions
25. Yes, it’s a fund set of solutions
26. Yes, it’s a fund set of solutions
27. No, it’s NOT a fund set of solutions
28. Yes, it’s a fund set of solutions
29. Yes, it’s a fund set of solutions
30. No, it’s NOT a fund set of solutions
184 Chapter 5 Review
Chapter 6
Techniques of Systems ofDi!erential Equations
6.1 A General Method, Part I: Solving Systems with Real,Distinct Eigenvalues
1. (a) Eigenvalues: &1 = !2, &2 = 1 " Saddle
Eigenvectors: v1 =
&21
', v2 =
&1!1
'
General Solution: x(t) = c1
&21
'e"2t + c2
&1!1
'et
(b) Eigenvalues: &1 = 1, &2 = !1 " Saddle
Eigenvectors: v1 =
&12
', v2 =
&11
'
General Solution: x(t) = c1
&12
'et + c2
&1!1
'e"t
2. (a) Eigenvalues: &1 = !2, &2 = !1 " Stable
Eigenvectors: v1 =
&!12
', v2 =
&1!3
'
General Solution: x(t) = c1
&!12
'e"2t + c2
&1!3
'e"t
(b) Eigenvalues: &1 = 3, &2 = 1 " Unstable
Eigenvectors: v1 =
&12
', v2 =
&13
'
General Solution: x(t) = c1
&12
'e3t + c2
&13
'et
3. (a) Eigenvalues: &1 = 4, &2 = 2 " Unstable
Eigenvectors: v1 =
&11
', v2 =
&13
'
General Solution: x(t) = c1
&11
'e4t + c2
&13
'e2t
(b) Eigenvalues: &1 = 5, &2 = !1 " Unstable
Eigenvectors: v1 =
&21
', v2 =
&!11
'
General Solution: x(t) = c1
&21
'e5t + c2
&!11
'e"t
185
186 Section 6.1
4. (a) Eigenvalues: &1 = !2, &2 = 3 " Saddle
Eigenvectors: v1 =
&11
', v2 =
&13
'
General Solution: x(t) = c1
&11
'e"2t + c2
&13
'e3t
(b) Eigenvalues: &1 = !4, &2 = 1 " Saddle
Eigenvectors: v1 =
&!21
', v2 =
&1!1
'
General Solution: x(t) = c1
&!21
'e"4t + c2
&1!1
'et
5. (a) Eigenvalues: &1 = 4, &2 = !2 " Saddle
Eigenvectors: v1 =
&11
', v2 =
&12
'
General Solution: x(t) = c1
&11
'e4t + c2
&12
'e"2t
(b) Eigenvalues: &1 = !3, &2 = !2 " Stable
Eigenvectors: v1 =
&11
', v2 =
&12
'
General Solution: x(t) = c1
&11
'e"3t + c2
&12
'e"2t
6. (a) Eigenvalues: &1 = !2, &2 = 1 " Saddle
Eigenvectors: v1 =
&21
', v2 =
&11
'
General Solution: x(t) = c1
&21
'e"2t + c2
&11
'et
(b) Eigenvalues: &1 =3 !
$5
2, &2 =
3 +$
52
" Unstable
Eigenvectors: v1 =
&1
!1.618
', v2 =
&1.618
1
'
General Solution: x(t) = c1
&1
!1.618
'e0.382t + c2
&1.618
1
'e2.618t
7. (a) Eigenvalues: &1 = 5, &2 = 1 " Unstable
Eigenvectors: v1 =
&11
', v2 =
&!13
'
General Solution: x(t) = c1
&11
'e5t + c2
&!13
'et
(b) Eigenvalues: &1 = 2 +$
3, &2 = 2 !$
3 " Unstable
Eigenvectors: v1 =
&1.732
1
', v2 =
&!1.732
1
'
General Solution: x(t) = c1
&1.732
1
'e3.732t + c2
&!1.732
1
'e0.268t
8. (a) Eigenvalues: &1 = !5, &2 = 1 " Saddle
Eigenvectors: v1 =
&31
', v2 =
&21
'
General Solution: x(t) = c1
&31
'e"5t + c2
&21
'et
(b) Eigenvalues: &1 = !2 +$
6, &2 = !2 !$
6 " Saddle
Eigenvectors: v1 =
&1
1.225
', v2 =
&1
!1.225
'
General Solution: x(t) = c1
&1
1.225
'e0.450t + c2
&1
!1.225
'e"4.450t
Section 6.1 187
9. (a) Eigenvalues: &1 = 3 + i, &2 = 3 ! i " Stable
(b) Eigenvalues: &1 = 0 + 2i, &2 = 0 ! 2i " Center
10. (a) Eigenvalues: &1 = 0 + i, &2 = 0 ! i " Center
(b) Eigenvalues: &1 = 0 + 2i, &2 = 0 ! 2i " Center
11. (a) Eigenvalues: &1 = 1 + 2i, &2 = 1 ! 2i " Unstable
(b) Eigenvalues: &1 = !1 + 2i, &2 = !1 ! 2i " Stable
12. (a) Eigenvalues: &1 = !3 + i, &2 = !3 ! i " Stable
(b) Eigenvalues: &1 = !3 + 2i, &2 = !3 ! 2i " Stable
13. (a) Eigenvalues: &1 = !6, &2 = !10 " Stable
(b) Eigenvalues: &1 = 3 + i, &2 = 3 ! i " Unstable
14. (a) Eigenvalues: &1 = 1, &2 = 5, &3 = 2 " Unstable
(b) Eigenvalues: &1 = 1, &2 = 2, &3 = !4 " Saddle
15. (a) Eigenvalues: &1 = !1 + 2i, &2 = !1 ! 2i, &3 = 1 " Stable
(b) Eigenvalues: &1 =3 +
$3
2, &2 =
3 !$
32
, &3 = 1 " Unstable
16. (a) Eigenvalues: &1 = !1, &2 = !1 + i, &3 = !1 ! i " Stable
(b) Eigenvalues: &1 = 11, &2 = 1, &3 = !1 " Saddle
17. (a) Eigenvalues: &1 =!3 +
$3
2, &2 =
!3 !$
32
, &3 = !1 " Stable
(b) Eigenvalues: &1 = !1 + i, &2 = !1 ! i, &3 = !1 " Stable
18. Eigenvalues are &1 =12
,(a + d) +
.(a ! d)2 + 4bc
-, &2 =
12
,(a + d) !
.(a ! d)2 + 4bc
-
(a) Must be real, so (a ! d)2 + 4bc & 0 and (a + d) ±.
(a ! d)2 + 4bc > 0
(b) Must be real, so (a ! d)2 + 4bc & 0 and (a + d) ±.
(a ! d)2 + 4bc < 0
(c) Must be complex, so (a ! d)2 + 4bc < 0 and a + d > 0
(d) Must be complex, so (a ! d)2 + 4bc < 0 and a + d < 0
(e) Must be real, so (a ! d)2 + 4bc & 0 and ad < bc
19. Eigenvalues: &1 = !1, &2 = !3 " Stable
Eigenvectors: v1 =
&1!1
', v2 =
&!13
'
General Solution: x(t) = c1
&1!1
'e"t + c2
&!13
'e"3t
188 Section 6.2
20. Eigenvalues: &1 = 2, &2 = !4 " Stable
Eigenvectors: v1 =
&12
', v2 =
&!14
'
General Solution: x(t) = c1
&12
'e2t + c2
&!14
'e"4t
6.2 A General Method, Part II: Solving Systems with Re-peated Real or Complex Eigenvalues
1. (a) Eigenvalues: &1 = 3 + i, &2 = 3 ! i
Eigenvectors: v1 =
&1!i
', v2 =
&1i
'"
a =
&10
', b =
&0!1
'
' = 3, " = 1
General Solution: x(t) = e3t
Oc1
!&10
'cos t !
&0!1
'sin t
"+ c2
!&10
'sin t +
&0!1
'cos t
"4
(b) Eigenvalues: &1 = 0 + 2i, &2 = 0 ! 2i
Eigenvectors: v1 =
&i1
', v2 =
&!i1
'"
a =
&01
', b =
&10
'
' = 0, " = 2
General Solution: x(t) =
Oc1
!&01
'cos 2t !
&10
'sin 2t
"+ c2
!&01
'sin 2t +
&10
'cos 2t
"4
2. (a) Eigenvalues: &1 = 0 + i, &2 = 0 ! i
Eigenvectors: v1 =
&2 ! i
5
', v2 =
&2 + i
5
'"
a =
&25
', b =
&!10
'
' = 0, " = 1
General Solution: x(t) =
Oc1
!&25
'cos t !
&!10
'sin t
"+ c2
!&25
'sin t +
&!10
'cos t
"4
(b) Eigenvalues: &1 = 0 + 2i, &2 = 0 ! 2i
Eigenvectors: v1 =
&i1
', v2 =
&!i1
'"
a =
&01
', b =
&10
'
' = 0, " = 2
General Solution: x(t) =
Oc1
!&01
'cos 2t !
&10
'sin 2t
"+ c2
!&01
'sin 2t +
&10
'cos 2t
"4
3. (a) Eigenvalues: &1 = 1 + 2i, &2 = 1 ! 2i
Eigenvectors: v1 =
&1
!3 ! i
', v2 =
&1
!3 + i
'"
a =
&1!3
', b =
&0!1
'
' = 1, " = 2
General Solution: x(t) = et
Oc1
!&1!3
'cos 2t !
&0!1
'sin 2t
"+ c2
!&1!3
'sin 2t +
&0!1
'cos 2t
"4
(b) Eigenvalues: &1 = !1 + 2i, &2 = !1 ! 2i
Eigenvectors: v1 =
&2
!3 ! i
', v2 =
&2
!3 + i
'"
Section 6.2 189
a =
&2!3
', b =
&0!1
'
' = !1, " = 2
General Solution: x(t) = e"t
Oc1
!&2!3
'cos 2t !
&0!1
'sin 2t
"+ c2
!&2!3
'sin 2t +
&0!1
'cos 2t
"4
4. (a) Eigenvalues: &1 = !3 + i, &2 = !3 ! i
Eigenvectors: v1 =
&!3 + i
5
', v2 =
&!3 ! i
5
'"
a =
&!35
', b =
&10
'
' = !3, " = 1
General Solution: x(t) = e"3t
Oc1
!&!35
'cos t !
&10
'sin t
"+ c2
!&!35
'sin t +
&10
'cos t
"4
(b) Eigenvalues: &1 = !3 + 2i, &2 = !3 ! 2i
Eigenvectors: v1 =
&2i1
', v2 =
&!2i1
'"
a =
&01
', b =
&20
'
' = !3, " = 2
General Solution: x(t) = e"3t
Oc1
!&01
'cos 2t !
&20
'sin 2t
"+ c2
!&01
'sin 2t +
&20
'cos 2t
"4
5. (a) Eigenvalues: &1 = !10, &2 = !6
Eigenvectors: v1 =
&12
', v2 =
&1!2
'"
a =
&12
', b =
&00
'
' = !10, " = 0
General Solution: x(t) = e"10t
&c1
2c1
'
(b) Eigenvalues: &1 = 3 + i, &2 = 3 ! i
Eigenvectors: v1 =
&1
3 + i
', v2 =
&1
3 ! i
'"
a =
&13
', b =
&01
'
' = 3, " = 1
General Solution: x(t) = e3t
Oc1
!&13
'cos t !
