Dif EQ 2015 Solutions

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Lecture Notes in Mathematics

Arkansas Tech UniversityDepartment of Mathematics

Introductory Notes in Ordinary Differential

Equations for Physical Sciences andEngineering

Solution Guide

Marcel B. FinancAll Rights Reserved



2



Contents

1 Basic Terminology 5

2 Existence and Uniqueness of Solutions to First Order LinearIVP 15

3 Analytical Solution: The Method of Integrating Factor 21

4 Existence and Uniqueness of Solutions to First Order Nonlin-

ear IVP 33

5 Separable Differential Equations 37

6 Exact Differential Equations 47

7 Substitution Techniques: Bernoulli and Riccati Equations 59

8 Graphical Solution: Direction Field of y = f (t, y) 67

9 Numerical Solutions to ODEs: Euler’s Method and its Vari-

ants 77

10 Second Order Linear Differential Equations: Existence and

Uniqueness Results 83

11 The General Solution of 2nd Order Linear Homogeneous Equa-tions 87

12 Second Order Linear Homogeneous Equations with Constant

Coefficients: Distinct Characterisitc Roots 95

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4 CONTENTS

13 Characteristic Equations with Repeated Roots 103

14 Characteristic Equations with Complex Roots 107

15 Series Solutions of Differential Equations 117

16 The Structure of the General Solution of Linear Nonhomo-

geneous Equations 129

17 The Method of Variation of Parameters 137

18 The Laplace Transform: Basic Definitions and Results 145

19 Further Studies of Laplace Transform 153

20 The Laplace Transform and the Method of Partial Fractions161

21 Laplace Transforms of Periodic Functions 171

22 Convolution Integrals 181



1 Basic Terminology

Problem 1.1

A car starts from rest and accelerates in a straight line at 1.6 m/sec 2 for 10seconds.

(a) What is its final speed?(b) How far has it travelled in this time?

Solution.

(a) The velocity is related to acceleration according to the equation

v(t) = v0 + at

where a stands for acceleration. Since v0 = 0 (car starting from rest) anda = 1.6 m/sec2 we can write

v(t) = 1.6t.

The final speed of the car is |v(10)| = 16 m/sec.(b) The displacement of the car is given by x(t) = 1

2at2 + v0t + x0 = 0.8t2.

Thus, the car traveled a distance of x(10) = 80 m

Problem 1.2

An object is thrown upward at time t = 0. After t seconds, its height is

y(t) = −4.9t2 + 7t + 1.5

meters above the ground.(a) From what height was the object thrown?(b) What is the initial velocity of the object?(c) What is the acceleration due to gravity?

5



6 1 BASIC TERMINOLOGY

Solution.

(a) The initial height is y(0) = 1.5 meters.(b) The velocity of the object at time t is given by v(t) = y (t) = −9.8t + 7.The initial velocity of the object is v(0) = 7 m/sec.(c) The acceleration due to gravity is y(t) = −9.8 m/sec2

Problem 1.3

An object thrown in the air on a planet in a distant galaxy is at heighty(t) = −25t2 + 72t + 40 feet at time t seconds after it is thrown. What isthe acceleration due to gravity on that planet? With what velocity was theobject thrown? From what height?

Solution.The acceleration due to gravity is y (t) = −50 ft/sec2. The initial velocity isv(0) = y (0) = 72 ft/sec. The initial height is y(0) = 40 ft

Problem 1.4

An object is dropped from a 400-foot tower. When does it hit the groundand how fast is it going at the time of the impact?

Solution.

The height of the object at any time t is given by y(t) = −16t2 + 400. Theobject hits the ground when y(t) = 0. Solving the equation

−16t2 + 400 = 0

we find t = 5 ft. The velocity of the object at the time of impact is v(5) =y(5) = −32(5) = −160 ft/sec

Problem 1.5

A ball that is dropped from a window hits the ground in five seconds. Howhigh is the window? (Give your answer in feet.)

Solution.

The height of the object at any time t is given by y(t) = −16t2 + y0. We aretold that y(5) = 0. Thus, y0 = 16(5)2 = 400 ft

Problem 1.6

A ball is thrown straight up from ground level and reaches its greatest heightafter 5 seconds. Find the initial velocity of the ball and the value of itsmaximum height above ground level.



7

Solution.

Let y(t) be the height of the ball above ground level at time t seconds afterit was thrown. We are given that y(0) = 0. We are also told that the ballreaches its maximum height after 5 seconds at which point the velocity iszero, i.e., v(5) = 0.The body’s position is governed by the differential equation y(t) = −32 ft/sec2.So y(t) = v(t) = −32t + C 1 for some constant C 1. Since v(5) = 0, solvingthe equation −32(5) + C 1 = 0 for C 1 we find C 1 = 160. Hence,

y(t) = v(t) = −32t + 160.

Using this equation we have now that the initial velocity of the ball wasv(0) = 160 ft/sec. We still need to find the position of the ball at time

5 seconds (when the ball was at its greatest height). By integrating theprevious equation we find

y(t) = −16t2 + 160t + C 2.

Since the ball was thrown from ground level, we have that y(0) = 0, so C 2 = 0and

y(t) = −16t2 + 160t.

We were told that the maximum height was reached after five seconds, sothe maximum height’s value is given by

y(5) = −16(5)2 + 160(5) = 400 ft

Problem 1.7

At time t = 0 an object having mass m is released from rest at a height y0above the ground. Let g represent the constant gravitational acceleration.Derive an expression for the impact time (the time at which the object strikesthe ground). What is the velocity with which the object strikes the ground?

Solution.

The motion satisfies the differential equation y = −g. Integrating twice andusing the facts that v(0) = 0 and y(0) = y0 we find

y(t) =

−

1

2

gt2 + y0.

The object strikes the ground when y(t) = 0. Thus, −12gt2 + y0 = 0. Solving

for t we find t =

2y0g

. The velocity with which the object strikes the ground

is v(

2y0g ) = −g(

2y0

g ) = −√ 2gy0




Problem 1.8

At time t = 0, an object of mass m is released from rest at a height of 252 ft above the floor of an experimental chamber in which gravitationalacceleration has been slightly modified. Assume (instead of the usual valueof 32 ft/sec2), that the acceleration has the form 32− sin

πt4

ft/sec2, where

is a positive constant. In addition, assume that the object strikes the groundexactly 4 sec after release. Can this information be used to determine theconstant ? If so, determine .

Solution.

The motion of the object satisfies the equation y = −(32 − sin

πt4

). The

velocity is given by v(t) = −32t − ( 4π

) cos πt4 + C. But v(0) = 0 so that

C = 4

π . Thus, v(t) = −32t− (4π ) cos πt

4 + 4

π .The displacement function is given by

y(t) = −16t2 −

4

π

2

sin

πt

4

+

4

π t + 252.

The object strikes the ground at t = 4 sec. In this case y(4) = 0. Solving for we find = π

4

Problem 1.9

Find the order of the following differential equations.

(a) ty + y = t3

.(b) y + y2 = 2.(c) sin y + 3t2y = 6t.

Solution.

(a) Since the highest derivative appearing in the equation is 2, the order of the equation is 2.(b) Order is 1(c) Order is 3

Problem 1.10

What is the order of the differential equation?(a) y (t) − 1 = 0.(b) y (t) − 1 = 0.(c) y (t) − 2ty(t) = 0.

(d) y (t)(y(t))12 − t

y(t) = 0.



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Solution.

(a) First order(b) , (c), and (d) second order

Problem 1.11

In the equation∂u

∂x − ∂ u

∂y = x − 2y

identify the independent variable(s) and the dependent variable.

Solution.

u is the depedent variable whereas x and y are the independent variables

Problem 1.12Classify the following equations as either ordinary or partial.

(a) (y)4 + t2

(y)2+4 = 0.

(b) ∂u∂x + y ∂u

∂y = y−xy+x .

(c) y − 4y = 0.

Solution.

(a) ODE.

(b) PDE.(c) ODE

Problem 1.13

Solve the equation y (t) − 2 = 0 by computing successive antiderivatives.

Solution.

Integrating for the first time we find y(t) = 2t + C 1. Integrating the lastequation we find y (t) = t2 + C 1t + C 2. Integrating for a third and final timewe get y(t) = t3

3 + C 1t2

2 + C 2t + C 3

Problem 1.14

Solve the initial-value problem

dy

dt = 3y(t), y(0) = 50.

What is the domain of the solution?




Solution.

The general solution is of the form y(t) = Ce

3t

. Since y(0) = 50, we find50 = C e3·0, and so C = 50. The solution is y(t) = 50e3t. The domain is theset of all real numbers

Problem 1.15

For what real value(s) of λ is y = cos λt a solution of the equation y+9y = 0?

Solution.

Finding the first and second derivatives, we find that y(t) = −λ sin λt andy(t) = −λ2 cos λt. By substitution, cos λt is a solution if and only if λ2−9 =0. This equation has roots λ = ±3

Problem 1.16

For what value(s) of m is y = emt a solution of the equation y+3y+2y = 0?

Solution.

Since ddt (emt) = memt and d2

dt2 (emt) = m2emt the requirement on m becomesm2 + 3m + 2 = 0. Factoring the left-hand side to obtain (m +2)(m + 1) = 0.Thus, m = −2 or m = −1

Problem 1.17

Show that y(t) = et is a solution to the differential equation

y − 2 + 2

t

y +

1 + 2

t

y = 0.

Solution.

Substituting y(t) = y (t) = y (t) = et into the equation we find

y −

2 + 2

t

y +

1 +

2

t

y =et −

2 +

2

t

et +

1 +

2

t

et

=et − 2et − 2

tet + et +

2

tet = 0

Problem 1.18Show that any function of the form y(t) = C 1 cos ωt + C 2 sin ωt satisfies thedifferential equation

d2y

dt2 + ω2y = 0.



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Solution.

Finding the first and the second derivatives of y we obtain

y(t) = −C 1ω sin ωt + C 2ω cos ωt

andy(t) = −C 1ω2 cos ωt − C 2ω2 sin ωt.

Substituting into the equation to obtain

d2y

dt2 + ω2y = − C 1ω2 cos ωt − C 2ω2 sin ωt + ω2(C 1 cos ωt + C 2 sin ωt)

=0

Problem 1.19

Suppose y(t) = 2e−4t is the solution to the initial value problem y + ky =0, y(0) = y0. Find the values of k and y0.

Solution.

Since y(0) = 2, we have y0 = 2. The given function satisfies the equationy + ky = 0, that is, −8e−4t + 2ke−4t = 0. Dividing through by 2e−4t > 0 toobtain −4 + k = 0. Thus, k = 4

Problem 1.20

Consider t > 0. For what value(s) of the constant n, if any, is y(t) = tn asolution to the differential equation

t2y − 2ty + 2y = 0?

Solution.

Since t2y − 2ty + 2y = t2(n(n − 1)tn−2) − 2t(ntn−1) + 2tn = 0 we findn(n− 1)− 2n + 2 = 0 or n2− 3n + 2 = 0. This last equation can be factoredas (n− 1)(n− 2) = 0. Solving we find n = 1 or n = 2

Problem 1.21(a) Show that y(t) = C 1e2t + C 2e−2t is a solution of the differential equationy − 4y = 0, where C 1 and C 2 are arbitrary constants.(b) Find the solution satisfying y(0) = 2 and y (0) = 0.(c) Find the solution satisfying y(0) = 2 and limt→∞ y(t) = 0




Solution.

(a) Finding the first and second derivatives of y(t) to obtain y (t) = 2C 1e

2t

−2C 2e−2t and y(t) = 4C 1e2t + 4C 2e−2t. Thus, y − 4y = 4C 1e2t + 4C 2e−2t −4(C 1e2t + C 2e−2t) = 0.(b) The condition y(0) = 2 implies that C 1 + C 2 = 2. The condition y (0) = 0implies that 2C 1 − 2C 2 = 0 or C 1 = C 2. But C 1 + C 2 = 2 and this impliesthat C 1 = C 2 = 1. In this case, the particular solution is y(t) = e2t + e−2t.(c) The first condition implies that C 1+C 2 = 2. The second condition impliesthat C 1 = 0 since limt→∞ et = ∞. Thus, C 2 = 2 and the particular solutionis given by y(t) = 2e−2t

Problem 1.22

Suppose that the graph below is the particular solution to the initial valueproblem y(t) = m + 1, y(1) = y0. Determine the constants m and y0 andthen find the formula for y(t).

Solution.

From the figure we see that y0 = y(1) = 1. Since y is the slope of the linewhich is −1 we have y(t) = −1 = m + 1. Solving for m we find m = −2.Hence, y(t) = −t + 2

Problem 1.23

Suppose that the graph below is the particular solution to the initial valueproblem y (t) = mt, y(t0) = −1. Determine the constants m and t0 and thenfind the formula for y(t).



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Solution.

From the graph we see that y(0) = −1 so that t0 = 0. Also, by integrationwe see that y = m

2 t2 + C. From the figure we see that C = −1. Finally, sincey(1) = −0.5 we find − 1

2 = m

2 − 1. Solving for m we find m = 1. Thus,

y(t) = t2

2 − 1

Problem 1.24

Show that y(t) = e2t is not a solution to the differential equation y + 4y = 0.

Solution.Finding the second derivative and substituting into the equation we find

y + 4y = 4e2t + 4e2t = 8e2t = 0.

Thus, y(t) = e2t is not a solution to the given equation

Problem 1.25

Consider the initial-value problem

y + 3y = 6t + 5, y(0) = 3.

(a) Show that y = Ce−3t + 2t + 1 is a solution to the above differentialequation.(b) Find the value of C.




Solution.

(a) Substituting y and y into the equation we find−3Ce−3t + 2 + 3[Ce−3t + 2t + 1] = −3Ce−3t + 3Ce−3t + 6t + 5 = 6t + 5

(b) Since y(0) = 3 we find C + 1 = 3. Solving for C we find C = 2. Thus,the solution to the initial value problem is y(t) = 2e−3t + 2t + 1



2 Existence and Uniqueness of

Solutions to First Order Linear

IVP

Problem 2.1

Find p(t) and y0 so that the function y(t) = 3et2 is the solution to the IVPy + p(t)y = 0, y(0) = y0.

Solution.

Since y(t) = 3et2, we have y(0) = y0 = 3e0 = 3. On the other hand, y(t)satisfies the equation y + p(t)y = 0 so that 6tet2 + p(t)3et2 = 0. Solving for

p(t), we find p(t) = −2t

Problem 2.2

For each of the initial conditions, determine the largest interval a < t < b onwhich Theorem 2.2 guarantees the existence of a unique solution.

y + 1

t2 + 1y = sin t

(a) y(0) = π (b) y(π) = 0.

Solution.

Here we have p(t) = 1t2+1 and g(t) = sin t.

(a) (−∞,∞).(b) (−∞,∞)

Problem 2.3For each of the initial conditions, determine the largest interval a < t < b onwhich Theorem 2.2 guarantees the existence of a unique solution

y + t

t2 − 4y =

et

t − 3

15



162 EXISTENCE AND UNIQUENESS OF SOLUTIONS TO FIRST ORDER LINEAR

(a) y(5) = 2 (b) y(− 32) = 1 (c) y(−6) = 2.

Solution.

Notice that p(t) and g(t) are defined for all t = −2, 2, 3(a) 3 < t < ∞.(b) −2 < t < 2.(c) −∞ < t < −2

Problem 2.4

(a) For what values of the constant C and the exponent r is y = Ctr thesolution of the IVP

2ty − 6y = 0, y(−2) = 8?

(b) Determine the largest interval of the form a < t < b on which Theorem2.2 guarantees the existence of a unique solution.(c) What is the actual interval of existence for the solution found in part (a)?

Solution.

(a) Substitution leads to 2trCtr−1 − 6Ctr = 0. Divide through by Ctr toobtain 2r − 6 = 0 or r = 3. Now, since y(−2) = 8 we have C (−2)3 = 8 orC = −1. Thus, y(t) = −t3.(b) Rewriting the equation in the form

y − 3

t y = 0

so that p(t) = − 3t and g(t) = 0. The largest interval of the form a < t < b

that guarantees the existence of a unique solution is the interval −∞ < t < 0since −2 is in that interval and both p(t) and g(t) are continuous over thatinterval.(c) By part (a) the actual interval of existence is the set of all real numberssince y(t) is a cubic polynomial

Problem 2.5

What information does Theorem 2.2 gives about the initial value problem

ty = y + t3 cos t, y(1) = 1?y(−1) = 1?

Solution.

We have p(t) = − 1t and g(t) = t2 cos t. The interval of existence is 0 < t < ∞

if y(1) = 1 and −∞ < t < 0 if y(−1) = 1



17

Problem 2.6

Consider the following differential equation

(t − 4)y + 3y = 1

t2 + 5t.

Without solving, find the interval over which a unique solution is guaranteedfor each of the following initial conditions:(a) y(−3) = 4 (b) y(1.5) = −2 (c) y(−6) = 0 (d) y(4.1) = 3.

Solution.

Rewriting the equation in the form

y + 3t − 4

y = 1(t − 4)(t2 + 5t)

we find that p(t) and g(t) are continuous for all t = −5, 0, 4.(a) −5 < t < 0.(b) 0 < t < 4.(c) −∞ < t < −5.(d) 4 < t < ∞

Problem 2.7

Without solving the initial value problem, (t−1)y+(ln t)y =

2

t−3 , y(t0) = y0,state whether or not a unique solution is guaranteed to exist for the y(t0) = y0

listed below. If a unique solution is guaranteed, find the largest interval forwhich the solution satisfies the differential equation and the initial condition.(a) y(2) = 4 (b) y(0) = 0 (c) y(4) = 2.

Solution.

Rewriting the equation in the form

y + ln t

t

−1

y = 2

(t

−3)(t

−1)

we find that p(t) and g(t) are continuous on (0, 1) ∪ (1, 3) ∪ (3,∞).(a) 1 < t < 3.(b) No such solution.(c) 3 < t < ∞




Problem 2.8

(a) State precisely the theorem (hypothesis and conclusion) for existence anduniqueness of a linear first order initial value problem.(b) Consider the equation y + t2y = et3 with initial conditions y(t0) = y0.Briefly discuss if this has a solution, if it is unique and why.

Solution.

(a) If p(t) and g(t) are continuous functions in the open interval I = (a, b)and t0 a point inside I then the IVP

y + p(t)y = g(t), y(t0) = y0

has a unique solution y(t) defined on I .(b) Since p(t) = t2 and g(t) = et3 are continuous in −∞ < t < ∞, the IVPhas a unique solution for any t0

Problem 2.9

Is the differential equation linear or non-linear? If the equation is linear,decide whether it is homogeneous or non-homogeneous.(a) y = ty2.(b) y = t2y.(c) (cos t)y + ety = sin t.(d) y

y + t3 = sin t, y > 0.

Solution.

(a) Non-linear because of y2.(b) Linear and homogeneous with p(t) = −t2 and g(t) = 0.(c) Linear and non-homogeneous with p(t) = et

cos t and g(t) = tan t.(d) y + (t3−sin t)y = 0. This is linear and homogeneous with p(t) = t3−sin tand g(t) = 0

Problem 2.10

Consider the initial value problem

y + p(t)y = g(t), y(3) = 1.

Suppose that p(t) and/or g(t) have discontinuities at t = −2, t = 0, and t = 5but are continuous for all other values of t. What is the largest interval (a, b)on which Theorem 2.2 is applied.



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Solution.

Because of the initial condition the largest interval of existence guaranteedby the existence and uniqueness theorem is 0 < t < 5

Problem 2.11

Determine α and y0 so that the graph of the solution to the initial-valueproblem

y + αy = 0, y(0) = y0

passes through the points (1, 4) and (3, 1).

Solution.

The general solution is given by y(t) = y(0)e−αt. Since y(1) = 4 and y(3) = 1we find

y(0)e−α

y(0)e−3α = 4.

Solving for α we find α = ln 42

= ln2. Thus, y(t) = y(0)e−(ln2)t = y(0)12

t.

Since y(1) = 4 we have y(0)2 = 4 so that y0 = 8

Problem 2.12

Match the following equations with the correct description. Every equationmatches exactly one description.(a) y = 3y

−5t.

(b) ∂yt

= ∂ 2y∂t2

+ ∂ 2y∂x2 .

(c) y − y2 = sin t.(d) y + 3y = 0.

(i) A partial differential equation.(ii) A homogeneous first order linear differential equation.(iii) A non-linear first order differential equation.(iv) A non-homogenous first order linear differential equation.

Solution.

(a) (iv) (b) (i) (c) (iii) (d) (ii)

Problem 2.13

Consider the differential equation y = −t2y +sin y. What is the order of thisequation? Is it linear or non-linear?




Solution.

A non-linear ( because of sin y) first order ordinary differential equation

Problem 2.14

Show that y = t+yt is a linear first order non-homogeneous equation.

Solution.

We have y = t+yt

= 1 + yt

. Thus,

y − 1

ty = 1.

This is a first-order linear non-homogeneous equation with p(t) =

−1t and

g(t) = 1



3 Analytical Solution: The

Method of Integrating Factor

Problem 3.1

Solve the IVPy = −2ty, y(1) = 1.

Solution.

First we rearrange the equation to the form recognizable as first-order linear:

y + 2ty = 0.

From this we see that p(t) = 2t so that

2tdt = t2. Thus, the general solution

to the DE is y(t) = Ce−t2 . But y(1) = 1 so that C = e. Hence, y(t) = e1−t2

Problem 3.2

Solve the IVPy + ety = 0, y(0) = 2.

Solution.

Since p(t) = et we find

etdt = et so that the general solution to the DE is

y(t) = Ce−et . But y(0) = 2 so that C = 2e. Hence, the unique solution isy(t) = 2e1−et

Problem 3.3

Consider the first order linear non-homogeneous IVP

y + p(t)y = αp(t), y(t0) = y0.

(a) Show that the IVP can be reduced to a first order linear homogeneousIVP by the change of variable z (t) = y(t) − α.(b) Solve this initial value problem for z (t) and use the solution to determiney(t).

21



223 ANALYTICAL SOLUTION: THE METHOD OF INTEGRATING FACTOR

Solution.

(a) Note that the given DE can be written as y + p(t)(y − α) = 0. Sincez (t) = y(t) − α, we get the IVP

z + p(t)z = 0, z (t0) = y(t0) − α.

(b)The general solution to the DE is z (t) = (y0 − α)e− t

t0 p(s)ds

. Thus, y(t) =

(y0 − α)e− t

t0 p(s)ds

+ α

Problem 3.4

Apply the results of the previous problem to solve the IVP

y + 2ty = 6t, y(0) = 4.

Solution.

Letting z (t) = y(t) − 3, the given IVP reduces to

z + 2tz = 0, z (0) = 1.

The unique solution to this IVP is z (t) = e−t2 . Hence, y(t) = e−t2 + 3

Problem 3.5

The unique solution to the IVP

ty − αy = 0, y(1) = y0

goes through the points (2, 1) and (4, 4). Find the values of α and y0.

Solution.

Rewriting the given IVP in the standard form

y − α

t y = 0, y(1) = y0,

we find p(t) = −α

t and −α

t dt = −α ln |t| = ln |t−α

|. Thus, the generalsolution to the DE is given by y(t) = C eln |t

−α| = C |t|−α. But y(2) = 1 andy(4) = 4 so that C 2−α = 1 and C 4−α = 4. Taking the ratio of these lastequations we find 2−α = 4 and thus α = −2. From C 2−α = 1 we find 4C = 1or C = 0.25. Hence, y0 = y(1) = 0.25(1)2 = 0.25



23

Problem 3.6

The table below lists values of t and ln[y(t)] where y(t) is the unique solutionto the IVPy + tny = 0, y(0) = y0.

t 1 2 3 4ln [y(t)] -0.25 -4.00 -20.25 -64.00

(a) Determine the values of n and y0.(b) What is y(−1)?

Solution.

(a) The general solution to the DE is y(t) = Ce−tn+1

n+1 . Since y(0) = y0 we

find C = y0 so that the unique solution is y(t) = y0e−tn+1

n+1 . Thus, ln [y(t)] =ln (y0) − tn+1

n+1. Since ln y(1) = − 1

4 and ln y(2) = −4 we find ln y0 − 1

n+1 −

ln y0 + 2n+1

n+1 = 4 − 14 = 15

4 . Thus, 2n+1−1n+1 = 15

4 . Using trial and error on the

values of n, one finds n = 3. Hence, ln y0 = − 14

+ 1n+1

= − 14

+ 14

= 0 so thaty0 = 1.

(b) y(−1) = 1 · e−(−1)4

4 = 4√

e−1

Problem 3.7

Consider the differential equation y + p(t)y = 0. Find p(t) so that y = ct is

the general solution.

Solution.

Substituting in the equation we find

− c

t2 +

c

t p(t) = 0.

Solving for p(t) we find p(t) = 1t

Problem 3.8

Consider the differential equation y + p(t)y = 0. Find p(t) so that y = ct3 isthe general solution.

Solution.Substituting in the equation we find

3ct2 + p(t)(ct3) = 0.

Solving for p(t) we find p(t) = − 3t




Problem 3.9

Solve the initial-value problem: y − 3

t y = 0, y(2) = 8.

Solution.

Using the method of integrating factor, we find y(t) = C t3. From y(2) = 8,we find C 23 = 8 and thus C = 1. The unique solution to the initial-valueproblem is y(t) = t3

Problem 3.10

Solve the differential equation y − 2ty = 0.

Solution.

Since p(t) = −2t, using the method of integrating factor, we find y(t) =Ce

2tdt = Cet2

Problem 3.11

Find the function f (t) that crosses the point (0, 4) and whose slope satisfiesf (t) = 2f (t).

Solution.

Solving the differential equation we find f (t) = Ce2t. Since f (0) = 4, we findC = 4. Thus, f (t) = 4e2t

Problem 3.12Find the general solution to the differential equation y − 2y = 0. Hint: Lety = z.

Solution.

Let z = y so that z = y . Thus, z − 2z = 0. Solving this DE by the methodof integrating factor, we find z (t) = Ce2t = y . Hence, y(t) = C e2t + C

Problem 3.13

Solve the IVP: y + 2ty = t, y(0) = 0.

Solution.

Since p(t) = 2t, we find µ(t) = e 2tdt = et2 . Multiplying the given equation

by et2 to obtain et2y

= tet2.



25

Integrating both sides with respect to t and using substitution on the right-

hand integral to obtainet2y =

1

2et2 + C.

Dividing the last equation by et2 to obtain

y(t) = Ce−t2 + 1

2.

Since y(0) = 0, we find C = − 12

. Thus, the unique solution to the IVP isgiven by

y = 1

2(1− e−t2)

Problem 3.14Find the general solution: y + 3y = t + e−2t.

Solution.

Since p(t) = 3, the integrating factor is µ(t) = e3t. Thus, the general solutionis

y(t) =e−3t

e3t(t + e−2t)dt + Ce−3t

=e−3t

(te3t + et)dt + Ce−3t

=e

−3t e3t

9 (3t − 1) + e

t + Ce

−3t

=3t − 1

9 + e−2t + Ce−3t

Problem 3.15

Find the general solution: y + 1t y = 3cos t, t > 0.

Solution.

Since p(t) = 1t , the integrating factor is µ(t) = e

dtt = eln t = t. Using the

method of integrating factor we find

y(t) =1

t 3t cos tdt + C

t=

3

t (t sin t + cos t) +

C

t

=3sin t + 3 cos t

t +

C

t




Problem 3.16

Find the general solution: y + 2y = cos(3t).

Solution.

We have p(t) = 2 so that µ(t) = e2t. Thus,

y(t) = e−2t

e2t cos (3t)dt + Ce−2t

But

e2t cos (3t)dt =e2t

3 sin(3t) − 2

3 e2t sin(3t)dt

=e2t

3 sin(3t) − 2

3

−e2t

3 cos (3t) +

2

3

e2t cos (3t)dt

13

9

e2t cos (3t)dt =

e2t

9 (3sin(3t) + 2 cos (3t))

e2t cos (3t)dt =e2t

13(3sin(3t) + 2 cos (3t)).

Hence,

y(t) = 1

13(3sin(3t) + 2 cos (3t)) + Ce−2t

Problem 3.17

Given that the solution to the IVP ty + 4y = αt2, y(1) = −13 exists on the

interval −∞ < t < ∞. What is the value of the constant α?

Solution.

Solving this equation with the integrating factor method with p(t) = 4t

wefind µ(t) = t4. Thus,

y(t) = 1

t4 t4(αt)dt +

C

t4

=α6

t2 + C t4

.

Since the solution is assumed to be defined for all t, we must have C = 0.Since y(1) = −1

3 we find α = −2



27

Problem 3.18

Suppose that y(t) = Ce−2t

+ t + 1 is the general solution to the equationy + p(t)y = g(t). Determine the functions p(t) and g(t).

Solution.