&01
'sin t
"+ c2
!&13
'sin t +
&01
'cos t
"4
6. (a) Eigenvalues: &1 = &2 = 1 "
Eigenvectors: v1 =
&10
', u1 =
&012
'
General Solution: x(t) = et
&c1 + c2t
c22
'
(b) Eigenvalues: &1 = &2 = !2 "
Eigenvectors: v1 =
&10
', u1 =
&013
'
General Solution: x(t) = e"2t
&c1 + c2t
c23
'
7. (a) Eigenvalues: &1 = &2 = 3 "
Eigenvectors: v1 =
&12
', u1 =
&0! 1
2
'
General Solution: x(t) = e3t
&c1 + c2t
2c1 + c2(2t ! 12 )
'
190 Section 6.2
(b) Eigenvalues: &1 = &2 = !1 "
Eigenvectors: v1 =
&12
', u1 =
&0!1
'
General Solution: x(t) = e"t
&c1 + c2t
2c1 + c2(2t ! 1)
'
8. (a) Eigenvalues: &1 = &2 = !1 "
Eigenvectors: v1 =
&11
', u1 =
&!10
'
General Solution: x(t) = e"t
&c1 + c2(t ! 1)
c1 + c2t
'
(b) Eigenvalues: &1 = &2 = 1 "
Eigenvectors: v1 =
&!11
', u1 =
&!10
'
General Solution: x(t) = et
&!c1 + c2(!t ! 1)
c1 + c2t
'
9. (a) Eigenvalues: &1 = &2 = !1 "
Eigenvectors: v1 =
&10
', u1 =
&014
'
General Solution: x(t) = e"t
&c1 + c2t
14c2
'
(b) Eigenvalues: &1 = &2 = !1 "
Eigenvectors: v1 =
&11
', u1 =
&120
'
General Solution: x(t) = e"t
&c1 + c2(t + 1
2 )c1 + c2t
'
10. (a) Eigenvalues: &1 = &2 = 1 "
Eigenvectors: v1 =
&!11
', u1 =
&120
'
General Solution: x(t) = et
&!c1 + c2(!t + 1
2 )c1 + c2t
'
(b) Eigenvalues: &1 = &2 = 2 "
Eigenvectors: v1 =
&!32
', u1 =
&! 1
20
'
General Solution: x(t) = e2t
&!3c1 + c2(!3t ! 1
2 )2c1 + 2c2t
'
11. Eigenvalues: &1 = !1 + 2i, &2 = !1 ! 2i, &3 = 1 " ' = !1, " = 2
Eigenvectors: v1 =
K
L!i!i0
M
N, v3 =
K
L012
M
N
a =
K
L0!10
M
N , b =
K
L!100
M
N
x1(t) = e"t
1
3
K
L0!10
M
N cos 2t !
K
L!100
M
N sin 2t
5
7
x2(t) = e"t
1
3
K
L0!10
M
N sin 2t +
K
L!100
M
N cos 2t
5
7
x3 = et
K
L012
M
N
12. Eigenvalues: &1 =3 +
$3
2, &2 =
3 !$
32
, &3 = 1 " ' = 1.5, " = 0.866
Section 6.2 191
Eigenvectors: v1 =
K
L0.3536 + .0.6124i
0.70710
M
N, v3 =
K
L!0.3015!0.90450.3015
M
N
a =
K
L0.35360.7071
0
M
N , b =
K
L0.6124
00
M
N
x1(t) = e1.5t
1
3
K
L0.35360.7071
0
M
N cos(0.866t) !
K
L0.6124
00
M
N sin(0.866t)
5
7
x2(t) = e1.5t
1
3
K
L0.35360.7071
0
M
N sin(0.866t) +
K
L0.6124
00
M
N cos(0.866t)
5
7
x3 = et
K
L!0.3015!0.90450.3015
M
N
13. Eigenvalues: &1 = !1 + i, &2 = !1 ! i, &3 = !1 " ' = !1, " = 1
Eigenvectors: v1 =
K
L10!i
M
N, v3 =
K
L011
M
N
a =
K
L100
M
N , b =
K
L00!1
M
N
x1(t) = e"t
1
3
K
L100
M
N cos(t) !
K
L00!1
M
N sin(t)
5
7
x2(t) = e"t
1
3
K
L100
M
N sin(t) +
K
L00!1
M
N cos(t)
5
7
x3 = e"t
K
L011
M
N
14. Eigenvalues: &1 = !1 + i, &2 = !1 ! i, &3 = !1 " ' = !1, " = 1
Eigenvectors: v1 =
K
L11i
M
N, v3 =
K
L100
M
N
a =
K
L110
M
N , b =
K
L001
M
N
x1(t) = e"t
1
3
K
L110
M
N cos(t) !
K
L001
M
N sin(t)
5
7
x2(t) = e"t
1
3
K
L110
M
N sin(t) +
K
L001
M
N cos(t)
5
7
x3 = e"t
K
L100
M
N
15. Eigenvalues: &1 = !1.5 + 0.866i, &2 = !1.5 ! 0.866i, &3 = !1 " ' = !1, " = 1
Eigenvectors: v1 =
K
L0.3536 + 0.6124i
00.7071
M
N, v3 =
K
L111
M
N
a =
K
L0.3536
00.7071
M
N , b =
K
L0.6124
00
M
N
192 Section 6.2
x1(t) = e"1.5t
1
3
K
L0.3536
00.7071
M
N cos(0.866t) !
K
L0.6124
00
M
N sin(0.866t)
5
7
x2(t) = e"t
1
3
K
L0.3536
00.7071
M
N sin(0.866t) +
K
L0.6124
00
M
N cos(0.866t)
5
7
x3 = e"t
K
L111
M
N
16. Eigenvalues: &1 = 2, &2 = 2, &3 = 1
Eigenvectors: v1 =
K
L!110
M
N, v2 =
K
L1!11
M
N
u1 =
K
L100
M
N
x(t) = c1(e2t ! te2t) ! c2e
2t + c3et
y(t) = c1te2t + c2e
2t ! c3et
z(t) = c3et
17. Eigenvalues: &1 = !1, &2 = 3, &3 = 3
Eigenvectors: v1 =
K
L001
M
N, v2 =
K
L1023
M
N v3 =
K
L01! 1
3
M
N
x(t) = c2e3t
y(t) = c3e3t
z(t) = c1e"t +
23c2e
3t ! 13c3e
3t
18. Eigenvalues: &1 = !1, &2 = 1, &3 = 1
Eigenvectors: v1 =
K
L100
M
N, v2 =
K
L! 1
21!2
M
N
u1 =
K
L3401
M
N
x(t) = c1e"t ! 1
2c2e
t + c3(34et ! 1
2tet)
y(t) = c2et + c3te
t
z(t) = !2c2et + c3(e
t ! 2tet)
19. & =
K
PPPPPPL
!1 + 2i!1 ! 2i
1222
M
QQQQQQN, v =
K
PPPPPPL
0 0 1 20 0 0 10 0 0 00 0 0 0!i i 0 01 1 0 0
M
QQQQQQN
x =
K
PPPPPPL
c3et + 2c4e2t + 2c5(te2t ! e2t) + c6(t2e2t ! 2te2t + 2e2t)c4e
2t + c5te2t + 1
2 c6t2e2t
c5e2t + c6te
2t
c6e2t
e"t(c1 sin(2t) + c2 cos(2t))e"t(c1 cos(2t) ! c2 sin(2t))
M
QQQQQQN
6.3. SOLVING LINEAR HOMOGENEOUS AND NONHOMOGENEOUS SYSTEMS OF EQUATIONS193
6.3 Solving Linear Homogeneous and Nonhomogeneous Sys-tems of Equations
1. Solution: &1 = !2, &2 = 3, v1 =
&!21
', v2 =
&1!3
'
x =
&!2c1e
"2t + c2e3t
c1e"2t ! 3c2e
3t
'
(a) !(t) =
&!2e"2t e3t
e"2t !3e3t
'
(b) " = V "1AV =
&3 00 !2
'" y! = "y " y =
&c1e
"2t
c2e3t
'" x = V y
(c) eAt = !(t)!"1(t0), !"1(0) =
15
&!3 !1!1 !2
'"
eAt =15
&6e"2t ! e3t 2e"2t ! 2e3t
3e3t ! 3e"2t !e"2t + 6e3t
'
2. Solution: &1 = !3, &2 = !4, v1 =
&!11
', v2 =
&!21
'
x =
&!c1e
"3t ! 2c2e"4t
c1e"3t + c2e
"4t
'
(a) !(t) =
&!e"3t !2e"4t
e"3t e"4t
'
(b) " = V "1AV =
&!3 00 !4
'" y! = "y " y =
&c1e
"3t
c2e"4t
'" x = V y
(c) eAt = !(t)!"1(t0) =
&2e"4t ! e"3t !2e"3t + 2e"4t
e"3t ! e"4t !e"4t + 2e"3t
'
3. Solution: &1 = 2, &2 = 4, v1 =
&01
', v2 =
&!21
'
x =
&!2c2e4t
c1e2t + c2e
4t
'
(a) !(t) =
&0 !2e4t
e2t e4t
'
(b) " = V "1AV =
&2 00 4
'" y! = "y " y =
&c1e2t
c2e4t
'" x = V y
(c) eAt = !(t)!"1(t0) =
&e4t 0
! 12e4t + 1
2e2t e2t
'
4. Solution: &1 = 1, &2 = 2, v1 =
&!11
', v2 =
&!21
'
x =
&!c1e
t ! 2c2e2t
c1et + c2e
t
'
(a) !(t) =
&!e"t !2e2t
et e2t
'
(b) " = V "1AV =
&1 00 2
'" y! = "y " y =
&c1e
t
c2e2t
'" x = V y
(c) eAt = !(t)!"1(t0) =
&!et + 2e2t 2e2t ! 2et
!e2t + et 2et ! e2t
'
194 Section 6.3
5. Solution: &1 = !1, &2 = 1, &3 = 2, v1 =
K
L1!21
M
N , v2 =
K
L!110
M
N , v3 =
K
L010
M
N
x =
K
Lc1e
"t ! c2et
!2c1e"t + c2e
t + c3e2t
c1e"t
M
N
(a) !(t) =
K
Le"t !et 0
!2e"t et e2t
e"t 0 0
M
N
(b) " = V "1AV =
K
L!1 0 00 1 00 0 2
M
N " y! = "y " y =
K
Lc1e
"t
c2et
c3e2t
M
N " x = V y
(c) eAt = !(t)!"1(t0) =
K
Let 0 e"t ! et
!et + e2t e2t e2t ! 2e"t + et
0 0 e"t
M
N
6. Solution: &1 = !1, &2 = 1, &3 = 3, v1 =
K
L!101
M
N , v2 =
K
L11!2
M
N , v3 =
K
L001
M
N
x =
K
L!c1e
"t + c2et
c2et
c1e"t ! 2c2e"t + c3e3t
M
N
(a) !(t) =
K
L!e"t et 0
0 et 0e"t !2et e3t
M
N
(b) " = V "1AV =
K
L!1 0 00 1 00 0 3
M
N " y! = "y " y =
K
Lc1e
"t
c2et
c3e3t
M
N " x = V y
(c) eAt = !(t)!"1(t0) =
K
Le"t !e"t + et 00 et 0
e3t ! e"t e3t + e"t ! 2et e3t
M
N
7. Solution: &1 = !1, &2 = 3, &3 = 1, v1 =
K
L!312
M
N , v2 =
K
L112
M
N , v3 =
K
L11!2
M
N
x =
K
L!3c1e
"t + c2e3t + c3e
t
c1e"t + c2e3t + c3et
2c1e"t + 2c2e
3t ! 2c3et
M
N
(a) !(t) =
K
L!3e"t e3t et
e"t e3t et
2e"t 2e3t !2et
M
N
(b) " = V "1AV =
K
L!1 0 00 3 00 0 1
M
N " y! = "y " y =
K
Lc1e
"t
c2e3t
c3et
M
N " x = V y
(c) eAt = !(t)!"1(t0) =14
K
L3e"t + e3t e3t ! 3e"t2et e3t ! et
e3t ! e"t 2et + e3t + e"t e3t ! et
2e3t ! 2e"t 2e3t + 2e"t ! 4et 2et + 2e3t
M
N
6.4. NONLINEAR EQUATIONS AND PHASE PLANE ANALYSIS 195
6.4 Nonlinear Equations and Phase Plane Analysis
1. Equilibria: (±2, 0)
J(!2, 0) =
&0 14 0
'" &1 = 2, &2 = !2 " Saddle
J(2, 0) =
&0 1!4 0
'" &1 = 2i, &2 = !2i " Linear center
2. Equilibria: (±1, 1)
J(1, 1) =
&0 12 !1
'" &1 = 1, &2 = !2 " Saddle
J(!1, 1) =
&0 1!2 !1
'" &1 =
!1 + i$
72
, &2 =!1 ! i
$7
2" Spiral
196 Section 6.4
3. Equilibria: (0, 0), (!2,!2)
J(0, 0) =
&!1 10 2
'" &1 = !1, &2 = 2 " Saddle
J(2, 0) =
&!1 1!4 2
'" &1 =
1 + i$
72
, &2 =1 ! i
$7
2" Unstable Spiral
4. Equilibria: (0, 0), (9, 3)
J(0, 0) =
&!1 01 !3
'" &1 = !1, &2 = !3 " Stable
J(9, 3) =
&!1 61 !3
'" &1 = !2 +
$7, &2 = 2 !