The integrating factor is µ(t) = e2t. Thus,

p(t)dt = 2t and this implies that p(t) = 2. On the other hand, the function t + 1 is the particular solutionto the non-homogeneous equation so that (t + 1) + 2(t + 1) = g(t). Hence,g(t) = 2t + 3

Problem 3.19

Suppose that y(t) = −2e−t + et + sin t is the unique solution to the IVPy + y = g(t), y(0) = y0. Determine the constant y0 and the function g(t).

Solution.

First, we find y0 : y0 = y(0) = −2 + 1 + 0 = −1. Next, we find g(t) : g(t) =y + y = (−2e−t + et +sin t) + (−2e−t + et +sin t) = 2e−t + et +cos t− 2e−t +et + sin t = 2et + cos t + sin t

Problem 3.20

Find the value (if any) of the unique solution to the IVP y + (1 + cos t)y =1 + cos t, y(0) = 3 in the long run, i.e., limr→∞ y(t)?

Solution.

The integrating factor is µ(t) = e (1+cos t)dt = et+sin t. Thus, the general solu-

tion is

y(t) =e−(t+sin t)

et+sin t(1 + cos t)dt + Ce−(t+sin t)

=1 + Ce−(t+sin t)

Since y(0) = 3 we find C = 2 and therefore y(t) = 1 + 2e−(t+sin t). Finally,

limt→∞

y(t) = limt→∞

(1 + 2e− sin te−t) = 1

Problem 3.21

Find the solution to the IVP

y + p(t)y = 2, y(0) = 1

where

p(t) =

0, if 0 ≤ t ≤ 11t if 1 < t ≤ 2.




Solution.

First, we solve the IVPy = 2, y(0) = 1, 0 ≤ t ≤ 1.

The general solution is y1(t) = 2t + C. Since y(0) = 1, we find C = 1. Hence,y1(t) = 2t + 1 and y(1) = 3.Next, we solve the IVP

y + 1

ty = 2, y(1) = 3, 1 < t ≤ 2.

The integrating factor is µ(t) = t and the general solution is y2(t) = t + C t .

Since y2(1) = 3, we find C = 2. Thus,

y(t) =

2t + 1, if 0 ≤ t ≤ 1

t + 2t if 1 < t ≤ 2

Problem 3.22


y + (sin t)y = g(t), y(0) = 3

where

g(t) =

sin t, if 0 ≤ t ≤ π− sin t if π < t ≤ 2π.

Solution.First, we solve the IVP

y + (sin t)y = sin t, y(0) = 3, 0 ≤ t ≤ π

The integrating factor is µ(t) = e− cos t and the general solution is y1(t) =1 + Cecos t. Since y1(0) = 3 we find C = 2e−1. Hence, y1(t) = 1+ 2ecos t−1 andy1(π) = 1 + 2e−2.Next, we solve the IVP

y + sin ty = − sin t, y(π) = 1 + 2e−2, π < t ≤ 2π

The integrating factor is µ(t) = e−cos t

and the general solution is y2(t) =−1 + Cecos t. Since y2(π) = 1 + 2e−2 we find C = 21e + e

. Thus,

y(t) =

1 + 2ecos t−1 if 0 ≤ t ≤ π−1 + 2

1e + e

ecos t if π < t ≤ 2π



29

Problem 3.23


y + y = g(t), t > 0, y(0) = 3

where

g(t) =

1, if 0 ≤ t ≤ 1

0 if t > 1.

Solution.

First, we solve the IVP

y + y = 1, y(0) = 3, 0 ≤ t ≤ 1.

The integrating factor is µ(t) = et and the general solution is y1(t) = 1+Ce−t.Since y1(0) = 3, we find C = 2. Hence, y1(t) = 1+ 2e−t and y1(1) = 1+ 2e−1.Next, we solve the IVP

y + y = 0, y(1) = 1 + 2e−1, t > 1.

The integrating factor is µ(t) = et and the general solution is y2(t) = C e−t

Since y2(1) = 1 + 2e−1, we find C = 2 + e. Thus,

y(t) = 1 + 2e−t, if 0 ≤ t ≤ 1

(2 + e)e−t if t > 1

Problem 3.24


y + p(t)y = 0, y(0) = 3

where

p(t) = 2t − 1, if 0 ≤ t ≤ 1

0 if 1 < t

≤3

−1t if 3 < t ≤ 4.

Sketch an accurate graph of the solution, and discuss the long-term behaviorof the solution. Is the solution differentiable on the interval t > 0? Explainyour answer.




Solution.

First, we solve the IVPy + (2t− 1)y = 0, y(0) = 3, 0 ≤ t ≤ 1.

The integrating factor is µ(t) = et2−t and the general solution is y1(t) =Ce−(t2−t). Since y1(0) = 3, we find C = 3. Hence, y1(t) = 3e−(t2−t) andy1(1) = 3.Next, we solve the IVP

y = 0, y(1) = 3, 1 < t ≤ 3.

The general solution is y2(t) = C. Since y2(1) = 3, we find C = 3 andy2(t)

≡3.

Next, we solve the IVP

y − 1

ty = 0, y(3) = 3, 3 < t ≤ 4.

The integrating factor is µ(t) = 1t

and the general solution is y3(t) = Ct.Since y3(3) = 3, we find C = 1. Hence, y3(t) = t. Hence,

y(t) =

3e−(t2−t), if 0 ≤ t ≤ 13, if 1 < t ≤ 3t if 3 < t ≤ 4.

The graph of y(t) is shown below

It follows that limt

→∞y(t) =

∞. The function y(t) is not differentiable at

t = 1 and t = 3 on the domain t > 0

Problem 3.25

Solve the initial-value problem y + y = ety2, y(0) = 1 using the substitutionu(t) = 1

y(t) .



31

Solution.

Substituting into the equation we findu − u = −et, u(0) = 1.

Solving this equation by the method of integrating factor with µ(t) = e−t,we find

u(t) = −tet + Cet.

Since u(0) = 1, we find C = 1 and therefore u(t) = −tet + et. The generalsolution is

y(t) = (−tet + et)−1






4 Existence and Uniqueness of

Solutions to First Order

Nonlinear IVP

For the given initial value problem in Problems 4.1 - 4.5,(a) Rewrite the differential equation, if necessary, to obtain the form

y = f (t, y), y(t0) = y0.

Identify the function f (t, y).(b) Compute ∂f

∂y . Determine where in the ty−plane both f (t, y) and ∂f ∂y

arecontinuous.

(c) Determine the largest open rectangle in the ty−plane that contains thepoint (t0, y0) and in which the hypotheses of Theorem 4.1 are satisfied.

Problem 4.1

3y + 2t cos y = 1, y(π

2) = −1.

Solution.

(a) y = 13(1− 2t cos y) = f (t, y).

(b) ∂f ∂y (t, y) = 2

3t sin y. The functions f (t, y) and f y(t, y) are both continuous

in the entire plane,

D = {(t, y) : −∞ < t < ∞, −∞ < y < ∞}.

(c) R = {(t, y) : −∞ < t < ∞, −∞ < y < ∞}

33



344 EXISTENCE AND UNIQUENESS OF SOLUTIONS TO FIRST ORDER NONLIN

Problem 4.2

3ty + 2 cos y = 1, y(π2

) = −1.

Solution.

(a) y = 13t

(1 − 2cos y) = f (t, y).(b) f y(t, y) = 2

3t sin y. Both f (t, y) and f y(t, y) are continuous in

D = {(t, y) : −∞ < t < 0, 0 < t < ∞, −∞ < y < ∞}.

(c) R = {(t, y) : 0 < t < ∞, −∞ < y < ∞}

Problem 4.3

2t + (1 + y3)y = 0, y(1) = 1.

Solution.

(a) y = − 2t1+y3

= f (t, y).

(b) f y(t, y) = 6ty2

(1+y3)2. Both f (t, y) and f y(t, y) are continuous in

D = {(t, y) : −∞ < t < ∞, −∞ < y < −1, − 1 < y < ∞}.

(c) R = {(t, y) : −∞ < t < ∞, − 1 < y < ∞}

Problem 4.4

(y2 − 9)y + e−y = t2, y(2) = 2.

Solution.

(a) y = t2−e−y

y2−9 = f (t, y).

(b) f y(t, y) = (y2+2y−9)e−y−2t2y2

y2−9 . Both f (t, y) and f y(t, y) are continuous in

D = {(t, y) : −∞ < t < ∞, −∞ < y < −3, − 3 < y < 3, 3 < y < ∞}.

(c) D = {(t, y) : −∞ < t < ∞, − 3 < y < 3}Problem 4.5

cos yy = 2 + tan t, y(0) = 0.



35

Solution.

(a) y = 2+tan t

cosy = f (t, y).(b) f y(t, y) = (2 + tan t)sec y tan y. Both f (t, y) and f y(t, y) are continuousin

D = {(t, y) : t = (2n + 1) π2 , y = (2m + 1) π

2 , where n and m are integers}.

(c) R = {(t, y) : −π2 < t < π

2 , − π2 < y < π

2}

Problem 4.6

Give an example of an initial value problem for which the open rectangle

R =

{(t, y) : 0 < t < 4,

−1 < y < 2

}represents the largest region in the ty−plane where the hypotheses of Theo-rem 4.1 are satisfied.

Solution.

An example is

y = 1

t(t − 4)(y + 1)(y − 2), y(2) = 0

Problem 4.7

Can we apply Theorem 4.1 to the following problem ? Explain what (if anything) we can conclude, and why (or why not):

y = y√

t, y(0) = 2.

Solution.

Since f (t, y) = y√ t

and f y(t, y) = 1√ t, both functions are continuous in the

regionD = {(t, y) : 0 < t < ∞, −∞ < y < ∞}.

Since (0, 2) is not in D, Theorem 4.1 can not be applied in this case

Problem 4.8

Consider the differential equation y = t−yt+y . For which of the following initial

value conditions does Theorem 4.1 apply?(a) y(0) = 0 (b) y(1) = −1 (c) y(−1) = −1.



364 EXISTENCE AND UNIQUENESS OF SOLUTIONS TO FIRST ORDER NONLIN

Solution.

The function f (t, y) =

t−y

t+y is continuous everywhere except along the linet + y = 0. (a) and (b): Since both (0, 0) and (1,−1) lie on this line, wecannot conclude existence from Theorem 4.1.(c) The point (−1,−1) is not on that line so we can find a small rectanglearound this point where Theorem 4.1 guarantees the existence of a solution.Furthermore, since f y(t, y) = − 2t

(t+y)2 is continuous at (−1,−1), the solution

is unique

Problem 4.9

Does the initial value problem y = yt + 2, y(0) = 1 satisfy the conditions of

Theorem 4.1?

Solution.

The equation is of the form y = f (t, y) = yt + 2. The function f is continuous

outside the line t = 0. The initial value point is (0,1), so there is no rectanglecontaining it in which f is continuous, and the conditions of Theorem 4.1 arenot satisfied

Problem 4.10

Does the initial value problem y = y sin y + t, y(0) = −1 satisfy the condi-tions of Theorem 4.1?

Solution.The equation is of the form y = f (t, y) = y sin y + t. the function f iscontinuous in the whole plane, and so is its partial derivative f y(t, y) =sin y + y cos y. In particular, any rectangle around the initial value point willsatisfy the conditions of Theorem 4.1



5 Separable Differential

Equations

Problem 5.1

Solve the (separable) differential equation

y = tet2−ln y2 .

Solution.

At first, this equation may not appear separable, so we must simplify theright hand side until it is clear what to do.

y =tet2−ln y2

=tet

2

· eln 1

y2

=tet2 · 1

y2

= t

y2et2.

Separating the variables and solving the equation we find

y2y =tet2

1

3 (y3)dt = tet2dt

1

3y3 =

1

2et2 + C

y3 =3

2et2 + C

37



38 5 SEPARABLE DIFFERENTIAL EQUATIONS

Problem 5.2


y = t2y − 4y

t + 2 , t = −2.

Solution.

Separating the variables and solving we find

y

y =

t2 − 4

t + 2 = t − 2

(ln |y|)dt = (t − 2)dt

ln |y| =t2

2 − 2t + C

y(t) =Cet2

2 −2t

Problem 5.3


ty = 2(y − 4).

Solution.


y

y − 4 =

2

t (ln |y − 4|)dt =

2

tdt

ln |y − 4| = ln t2 + C

ln |y − 4

t2 | =C

y(t) =Ct2 + 4

Problem 5.4


y = 2y(2 − y).



39

Solution.

Separating the variables and solving (using partial fractions in the process)we find

y

y(2− y) =2

y

2y +

y

2(2− y) =2

y

y +

y

(2− y) =4

(ln |y|)dt− 1

2 (ln |2 − y|)dt =

4dt

ln y

2 − y

=4t + C y

2 − y

=Ce4t

y(t) = 2Ce4t

1 + Ce4t.

Note that y(t) ≡ 2 is a singular solution

Problem 5.5

Solve the IVP:

y = 4 sin (2t)y

, y(0) = 1.

Solution.


yy =4 sin (2t)

(y2) =8 sin (2t) (y2)dt =

8sin(2t)dt

y2 =

−4cos(2t) + C

y(t) =± C − 4cos(2t).

Since y(0) = 1, we find C = 5 and hence

y(t) =

5 − 4cos(2t)




Problem 5.6

Solve the IVP: yy = sin t, y(π2

) = −2.

Solution.


y2

2

dt =

sin tdt

y2

2 = − cos t + C

y2 = − 2cos t + C.

Since y( π2

) = −2, we find C = 4. Thus, y(t) = ±

(−2cos t + 4). Since

y( π2

) = −2, y(t) = −

(−2cos t + 4)

Problem 5.7

Solve the IVP:

y + 1

y + 1 = 0, y(1) = 0.

Solution.Separating the variables and solving we find

d

dt[(1 + y)2] = − 2

(y + 1)2 = − 2t + C

y + 1 = ±√ −2t + C

y(t) = ±√ −2t + C − 1.

Since y(1) = 0, we find C = 3. Thus, y(t) =√

−2t + 3

−1

Problem 5.8

Solve the IVP:

y − ty3 = 0, y(0) = 2.



41

Solution.

Separating the variables and solving we find yy−3dt =

tdt

y−2

−2

dt =

t2

2 + C

− 1

y2 =t2 + C

y2 = 1

−t2 + C .

Since y(0) = 2, we find C = 14

. Thus, y(t) = ±

4−4t2+1

. Since y(0) = 2, we

have y(t) = 2√ −4t2+1

Problem 5.9

Solve the IVP:

y = 1 + y2, y(π

4) = −1.

Solution.Separating the variables and solving we find

y

1 + y2 =1

arctan y =t + C

y(t) =tan (t + C ).

Since y( π4

) = −1, we have C = −π2

. Hence, y(t) = tan

t− π2

Problem 5.10

Solve the IVP:

y = t − ty2, y(0) = 1

2.




Solution.

Separating the variables and solving we findy

y2 − 1 = − t

y

y − 1 − y

y + 1 = − 2t

ln

y − 1

y + 1

= − 2t + C

y − 1

y + 1 =Ce−2t

y(t) =

1 + Ce−2t

1 − Ce−2t .

Since y(0) = 12 , we find C = −1

3 . Thus,

y(t) = 3 − e−2t

3 + e−2t

Problem 5.11

Solve the IVP:(2y − sin y)y = sin t − t, y(0) = 0.

Solution.

Separating the variables and solving we find (2y − sin y)ydt =

(sin t − t)dt

y2 + cos y = − cos t − t2

2 + C.

Since y(0) = 0, we find C = 2. Thus,

y2 + cos y + cos t + t2

2 = 2

Problem 5.12For what values of the constants α, y0, and integer n is the function y(t) =

(4 + t)−12 a solution of the initial value problem

y + αyn = 0, y(0) = y0?



43

Solution.

We have y0 = y(0) = (4 + 0)−12

=

1

2 . Also, y = −1

2(4 + t)−32

= −1

2y

3

. Thus,

y + 1

2y3 = 0

so that α = 12 and n = 3

Problem 5.13

State an initial value problem, with initial condition imposed at t0 = 2,having implicit solution y3 + t2 + sin y = 4.

Solution.

Differentiating both sides of the given equation we find

3y2y + (cos y)y + 2t = 0, y(2) = 0

Problem 5.14


y = 2y2, y(0) = y0.

For what value(s) of y0 will the solution have a vertical asymptote at t = 4,where the t−interval of existence is −∞ < t < 4?

Solution.

Solving the differential equation by the method of separating the variableswe find

y

y2 =2

y

y2dt =

2dt

−1

y =2t + C

y(t) = 1C − 2t

.

Since y(0) = y0, we find C = 1y0

. Thus, y(t) = y01−2y0t

. This function will have

a vertical asymptote at t = 4 when 1 − 2(4)y0 = 0 or y0 = 18




Problem 5.15

Assume that y sin y − 3t + 3 = 0 is an implicit solution of the initial valueproblem y = f (y), y(1) = 0. What is f (y)? What is an implicit solution tothe initial value problem y = t2f (y), y(1) = 0?

Solution.

Taking the derivative of the given equation with respect to t we find

y sin y + yy cos y − 3 = 0.

Thus,

y = 3

sin y + y cos y = f (y).

If y = t2f (y) then

y = 3t2

sin y + y cos y.

Solving this equation by the method of separation of variables we find

y sin y + yy cos y =3t2

(y sin y) =3t2

y sin y =t3 + C.

Since y(1) = 0, we find C = −1. Hence, the implicit solution is given by

y sin y − t3 + 1 = 0

Problem 5.16

Find all the solutions to the differential equation y = 2ty1+t

.

Solution.

Separating the variables to obtain

y

y =

2t

1 + t = 2 − 2

t + 1

ln |y| =2t − ln (t + 1)2 + C

ln |(t + 1)2y| =2t + C (t + 1)2y =Ce2t

y(t) = Ce2t

(t + 1)2



45

Problem 5.17

Solve the initial-value problem y = cos

2

y cos

2

t, y(0) =

π

4 .Solution.

Solving by the method of separation of variables we find

y

cos2 y =cos2 t =

1 + cos (2t)

2

tan y =t

2 +

1

4 sin 2t + C.

Since y(0) = π4

, we find C = 1. Hence,

tan y = t

2 + 1

4 sin 2t + 1

Problem 5.18

Solve the initial-value problem y = et+y, y(0) = 0 and determine the intervalon which the solution y(t) is defined.

Solution.

Separating the variables we obtain

ye−y = et.

Integrating both sides to obtaine−y = −et + C.

But y(0) = 0 so that C = 2. Hence, e−y = −et + 2. Solving for y we find

y(t) = − ln(2− et).

This function is defined for 2 − et > 0, that is, for t < ln 2

Problem 5.19

(a) Solve the initial-value problem

y = t2

e−y − ey

t2.

State the name of the method you are using.(b) Find the solution which satisfies the condition y(1) = 1.




Solution.

(a) Using the method of separation of variables we find

ye−y =t2 − 1

t2

e−y = − t3

3 − 1

t + C.

(b) Since y(1) = 1, we find C = e−1 + 43 . Thus, the unique solution is defined

implicitly by the expression

e−y + t3

3 +

1

t = e−1 +

4

3



6 Exact Differential Equations

Problem 6.1

Find ∂f ∂t and ∂f

∂y if f (t, y) = y ln y − e−ty.

Solution.

∂f ∂t = ye−ty.

∂f ∂y = ln y + 1 + te−ty

Problem 6.2

Find ∂f ∂t

and ∂f ∂y

if f (t, y) = ln (ty) + t2+1y−5

.

Solution.

∂f ∂t

= 1t + 2t

y−5.

∂f ∂y = 1

y − t2+1(y−5)2

Problem 6.3

Let f (u, v) = 2u − 3uv where u(t) = 2 cos t and v(t) = 2 sin t. Find df dt .

Solution.

By the Chain Rule

df

dt =

∂f

∂u

du

dt +

∂ f

∂v

dv

dt=(2− 3v)(−2sin t) − 3u(2cos t) = (2− 6sin t)(−2sin t) − 6cos t(2 cos t)

=12 sin2 t − 12cos2 t − 4sin t = 12sin2 t − 12(1− sin2 t) − 4sin t

=24 sin2 t − 4sin t − 12

47



48 6 EXACT DIFFERENTIAL EQUATIONS

In Problems 6.4 - 6.8, determine whether the given differential equation is

exact. If the equation is exact, find an implicit solution and (where possible)an explicit solution.

Problem 6.4

yy + 3t2 − 2 = 0, y(−1) = −2.

Solution.

We have M (t, y) = 3t2− 2 and N (t, y) = y. Thus, ∂M ∂y (t, y) = 0 = ∂N

∂t (t, y) sothat the equation is exact.

∂H

∂t (t, y) = 3t2

−2 =

⇒H (t, y) = (3t2

−2)dt = t3

−2t + h(y).

But ∂H ∂y (t, y) = y so that h(y) = y and hence h(y) = y2

2 + C. Therefore

t3 − 2t + y2

2 = C.

Since y(−1) = −2, we find C = 3. It follows

t3 − 2t + y2

2 = 3.

Solving for y we find y(t) = ±√

4t−

2t3 + 6. Since y(−

1) = −

2, we findy(t) = −√ 4t − t3 + 6

Problem 6.5

y = (3t2 + 1)(y2 + 1), y(0) = 1.

Solution.

Since the equation is separable, it is exact. Integrating ∂H ∂t (t, y) = 3t2 + 1

with respect to t we find H (t, y) = t3 + t + h(y). But ∂H ∂y

(t, y) = −(y2 + 1)−1

which implies that h(y) = −(y2 + 1)−1. Thus, h(y) = − arctan y + C. Hence,

t3 + t − arctan y = C.

Since y(0) = 1 we find C = −π4 . It follows

t3 + t − arctan y = −π

4.



49

Solving for y(t) we find

y(t) = tan

t3 + t + π4

Problem 6.6

(6t + y3)y + 3t2y = 0, y(1) = 2.

Solution.

We have M (t, y) = 3t2y and N (t, y) = 6t + y3. Since ∂M ∂y (t, y) = 3t2 and

∂N ∂t

(t, y) = 6, the given differential equation is not exact

Problem 6.7

(et+y + 2y)y + (et+y + 3t2) = 0, y(0) = 0.

Solution.

We have M (t, y) = et+y +3t2 and N (t, y) = et+y +2y. Since ∂M ∂y

(t, y) = et+y =∂N ∂t (t, y), the given differential equation is exact.

∂H

∂t (t, y) = et+y + 3t2 =⇒ H (t, y) = (et+y + 3t2)dt = et+y + t3 + h(y).

Also

∂H

∂y (t, y) = et+y + 2y = h(y) + et+y =⇒ h(y) = 2y =⇒ h(y) = y2 + C.

Hence,

et+y + t3 + y2 = C.

Since y(0) = 0 we find C = 1. Therefore,

et+y + t3 + y2 = 1

Problem 6.8

(sin (t + y) + y cos(t + y) + t + y)y + (y cos(t + y) + y + t) = 0, y(1) = −1.




Solution.

We have M (t, y) = y cos(t + y)+t+y and N (t, y) = sin (t + y)+y cos(t + y)+t+y. Since ∂M ∂y (t, y) = cos (t + y)−y sin(t + y)+1 = ∂N

∂t (t, y), the differentialequation is exact. Thus,

∂H

∂t (t, y) =y cos(t + y) + t + y =⇒ H (t, y) =

(y cos(t + y) + t + y)dt

=y sin(t + y) + t2

2 + yt + h(y).

Also,

∂H

∂y (t, y) =sin (t + y) + y cos(t + y) + t + y = y cos(t + y) + sin (t + y) + t + h(y)

h(y) =y =⇒ h(y) = y2

2 + C.

Hence,

y sin(t + y) + t2

2 + ty +

y2

2 = C.

Since y(1) = −1 we find C = 0. Therefore,

y sin(t + y) + t2

2

+ ty + y2

2

= 0

Problem 6.9

For what values of the constants m,n, and α (if any) is the following differ-ential equation exact?

tmy2y + αt3yn = 0.

Solution.

We have M (t, y) = αt3yn and N (t, y) = tmy2. Thus, ∂M ∂y (t, y) = nαt3yn−1

and ∂N ∂t

(t, y) = mtm−1y2. For the differential equation to be exact we musthave ∂M

∂y (t, y) = ∂N ∂t (t, y), i.e.,

nαt3yn−1 = mtm−1y2.

This shows that m − 1 = 3 so that m = 4. Also, n − 1 = 2 so that n = 3.Finally, 3α = 4 so that α = 4

3



51

Problem 6.10

Assume that N (t, y)y + t

2

+ y

2

sin t = 0 is an exact differential equation.Determine the general form of N (t, y).

Solution.

We have M (t, y) = t2 + y2 sin t. Since the differential equation is exact,∂N ∂t

(t, y) = ∂M ∂y

(t, y) = 2y sin t. Hence,

N (t, y) =

2y sin tdt = −2y cos t + h(y)

Problem 6.11

Assume that t3y + et + y2 = 5 is an implicit solution to the differential

equationN (t, y)y + M (t, y) = 0, y(0) = y0.

Determine possible functions M (t, y), N (t, y), and the possible value(s) fory0.

Solution.

Replacing y by y0 and t by 0 to obtain y0 = ±2. Differentiating the givenequation with respect to t we find 3t2y + et + (t3 + 2y)y = 0. Thus, M (t, y) =3t2 + et and N (t, y) = t3 + 2y

Problem 6.12Assume that y = −t −√ 4− t2 is an explicit solution of the following initialvalue problem

(y + at)y + (ay + bt) = 0, y(0) = y0.

Determine values for the constants a, b and y0.

Solution.

We have y0 = −0 − √ 4 − 02 = −2. Since ∂N

∂t (t, y) = ∂M ∂y (t, y) = a, the

differential equation is exact. From this we have

∂H

∂y (t, y) = y + at =⇒ H (t, y) =

y2

2 + aty + h(t)

and

∂H

∂t (t, y) = ay + bt = ay + h(t) =⇒ h(t) = bt =⇒ h(t) =

b

2t2 + C.




Hence,y2

2 + aty + b

2t2 = C.

Since y(0) = −2 we find C = 2. Therefore,

y2 + 2aty + bt2 = 4.

Solving this quadratic equation for y we find

y = −2at±

4a2t2 − 4(bt2 − 4)

2 .

Thus,y(t) = −at±

a2t2 − (bt2 − 4).

Since y(0) = −2 we obtain y(t) = −at−√ t2a2 − bt2 + 4. Finally, a = 1, a2

−b = −1, b = 2

Problem 6.13

Let k be a positive constant. Use the exactness criterion to determinewhether or not the population equation dP

dt = kP is exact. Do NOT tryto solve the equation or carry out any further calculation.

Solution.

Rewriting the equation in the form kP − dP dt = 0 we find that M (t, P ) = kP

and N (t, P ) = −1. Since ∂M ∂t

(t, P ) = ∂N ∂P

(t, P ) = 0, the differential equationis exact

Problem 6.14

Consider the differential equation (2t+3)+(2y−2)y = 0. Determine whetherthis equation is exact or not. If it is, solve it.

Solution.

We have M (t, y) = 2t+3 and N (t, y) = 2y−2. Since ∂M ∂y

(t, y) = ∂N ∂t

(t, y) = 0,the differential equation is exact. Now,

∂H

∂t (t, y) = 2t + 3 =⇒ H (t, y) =

(2t + 3)dt = t2 + 3t + h(y).

Also

∂H ∂y

(t, y) = 2y − 2 = h(y) =⇒ h(y) = 2y − 2 =⇒ h(y) = y2 − 2y + C.

Hence,t2 + 3t + y2 − 2y = C



53

Problem 6.15

Consider the differential equation (ye

2ty

+ t) + bte

2ty

y = 0. Determine forwhich value of b this equation is exact, and then solve it with this value of b.

Solution.

We have M (t, y) = ye2ty + t and N (t, y) = bte2ty. For the equation to beexact we must have ∂M

∂y (t, y) = ∂N ∂t (t, y), that is,

e2ty + 2tye2ty = be2ty + 2ybte2ty.

Dividing through by e2ty to obtain

1 + 2ty = b + 2byt = b(1 + 2ty).

This implies b = 1. Hence, the equation is

(ye2ty + t) + te2tyy = 0.

Now,

∂H

∂t (t, y) = ye2ty + t =⇒ H (t, y) =

(ye2ty + t)dt =

1

2e2ty +

t2

2 + h(y).

Also

∂H ∂y

(t, y) = te2ty = te2ty + h(y) =⇒ h(y) = 0 =⇒ h(y) = C.

Hence,1

2e2ty +

t2

2 = C

Problem 6.16

Consider the differential equation y + (2t − yey)y = 0. Check that thisequation is not exact. Now multiply the equation by y. Check that the newequation is exact, and solve it.

Solution.

If we let M (t, y) = y and N (t, y) = 2t − yey we see that ∂M ∂y (t, y) = 1 and

∂N ∂t

(t, y) = 2 so that the equation is not exact. If we multiply the givenequation by y then M (t, y) = y2 and N (t, y) = 2ty − y2ey. In this case,




∂M ∂y (t, y) = ∂N

∂t (t, y) = 2y so that the equation is exact.