$7 " Saddle
Section 6.4 197
5. Equilibria: (0, 0), (1, 0), (!1, 0)
J(0, 0) =
&0 1!1 0
'" &1 = i, &2 = !i " Linear center
J(1, 0) =
&0 12 0
'" &1 =
$2, &2 = !
$2 " Saddle
J(!1, 0) =
&0 12 0
'" &1 =
$2, &2 = !
$2 " Saddle
6. Equilibria: (#6
,12), (!#
6,!1
2)
J(#6
,12) =
& #3
2 !10 1
'" &1 = 1, &2 =
$3
2" Unstable
J(!#6
,!12) =
& #3
2 !10 !1
'" &1 = !1, &2 =
$3
2" Saddle
198 Section 6.4
7. Equilibria: (±1, 1)
J(!1, 1) =
&0 2!2 !1
'" &1 = !1
2+
12
$15, &2 = !1
2!
$15 " Stable
J(1, 1) =
&0 22 !1
'" &1 = !1
2+
12
$17, &2 = !1
2!
$17 " Stable
8. Equilibria: (0, 0)
J(0, 0) =
&1 10 !8
'" &1 = 1, &2 = !8 " Saddle
Section 6.4 199
9. Equilibria: (0, 0), (0, 2), (0, 3)
J(0, 0) =
&3 00 2
'" &1 = 3, &2 = 2 " Unstable
J(0, 2) =
&1 0!2 !2
'" &1 = 1, &2 = !2 " Saddle J(0, 3) =
&!3 !30 !1
'" &1 = !3, &2 = !1 "
Stable
10. Equilibria: (0, 0), (0, 2), (3, 0), (1, 1)
J(0, 0) =
&3 00 2
'" &1 = 3, &2 = 2 " Unstable
J(0, 2) =
&!1 0!2 !2
'" &1 = !1, &2 = !2 " Stable
J(3, 0) =
&3 !60 !1
'" &1 = !3, &2 = !1 " Stable
J(1, 1) =
&!1 !2!1 !1
'" &1 = !1 +
$2, &2 = !1 !
$2 " Saddle
200 Section 6.4
11. Equilibria: (0, 0), (3, 0), (2, 1)
J(0, 0) =
&3 00 !2
'" &1 = 3, &2 = !2 " Saddle
J(3, 0) =
&!3 !30 1
'" &1 = !3, &2 = 1 " Saddle
J(2, 1) =
&!2 !21 0
'" &1 = !1 + i, &2 = !1 ! i " Stable spiral
12. u1 = x, u2 = x! " u!1 = u2, u!
2 = !(u2(u21 ! 1) ! u1 " x! = y, y! = !(y(x2 ! 1) ! x " Equilibria: (0, 0)
J(0, 0) =
&0 1!1 (
'" &1 =
(+$(2 ! 42
, &2 =(!
$(2 ! 42
"
Stable for ( < 0, spiral for !2 < ( < 0, node for ( ' !2Unstable for ( > 0, spiral for 0 < ( < 2, node for ( & 2Center for ( = 0Never a saddle.
(a) ! = 0.1 (b) ! = 10
Section 6.4 201
13. u1 = x, u2 = x! # u!1 = u2, u!
2 = "u1 " !u31 # x! = y, y! = "x " !x3 #
Equilibria: (0, 0),
!"1$"!
, 0
",
!1
$"!
, 0
"
J(0, 0) =
&0 1"1 0
'# "1 = i, "2 = "i # Center, %!
J
!"1$"!
, 0
"=
&0 12 0
'# "1 =
$2, "2 = "
$2 # Saddle, %!
J
!1
$"!
, 0
"=
&0 12 0
'# "1 =
$2, "2 = "
$2 # Saddle, %!
(a) # = 0.25 (b) # = 1
14. $! = v, v! ="g
msin $
Equilibria: centers at ($, $!) = (±2n%, 0), n = 0, 1, 2, · · ·saddles at ($, $!) = (±(2n + 1)%, 0), n = 0, 1, 2, · · ·
The closed orbits around the centers correspond to the typical back-and-forth pendulum motion. The sad-dles correspond to the pendulum standing on end. The trajectories outside the ones connecting the saddlescorrespond to a pendulum whirling over the top. These do not stop because there is no friction present.
202 Section 6.4
15. $! = v, v! = "0.3$! " $Equilibria: spiral sinks at ($, $!) = (±2n%, 0), n = 0, 1, 2, · · ·saddles at ($, $!) = (±(2n + 1)%, 0), n = 0, 1, 2, · · ·
Because friction is present, almost all trajectories eventually go to the rest position of the hanging pendulum.Some trajectories will end up with the pendulum standing on end.
16. r < 0 =# no equilibriar = 0 =# (0, 0) is degenerate node (“half-stable”)r > 0 =# ("
$r, 0) is unstable node, (
$r, 0) is stable node
This is a saddlenode bifurcation.
r < 0 r = 0 r > 0
17. r < 0 =# (r, 0) is unstable node, (0, 0) is stable noder = 0 =# (0, 0) is degenerate node (“half-stable”)r > 0 =# (0, 0) is unstable node, (r, 0) is stable nodeThis is a transcritical bifurcation.
r < 0 r = 0 r > 0
Section 6.4 203
18. r < 0 =# (0, 0) is stable noder = 0 =# (0, 0) is stable noder > 0 =# (0, 0) is a saddle, (±r, 0) is stable nodeThis is a supercritical pitchfork bifurcation.
r < 0 r = 0 r > 0
19. r < 0 =# (0, 0) is stable node, (±r, 0) are saddlesr = 0 =# (0, 0) is a saddler > 0 =# (0, 0) is a saddleThis is a subcritical pitchfork bifurcation.
r < 0 r = 0 r > 0
20. Equilibria: (0, 0)
J(0, 0) =
&r 2"2 r
'# "1 = r + 2i, "2 = r " 2i # (0, 0) is a stable spiral when r < 0 and an unstable spiral
when r > 0.(a) r = "0.25, no limit cycle (b) r = 0.25, stable limit cycle
204 Section 6.5
21. Equilibria: (0, 0)
J(0, 0) =
&r 2"2 r
'# "1 = r + 2i, "2 = r " 2i # (0, 0) is a stable spiral when r < 0 and an unstable spiral
when r > 0.(a) r = "0.25, unstable limit cycle (b) r = 0.25, no limit cycle
6.5 Epidemiological Models
1. (a)d)ds
=")s ! ')!"s) = !1 +
'"s
" )! '"
ln s + s = C
(b) IC " )! '"
ln s + s = )(0) ! '"
ln s(0) + s(0)
Evaluate at (s$, )$) " )$ ! '"
ln s$ + s$ = )(0) ! '"
ln s(0) + s(0)
(c) )$ ! )(0) +'"
ln
!s(0)s$
"+ s$ ! s(0) = 0
Substitution gives'"
ln
!s(0)s$
"+ s$ ! 1 = 0 " "
'=
ln,
s(0)s#
-
1 ! s$
2. Disease spreads when R0 > !" > 1. Thus we need 3.5" > 1 " " >
13.5
= .285.
3. If we let v = %vaccinated, then 1! v are not vaccinated. We replace s(0) and s$ with (1! v)s(0) and
(1 ! v)s$ respectively. This gives us 1 >
,(1"v)s(0)(1"v)s#
-
1 ! (1 ! v)s$" v >
,s(0)s#
-
s$= 1.94, which does not make
sense. If we go back to 1c, we would obtain 1 >"'
,(1"v)s(0)(1"v)s#
-
(1 ! v)(1 ! s$)" v < 1 !
,s(0)s#
-
1 ! s$= !.41, which
again doesn’t make sense. Our system is nonlinear and thus we cannot make the assumption statedin this problem because a percentage change in the initial condition does not necessarily correspondto the same change in the final solution.
4. (a) s = 0 " s! = 0)! = !')
4" s(t) = 0 (because s(0) = 0), and )(t) ( 0
(b) ) = 0 " s! = 0)! = 0
4" s(t) = s(0) and )(t) = 0
(c) If s + ) = 1, then
O) = 1 ! ss = 1 ! ) "
Os! = !"s(1 ! s) < 0)! = !")(1 ! )) ! ') < 0
Thus, solutions cannot leave the closed triangle, which makes the problem well-posed.
Section 6.5 205
#4 #5
5. (a) s = 0 # s! = µ&! = "(' + µ)&
4# s(t) increases (not always 0, but doesn’t leave set) and &(t) decreases
(b) & = 0 # s! = µ(1 " s)&! = 0
4# s(t) = increases and &(t) = 0
(c) If s + & = 1, then
O& = 1 " ss = 1 " &
#
ABBC
BBD
s! = µ(1 " s) " (s(1 " s)< 0 for s & 1> 0 for s '
&! = "(&(1 " &) " (' + µ)& < 0Thus, solutions cannot leave the closed triangle, which makes the problem well-posed.
6. Let s = S/N, & = I/N, r = R/N . Substitute
dSdt = "( I
N SdIdt = ( I
N S " 'IdRdt = 'I,
FG
H =#
ABC
BD
d(sN)dt = "( #N
N (sN)d(#N)
dt = ( #N (sN) " '(&N)
d(rN)dt = '(&N),
which gives the desired result after cancelling the constant N .
7. Again substituting s = S/N, & = I/N, r = R/N , we haved(sN)
dt = "( #NN (sN)
d(#N)dt = ( #
N (sN) " '(&N)d(rN)
dt = '(&N),
FBG
BH=#
dsdt = µ " µs " (&sd#dt = (&s " (' + µ)&drdt = '& " µr.
Now we let s+ &+r = 1 (and thus r = 1"s" &) to get the desired results, where the r! equation can be ignoredbecause knowing s(t) and &(t) gives r(t).
8. (a) N is time dependent; N is constant; s = SN , & = I
N give proportions of the students still in school while
r = R/N is the drop out proportion of the original population.
(b)N
N=
S + I
S + I + R=
S + I + R " R
S + I + R= 1 "
R
S + I + R= 1 " r
(c)
d(sN)dt = "(&sN
d(#N)dt = (&sN " '&N
d(rN)dt = '&N
FBG
BH=#
ABC
BD
dsdt = "(&sd#dt = (&s " '&drdt = '&N
N= '&(1 " r)
9. The equation satisfied by N !(t) isdN
dt= !" µN.
Solving this for N(t) gives
N(t) =
!N(0) "
!
µ
"e"µt +
!
µ.
Note thatlim
t%&e"µt = 0.
It is thus clear that N(t) ( !µ as t ( ).
10. (a) dNdt = 0 implies that the total population size remains constant for all time t * 0.
(b) It is su"cient to study the first two equations since they are independent of the third variable R andR = N " S " I. That is, there are 3 equations and one constraint, so there are 2 independent equations.
The DFE is given by (S$, I$) =,
µNµ+v , 0
-.
206 Section 6.5
(c) The Jacobian is
J(DFE) =
/"(µ + v) !µ
µ+v
0 !µµ+v " ' " µ
0
,
Because the matrix is upper triangular, the eigenvalues are immediate: "1 = "(µ + v), "2 =(µ
µ + v" ' " µ
Thus the DFE is locally asymptotically stable if !µµ+v " ' " µ < 0 and is unstable if "2 > 0
(d) Manipulating the only eigenvalue that can change sign, "2, we see that R0 =(µ
(µ + v)(' + µ), which can
be rewritten as
R0(v) =(
µ + '
µ
µ + v.
With no vaccination, this becomes R0(0) = !µ+" . R0 depends on two factors: the !
µ+" factor gives a
measure of those entering the I-class versus those leaving, while the µµ+v factor gives a factor that is
always less than or equal to one and is farther from one when v is bigger.
11. (a) Adding the equations, we see that
dN
dt=
dS
dt+
dE
dt+
dI
dt+
dR
dt= 0.
Thus N is a constant.
(b) The reduced 3-D system is thus
dS
dt= "(S
I
N+ µN " µS
dE
dt= (S
I
N" µE " #E
dI
dt= #E " µI " )I.
(6.1)
(c) At the DFE (N, 0, 0), the Jacobian is1
3"µ 0 " (0 "(µ + #) (0 # "(µ + ))
5
7
One of the eigenvalues, "1 = "µ (always less than zero) can be read immediately. Its location allows usto separate the matrix into it’s block form and now just consider
J(DFE) =
1
3"(µ + #) (# "(µ + ))
5
7
to determine the stability of the DFE. We can calculate the trace and determinant as
trJ = "(µ + # + µ + )) < 0,
DetJ = (µ + ))(µ + #) " #(.
We require Det J > 0 to ensure that the disease-free equilibrium is asymptotically stable, so that thedisease will die out under the condition of Det J > 0. Intuitively, the extinction of a disease meansthat new infections cannot replace the old/current ones. It turns out that the basic reproductive numbershould be less than 1. Therefore, because Det J > 0, we have
(#
(# + µ)(µ + ))< 1.
So,
R0 =(
(µ + ))
#
(# + µ).
We have thus shown that if R0 < 1, the disease-free equilibrium is asymptotically stable, i.e., that thedisease will die out. We know that the average life-span for the infectious class is 1
µ+$ , so !µ+$ represents
the total e#ective contact number of a typical infection. The quantity %%+µ is the fraction or probability
that an individual in the latent class becomes infected. Therefore,
R0 =(
(µ + ))
#
(µ + #)
represents the total number of new cases generated by a typical infection during its lifetime.
6.6. CHAPTER 6: ADDITIONAL PROBLEMS 207
6.6 Chapter 6: Additional Problems
1. False. The systems must also be homogeneous, constant coe#cient and with a full set of eigenvectors.
2. True.
3. False. The method of solution combines the complex (non-real) method with the repeated eigenvaluemethod.
4. True.
5. True. See a linear algebra text for more details.
6. (a) i. &1 = 2, &2 = 1, v1 =
&11
', v2 =
&01
'
ii. Unstable node
iii. x1 = e2tv1, x2 = etv2
iv. x =
&c1e
2t
c1e2t + c2e
t
'
(b) i. &1 = 4i, &2 = !4i, v1 =
&1i2
', v2 =
&1"i2
'
ii. Center
iii. x1 =
&10
'cos 4t !
&0
1/2
'sin 4t, x2 =
&10
'sin 4t +
&0
1/2
'cos 4t
iv. x =
&c1 cos 4t + c2 sin 4t
12c2 cos 4t ! 1
2 c1 sin 4t
'
7. (a) i. &1 = 2, &2 = !2, v1 =
&11
', v2 =
&1!3
'
ii. Saddle
iii. x1 = e2tv1, x2 = e"2tv2
iv. x =
&c1e2t + c2e"2t
c1e2t ! 3c2e
"2t
'
(b) iv. x(t) = 2c1 cos 4t + 2c2 sin 4t,y(t) = !c1 sin 4t + c2 cos 4t
8. (a) i. &1 = 1, &2 = 5, v1 =
&15
', v2 =
&11
'
ii. Unstable node
iii. x1 = etv1, x2 = e5tv2
iv. x =
&c1e
t + c2e5t
5c1et + c2e5t
'
(b) i. &1 = 7 + 4i, &2 = 7 ! 4i, v1 =
&1!i
', v2 =
&1i
'
ii. Unstable spiral
iii. x1 = e7t
!&10
'cos 4t !
&0!1
'sin 4t
", x2 = e7t
!&10
'sin 4t +
&01
'cos 4t
"
iv. x =
&e7t (c1 cos 4t + c2 sin 4t)e7t (c2 cos 4t + c1 sin 4t)
'
9. (a) i. &1 = !3, &2 = !2, v1 =
&1!2
', v2 =
&1! 3
2
'
ii. Stable node
iii. x1 = e"3tv1, x2 = e"2tv2
iv. x =
&c1e
"3t + c2e"2t
!2c1e"3t ! 32 c2e"2t
'
(b) i. &1 = 4i, &2 = !4i, v1 =
&1
!4i
', v2 =
&14i
'
208 Chapter 6 Review
ii. Center
iii. x1 =
&10
'cos 4t !
&0!4
'sin 4t, x2 =
&10
'sin 4t +
&04
'cos 4t
iv. x =
&c1 cos 4t + c2 sin 4t
4c1 sin 4t + 4c2 cos 4t
'
10. (a) i. &1 = !2, &2 = 1, v1 =
&132
', v2 =
&12
'
ii. Saddle
iii. x1 = e"2tv1, x2 = etv2
iv. x =
&c1e
"2t + c2et
32c1e
"2t + 2c2et
'
(b) i. &1 = 1 + 2i, &2 = 1 ! 2i, v1 =
&1
3 ! 2i
', v2 =
&1
!3 + 2i
'
ii. Unstable spiral
iii. x1 = et
!&13
'cos 2t !
&0!2
'sin 2t
", x2 = et
!&1!3
'sin 2t +
&02
'cos 2t
"
iv. x =
&et (c1 cos 2t + c2 sin 2t)
et (3c1 cos 2t + 2c2 sin 2t ! 3c1 sin 2t + 2c2 cos 2t)
'
11. (a) i. &1 = !1, &2 = !2, v1 =
&!11
', v2 =
&1!2
'
ii. Stable node
iii. x1 = e"tv1, x2 = e"2tv2
iv. x =
&c1e
"t + c2e"2t
c1e"t ! 2c2e
"2t
'
(b) i. &1 = 2 + i, &2 = 2 ! i, v1 =
&!3 ! i
1
', v2 =
&!3 + i
1
'
ii. Unstable spiral
iii. x1 = e2t
!&!31
'cos t !
&!10
'sin t
", x2 = e2t
!&!31
'sin t +
&10
'cos t
"
iv. x =
&e2t (!3c1 cos t + c1 sin t ! 3c2 sin t + c2 cos t)
e2t (c1 cos t + c2 sin t)
'
12. (a) i. &1 = !8, &2 = !12, v1 =
&1!2
', v2 =
&12
'
ii. Stable node
iii. x1 = e"8tv1, x2 = e"12tv2
iv. x =
&c1e
"8t + c2e"12t
!2c1e"8t + 2c2e
"12t
'
(b) i. &1 = 1 + 2i, &2 = 1 ! 2i, v1 =
&1 ! i
1
', v2 =
&1 + i
1
'
ii. Unstable spiral
iii. x1 = et
!&11
'cos 2t !
&!10
'sin 2t
", x2 = et
!&11
'sin 2t +
&10
'cos 2t
"
iv. x =
&et (c1 cos 2t + c2 sin 2t + c1 sin 2t + c2 cos 2t)
et (c1 cos 2t + c2 sin 2t)
'
13. (a) i. &1 = !1, &2 = !5, v1 =
&15
', v2 =
&11
'
ii. Stable node
iii. x1 = e"tv1, x2 = e"5tv2
iv. x =
&c1e
"t + c2e"5t
5c1e"t + c2e
"5t
'
Chapter 6 Review 209
(b) i. &1 = 4 +$
3i, &2 = 4 !$
3i, v1 =
&1
1+#
3i2
', v2 =
&1
1"#
3i2
'
ii. Unstable spiral
iii. x1 = e4t
!&1
1/2
'cos
$3t !
&0$3/2
'sin
$3t
", x2 = e4t
!&1
1/2
'sin
$3t +
&0
!$
3/2
'cos
$3t
"
iv. x =
=e4t#c1 cos
$3t + c1 sin
$3t + c2 cos
$3t$
e4t,
12 c1 cos
$3t !
#3
2 c2 sin$
3t + 12c1 sin
$3t !
#3
2 c2 cos$
3t->
14. (a) i. &1 = !1, v1 =
&10
'
ii. Stable node
iii. u1 =
&01
'
iv. x =
&c1e
"t + c2te"t
c2e"t
'
(b) i. &1 = !2, v1 =
&1!3
'
ii. Stable node
iii. u1 =
&01
'
iv. x =
&c1e
"2t + c2te"2t
!3c1e"2t + c2(!3te"2t + e"2t)
'
15. (a) i. &1 = 1, v1 =
&132
'
ii. Unstable node
iii. u1 =
&0
!1/4
'
iv. x =
&c1e
t + c2tet
32c1et + 3
2 c2tet ! 14c2et
'
(b) i. &1 = 2, v1 =
&21
'
ii. Unstable node
iii. u1 =
&!1/3
0
'
iv. x =
&2c1e
2t + c2(2te2t ! 13e2t)
c1e2t + c2te
2t
'
16. (a) i. &1 = 3, v1 =
&10
'
ii. Unstable node
iii. u1 =
&0
1/2
'
iv. x =
&c1e3t + c2te3t
12c2e
3t
'
(b) i. &1 = 3, v1 =
&21
'
ii. Stable node
iii. u1 =
&!10
'
iv. x =
&2c1e3t + c2(2te3t ! e3t)
c1e3t + c2te
3t
'
210 Chapter 6 Review
17. x =
&c1e
2t
c1e2t + c2et
'
18. x =
&c1e
"2t + c2e2t
!3c1e"2t + c2e2t
'
19. x =
&15c1et + c2e5t
c1et + c2e
5t
'
20. x =
K
Lc1e2t + c2e3t + c3e"2t
!c1e2t ! 1
2c2e3t ! 1
2c3e"2t
c1e2t + 1
2c2e3t + c3e
"2t
M
N
21. x =
K
Lc1e
5t + c2et ! c3e
"t
!c1e5t + c2e
t ! c3e"t
c1e5t + c2e
t + c3e"t
M
N
23. a. i. &1 = !4,&2,3 = 1,
v1 =
1
3!101
5
7 ,v2 =
1
3!213
5
7; ii. saddle;
b. i. &1 = 3, &2 = 3 + 6i, &3 = 3 ! 6i,
v1 =
1
31!2!3
5
7 ,v2 =
1
3!21
92 + 3
2 i
5
7 v3 =
1
3!21
92 ! 3
2 i
5
7; ii. unstable;
26. a. i. &1 = 1, &2 = !1, &3 = 5,
v1 =
1
3111
5
7 ,v2 =
1
311!1
5
7 ,v2 =
1
31!11
5
7; ii. saddle;
b. i. &1 = !3, &2 = 5, &3 = 1,
v1 =
1
3!1!11
5
7 ,v2 =
1
31!11
5
7 v3 =
1
3111
5
7; ii. saddle
28. (!1, 1) unstable node; (!1,!1) saddle
Chapter 6 Review 211
29. (0, 1) saddle; (0,!1) linear center
30. (0, 0) stable spiral; (!1,!1) saddle; (1, 1) saddle
212 Chapter 6 Review
31. i. (x$, y$) = (0, 0), (1, 1);ii. (1, 1) is a saddle, (0, 0) is an indeterminate form since linearization gives a zero eigenvalue; graph-ically, (0, 0) is stable from above but unstable from below.