Now, ∂H

∂t (t, y) = y2 =⇒ H (t, y) =

y2dt = ty2 + h(y).

Also∂H

∂y (t, y) = 2ty − y2ey = 2ty + h(y) =⇒ h(y) = −y2ey.

Using integration by parts twice we find

h(y) = −y2ey + 2yey − 2ey + C.

Hence,

ty2 − y2ey + 2yey − 2ey = C

Problem 6.17

(a) Consider the differential equation

y + p(t)y = g(t)

with p(t) = 0. Show that this equation is not exact.(b) Let µ(t) = e

p(t)dt. Show that the equation

µ(t)(y + p(t)y) = µ(t)g(t)

is exact and solve it.

Solution.

(a) We have M (t, y) = p(t)y−g(t) and N (t, y) = 1. Since ∂M ∂y

(t, y) = p(t) = 0

and ∂N ∂t (t, y) = 0, the differential equation is not exact.

(b) Here, we have M (t, y) = µ(t) p(t)y − µ(t)g(t) and N (t, y) = µ(t). Thus,∂M ∂y (t, y) = ∂N

∂t (t, y) = p(t)e

p(t)dt. That is, the new differential equation isexact.Now,

∂H

∂t (t, y) =µ(t) p(t)y − µ(t)g(t)

H (t, y) =

(µ(t) p(t)y − µ(t)g(t))dt = µ(t)y −

µ(t)g(t)dt + h(y).



55

Also

∂H ∂y

(t, y) = µ(t) = µ(t) + h(y) =⇒ h(y) = 0 =⇒ h(y) = C.

Hence,

µ(t)y −

µ(t)g(t) = C

and so

y(t) = e−

p(t)dt

e

p(t)dtg(t)dt + Ce−

p(t)dt

Problem 6.18

Use the method of the previous problem to solve the linear, first-order equa-tion y− y

t = 1, with initial condition y(1) = 7. First, check that this equationis not exact. Next, find µ(t). Multiply the equation by µ(t) and check thatthe new equation is exact. Solve it, using the method of exact equations.

Solution.

For the given equation we have M (t, y) = 1 + yt

and N (t, y) = −1. Since∂M ∂y (t, y) = 1

t and ∂N ∂t (t, y) = 0, the equation is not exact. Let µ(t) = e−

dtt =

1t . Multiply the given equation by µ(t) to obtain

(1 + y

t

)(1

t

)

− 1

t

y = 0.

In this equation, M (t, y) = (1 + yt )(1t ) and N (t, y) = − 1

t . Also, ∂M ∂y (t, y) =

∂N ∂t (t, y) = 1

t2 so that the new equation is exact. By the previous exercise thesolution is given by

y(t) = t

1

tdt + Ct = t ln t + Ct.

Since y(1) = 7 we find C = 7. Hence, y(t) = t ln t + 7t

Problem 6.19

Put the following differential equation in the “Exact Differential Equation”form and find the general solution

y = y3 − 2ty

t2 − 3ty2.




Solution.

Rewriting this equation in the form(y3 − 2ty) + (3ty2 − t2)y = 0

we find M (t, y) = y3−2ty and N (t, y) = 3ty2−t2. Also, notice that ∂M ∂y (t, y) =

∂N ∂t

(t, y) = 3y2 − 2t. Now,

∂H

∂t (t, y) = y3 − 2ty =⇒ H (t, y) =

(y3 − 2ty)dt = ty3 − t2y + h(y).

Also

∂H

∂y

(t, y) = 3ty2

−t2 = 3ty2

−t2 + h(y) =

⇒h(y) = 0 =

⇒h(y) = C.

Hence,ty3 − t2y = C

Problem 6.20

The following differential equations are exact. Solve them by that method.(a) (4t3y + 4t + 4)y = 8− 4y − 6t2y2, y(−1) = 1.(b) (6− 4y + 16t) + (10y − 4t + 2)y = 0, y(1) = 2.

Solution.

(a) We have M (t, y) = 6t2y2 + 4y − 8 and N (t, y) = 4t3y + 4t + 4. Notice

that ∂M

∂y (t, y) = ∂N

∂t (t, y) = 12t2

y + 4. Now,∂H

∂t (t, y) = 6t2y2+4y−8 =⇒ H (t, y) =

(6t2y2+4y−8)dt = 2t3y2+4ty−8t+h(y).

Also

∂H

∂y (t, y) = 4t3y + 4t + 4 = 4t3y + 4t + h(y) =⇒ h(y) = 4 =⇒ h(y) = 4y + C.

Hence,2t3y2 + 4ty − 8t + 4y = C.

Since y(−

1) = 1 we find C = 6. Hence, 2t3y2 + 4ty−

8t + 4y = 6(b) We have M (t, y) = 6− 4y + 16t and N (t, y) = 10y − 4t + 2. Notice that∂M ∂y (t, y) = ∂N

∂t (t, y) = −4. Now,

∂H

∂t (t, y) = 6−4y+16t =⇒ H (t, y) =

(6−4y+16t)dt = 6t−4ty+8t2+h(y).



57

Also

∂H ∂y

(t, y) = 10y−4t+2 = −4t+h(y) =⇒ h(y) = 10y+2 =⇒ h(y) = 5y2+2y+C.

Hence,6t− 4ty + 8t2 + 5y2 + 2y = C.

Since y(1) = 2 we find C = 30. Hence, 6t− 4ty + 8t2 + 5y2 + 2y = 30






7 Substitution Techniques:

Bernoulli and Riccati Equations

Problem 7.1

Solve the Bernoulli equation

y = t2 + 3y2

2ty , t > 0.

Solution.

The given equation can be written in the form

y − 3

2ty =

1

2ty−1.

Divide through by y−1 to obtain

yy − 3

2ty2 =

t

2.

Let z = y2. Then the last equation becomes

z − 3

tz = t

and this is a linear first order differential equation.To solve this equation, we use the integrating factor method. Let µ(t) = t−3.Then

z (t) = t3 t−3tdt + Ct3 = −t2 + Ct3.

The general solution to the initial problem is implicitly defined by

y2 = −t2 + Ct3

59



607 SUBSTITUTION TECHNIQUES: BERNOULLI AND RICCATI EQUATIONS

Problem 7.2

Find the general solution of y + ty = te−t2

y−3

.Solution.

Divide the given equation by y−3 to obtain

y3y + ty4 = te−t2.

Let z = y4 so that the previous equation becomes

z + 4tz = 4te−t2.

The integrating factor is µ(t) = e2t2. Thus,

z (t) = e−2t2 e

2t2

4te−t2

dt + Ce−2t2

= 2e−t2

+ Ce−2t2

.

Finally, the general solution to the original equation is defined implicitly bythe equation

y4 = 2e−t2 + Ce−2t2

Problem 7.3

Solve the IVP ty + y = t2y2, y(0.5) = 0.5.

Solution.

Divide through by y2 to obtain

ty−2

y + y−1

= t2

or

y−2y + 1

ty−1 = t.

Let z = y−1 so that

z − 1

tz = −t, z

1

2

= 2.

Solving this equation by the integrating factor method with µ(t) = 1t we find

z (t) = t 1

t · (−t)dt + Ct = −t2 + Ct = t(C − t).

Hence, y(t) = 1t(C −t) . But y(1

2) = 1

2 so that C = 4.5. Thus,

y(t) = 1

t(4.5 − t)



61

Problem 7.4

Solve the IVP y − 1

t y = −y

2

, y(1) = 1, t > 0.

Solution.

Divide through by y2 to obtain

y−2y − 1

ty−1 = −1.

So let z = y−1. Thus,

z + 1

tz = 1, z (1) = 1.

Solving this equation using the integrating factor method with µ(t) = t wefind

z (t) = 1

t

tdt + Ct−1 =

t

2 + Ct−1.

Since z (1) = 1, we find C = 12 . Hence, z = 1

2(t + 1t ) and y(t) = 2t

t2+1

Problem 7.5

Solve the IVP y = y(1− y), y(0) = 12

.

Solution.

Rewriting the given equation in the form y − y = −y2

. Divide through byy2 to obtain

y−2y − y−1 = −1.

Let z = y−1. Then

z + z = 1, z (0) = 2.

Solving this equation using the integrating factor method with µ(t) = et weobtain

z (t) = e−t

etdt + Ce−t = 1 + Ce−t.

But z (0) = 2 so that C = 1 and thus z (t) = 1 + e−t

. Finally, y(t) =(1 + e−t)−1

Problem 7.6

Solve y + y = ty4.




Solution.

Divide through by y

4

to obtainy−4y + y−3 = t.

Let z = y−3 so thatz − 3z = −3t.

Solving this equation using the integrating factor method with µ(t) = e−3t

we find

z (t) = e3t

e−3t(−3t)dt + Ce3t = t +

1

3 + Ce3t.

Hence, y(t) = (t + 13 + Ce3t)−

13

Problem 7.7

Solve the differential equation y = 1 + t2 − y2 given that y1(t) = t is aparticular solution.

Solution.

Let 1z

= y − t. Then − z

z2 = y − 1. Substituting we find

− z

z 2 + 1 = 1 + t2 −

1

z + t

2

.

Simplifying this last equation to obtain

z − 2tz = 1.

Solving this equation by the method of integrating factor with µ(t) = e−t2

we find

z (t) = et2 t

0

e−s2ds + Cet2.

The general solution to the differential equation is

y(t) = (et2

t

0

e−s2ds + Cet2)−1 + t

Problem 7.8

Perform a change of variable that changes the Bernoulli equation y+y+y2 =0 into a linear equation in the new variable. Do NOT try to solve the equationor proceed further than with any calculations.



63

Solution.

Dividing through by y

2

to obtainy−2y + y−1 = −1.

Letting z = y−1 to obtainz − z = 1

Problem 7.9

Consider the equation

y = y − σy3, > 0, σ > 0.

(a) Use the Bernoulli transformation to change this nonlinear equation intoa linear equation.(b) Solve the resulting linear equation in part (a) and use the solution to findthe solution of the given differential equation above.

Solution.

(a) Dividing through by y3 to obtain

y−3y − y−2 = −σ.

Letting z = y−2 to obtainz + 2z = 2σ.

(b) Using the method of integrating factor with µ(t) = e2t we find

z (t) = e−2t

e2t2σdt + Ce−2t = σ

+ Ce−2t.

Hence,

y(t) = (σ

+ Ce−2t)−

12

Problem 7.10

Consider the differential equation

y = f y

t

.

(a) Show that the substitution z = yt leads to a separable differential equation

in z.(b) Use the above method to solve the initial-value problem

y = t + y

t − y, y(1) = 0.




Solution.

(a) Letting z =

y

t then y = tz . Thus, y = z + tz . Hence,tz + z = f (z )

orz

f (z ) − z =

1

t

which is a separable differential equation.(b) Note first that

t + y

t − y =

1 + yt

1 − yt

.

Letting z =

y

t we obtain

tz + z =1 + z

1 − z

tz =1 + z 2

1− z 1− z

1 + z 2z =

1

t z

1 + z 2dt −

zz

1 + z 2dt =

dt

t

arctan z −

1

2 ln (1 + z 2) = ln

|t|

+ C

2 arctan z = ln t2(1 + z 2) + C

2 arctany

t

= ln t2

1 +

y

t

2 + C

Since y(1) = 0, C = 0. Hence,

2 arctany

t

= ln t2

1 +

y

t

2

Problem 7.11

Solve: y + y

3 = ety4.

Solution.

Divide through by y4 to obtain y−4y + 13y−3 = et. Letting z = y−3 to obtain

z − z = −3et.



65

Solving this equation by the method of integrating factor with µ(t) = e−t we

findz (t) = et

e−t(−3et)dt + Cet = −3tet + Cet.

Hence,

y(t) = (−3tet + Cet)−13

Problem 7.12

Solve: ty + y = ty3.

Solution.

Dividing through by ty

3

to obtain y−3

y + 1

t y−2

= 1. Letting z = y−2

toobtain

z − 2

tz = −2.

Solving this equation by the method of integrating factor with µ(t) = 1t2 we

find

z (t) = t2

1

t2(−2)dt + Ct2 = 2t + Ct2.

Hence,

y(t) = (2t + Ct2)−12

Problem 7.13

Solve: ty + y = t2y2 ln t.

Solution.

Dividing through by ty2 to obtain y−2y + 1t y−1 = t ln t. Letting z = y−1 to

obtain

z − 1

tz = −t ln t.

Solving this equation by the method of integrating factor with µ(t) = 1t we

find

z (t) = t

(− ln t)dt + Ct = −t2 ln t + t2 + Ct.

Hence,

y(t) = (−t2 ln t + t2 + Ct)−1




Problem 7.14

Verify that y1(t) = 2 is a particular solution to the Riccati equationy = −2 − y + y2,

and then find the general solution.

Solution.

Since y 1 = 0 and −2− y1 + y21 = −2− 2 + 4 = 0, we have y 1 = −2− y1 + y2

1.Now, to solve the equation we let 1

z = y − 2. Substituting this into the aboveequation to obtain

z + 3z = −1.

Solving this equation by the method of integrating factor with µ(t) = e3t we

findz (t) = e−3t

−e3tdt + Ce−3t = −1

3 + Ce−3t.

Hence,

y(t) = (−1

3 + Ce−3t)−1 + 2

Problem 7.15

Verify that y1(t) = 2t is a particular solution to the Riccati equation

y = − 4

t2 − 1

ty + y2,

and then find the general solution.

Solution.

Since y 1 = − 2t2 and − 4

t2 − 1

t y1 + y21 = − 2

t2 , y1 is a solution to the differentialequation. Next, let z −1 = y − 2

t then substituting into the previous equation

we find

z + 3

tz = −1.

Solving this equation by the method of integrating factor with µ(t) = t3 wefind

z (t) = t−3 −t3dt + Ct−3 =−

t

4 + Ct−3.

Hence,

y(t) = (− t

4 + Ct−3)−1 +

2

t



8 Graphical Solution: Direction

Field of y = f (t, y)

Problem 8.1Sketch the direction field for the differential equation in the window −3 ≤t ≤ 3,−3 ≤ y ≤ 3.(a) y = y (b) y = t − y.

Solution.

(a)

(b)

67



68 8 GRAPHICAL SOLUTION: DIRECTION FIELD OF Y = F (T, Y )

Problem 8.2

Match each direction field with the equation that the slope field could repre-sent. Each direction field is drawn in the portion of the ty-plane defined by−6 ≤ t ≤ 6,−4 ≤ y ≤ 4.(a) y = −t (b) y = sin t (c) y = 1 − y (d) y = y(2− y).