Chapter 7
Laplace Transforms
7.1 Fundamentals of the Laplace Transform
1.2s3
, s > 0
2.1 ! 3s
s2, s > 0
3.1
(s + 1)2, s > !1
4.2 ! 2e"4s
s, s > 0
5.2s + 1
s2, s > 0
6.1 ! e"s ! se"2s
s2, s > 0
7.e"2s(se2s + es(2 + s) ! 2(1 + 2s + 2s2))
s3, s > 0
8.1
s2 ! 1, s > 1
9. Laplace transform does not exist (integral does not converge).
10.28
s ! 13, s > 13
11. L{t"} =
% &
0
e"stt" dt
!Let st = u " t" =
u"
s", dt =
dus
"=
% &
0
e"u u"
s"dus
=1
s"+1
% &
0
u" e"u du
=$('+ 1)
s"+1which converges only if ' > !1
12. (a) LO
1$t
4=$(1/2)s1/2
=
9#s
(b) LS$
tT
=$(3/2)s3/2
=
$#
2s3/2
(c) LS
t$
tT
=$(5/2)s5/2
=2$#
4s5/2
13. )',", M, N such that e""t |f(t)| < M, e"!t |g(t)| < N and let ! = max {', "} "M + N > e""t |f(t)| + e"!t |g(t)| > e"$t [|f(t)| + |g(t)|] > e"$t |f(t) + g(t)| " f(t) + g(t) is exponen-tial.
14. Let I1 be the partition of [a, b] for f(t) and I2 be the partition of [a, b] for g(t). With I = I1 * I2, bothf and g are piecewise continuous on [a, b] with partition I . Since f and g both continuous impliesf + g is continuous, then f + g is piecewise continous on [a, b] with partition I .
213
214 Section 7.3
15. L{f(t) + g(t)} =
% &
0
e"st [f(t) + g(t)] dt =
% &
0
e"stf(t) dt +
% &
0
e"stg(t) dt = L{f(t)} + L{g(t)} and
this transform exists for s > ! (see problem #13 for !) by theorem 7.1.3 and problems 13 and 14.
16. (a) e""t8883 sin(et2)
888 ' 3 for t > 0 and ' = 1 " f(t) is of exponential order.
(b) f !(t) = 6tet2 cos(et2) and limt%&
e""t6tet2 cos(et2) = + " f !(t) is not of exponential order.
7.2 Properties of the Laplace Transform
1.1 ! 5s
s2
2.1 + 5s
s2
3.1
(s ! 2)2
4.5s ! 14(s ! 3)2
5.12s
+s
s2 + 4
6.2a2 + s2
s(s2 + 4a2)
7.4s
(s2 + 9)(s2 + 1)
8.2abs
(s2 + (b + a)2)(s2 + (a ! b)2)
9.6
(s2 + 9)(s2 + 1)
10.2
s(s2 + 4)+
2s2 + 1
11.'
s2 + '2
12.s
s2 + '2
13.' cos%! s sin%
s2 + '2
14. (a)
% &
s
F (z) dz =
% &
s
!% &
0
e"ztf(t) dt
"dz =
% &
0
!% &
s
e"ztf(t) dz
"dt
=
% &
0
!!1
te"ztf(t)
8888z=&
z=s
"dt =
% &
0
1te"stf(t) dt
(b) Di#culty comes from t in the denominator, but limt%0
sin tt
= 1.
(c) LO
sin tt
4=
% &
s
!% &
0
e"zt sin(t) dt
"dz =
% &
s
1z2 + 1
dz = arctan z|&s =#2! arctan s
7.3 Step Functions, Translated Functions, and Periodic Func-tions
1.3e"4s
s
2.5e"6s
s
7.4. THE INVERSE LAPLACE TRANSFORM 215
3.4 ! 4e"10s
s
4.25
#e"5s ! e"7s$
5.6e"9s ! 6e"3s
s6. NOTE: t > 5 for piecewise definition of function
1 + e"2s + e"3s ! 3e"5s
s
7.1 ! e"2s + e"6s
s
8.1 ! e"3s
s2
9.2 ! 2e"5s
s2
10. e"s 1e + es
=e"s"1
s + 1
11.3e"7s
s! 4e"4s
s+
e"4s(1 + 4s)s2
12.6s
+2e"s
s! 2e"s(1 + s)
s2+
2e"3s(1 + 3s)s2
13.1 + e2
e2
14. 0
15. 1
16.1 ! 2e4
e4
17.
34
=
% 1
0
sin2[#(t ! t0)]*(t !12) dt
= sin2[#(12! t0)]
=12!
cos(2#( 12 ! t0))
2
!12
= cos(2#(12! t0))
= cos(# ! 2#t0)
= cos #(cos(2#t0)) + sin #(sin(2#t0))
12
= cos(2#t0)
t0 =cos"1( 1
2 )
2#
=16
18. Integrand only has value at t = 2 " n = 3
19. !1 + U0(t)
7.4 The Inverse Laplace Transform
1.32
sin(2t)
216 Section 7.5
2. 5 cosh(3t)
3. 5te3t
4. 5e"2t ! 10te"2t
5. e"at cos($
3t)
6. e"4t(cos(2t) + sin(2t))
7.12
,1 ! cos
$2 +
$2 sin(
$2t)-
8.t6
sin(3t) +154
sin(3t) ! t18
cos(3t)
9. F (s) =2s + 7
(s + 3)4=
2(s + 3)3
+1
(s + 3)4" f(t) = t2e"3t +
16t3e"3t
10. F (s) =8(s + 1)
(2s ! 1)3=
(s ! 12 )
(s ! 12 )3
+32
(s ! 12 )3
" f(t) = tet/2 +34t2et/2
11. F (s) =s ! 2
s2 + 5s + 6=
5s + 3
+!4
s + 2" f(t) = 5e"3t ! 4e"2t
12. F (s) =5s + 8
s2 + 3s ! 10=
177
s + 5+
187
s ! 2" f(t) =
177
e"5t +187
e2t
13. L"1 {F (s)} = U&(t)f(t ! #) where f(t ! #) = !5 cos(3t) ! 2 sin(3t) =
O0 0 < t < #
!5 cos(3t) ! 2 sin(3t) t > #
14. L"1 {F (s)} = U3(t)f(t ! 3) where f(t ! 3) = 2e"2t+6 cos(3t ! 9) +53e"2t+6 sin(3t ! 9)
=
O0 0 < t < 3
2e"2t+6 cos(3t ! 9) + 53e"2t+6 sin(3t ! 9) t > 3
15. L"1 {F (s)} = U4(t)f(t ! 4) ! U7(t)f(t ! 7) =
AC
D
0 0 < t < 4t ! 4 4 < t < 7
3 t > 7
16.13
[sin(9 ! 3t)U3(t ! 3) + 2 sin(3t)U0(t)]
7.5 Laplace Transform Solution of Linear Di!erential Equa-tions
1.12e3t +
32et
2.12
#3 cos t + 3 sin t ! 5e"t$
3. 2 cos(2t) +32
sin(2t)
4. 7 cos(4t)
5. 2e3t
6. e3t
7.137
e"4t +157
e3t
8.5 sin(2
$2t)$
2
9. 3e"t sin(2t) + 2e"t cos(2t)
10.74e"t +
9t4
e"t +14e"3t +
t4e"3t
11. !52e5t +
152
e3t ! 6te4t
7.6. SOLVING LINEAR SYSTEMS USING LAPLACE TRANSFORMS 217
12.3118
e2t ! 229
e"t +1318
e"t (cos(3t) ! sin(3t))
13. !e2t (! cos t + 7 sin t) +47
#e"5t ! e2t$
15.!20936
e4t +11436
te4t +112
e3t +174
e2t ! 8918
et
7.6 Solving Linear Systems using Laplace Transforms
1. x(t) = !e2t + 2et + 2y(t) = !e2t + et
2. x(t) = 2e2t +12e"t ! 1
2et
y(t) = !e2t +12e"t +
12et
3. x(t) = !3t ! e"t
y(t) = !1 + e"t + 2t
4. x(t) = 2t + 3 +43t3
y(t) =83t3 ! 2t2 + 5t + 5
5. !2et + 5e4t
!4et + 4e4t
6. x(t) = ! sin(2t) + 2 cos(2t) + e2t
y(t) = 5 sin(2t) + 2e2t
7. x(t) = 8 sin t + 2 cos t
y(t) = cos t ! 13 sin t +12et ! 1
2e"t
8. x(t) = !e"2t + 3e4t + 1 ! ty(t) = !e"2t ! 3e4t + t
9. x(t) = 2e"t + 2e"2t
y(t) = 2e"t + 3 ! 8e"2t
10. x(t) =47
,!1 + e
7t5
-
y(t) = !17! 16
49e
7t5 +
1649
11. x(t) = !3et + 2y(t) = !2et ! 8
12. x(t) = 4 sin(2t) ! 4 cos(2t)y(t) = cos(2t) + sin(2t)
13. x(t) = !4 +t3
3+ 5t ! 2t2 + 5e"t
y(t) = 5 ! 5t + 2t2 ! 5e"t
7.7 The Convolution
1.t3
3
2. !25
cos(2t) +15
sin(2t) +25e"t
3.3
128sin(4t) +
14t3 ! 3
32t
4.s cos(2t) + 2 sin(2t) ! se"st
s2 + 4
218 Chapter 7 Review
5. t
6. ! cos t + cos tU0(# ! 2t)
7. f(t) = e"2t ! e"3t
8. f(t) = !15e"4t +
15et
9. f(t) = !19
cos(3t) +19
10. f(t) =113
!1 ! e"2t cos(3t) ! 2
3e"2t sin(3t)
"
11. f(t) =127
! 19t +
16t2 ! 1
27e"3t
12. f(t) =153
!e"7t ! cos(2t) +
72
sin(2t)
"
13.
(f , g) , h =
% t
0
[(f , g)(t! + )] h(+ ) d+
=
% t
0
&% t"'
0
f((t ! + ) ! z)g(z) dz
'h(+ ) d+
f , (g , h) =
% t
0
f(t ! + )(g , h)(+ ) d+
=
% t
0
f(t ! + )&% '
0
g(+ ! z)h(z) dz
'd+
(interchange order of integration)
=
% t
0
h(z)
&% t
z
f(t ! + )g(+ ! z) d+
'dz
(let u = + ! z, du = d+ )
=
% t
0
h(z)
&% t"z
0
f(t ! (z + u))g(u) du
'dz
=
% t
0
h(z)
&% t"z
0
f((t ! z) ! u)g(u) du
'dz
=
% t
0
f , g(t ! z)h(z) dz
= (f , g) , h
16. LS
t"1/2T
=
9#s" L
St"1/2 , t"1/2
T=#s" L"1
SLS
t"1/2 , t"1/2TT
= #
17. (a) sin kt (cos kt sin t + sin kt ! cos t sin kt)
(b) sin kt (! cos kt + cos t cos kt + sin t sin kt)
(c) cos kt (cos kt sin t + sin kt ! cos t sin kt)
(d)12t sin kt
7.8 Chapter 7: Additional Problems
1. False, f(t) must be a real-valued function for which
% &
0
e"stf(t) dt exists.