Solution.(A) y = sin t (B) y = y(2 − y) (c) y = −t (D) y = 1 − y

Problem 8.3

State whether or not the equation is autonomous.



69

(a) y = −t (b) y = sin t (c) y = 1 − y (d) y = y(2− y).

Solution.

(a) No (b) No (c) Yes (d) Yes

Problem 8.4

Find all the equilibrium solutions of each of the autonomous differential equa-tions below(a) y = (y − 1)(y − 2).(b) y = (y − 1)(y − 2)2.(c) y = (y − 1)(y − 2)(y − 3).

Solution.

(a) y(t) ≡ 1, y(t) ≡ 2.(b) y(t) ≡ 1, y ≡ 2.(c) y ≡ 1, y ≡ 2, y ≡ 3

Problem 8.5

Find an autonomous differential equation with an equilibrium solution aty = 1 and satisfying y < 0 for −∞ < y < 1 and 1 < y < ∞.

Solution.One answer is the differential equation: y = −(y − 1)2

Problem 8.6

Find an autonomous differential equation with no equilibrium solutions andsatisfying y > 0.

Solution.

Consider the differential equation y = C where C is a positive constant

Problem 8.7

Find an autonomous differential equation with equilibrium solutions y = n2 ,

where n is an integer.

Solution.

One answer is the DE y = sin (2πy)




Problem 8.8

Find an autonomous differential equation with equilibrium solutions y = 0and y = 2 and satisfying the properties y > 0 for 0 < y < 2; y < 0 for y < 0or y > 2.

Solution.

An answer is y = y(2− y)

Problem 8.9Classify whether the equilibrium solutions are stable, unstable, or neither.(a) y = 1 − y2.(b) y = (y + 1)2.

Solution.

Using the direction fields shown below we find(a) y = 1 stable, y = −1 unstable

(b) y = −1 is neither. This is a semi-stable equilibrium



71

Problem 8.10

Consider the direction field below. Classify the equilibrium points, as asymp-totically stable, semi-stable, or unstable.




Solution.

The equilibrium solution at y = 1 is asymptotically stable where as theequilibrium solution at y = 0 is unstable

Problem 8.11

Sketch the direction field of the equation y = y3. Sketch the solution satis-fying the condition y(1) = −1.

Solution.

As shown in the figure below, the domain of the solution is the interval−∞ < t < 2

Problem 8.12

Find the equilibrium solutions and determine their stability

y = y

2

(y

2

− 1), y(0) = y0.

Solution.

The direction field is given below.



73

The equilibrium solution y = 1 is unstable; y = 0 is semi-stable; y = −1 isasymptotically stable

Problem 8.13

Find the equilibrium solutions of the equation

y = y2 − 4y

then decide whether they are asymptotically stable, semi-stable, or unstable.What is the long-run behavior if y(0) = 5?y(0) = 4?y(0) = 3?

Solution.

The direction field is given below.




The equilibrium solution y = 4 is unstable while y = 0 is asymptoticallystable. If y(0) = 5 then limt→∞ y(t) = ∞. If y(0) = 4 then limt→∞ y(t) = 4.If y(0) = 3 then limt→∞ y(t) = 0

Problem 8.14What is limt→∞ y(t) for the initial-value problem

y = sin (y(t)), y(0) = π

2.

Solution.

y(t) = 0 and y(t) = π are two equilibrium solutions. According to thedirection field shown below we conclude that

limt→∞

y(t) = π



75






9 Numerical Solutions to

ODEs: Euler’s Method and its

Variants

In Problems 9.1 - 9.3, answer the following questions:(a) Solve the initial value problem analytically, using an appropriate solutiontechnique.(b) For the given initial value problem express yn+1 in terms of yn usingHeun’s method.(c) For the given initial value problem express yn+1 in terms of yn using theModified Euler’s method.(d) Use a step size h = 0.1. Compute the first three approximations y1, y2, y3

using the method in part (b).(e) Use a step size h = 0.1. Compute the first three approximations y1, y2, y3using the method in part (c).(f) For comparison, calculate and list the exact solution values y(t1), y(t2), y(t3).

Problem 9.1

y = 2t− 1, y(1) = 0.

Solution.

(a) Integrating y = 2t − 1 we find y(t) = t2 − t + C. Imposing the initial

condition y(1) = 0, we obtain y(t) = t2

− t.(b) Since f (t, y) = 2t− 1, it follows that f (t + h, y + hf (t, y)) = 2(t + h)− 1.Therefore, Heun’s method takes the form

yn+1 = yn + h

2[(2tn − 1) + (2tn+1 − 1)].

77



789 NUMERICAL SOLUTIONS TO ODES: EULER’S METHOD AND ITS VARIANT

(c) As in part (b), we find the modified Euler’s method takes the form

yn+1 = yn + h[2(tn + h2

) − 1].

(d)

n tn yn

0 1.0000 01 1.1000 0.11002 1.2000 0.24003 1.3000 0.3900

(e)

n tn yn

0 1.0000 01 1.1000 0.11002 1.2000 0.24003 1.3000 0.3900

(f)

n tn yn

0 1.0000 01 1.1000 0.11002 1.2000 0.2400

3 1.3000 0.3900

Problem 9.2

y = −y, y(0) = 1.

Solution.

(a) Integrating y = −y and imposing the initial condition we find y(t) = e−t.(b) Heun’s method takes the form yn+1 = (1− h + 0.5h2)yn.(c) The modified Euler’s method takes the form yn+1 = (1− h + 0.5h2)yn.(d)

n tn yn

0 0.0000 1.00001 0.1000 0.90502 0.2000 0.81903 0.3000 0.7412



79

(e)

n tn yn

0 0.0000 1.00001 0.1000 0.90502 0.2000 0.81903 0.3000 0.7412

(f)

n tn yn

0 0.0000 1.00001 0.1000 0.9048

2 0.2000 0.81873 0.3000 0.7408

Problem 9.3

y2y + t = 0, y(0) = 1.

Solution.

(a) Solving the separable equation y2y + t = 0 we find y3 = −1.5t2 + C.

Imposing the initial condition y(0) = 1, we obtain y(t) = (1 − 1.5t2)13 .

(b) Since f (t, y) = −ty−2, it follows that f (t + h,hf (t, y)) = −(t + h)[y +

h(−ty)]−2

. Therefore, Heun’s method takes the form

yn+1 = yn + h

2[−tny−2

n − tn+1(yn − htny−2n )−2].

(c) The modified Euler’s method takes the form

yn+1 = yn − h(tn + 0.5h)(yn − 0.5htny−2n )−2.

(d)

n tn yn

0 0.0000 1.0000

1 0.1000 0.99502 0.2000 0.97963 0.3000 0.9529

(e)




n tn yn

0 0.0000 1.00001 0.1000 0.99502 0.2000 0.97973 0.3000 0.9531

(f)

n tn yn

0 0.0000 1.00001 0.1000 0.99502 0.2000 0.97963 0.3000 0.9528

Problem 9.4


y = 1 + y2, y(0) = −1.

(a) Find the exact solution of the given initial value problem.(b) Use step size h = 0.05. Compute 20 steps of Euler’s method, Heun’smethod, and modified Euler’s method. Compare the numerical values ob-tained at t = 1 by calculating the error |y(1) − y20|.

Solution.

(a) Solving the separable equation y = 1 + y2 and imposing the initialcondition we obtain y(t) = tan

t − π

4

.

(b)

n tn Euler Heun Modified Euler0 0.0000 −1.0000 −1.0000 −1.00001 0.0500 −0.9000 −0.9047 −0.90492 0.1000 −0.8095 −0.8177 −0.81793 0.1500 −0.7267 −0.7375 −0.7378

4 0.2000 −0.6503 −0.6630 −0.6634... ...

... ...

...20 1.0000 0.2355 0.2181 0.2173

The errors at t = 1 are, respectively, 0.0175, 1.419×10−4, and 6.581×10−4



81

Problem 9.5

Consider the initial value problemy + 2y = 4, y(0) = 3.

(a) Find the exact solution of the given initial value problem.(b) Use step size h = 0.05. Compute 20 steps of Euler’s method, Heun’smethod, and modified Euler’s method. Compare the numerical values ob-tained at t = 1 by calculating the error |y(1) − y20|.

Solution.

(a) Solving the equation y + 2y = 4 and imposing the initial condition weobtain y(t) = 2 + e−2t.(b)

n tn Euler Heun Modified Euler0 0.0000 3.0000 3.0000 3.00001 0.0500 2.9000 2.9050 2.90502 0.1000 2.8100 2.8190 2.81903 0.1500 2.7290 2.7412 2.74124 0.2000 2.6561 2.6708 2.6708...

... ...

... ...

20 1.0000 2.1216 2.1358 2.1358

The errors at t = 1 are, respectively, 1.3758 × 10−2, 4.8717 × 10−4, and4.8717 × 10−4

In Problems 9.6 - 9.8, the given iteration is the result of applying N steps of Euler’s method, Heun’s method, or the modified Euler’s method to an initialvalue problem of the form

y = f (t, y), y(t0) = y0, t0 ≤ t ≤ t0 + T.

Identify the numerical method and determine t0, N , T , and f (t, y).

Problem 9.6

yn+1 = yn + h(yn + t2ny3n), y0 = 1

tn = 2 + nh, h = 0.02, n = 0, 1, 2, · · · , 49.




Solution.

Since tn = 2 + nh, h = 0.02, n = 0, 1, · · · , 49, it follows that t0 = 2 andN − 1 = 49. Thus, N = 50 and T = tN − t0 = 2 + N h − 2 = 50(0.02) =1. From the form of the iteration, it must be Euler’s method. Thereforef (t, y) = y + t2y3

Problem 9.7

yn+1 = yn + h

tn +

h

2

sin2

yn +

h

2tn sin2 yn

, y0 = 1

tn = nh, h = 0.01, n = 0, 1, 2, · · · , 199.

Solution.Since tn = nh, h = 0.01, n = 0, 1, · · · , 199, it follows that t0 = 0 andN − 1 = 199. Thus, N = 200 and T = tN − t0 = N h = 200(0.01) = 2. Fromthe form of the iteration, it must be the modified Euler’s method. Therefore,f (t, y) = t sin2 y

Problem 9.8

yn+1 = yn + h

2[tny2

n + 2 + (tn + h)(yn + h(tny2n + 1))2], y0 = 2

tn = 1 + nh, h = 0.05, n = 0, 1, 2,

· · · , 99.

Solution.

Since tn = 1 + nh, h = 0.05, n = 0, 1, · · · , 99, it follows that t0 = 1 andN − 1 = 99. Thus, N = 100 and T = tN − t0 = 1 + N h − 1 = 100(0.05) =5. From the form of the iteration, it must be Heun’s method. Thereforef (t, y) = ty2 + 1



10 Second Order Linear

Differential Equations:

Existence and Uniqueness

Results

In Problems 10.1 - 10.6, determine the largest t−interval on which Theorem15.1 guarantees the existence of a unique solution.

Problem 10.1

y + y + 3ty = tan t, y(π) = 1, y(π) = −1.

Solution.

In this equation p(t) = 1, q (t) = 3t and g(t) = tan t. All three functions arecontinuous for all t = (2n + 1) π

2 , where n is an integer. With t0 = π then thelargest interval of existence guaranteed by Theorem 15.1 is π

2 < t < 3π

2

Problem 10.2

ety + 1t2−1

y = 4t

, y(−2) = 1, y(−2) = 2.

Solution.

In this equation p(t) = 0, q (t) = 1et(t2−1) , and g(t) = 4

t e−t. All three functionsare continuous for all t = −1, 0, 1. With t0 = −2 then the largest interval of existence guaranteed by Theorem 15.1 is −∞ < t < −1

83



8410 SECOND ORDER LINEAR DIFFERENTIAL EQUATIONS: EXISTENCE AND

Problem 10.3

ty + sin2tt2−9

y + 2y = 0, y(1) = 0, y(1) = 1.

Solution.

In this equation p(t) = sin2tt(t2−9) , q (t) = 2

t , and g(t) = 0. All three functionsare continuous for all t = −3, 0, 3. With t0 = 1 then the largest interval of existence guaranteed by Theorem 15.1 is 0 < t < 3

Problem 10.4

ty − (1 + t)y + y = t2e2t, y(−1) = 0, y(−1) = 1.

Solution.

In this equation p(t) = −1+tt , q (t) = 1

t , and g(t) = te2t. All three functionsare continuous for all t = 0. With t0 = −1 then the largest interval of existence guaranteed by Theorem 15.1 is −∞ < t < 0

Problem 10.5

(sin2 t)y

−(2sin t cos t)y + (cos2 t + 1)y = sin3 t, y( π

4 ) = 0, y( π4 ) =

√ 2.

Solution.

In this equation p(t) = −2 cos tsin t

, q (t) = cos2 t+1sin2 t

, and g(t) = sin t. All threefunctions are continuous for all t = nπ, where n is an integer. With t0 = π

4

then the largest interval of existence guaranteed by Theorem 15.1 is 0 < t <π

Problem 10.6

t2y + ty + y = sec (ln t), y( π3 ) = 0, y( π

3 ) = −1.

Solution.

In this equation p(t) = 1t , q (t) = 1

t2 , and g(t) = sec(ln t)t2 . All three functions

are continuous for all t > 0 and t = e(2n+1)π2 , where n is an integer. Note that

secln t = 0 only if cosln t = 0 which implies that ln t = (2n + 1)) π2 . With



85

t0 = π3 then the largest interval of existence guaranteed by Theorem 15.1 is

e−π2

< t < e

π2

In Problems 10.7 - 10.8, give an example of an initial value problem of theform y + p(t)y + q (t)y = 0, y(t0) = y0, y(t0) = y0 for which the givent−interval is the largest on which Theorem 15.1 guarantees a unique solu-tion.

Problem 10.7

−∞< t <

∞.

Solution.

One such an answer is

y + y + y = 0, y(0) = 0, y(0) = 1

Problem 10.8

3 < t < ∞.

Solution.

One such an answer is

y + 1t−3y + y = 1, y(4) = 0, y(4) = −1

Problem 10.9

Determine the largest interval in which the initial-value problem

(t − 3)y + ty + (ln |t|)y = 0, y(1) = 0, y(1) = 1

is certain to have a unique solution.

Solution.