2. True. Theorem 7.3.1
3. True. Problem 7.2.14
Chapter 7 Review 219
4. False. L{C1f1(t) + C2f2(t)} = C1L{f1(t)} + C2L{f2(t)}.5. True. Uniqueness of Laplace transforms.
6. !5e"7s
s+
5s
7.48s5
8.2
(s + 1)3
9.2s
+1s2
10.s ! 1
(s ! 1)2 + 1
11.1
s ! (a + bi)
16. 1 ! e"3
17. 0
18. 0
19. tan(#/4) = 1
20.53
sin 3t
21. 7 cos 2t
22. 6te2t
23. 5e"3t ! 15te"3t
24.115
e"5t/2
&15 cos
!t$
152
"! 11
$15 sin
!t$
152
"'
25.13e4t +
173
et
26. 2 cos 3t + sin 3t
27. cos 9t +29
sin 9t
28.37e4t +
47e"3t
29. 6e"t cos 2t + 3e"t sin 2t
30. 2e"2t + te"2t ! e"t + 2te"t
31.209
e2t +159
te2t ! 479
e5t + 8e3t
32.3825
e2t ! e"t ! 1325
cos 4t e"t ! 39100
sin 4t e"t
33. !16et ! 17
42e"5t ! 3
7e2t
34.9136
! 136
t ! 2320
e"2t ! 34e"t +
67180
e3t
37. x(t) = !119
e"t +5963
e"10t +167
e"3t
y(t) = !119
e"t ! 236315
e"10t ! 247
e"3t +125
38. x(t) =875
e2t ! 14e3t ! 125
cos t ! 245
sin t
y(t) = !875
e2t +563
e3t +245
sin t +125
cos t ! 23
220 Chapter 7 Review
39. x(t) = !52e2t +
32
y(t) = !2e2t ! 8
40. x(t) = !14e3t/2
&2 cosh(3t/2) ! 1
3sinh(3t/2)
'
y(t) = !10e3t/2 cosh(3t/2)
41.34t4
42.113
#!3 cos 3t + 2 sin 3t + 3e"2t$
43.!316
t +12t3 +
364
sin 4t
44.b cos at + a sin at ! be"bt
b2 + a2
45. sin t ! sin tU0(# ! 2t)
46. e"3t ! e"4t
47.113
#e5t ! e"8t$
48.!14
cos 2t +14
49.16t ! 1
36sin 6t
50.1
128
#e"8t ! cos 8t + sin 8t
$
Chapter 8
Series Methods
8.1 Power Series Representations of Functions
1. (a) limn%&
8888xn+1
xn
8888 = limn%&
|x| = |x| < 1 for convergence.
For x = 1,&U
n=0
1n diverges
For x = !1,&U
n=0
(!1)n diverges (the sequence of partial sums oscillates between 0 and 1...they
don’t converge)
(b)U
xn is a geometric series which converges to1
1 ! x.
To see this, observe Sk =kU
n=0
xn = 1 + x + x2 + · · · + xk and xSk = x + x2 + x3 + · · · + xk + xk+1 "
Sk ! xSk = 1 ! xk+1 " Sk =1 ! xk+1
1 ! x. From above, we know that |x| < 1, so
S(x) = limk%&
Sk = limk%&
1 ! xk+1
1 ! x=
11 ! x
.
2. (a) f(x) = sin x f(0) = 0f !(x) = cos x f !(0) = 1
f !!(x) = ! sin x f !!(0) = 0f !!!(x) = ! cos x f !!!(0) = !1
......
"
f(x) = sin x = 0 +1(x)1!
+0(x2)
2!+
(!1)(x3)3!
+0(x4)
4!+ · · · = x ! x3
3!+
x5
5!+ · · ·
(b) f(x) = cos x f(0) = 1f !(x) = ! sin x f !(0) = 0f !!(x) = ! cos x f !!(0) = !1f !!!(x) = sin x f !!!(0) = 0
......
"
f(x) = cos x = 1 +0(x)1!
+(!1)(x2)
2!+
(0)(x3)3!
+1(x4)
4!+ · · · = 1 ! x2
2!+
x4
4!+ · · ·
221
222 Section 8.1
(c) f(x) = sinh x f(0) = 0f !(x) = cosh x f !(0) = 1f !!(x) = sinhx f !!(0) = 0f !!!(x) = cosh x f !!!(0) = 1
......
"
f(x) = sinh x = 0 +1(x)1!
+0(x2)
2!+
(1)(x3)3!
+0(x4)
4!+ · · · = x +
x3
3!+
x5
5!+ · · ·
(d) f(x) = cosh x f(0) = 1f !(x) = sinhx f !(0) = 0f !!(x) = cosh x f !!(0) = 1f !!!(x) = sin x f !!!(0) = 0
......
"
f(x) = cosh x = 1 +0(x)1!
+(1)(x2)
2!+
(0)(x3)3!
+1(x4)
4!+ · · · = 1 +
x2
2!+
x4
4!+ · · ·
To show analytic, sin x =&U
n=0
(!1)nx2n+1
(2n + 1)!" lim
n%&
888 ("1)n+1x2(n+1)+1
(2(n+1)+1)!
888888 x2n+1
(2n+1)!
888= lim
n%&
888 ("1)n+1x2n+3
(2n+3)!
888888 x2n+1
(2n+1)!
888=
limn%&
8888x2
(2n + 3)(2n + 2)
8888 = 0, -x. The argument is similar for the other functions.
3. (a) f !(0) = limx%0
e"1/x2
x= lim
x%0
1x
e1/x2 = limx%0
"1x2
e1/x2 "2x3
= limx%0
x
2e1/x2 = 0
f !!(0) = limx%0
f !(x) ! f !(0)x
= limx%0
2
x4e1/x2 = 0
4. limn%&
88888
x2n+3
2n+3
x2n+1
2n+1
88888 = limn%&
8888x2(2n + 1)
2n + 3
8888 = |x2| < 1 for convergence " |x| < 1
For x = 1,&U
n=0
(!1)n(1)2n+1
2n + 1which is an alternating series which converges by the Alternating Series
Test.
For x = !1,&U
n=0
(!1)n(!1)2n+1
2n + 1=
&U
n=0
(!1)3n+1
2n + 1which is an alternating series which converges by
the Alternating Series Test.Therefore, the interval of convergence is |x| ' 1 and radius of convergence is R = 1.
5. ex = 1 + x + x2/2! + x3/3! + · · · , e"x = 1 ! x + x2/2! ! x3/3! + · · ·" exe"x = (1 + x + x2/2! + x3/3! + · · · )(1 ! x + x2/2! ! x3/3! + · · · )= (1 ! x + x2/2! ! x3/3! + · · · ) + x(1 ! x + x2/2! ! x3/3! + · · · )
+ (x2/2!)(1 ! x + x2/2! ! x3/3! + · · · ) + · · ·= 1
6. sin 2x =&U
n=0
(!1)n(2x)2n+1
(2n + 1)!= 2x ! (2x)3
3!+
(2x)5
5!+ · · ·
= 2x ! 8x3
3!+
32x5
5!+ · · ·
2 sin x cos x = 2(x ! x3/3! + x5/5! + · · · )(1 ! x2/2! + x4/4! + · · · )= 2(x ! x3/2! + x5/4! + · · ·! x3/3! + x5/(2!3!) + · · · + x5/5! + · · · )= 2(x ! 3x3/6 ! x3/6 + x5/24 + x5/12 + x5/120 + · · · )
= 2x ! 4x3
3+
32x5
120+ · · · = 2x ! 8x3
3!+
32x5
120+ · · · = sin 2x
8.2. THE POWER SERIES METHOD 223
7.
eix = 1 + (ix) + (ix)2/2! + (ix)3/3! + (ix)4/4! + · · ·= 1 + ix ! x2/2! ! ix3/3! + x4/4! + · · ·= (1 ! x2/2! + x4/4! + · · · ) + i(x ! x3/3! + x5/5! + · · · )= cos x + i sin x
8. limx%0
1 ! cos xx
= limx%0
1 ! (1 ! x2/2! + x4/4! + · · · )x
= limx%0
x/2! ! x3/4! + x5/6! + · · · = 0
9. (a)
W (x) =
8888f(x) g(x)f !(x) g!(x)
8888
= f(x)g!(x) ! f !(x)g(x)
= (a0 + a1(x ! c) + a2(x ! c)2 + · · · )(b1 + 2b2(x ! c) + · · · )!(a1 + 2a2(x ! c) + · · · )(b0 + b1(x ! c) + b2(x ! c)2 + · · · )
= a0b1 + 2a0b2(x ! c) + · · · + a1b1(x ! c) + 2a1b2(x ! c)2
+ · · · + a2b1(x ! c)2 + 2a2b2(x ! c)3 + · · ·!a1b0 ! aab1(x ! c) ! a1b2(x ! c)2 ! 2a2b0(x ! c) ! 2a2b1(x ! c)2 ! 2a2b2(x ! c)3
= a0b1 ! a1b0
(b) See Theorem 3.2.4
8.2 The Power Series Method
1. y = 1 + x +x2
2+
2x3
3+
7x4
12+ · · ·
2. y = 1 + 3(x ! 1) + 4(x ! 1)2 + 3(x ! 1)3 +32(x ! 1)4 + · · ·
3. y =x2
2+
x3
6+
x4
6+
x5
15+ · · ·
4. y = 1 + x + x2 +x3
3+
x4
12+ · · ·
5. y = 1 + x + x2 +4x3
3+
7x4
6+ · · ·
6. y =x2
2+
x3
6+
x4
24+
x5
120+ · · ·
7. y = x + x2 +x3
2+
x4
8! x5
15+ · · ·
8. y = 1 + 2x + 4x2 +25x3
3+
49x4
4+ · · ·
9. y = 1 + x +x3
3! x4
3+ · · ·
10. y = x +x3
3+
x5
30! x7
56+ · · ·
11. y = 1 + 2(x ! 1) + 4(x ! 1)2 +253
(x ! 1)3 +814
(x ! 1)4 + · · ·
12. y = 1 + 2(x ! 1) +72(x ! 1)2 +
143
(x ! 1)3 +7312
(x ! 1)4 + · · ·
13. y = 1 ! x +x3
3+
x4
12! x5
60+ · · ·
224 Section 8.3
14. y = 1 ! x ! x2
2+
x3
6! x4
12! x5
24+ · · ·
15. y = c0
!1 ! x3
6+
3x5
40
"+ c1
!x ! x3
6! x4
12+
3x5
40
"+ · · ·
16. y = x + x2 +x3
2+
x4
6+ · · ·
17. y = 1 ! x4
+ x2 +x4
2+ · · ·
18. y = c0
!1 +
x3
6
"+ c1
!x +
x4
12
"
19. c0
!1 ! px2 +
16(p ! 2)px4
"+ c1
!x +
x3(1 ! p)3
+130
(3 ! p)x5 +130
(p ! 3)px5
"
8.3 Ordinary and Singular Points
1.
x0 = Singular point p(x) q(x) (x ! x0)p(x) (x ! x0)2q(x) Classify
x0 = 0x + 2
x2 ! 3x1
x2 ! 3xx + 2x ! 3
xx ! 3
Regular
x0 = 3x + 2
xx ! 3
xRegular
2.
x0 = Singular point p(x) q(x) (x ! x0)p(x) (x ! x0)2q(x) Classify
x0 = 0x2 ! 2xx3 + x2
4x3 + x2
x2 ! 2xx2 + x
4x + 1
Irregular
x0 = !1x2 ! 2x
x2
4x + 4x2
Regular
3.
x0 = Singular point p(x) q(x) (x ! x0)p(x) (x ! x0)2q(x) Classify
x0 = 01
x(x + 2)2x ! 1
x2(x + 2)1
x + 22x ! 1x + 2
Regular
x0 = !21x
(2x ! 1)(x + 2)x2
Regular
4.
x0 = Singular point p(x) q(x) (x ! x0)p(x) (x ! x0)2q(x) Classify
x0 = 01
x3(x + 3)(x ! 2)1
x3(x + 3)1
x2(x + 3)(x ! 2)1
x(x + 3)Irregular
x0 = 21
x(x + 3)x ! 2
x3(x + 3)Regular
x0 = !31
x(x ! 2)x + 3x3
Regular
5.
x0 = Singular point p(x) q(x) (x ! x0)p(x) (x ! x0)2q(x) Classify
x0 = 0!(x + 2)
x2(2x + 1)2ex
x2(2x + 1)!(x + 2)x(2x + 1)
2ex
2x + 1Irregular
x0 = ! 12
!(x + 2)2x2
ex(2x + 1)2x2
Regular
6.
x0 = Singular point p(x) q(x) (x ! x0)p(x) (x ! x0)2q(x) Classify
x0 = 02
x2(x ! 1)1
(x ! 1)22
x(x ! 1)x2
(x ! 1)2Irregular
x0 = 12x2
1 Regular
7. T5(x) = 5x ! 20x3 + 16x5
T6(x) = !1 + 18x2 ! 48x4 + 32x6
8. (a) y = cos(n arccos x) " y! = n sin(n arccos x)
!1$
1 ! x2
""
y!! =nx sin(n arccos x)
(1 ! x2)$
1 ! x2! n2 cos(n arccos x)
1 ! x2"
8.4. THE METHOD OF FROBENIUS 225
(1 ! x2)y!! =nx sin(n arccos x)$
1 ! x2! n2 cos(n arccos x)
!xy! =!nx sin(n arccos x)$
1 ! x2, n2y = n2 cos(n arccos x) " (1 ! x2)y!! ! xy! + n2y = 0
(b) • n = 0; y(x) = cos 0 = 1 = T0(x)
• n = 1; y(x) = cos(arccos x) = x = T1(x)
• n = 2; y(x) = cos(2 arccos x) = 2 cos2(arccos x) ! 1 = 2x2 ! 1 = T2(x)
8.4 The Method of Frobenius
The Mathematica code may need to be modified as follows:
yr[x_] = c0*x^r + c1*x^(r + 1) + c2*x^(r + 2) + c3*x^(r + 3) + c4*x^(r + 4)eqODE[x_] = (x^2)*y’’[x] + (3 x - x^2)*y’[x] + (5 - x)*y[x]
(*your differential equation*)eq1 = Collect[ReplaceAll[eqODE[x], {y[x] -> yr[x], y’[x] -> yr’[x],
y’’[x] -> yr’’[x]}],x^(r - 3)]eq4a = Coefficient[Coefficient[eq1, x^(r - 0)], x, 0](*In the above equation, the x^(r - 0) term needs to be adjusted based
on the result from eq1. If eq1 has a x^(r - 1),then r - 1 needs to replace r - 0 in eq4a and eq5a, r needs toreplace r + 1, r + 1 needs to replace r + 2,and r + 2 needs to replace r + 3 in the equations eq5b - 5d below*)
eq4b = Solve[eq4a == 0, r]eq5a = Coefficient[Expand[eq1], x^(r - 0)]eq5b = Coefficient[Expand[eq1], x^(r + 1)]eq5c = Coefficient[Expand[eq1], x^(r + 2)]eq5d = Coefficient[Expand[eq1], x^(r + 3)]eqsoln1 = Solve[ReplaceAll[{eq5a == 0, eq5b == 0, eq5c == 0, eq5d == 0},
eq4b[[1]]], {c1, c2, c3, c4}]eqsoln2 = Solve[ReplaceAll[{eq5a == 0, eq5b == 0, eq5c == 0, eq5d == 0},
eq4b[[2]]], {c1, c2, c3, c4}]y1[x_]=Collect[ReplaceAll[ReplaceAll[ReplaceAll[yr[x], eqsoln1[[1]]],
c4 -> 0],eq4b[[1]]], c0]y2[x_]=Collect[ReplaceAll[ReplaceAll[ReplaceAll[yr[x], eqsoln2[[1]]],
c4 -> 0],eq4b[[2]]], c0]
1. y1(x) = c0
!1$x! x3/2
2
"
y2(x) = c0
!x ! x3
14
"
4. y1(x) = c0
!x1/3 ! 3
2x7/3
"
y2(x) = c0
!x5/3 ! 3
10x11/3
"
5. y = 3c1x12
&U
n=0
2n+1(n + 1)(2n + 3)!
xn + c2x"1
/1 !
&U
n=1
2n"1(n ! 1)!(2n ! 2)!
xn
0
7. y = 3c1x13
!1 +
2x9
!2&U
n=2
(!1)n 1 · 4 · 7 . . . (3n ! 5)9nn!
xn + c2x
0
10. y1(x) = c0
!!1 +
1x
"
y2(x) = c0
!x ! x2
3+
x3
12! x4
60
"
226 Section 8.5
13. y = c1x3
1 + x+ c2(1 ! x + x2)
16. y = c1
&U
n=0
xn
(n + 1)!
19. y1(x) = c0$
x + c1x3/2
y2(x) = c0x3/2
21. y1(x) = c1
&U
n=0
(!1)n xn
(n!)2,
y = c1y1(x) + c2
/y1(x) ln x ! 2
&U
n=2
(!1)n 1 + 12 + . . . + 1
n
(n!)2xn
0
22. y1(x) = y2(x) = c0
#x ! x2 + x3 ! x4$
23. y1(x) = x + x2,
y = c1y1(x) + c2
/y1(x) ln x ! 2x2 !
&U
n=2
(!1)n xn+1
n(n ! 1)
0
24. Same as #16
27. y1(x) = xex,
y = c1y1(x) + c2
/y1(x) ln x1 !
&U
n=2
1 + 12 + . . . + 1
n
(n ! 1)!xn
0
28. p(x) = 1 ! 35x +
15x2 + . . .,
q(x) =15x ! 3
20x2 + . . .,
y = c1x[p(x) cos ln x ! q(x) sin lnx] + c2x[q(x) cos lnx + p(x) sin ln x]
8.5 Bessel Functions
1. limn%&
8888888
( x2 )2n+2
((n+1)!)2
( x2 )2n
(n!)2
8888888= lim
n%&
88888
#x2
$2n!n!
(n + 1)!(n + 1)!
88888 = limn%&
88888
#x2
$2
(n + 1)2
88888 = 0, -x " converges.
2. (a) limn%&
8888888
( x2 )2n+3
(n+1)!(n+2)!
( x2 )2n+1
(n!)(n+1)!
8888888= lim
n%&
88888
#x2
$2n!(n + 1)!
(n + 1)!(n + 2)!
88888 = limn%&
88888
#x2
$2
(n + 1)(n + 2)
88888 = 0, -x " converges.
(b) J !0(x) =
&U
n=1
(!1)n(2n)x2n"1
(n!)2(22n)=
&U
n=0
(!1)n+1x2n+1
(n + 1)!n!(22n+1)= (!1)
&U
n=0
(!1)nx2n+1
(n + 1)!n!(22n+1)= !J1(x)
3. (a) kxpJp"1(kx) =&U
n=0
(!1)n
n!$(n + p)k2n+px2n+2p"1
22n+p"1
Section 8.5 227
ddx
(xpJp(kx)) =ddx
/ &U
n=0
(!1)n k2n+p x2n+2p
n!$(n + p + 1) 22n+p
0
=&U
n=0
(!1)n k2n+p(2n + 2p)x2n+2p"1
n!$(n + p + 1) 22n+p
=&U
n=0
(!1)n k2n+p 2(n + p)x2n+2p"1
n!$(n + p + 1) 22n+p
=&U
n=0
(!1)n k2n+p (n + p)x2n+2p"1
n! (n + p)$(n + p) 22n+p"1
=&U
n=0
(!1)n k2n+p x2n+2p"1
n!$(n + p) 22n+p"1
= kxp Jp"1(kx)
(b) !kx"pJp+1(kx) =&U
n=0
(!1)n+1
n!$(n + p + 2)k2n+p+2x2n+1
22n+p+1
ddx
#x"pJp(kx)
$=
ddx
/ &U
n=0
(!1)n k2n+p x2n
n!$(n + p + 1) 22n+p
0
=&U
n=0
(!1)n k2n+p(2n)x2n"1
n!$(n + p + 1) 22n+p
=&U
n=0
(!1)n k2n+p x2n"1
(n ! 1)!$(n + p + 1) 22n+p"1
(n = m + 1)
=&U
m="1
(!1)m+1 k2m+p+2 x2m+1
m!$(m + p + 2) 22m+p+1
=(!1)0x"1kp
(!1)!$(p + 1)2p"1
&U
m=0
(!1)m+1 k2m+p+2 x2m+1
m!$(m + p + 2) 22m+p+1
=&U
m=0
(!1)m+1 k2m+p+2 x2m+1
m!$(m + p + 2) 22m+p+1
= !kx"p Jp+1(kx)
To see this,(!1)0x"1kp
(!1)!$(p + 1)2p"1=
x"1kp
$(0)$(p + 1)2p+1= 0
because limx%0!
$(x) = !+and limx%0+
$(x) = +
228 Section 8.5
4.
J0(kx) =&U
n=0
(!1)n k2n x2n
(n!)2 22n
" k2x J0(kx) =&U
n=0
(!1)n k2n+2 x2n+1
(n!)2 22n
= k2x +&U
n=1
(!1)n k2n+2x2n+1
(n!)2 22n
J !0(kx) =
&U
n=1
(!1)n k2n 2n x2n"1
(n!)2 22n
J !!0 (kx) =
&U
n=1
(!1)n k2n 2n (2n ! 1) x2n"2
(n!)2 22n
xJ !!0 (kx) =
&U
n=1
(!1)n k2n 2n (2n ! 1) x2n"1
(n!)2 22n
" xy!! + y! + k2xy = k2x +&U
n=1
(!1)n k2n x2n"1
(n!)2 22n
;k2 x2 + 4n2<
= k2x + k2x&U
n=1
(!1)n k2nx2n
(n!)2 22n+
&U
n=1
(!1)n k2nx2n"1 4n2
(n!)2 22n
= k2x + k2x J0(kx) +&U
n=1
(!1)n k2nx2n"1
((n ! 1)!)2 22n"2
= k2x + k2x (J0(kx) ! 1) + k2x&U
n=1
(!1)n k2n"2x2n"2
((n ! 1)!)2 22n"2
= k2x + k2x (J0(kx) ! 1) + k2x&U
n=0
(!1)n"1 k2nx2n
(n!)2 22n
= k2x + k2xJ0(kx) ! k2x + k2x (!J0(kx))
= 0
5. (a) y =u(x)$
x" y! =
$xu! ! u(1/2)x"1/2
x
x2 y!! = x3/2u!! ! x1/2u! +34u x"1/2
xy! =$
xu! ! u
2$
x#x2 ! p2$ y = x3/2u ! p2x"1/2u
x2y!! + xy! + (x2 ! p2)y = x3/2u!! ! x1/2u! +34ux"1/2 + x1/2u! ! 1
2ux"1/2 + x3/2u ! p2x"1/2u
= x3/2u!! +
!14! p2
"ux"1/2 + ux3/2
= x3/2
&u!! +
!14! p2
"ux"2 + u
'= 0
(assuming x #= 0) " u!! + u
&1 +
!14! p2
"x"2
'= 0
(b) p =12, y1 =
cos x$x
" u = cos x " u!! = ! cos x "! cos x + cos x
&1 +
!14! 1
4
"x"2
'= 0 Check!