We have p(t) = tt−3

and q (t) = ln |t|t−3

. Both functions are continuous for allt = 0, 3. Since t0 = 1, the largest t-interval is 0 < t < 3



8610 SECOND ORDER LINEAR DIFFERENTIAL EQUATIONS: EXISTENCE AND

Problem 10.10

Theorem 15.1 tells us that the initial-value problemy + t2y = 0, y(0) = y (0) = 0

defines exactly one function y(t). Using only Theorem 15.1, show that thisfunction has the additional property y(−t) = y(t). That is, y(t) is an evenfunction.

Solution.

Let Y (t) = y(−t). Then Y + t2Y = y + t2y = 0, Y (0) = Y (0) = 0 so thatY (t) is a solution to the given initial-value problem. By Theorem 15.1, wemust have Y (t) = y(t), i.e., y(

−t) = y(t) for all real numbers t



11 The General Solution of 2nd

Order Linear Homogeneous

Equations

In Problems 11.1-11.7, the t-interval of solution is −∞ < t < ∞ unless indi-cated otherwise.(a) Determine whether the given functions are solutions to the differentialequation.(b) If both functions are solutions, calculate the Wronskian. Does this cal-culation show that the two functions form a fundamental set of solutions?(c) If the two functions have been shown in (b) to form a fundamental set,construct the general solution and determine the unique solution satisfying

the initial value problem.

Problem 11.1

y − 4y = 0, y1(t) = e2t, y2(t) = 2e−2t, y(0) = 1, y(0) = −2.

Solution.

(a)

y1 − 4y1 = 4e2t − 4e2t = 0

y2 −

4y2 = 8e−2t

−8e−2t = 0

so both functions are solutions.(b)

W (y1(t), y2(t)) =

e2t 2e−2t

2e2t −4e−2t

= −8 = 0

87



8811 THE GENERAL SOLUTION OF 2ND ORDER LINEAR HOMOGENEOUS EQU

so {y1, y2} is a fundamental set of solutions.

(c) We have y(t) = c1e

2t

+ 2c2e−2t

and y(t) = 2c1e

2t

− 4c2e−2t

. The initialconditions imply c1 + 2c2 = 1 and 2c1 − 4c2 = −2. Solving we find c1 = 0and c2 = 1

2. Hence, y(t) = e−2t

Problem 11.2

y + y = 0, y1(t) = sin t cos t, y2(t) = sin t, y( π2 ) = 1, y( π

2 ) = 1.

Solution.

(a)

y1 + y1 = −2cos t sin t + 2 cos t sin t + sin t cos t = 0

so y1 is not a solution.

y2 + y2 = − sin t + sin t = 0

so y2 is a solution

Problem 11.3

y

−4y + 4y = 0, y1(t) = e2t, y2(t) = te2t, y(0) = 2, y(0) = 0.

Solution.

(a)

y1 − 4y1 + 4y1 = 4e2t − 8e2t + 4e2t = 0.

y2 − 4y2 + 4y2 = 4e2t + 4te2t − 4e2t − 8te2t + 4te2t = 0


W (y1(t), y2(t)) = e2t te2t

2e2t (2t + 1)e2t = e4t

= 0

so {y1, y2} is a fundamental set of solutions.(c) We have y(t) = c1e2t + c2te2t and y (t) = 2c1e2t + (c2 + 2c2t)e2t. The initialconditions imply c1 = 2 and c2 = −4. Hence, y(t) = 2e2t − 4te2t



89

Problem 11.4

ty + y = 0, y1(t) = ln t, y2(t) = ln 3t, y(3) = 0, y(3) = 3, 0 < t < ∞.

Solution.

(a)

ty1 + y1 = − tt2

+ 1t = 0

ty2 + y2 = − tt2 + 1

t = 0


W (y1(t), y2(t)) = ln t ln(3t)

1t

1t

= −1t

ln 3 = 0

so {y1, y2} is a fundamental set of solutions.(c) We have y(t) = c1 ln t+c2 ln(3t) and y(t) = c1

t + c2t . The initial conditions

imply c1 + 2c2 = 0 and c1 + c2 = 9. Solving we find c1 = 18 and c2 = −9.Hence, y(t) = 18ln t − 9ln(3t), t > 0

Problem 11.5

t2y

−ty

−3y = 0, y1(t) = t3, y2(t) =

−t−1, y(

−1) = 0, y(

−1) =

−2, t < 0.

Solution.

(a)

t2y1 − ty1 − 3y1 = t2(6t) − t(3t2) − 3t3 = 0

t2y2 − ty2 − 3y2 = t2(−2t−3) − t(t−2) − 3(−t−1) = 0


W (y1(t), y2(t)) = t3 −t−1

3t2 t−2 = 4t

= 0, t < 0

so {y1, y2} is a fundamental set of solutions.(c) We have y(t) = c1t3−c2t−1 and y(t) = 3c1t2+c2t−2. The initial conditionsimply −c1 + c2 = 0 and 3c1 + c2 = −2. Solving we find c1 = − 1

2 and c2 = −1

2.

Hence, y(t) = 12(t−1 − t3), t > 0




Problem 11.6

y = 0, y1(t) = t + 1, y2(t) = −t + 2, y(1) = 4, y(1) = −1.

Solution.

(a) Since y1 = y 2 = 0, both functions are solutions.(b)

W (y1(t), y2(t)) =

t + 1 −t + 21 −1

= −3 = 0

so {y1, y2} is a fundamental set of solutions.(c) We have y(t) = c1(t + 1) + c2(

−t + 2) and y(t) = c1

−c2. The initial

conditions imply 2c1 + c2 = 4 and c1 − c2 = −1. Solving we find c1 = 1 andc2 = 2. Hence, y(t) = −t + 5

Problem 11.7

4y + 4y + y = 0, y1(t) = et2 , y2(t) = te

t2 , y(1) = 1, y(1) = 0.

Solution.

(a)

4y1 + 4y1 + y1 = 4e t2 = 0

so y1 is not a solution.

4y2 + 4y2 + y2 = 8et2 + 4te

t2 = 0

so y2 is not a solution

Problem 11.8

The functions y1(t) = t and y2(t) = t ln t form a fundamental set of solutionsto the differential equation

t2y − ty + y = 0, 0 < t < ∞.

(a) Show that y(t) = 2t + t ln 3t is a solution to the differential equation.(b) Find c1 and c2 such that y(t) = c1y1(t) + c2y2(t).



91

Solution.

(a) t

2

y − ty + y = t

2

t−1

− t(3 + ln (3t)) + 2t + t ln(3t) = 0.(b) We havec1t + c2t ln t = 2t + t ln(3t)

c1 + c2(1 + ln t) = 3 + ln (3t)

Using the elimination method we find c1 = 2 + ln3 and c2 = 1. Thus,y(t) = (2 + ln 3)t + t ln t

Problem 11.9

The functions y1(t) = e3t and y2(t) = e−3t are known to be solutions of y + αy + βy = 0, where α and β are constants. Determine α and β .

Solution.Since y1 + αy 1 + βy1 = 0, 3α + β = −9. Since y2 + αy2 + βy2 = 0 we find−3α + β = −9. Hence, α = 0 and β = −9

Problem 11.10

The functions y1(t) = t and y2(t) = et are known to be solutions of y + p(t)y + q (t)y = 0.(a) Determine the functions p(t) and q (t).(b) On what t−intervals are the functions p(t) and q (t) continuous?(c) Compute the Wronskian of these two functions. On what t−intervals isthe Wronskian nonzero?

(d) Are the observations in (b) and (c) consistent with Abel’s Theorem?Solution.

(a) Since y1 + p(t)y1+q (t)y1 = 0, p(t)+tq (t) = 0. Since y2 + p(t)y2+q (t)y2 = 0we find p(t) + q (t) = −1. Solving for p(t) and q (t) we find p(t) = −t

t−1 and

q (t) = 1t−1

.(b) Both p(t) and q (t) are continuous on (−∞, 1) ∪ (1,∞).(c)

W (y1(t), y2(t)) =

t et

1 et

= et(t − 1).

The Wronskian is nonzero for all t

= 1.

(d) Yes. W = 0 on the two intervals on which p and q are both continuous

Problem 11.11

It is known that two solutions of y+ty+2y = 0 has a Wronskian W (y1(t), y2(t))that satisfies W (y1(1), y2(1)) = 4. What is W (y1(2), y2(2))?




Solution.

From Abel’s Theorem, we have

W (y1(t), y2(t)) = W (y1(1), y2(1))e− t1

sds = 4e−t2

2 + 1

2 .

Hence, W (y1(2), y2(2)) = 4e−1.5

Problem 11.12

The pair of functions {y1, y2} is known to form a fundamental set of solutionsof y +αy+βy = 0, where α and β are constants. One solution is y1(t) = e2t,and the Wronskian formed by these two solutions is W (y1(t), y2(t)) = e−t.Determine the constants α and β .

Solution.Since y 1 + αy1 + βy1 = 0, we have 2α + β = −4. Since W (y1(t), y2(t)) = e−t,we find W (t) = −e−t. But W + pW = 0 so that −e−t + pe−t = 0. Hence,

p(t) = 1 = α. Thus, β = −4 − 2α = −6

Problem 11.13

The Wronskian of a pair of solutions of y + p(t)y + 3y = 0 is W (t) = e−t2 .What is the coefficient function p(t)?

Solution.

Since W = − pW, we have −2te−t2 = − p(t)e−t2 so that p(t) = 2t

Problem 11.14

Prove that if y1 and y2 have maxima or minima at the same point in aninterval I, then they cannot be a fundamental set of solutions on that interval.

Solution.

Suppose for example that both functions have a same maximum at t0. Theny1(t0) = y 2(t0) = 0. But

W (y1(t0), y2(t0)) = y1(t0)y2(t0) − y1(t0)y2(t0) = 0.

Thus, {y1, y2} is not a fundamental set

Problem 11.15

Without solving the equation, find the Wronskian of two solutions of Bessel’sequation

t2y + ty + (t2 − µ2)y = 0.



93

Solution.

By Abel’s Theorem

W (y1(t), y2(t)) = W (y1(t0), y2(t0))e− t

t0

dss = t0

W (y1(t0), y2(t0))

t

Problem 11.16

If W (y1, y2) = t2et and y1(t) = t then find y2(t).

Solution.

By the quotient rule

y2y1

=

W

y21

= et.

Thus, one possible answer is

y2(t) = tet

Problem 11.17

The functions t2 and 1/t are solutions to a 2nd order, linear homogeneousODE on t > 0. Verify whether or not the two solutions form a fundamentalsolution set.

Solution.

Finding the Wronskian

W (t2, 1

t) =

t2 t−1

2t −t−2

= −3 = 0

so that {y1, y2} is a fundamental set

Problem 11.18

Show that t3 and t4 can’t both be solutions to a differential equation of theform y + p(t)y + q (t)y = 0 where p and q are continuous functions definedon the real numbers.

Solution.Suppose that t3 and t4 are both solutions. Since W (t) = t6, we find W (1) = 1and so {y1, y2} is a fundamental set. By Abel’s Theorem, W (t) = 0 for all−∞ < t < ∞. But W (0) = 0, a contradiction. Hence, t3 and t4 can’t beboth solutions for the differential equation for −∞ < t < ∞




Problem 11.19

Suppose that t

2

+ 1 is the Wronskian of two solutions to the differentialequation y + p(t)y + q (t)y = 0. Find p(t).

Solution.

Since W = − p(t)W, we have 2t = − p(t)(t2 + 1). Thus, p(t) = − 2tt2+1



12 Second Order Linear

Homogeneous Equations with

Constant Coefficients: Distinct

Characterisitc Roots

Problem 12.1

Solve the initial value problem

y + y − 2y = 0, y(0) = 3, y(0) = −3.

Describe the behavior of the solution y(t) as t → −∞ and t →∞.

Solution.

The characteristic equation r

2

+ r − 2 = 0 has roots r = 1 and r = −2 sothat the general solution is given by

y(t) = c1et + c2e−2t.

The initial conditions and y (t) = c1et−2c2e−2t lead to the system c1+c2 = 3and c1 − 2c2 = −3. Solving this system, we find c1 = 1 and c2 = 2. Hence,the unique solution to the initial value problem is

y(t) = et + 2e−2t.

limt→−∞ y(t) = ∞ and limt→∞ y(t) = ∞

Problem 12.2Solve the initial value problem

y − 4y + 3y = 0, y(0) = −1, y(0) = 1.


95



9612 SECOND ORDER LINEAR HOMOGENEOUS EQUATIONS WITH CONSTAN

Solution.


2

−4r + 3 = 0 has roots r = 1 and r = 3 so thatthe general solution is given by

y(t) = c1et + c2e3t.

The initial conditions and y(t) = c1et +3c2e3t lead to the system c1+c2 = −1and c1 + 3c2 = 1. Solving this system, we find c1 = −2 and c2 = 1. Hence,the unique solution to the initial value problem is

y(t) = −2et + e3t.

limt

→−∞y(t) = 0 and limt

→∞y(t) = limt

→∞e3t 1

− 2e2t =

∞Problem 12.3


y − y = 0, y(0) = 1, y(0) = −1.


Solution.

The characteristic equation r2 − 1 = 0 has roots r = −1 and r = 1 so thatthe general solution is given by

y(t) = c1et + c2e−t.

The initial conditions and y (t) = c1et − c2e−t lead to the system c1 + c2 = 1and c1− c2 = −1. Solving this system, we find c1 = 0 and c2 = 1. Hence, theunique solution to the initial value problem is

y(t) = e−t.

limt→−∞ y(t) = ∞ and limt→∞ y(t) = 0

Problem 12.4


y + 5y + 6y = 0, y(0) = 1, y(0) = −1.




97

Solution.


2

+ 5r + 6 = 0 has roots r = −2 and r = −3 sothat the general solution is given by

y(t) = c1e−2t + c2e−3t.

The initial conditions and y(t) = −2c1e−2t − 3c2e−3t lead to the systemc1 + c2 = 1 and 2c1 + 3c2 = 1. Solving this system, we find c1 = 2 andc2 = −1. Hence, the unique solution to the initial value problem is

y(t) = 2e−2t − e−3t.

limt→−∞ y(t) = limt→−∞ e−3t

(2e

t

− 1) = −∞ and limt→∞ y(t) = 0

Problem 12.5


y − 4y = 0, y(3) = 0, y(3) = 0.


Solution.

The characteristic equation r2 − 4 = 0 has roots r = −2 and r = 2 so that

the general solution is given by

y(t) = c1e2t + c2e−2t.