(c) p =12, y1 =
sin x$x
" u = sin x " u!! = ! sin x " ! sin x + sin x
&1 +
!14! 1
4
"x"2
'= 0 Check!
8.6. CHAPTER 8: ADDITIONAL PROBLEMS 229
6.
y =$
xf('x!)
y! = f !('x!)('"x!"1/2) +f('x!)
2$
x
y!! = f !('x!)('"(" ! 12)x!"3/2) + f !!('x!)('"x!"1)('"x!"1/2)
+2$
xf !('x!)('"x!"1) ! f('x!)( 1#x)
4x
x2y!! +
!'2"2x2! +
14! p2"2
"y = '"2f !('x!)x!
$x ! 1
2f !('x!)'"x!
$x + f !!('x!)('2"2x2!)
$x
+12f !('x!)('")x!
$x ! 1
4
$xf('x!) + f('x!)'2"2x2!$x
+14
$xf('x!) ! p2"2f('x!)
$x
=$
x"2If !!('x!)'2x2! + f !('x!)'x! + f('x!)
,'2x2! ! p2
-J
,if u = 'x! , u2 = '2x2!
-
=$
x"2 ;u2f !!(u) + uf !(u) + f(u)#u2 ! p2$<
= 0 since f('x!) is a solution to a Bessel function of order p
8. Using the chain rule (from calculus) to convert the variables, we see thatdydt
=dydx
dxdt
= !e"t dydx
and
d2ydt2
=ddt
!!e"t dy
dx
"= e"2t d2y
dx2+ e"t dy
dx.
Substitution of this (together with x = e"t) into the ODE gives the desired result.
8.6 Chapter 8: Additional Problems
1. False. See Theorem 8.1.1.
2. True, this is just one reason they are useful.
3. If you think this is true, then you have missed the point of this chapter.
4. False. See Definition 8.1.
5. False, check out Taylor’s theorem.
9. y = 3 + 12x + 12x2 + 8x3 + 4x4 + · · ·
10. y = 1 + 4(x ! 1) + 7(x ! 1)2 +203
(x ! 1)3 +236
(x ! 1)4 + · · ·
11. y = 6 ! 2x ! 4x2 + x3 +3x4
4+ · · ·
12. y = 3 + 6x +13x2
2+
13x3
3+
13x4
6+ · · ·
13. y = 5 + 3x +25x2
2+ 10x3 +
67x4
2+ · · ·
14. y = 1 + 2x + x2 +x3
6+
x4
24+ · · ·
15. y = x + x2 + x3 +7x4
12+
3x5
10+ · · ·
16. y = 6 + 30x + 75x2 +376x3
3+
470x4
3+ · · ·
17. y = 2 + 16x + 256x2 +14338
3x3 +
2867363
x4 + · · ·
230 Chapter 8 Review
18. y = 2 + x +cos(2)
2x2 ! sin(2)
3x3 + · · ·
19. y = 4 + 16x + 64x2 + 256x3 +4097
4x4 + · · ·
20. y = 6 + 432x + 46656x2 +16796164
3x3 + 705438792x4 + · · ·
22. y = c1x43
,1 ! 3x2
16 + 9x4
896 ! . . .-
+c2x23
,1 ! 3x2
8 + 9x4
320 ! . . .-
25. y = c1x13
,1 ! 3x2
16 + 9x4
896 ! . . .-
+c2x!13
,1 ! 3x2
8 + 9x4
320 ! . . .-
29. y = c1
,1 + x + 3x2
10 + . . .-
+c2x13
,1 + 7x
12 + 5x2
36 ! . . .-
Appendix B
Graphing Factored Polynomials
1. f(x) = x2(x ! 2)(4 ! 3x)
2. f(x) = (x2 ! 9)2(x ! 1)
3. g(x) = (1 + x)(3 + 2x)(5 ! x)7
231
232 Appendix B
4. h(x) = x4(2x ! 3)3(x2 + 4)(x2 + 2x + 1)
5. P (x) = (x4 ! 16)2
6. x(t) = (16 ! 3t2)(t2 ! 4)3
7. f(x) = (4 ! x)8(2 ! x)(4 ! 3x2)4(2x + 2)(x + 5)
8. f(x) = x(4! 3x)3(2x + 1)4(x3 ! 1)
Appendix C
Selected Topics from LinearAlgebra
C.1 A Primer on Matrix Algebra
1. (a)
!3 51 1
"!xy
"=
!7!1
"
(b)
1
32 5 !1!1 2 01 0 4
5
7
1
3xyz
5
7 =
1
32!10
5
7
(c)
!x!
y!
"=
!1 !52 1
"!xy
"
2.
AB =
!2 !3!1 4
"!!2 21 !5
"
=
!!4 ! 3 4 + 152 + 4 !2 ! 20
"
=
!!7 196 !22
"
BA =
!!2 21 !5
"!2 !3!1 4
"
=
!!4 ! 2 6 + 82 + 5 !3 ! 20
"
=
!!6 147 !23
"
3.
AB =
1
31 2 !30 !1 43 0 2
5
7
1
33 !2 2!1 1 !50 2 2
5
7
=
1
33 ! 2 + 0 !2 + 2 + 6 2 ! 10 ! 60 + 1 + 0 0 ! 1 ! 8 0 + 5 + 89 + 0 + 0 !6 + 0 + 4 6 + 0 + 4
5
7
=
1
31 !6 !141 7 139 !2 10
5
7
233
234 Appendix C.1
BA =
1
33 !2 2!1 1 !50 2 2
5
7
1
31 2 !30 !1 43 0 2
5
7
=
1
33 + 0 + 6 6 + 2 + 0 !9 ! 8 + 4
!1 + 0 ! 15 !2 ! 1 + 0 3 + 4 ! 100 + 0 + 6 0 ! 2 + 0 0 + 8 + 4
5
7
=
1
39 8 !13
!16 !3 !36 !2 12
5
7
4. (a)
1
32 2 64 1 47 4 2
5
7
(b)
1
30 8 !2!6 !1 21 0 0
5
7
(c)
1
35 1 1613 3 917 10 5
5
7
(d)
1
36 0!2 42 2
5
7
5. (a) Not possible: Matrices are of di%erent order.
(b)
1
32 !22 1222 4 !24 4 2
5
7
(c) 3
(d) 6
6. (a) 6
(b) Not possible: Matrices are of di%erent order.
(c)
!3 !1 10 2 1
"
(d)
1
31 !1 45 0 22 3 1
5
7
7. (a)
1
32 4 72 1 46 4 2
5
7
(b) Not possible: Number of columns of A (2) does not equal the number of rows of A (3).
(c)
1
33 12 !611 2 47 7 !1
5
7
(d)
1
332 6 118 9 !117 !8 19
5
7
8. (a)
1
320 13 !38 27 145 17 13
5
7
C.2. GAUSSIAN ELIMINATION, MATRIX INVERSES, AND CRAMER’S RULE 235
(b)
1
312 !3!4 54 1
5
7
(c) Not possible: Number of columns of B (2) does not equal the number of rows of A (3).
(d)
!12 !6 30 4 2
"
9. (a)
1
310 !215 38 5
5
7
(b) Not possible: Number of columns of A (2) does not equal the number of rows of E (3).
(c)
!1 !8 1213 4 3
"
(d)
1
3!6 39 !2726 !1 1310 19 !7
5
7
10. (a) 8
(b) 55
(c) 33
(d) 64
11. (a) 1815
(b) 1815
(c) 1815
(d) 55
C.2 Gaussian Elimination, Matrix Inverses, and Cramer’sRule
1. (a) Yes
(b) Yes
(c) No
2. (a) Yes
(b) Yes
(c) No
3. (a)
x1 =18
x2 =34
(b)
x1 = 3
x2 = 1
x3 = 2
236 Appendix C.2
4. (a) There are an infinite number of solutions.
x1 = !17(1 + 3x3)
x2 =17(1 ! 4x3)
(b) No solution.
5. (a)
x = 6
y = 8
z = 3
(b)
x = 5
y = !6
z = 5
6. (a)
x =
8888!2 !31 1
88888888
2 !32 1
8888
=!2 ! (!3)2 ! (!6)
=18
y =
88882 !22 1
88888888
2 !32 1
8888
=!2 ! (!4)2 ! (!6)
=34
(b)
x =
888888
2 0 43 !3 00 !4 6
888888888888
1 0 42 !3 03 !4 6
888888
=!84!14
= 6
Appendix C.2 237
y =
888888
1 2 42 3 03 0 6
888888888888
1 0 42 !3 03 !4 6
888888
=!42!14
= 3
z =
888888
1 0 22 !3 33 !4 0
888888888888
1 0 42 !3 03 !4 6
888888
=14!14
= 1
(c) Since
888888
2 3 !21 !2 34 !1 4
888888= 0, there is no solution.
7. (a)
x1 = 1
x2 = 2
(b)
x =311
y =211
z = ! 111
8. (a)
x = !14455
y = !6155
z =4611
(b)
x = 1
y = 0
z = 2
w = 0
9.
x! =
8888x ! sin $y cos $
88888888
cos $ ! sin $sin $ cos $
8888
= x cos $ + y sin $
238 Appendix C.3
y! =
8888cos $ xsin $ y
88888888
cos $ ! sin $sin $ cos $
8888
= y cos $ ! x sin $
10. The formula for the 2 . 2 inverse:
A =
!a bc d
"then A! =
1detA
!d !b!c a
".
A!1 =
!!1 !21 1
"
A!2 =
119
!5 4!1 3
"
11. A!1 =
13
!7 !2!2 1
"
A!2 =
121
!6 53 !1
"
12. A! =12
1
3!32 !22 67 5 !1!5 !3 1
5
7
B! =134
1
317 34 !17!9 !6 7!11 4 1
5
7
C.3 Coordinates and Change of Basis
1. (a)111
!4 3!1 2
"
(b)
!2 !31 4
"
(c) [w]B =
!3!5
"
(d) [w]B" =
!! 3
11! 13
11
"
2. (a)12
!1 !5!1 !13
"
(b)19
!13 !5!1 !1
"
(c) [w]B =
!! 17
9119
"
(d) [w]B" =
!!4!7
"
3. (a)
1
334
34 0
! 34 ! 17
12 ! 32
0 23 1
5
7
(b) [w]B =
1
3! 31
95! 1
3
5
7
4. (a)
1
372 2 5
2! 5
2 !3 ! 32
6 1 5
5
7
Appendix C.3 239
(b) [w]B =
1
37!6!6
5
7
5. (a) Matrix of eigenvectors P =
!0 10 0
"which is not invertible. Hence P"1 does not exist.
(b) Matrix of eigenvectors P =
!0 21 7
", P"1 =
!! 7
2 112 0
"
P"1AP =
!!1 00 1
"
6. (a) Matrix of eigenvectors P =
!3+i
#3
23"i
#3
21 1
",
P"1 =
/! i#
31+i
#3
2i#3
1"i#
32
0
P"1AP =
/1+i
#3
2 0
0 1"i#
32
0
(b) Matrix of eigenvectors P =
1
31 0 00 0 00 1 0
5
7, which is not invertible. Hence P"1 does not exist.
7. (a) Matrix of eigenvectors P =
1
3!2 0 10 1 03 0 0
5
7, P"1 =
1
30 0 1
30 1 01 0 2
3
5
7
P"1AP =
1
35 0 00 3 00 0 2
5
7
(b) Matrix of eigenvectors P =
1
3!1 0 !20 1 11 0 1
5
7,
P"1 =
1
31 0 21 1 1!1 0 !1
5
7
P"1AP =
1
32 0 00 2 00 0 1
5
7