The initial conditions and y(t) = 2c1e2t − 2c2e−2t lead to the system c1e6 +c2e−6 = 0 and 2c1e6 − 2c2e−6 = 0. Solving this system, we find c1 = 0 andc2 = 0. Hence, the unique solution to the initial value problem is y(t) ≡ 0.

limt→−∞ y(t) = 0 and limt→∞ y(t) = 0

Problem 12.6


2y − 3y = 0, y(−2) = 3, y(−2) = 0.





Solution.

The characteristic equation 2r

2

− 3r = 0 has roots r = 0 and r = 1.5 so thatthe general solution is given by

y(t) = c1e1.5t + c2.

The initial conditions and y (t) = 1.5c1e1.5t lead to the system c1e−3 + c2 = 3and c1 = 0. Solving this system, we find c2 = 3. Hence, the unique solutionto the initial value probem y(t) ≡ 3.

limt→−∞ y(t) = 3 and limt→∞ y(t) = 3

Problem 12.7


y + 4y + 2y = 0, y(0) = 0, y(0) = 4.


Solution.

The characteristic equation r2 + 4r + 2 = 0 has roots r = −2 − √ 2 and

r = −2 +√

2 so that the general solution is given by

y(t) = c1e(−2−√ 2)t + c2e(−2+√ 2)t.

The initial conditions and y(t) = c1(−2−√ 2)e(−2−√ 2)t+c2(−2+√ 2)e(−2+√ 2)t

lead to the system c1 + c2 = 0 and (−2 −√ 2)c1 + (−2 +√

2)c2 = 4. Solvingthis system, we find c1 = −√ 2 and c2 =

√ 2. Hence, the unique solution to

the initial value problem is

y(t) = −√

2e(−2−√ 2)t +√

2e(−2+√ 2)t.

limt→−∞ y(t) = limt→−∞ e(−2−√ 2)t[−√ 2 +√

2e2√ 2t] = −∞ and

limt→∞ y(t) = 0

Problem 12.8


2y − y = 0, y(0) = −2, y(0) =√

2.




99

Solution.

The characteristic equation 2r

2

− 1 = 0 has roots r = −√ 2

2 and r =

√ 2

2 sothat the general solution is given by

y(t) = c1e√ 22

t + c2e−√ 22

t.

The initial conditions and y(t) =√ 22 c1e

√ 22

t −√ 22 c2e−

√ 22

t lead to the systemc1+c2 = −2 and c1−c2 = 2. Solving this system, we find c1 = 0 and c2 = −2.Hence, the unique solution to the initial value problem is

y(t) = −2e−√ 22

t.

limt→−∞

y(t) =−∞

and limt→∞

y(t) = 0

Problem 12.9

Obtain the general solution to the differential equation y − 5y + 6y = 0.

Solution.

Let u = y . Then u = y and u = y so that the given equation becomes

u − 5u + 6u = 0.

The characteristic equation r2−5r + 6 = 0 has roots r = 2 and r = 3 so thatthe general solution is given by

u(t) = c1e2t + c2e3t.

But y (t) = u(t) so that

y(t) =

u(t)dt = C 1e2t + C 2e3t + C 3

Problem 12.10

A particle of mass m moves along the x-axis and is acted upon by a dragforce proportional to its velocity. The drag constant is denoted by k. If x(t)represents the particle position at time t, Newton’s law of motion leads to

the differential equation mx(t) = −kx(t).(a) Obtain the general solution to this second order linear differential equa-tion.(b) Solve the initial value problem if x(0) = x0 and x(0) = v0.(c) What is limt→∞ x(t)?




Solution.

(a) The characteristic equation is mr

2

+kr = 0 with roots r = 0 and r = −k

m .Thus, the general solution is

x(t) = c1 + c2e−km

t.

(b) The initial conditions and x(t) = − km

c2e−km

t lead to c1 = x0 + mk v0 and

c2 = −mk v0. Hence,

x(t) =

x0 + m

k v0

− m

k v0e−

km

t.

(c) limt→∞ x(t) = x0 + mk v0

Problem 12.11

Find a homogeneous second-order linear ordinary differential equation whosegeneral solution is y(t) = c1e2t + c2e−t.

Solution.

The roots for the characteristic equation are r = 2 and r = −1 so that(r − 2)(r + 1) = 0 and hence r2 − r − 2 = 0. The homogeneous equation isthen y − y − 2y = 0

Problem 12.12

Find the general solution of the differential equation y

−3y

−4y = 0.

Solution.

The characteristic equation r2 − 3r − 4 = 0 has roots r = −1 and r = 4.Thus,

y(t) = c1e−t + c2e4t

Problem 12.13

Find the general solution of the differential equation y + 4y − 5y = 0.

Solution.

The characteristic equation r2 + 4r

−5 = 0 has roots r = 1 and r =

−5.

Thus,y(t) = c1et + c2e−5t

Problem 12.14

Find the general solution of the differential equation −3y + 2y + y = 0.



101

Solution.

The characteristic equation −3r

2

+ 2r + 1 = 0 has roots r = 1 and r = −1

3 .Thus,y(t) = c1et + c2e

−t3

Problem 12.15

Solve the initial-value problem: y + 3y − 4y = 0, y(0) = −1, y(0) = 1.

Solution.

The characteristic equation r2 + 3r − 4 = 0 has roots r = 1 and r = −4.Thus,

y(t) = c1et + c2e−4t.

The initial conditions and y(t) = c1et−4c2e−4t lead to the system c1+c2 = −1and c1 − 4c2 = 1. Solving this system, we find c1 = −3

5 and c2 = − 2

5 . Thus,

y(t) = −1

5(3et + 2e−4t)

Problem 12.16

Solve the initial-value problem: 2y + 5y − 3y = 0, y(0) = 2, y(0) = 1.

Solution.

The characteristic equation 2r2 + 5r − 3 = 0 has roots r = −3 and r = 12 .

Thus,y(t) = c1e−3t + c2e

t2 .

The initial conditions and y(t) = −3c1e−3t + c22 e

t2 lead to the system c1+c2 =

2 and −3c1 + c22

= 1. Solving this system, we find c1 = 0 and c2 = 2. Thus,

y(t) = 2et2






13 Characteristic Equations

with Repeated Roots

In Problems 13.1 - 13.5 answer the following questions.(a) Obtain the general solution of the differential equation.(b) Impose the initial conditions to obtain the unique solution of the initialvalue problem.(c) Describe the behavior of the solution as t → −∞ and t →∞.

Problem 13.1

9y − 6y + y = 0, y(3) = −2, y(3) = −5

3.

Solution.

(a) The characteristic equation 9r2 − 6r + 1 = 0 has the roots r1 = r2 = 13 .

The general solution is then

y(t) = c1et3 + c2te

t3 .

(b) The initial conditions and y (t) = c13 e

t3 + c2e

t3 + c2

3 te

t3 lead to the system

c1 + 3c2 = −2e−1 and c1 + 6c2 = −5e−1. Solving this system, we find c1 = e−1

and c2 = −e−1. Thus, the unique solution is

y(t) = et3−1(1− t).

(c)

limt→−∞ y(t) = limt→−∞ 1−t

e1−t3

= limt→−∞ −1

−1/3e1−t3

= 0

Now, for large t we have t−1 ≥ 1 so that et3−1(t−1) ≥ e

t3−1. Since e

t3−1 →∞

as t →∞, we find et3−1(t − 1) →∞ as t →∞. Hence,

103



104 13 CHARACTERISTIC EQUATIONS WITH REPEATED ROOTS

limt→∞ y(t) = − limt→∞ et3−1(t − 1) = −∞

Problem 13.2

25y + 20y + 4y = 0, y(5) = 4e−2, y(5) = −3

5e−2.

Solution.

(a) The characteristic equation 25r2+20r+4 = 0 has the roots r1 = r2 = − 25 .

The general solution is then

y(t) = c1e−2t5 + c2te−

2t5

(b) The initial conditions and y (t) =

−2c15

e−2t5 + c2e−

2t5

−2c25

te−2t5 lead to the

system c1 + 5c2 = 4 and 2c1 + 5c2 = 3. Solving this system, we find c1 = −1and c2 = 1. Thus, the unique solution is

y(t) = e−2t5 (t− 1).

(c)

limt→−∞ y(t) = − limt→−∞ e−2t5 (1 − t) = −∞

and

limt→∞ y(t) = limt→∞ e−2t5 (t − 1) = limt→∞ 1

25

e2t5

= 0

Problem 13.3

y − 4y + 4y = 0, y(1) = −4, y(1) = 0.

Solution.

(a) The characteristic equation r2 − 4r + 4 = 0 has the roots r1 = r2 = 2.The general solution is then

y(t) = c1e2t + c2te2t.

(b) The initial conditions and y(t) = 2c1e2t +c2e2t +2c2te2t lead to the system

c1 + c2 = −4e−2

and 2c1 + 3c2 = 0. Solving this system, we find c1 = −12e−2

and c2 = 8e−2. Thus, the unique solution is

y(t) = 4e2t−2(2t − 3).

(c)



105

limt→−∞ y(t) = limt→−∞ 2t−3e2−2t = limt→−∞ 2

−2e2−2t = 0

and

limt→∞ y(t) = ∞Problem 13.4

y + 2√

2y + 2y = 0, y(0) = 1, y(0) = 0.

Solution.

(a) The characteristic equation r2 + 2√

2r + 2 = 0 has the roots r1 = r2 =−√ 2. The general solution is then

y(t) = c1e−

√ 2t

+ c2te−

√ 2t

.(b) The initial conditions and y(t) = −√ 2c1e−

√ 2t + c2e−

√ 2t − √

2c2te−√ 2t

lead to c1 = 1 and c2 =√

2.Thus, the unique solution is

y(t) = e−√ 2t +

√ 2te−

√ 2t = e−

√ 2t(1 +

√ 2t).

(c) We havelim

t→−∞y(t) = ∞×−∞ = −∞.

Also,

limt→∞

y(t) = limt→∞

1 +√

2t

e

√ 2t

= limt→∞

√ 2

√ 2e

√ 2t

= 0

Problem 13.5

3y + 2√

3y + y = 0, y(0) = 2√

3, y(0) = 3.

Solution.

(a) The characteristic equation 3r2 + 2√

3r + 1 = 0 has the roots r = r1 =r2 = − 1√

3. The general solution is then

y(t) = c1ert + c2tert.

(b) The initial conditions and y (t) = rc1e

rt

+ c2e

rt

+ rc2te

rt

lead to c1 = 2

√ 3and c2 = 5. Thus, the unique solution is

y(t) = e− t√

3 (5t + 2√

3).

(c)



106 13 CHARACTERISTIC EQUATIONS WITH REPEATED ROOTS

limt→−∞ y(t) = ∞×−∞ = −∞

and

limt→∞ y(t) = limt→∞ 2√ 3+5t

et√ 3

= limt→∞ 5

(1/√ 3)e

t√ 3

= 0

Problem 13.6

Find the general solution of y − 6y + 9y = 0.

Solution.

The characteristic equation r2 − 6r + 9 = 0 has double roots r1 = r2 = 3 sothe general solution is

y(t) = c1e

3t

+ c2te

3t

Problem 13.7

Find the general solution of 4y − 4y + y = 0.

Solution.

The characteristic equation 4r2− 4r + 1 = 0 has double roots r1 = r2 = 12

sothe general solution is

y(t) = c1et2 + c2te

t2

Problem 13.8

Solve the initial-value problem: y + y + y

4 = 0, y(0) = 2, y(0) = 0.

Solution.

The characteristic equation r2 + r + 14 = 0 has double roots r1 = r2 = − 1

2 sothe general solution is

y(t) = c1e−t2 + c2te−

t2 .

Since y(0) = 2, we find c1 = 2. Since y (0) = 0, we find c1 − 2c2 = 0. Solvingfor c2, we find c2 = 1. Hence, the unique solution is

y(t) = 2e−t2 + te−

t2



14 Characteristic Equations

with Complex Roots

Problem 14.1

For any z = α + iβ we define the conjugate of z to be the complex numberz = α − iβ. show that α = 1

2(z + z ) and β = 1

2i(z − z ).

Solution.

Adding z and z , we find 2α = z + z. Hence, α = 12(z + z ). Next, subtracting

z from z , we find 2iβ = z − z. Therefore, β = 12i

(z − z )

Problem 14.2

Write each of the complex numbers in the form α + iβ, where α and β arereal numbers.1. 2ei π

3 .

2. (2 − i)ei 3π2 .3. (

√ 2eiπ

6 )4.

Solution.

Recall Euler’s identity: eα+iβ = eα(cos β + i sin β ). Thus,1. 2ei π

3 = 2 cos ( π3

) + 2i sin( π3

) = 1 + i√

3.2. We have

(2− i)ei 3π2 =2ei 3π

2 − iei 3π2

=2 cos3π

2 + i sin3π

2 − i cos3π

2 + i sin3π

2 =2(0− i) − i(0− i)

= − 1− 2i.

3. (√

2eiπ6 )4 = (

√ 2)4ei 2π

3 = 4(− 12 + i

√ 32 ) = −2 + 2i

√ 3

107



108 14 CHARACTERISTIC EQUATIONS WITH COMPLEX ROOTS

Problem 14.3

Write each function in the form Ae

αt

cos βt + iBe

αt

sin βt, where α, β,A, andB are real numbers.1. 2ei

√ 2t.

2. − 12e2t+i(t+π).

3. (√

3e(1+i)t)3.

Solution.

Using Euler’s formula we find1. 2ei

√ 2t = 2cos

√ 2t + 2i sin

√ 2t.

2. − 12

e2t+i(t+π) = −12

e2t cos(t + π) − 12

e2ti sin(t + π) = 12

e2t cos t + 12

ie2t sin t.

3. (√

3e(1+i)t)3 = 3√

3e3(1+i)t = 3√

3e3t cos (3t) + 3√

3ie3t sin(3t)

In Problems 14.4 - 14.8(a) Determine the roots of the characteristic equation.(b) Obtain the general solution as a linear combination of real-valued solu-tions.(c) Impose the initial conditions and solve the initial value problem.

Problem 14.4

y + 2y + 2y = 0, y(0) = 3, y(0) = −1.

Solution.(a) The characteristic equation r2 + 2r + 2 = 0 has roots r1 = −1 − i andr2 = −1 + i.(b) y(t) = e−t(c1 cos t + c2 sin t).(c) The initial conditions and y (t) = e−t cos t(c2− c1)− e−t sin t(c1 + c2) leadto the equations c1 = 3 and −c1 + c2 = −1. Solving, we find c1 = 3 andc2 = 2. Hence, the unique solution to the initial value problem is

y(t) = 3e−t cos t + 2e−t sin t

Problem 14.5

2y − 2y + y = 0, y(−π) = 1, y(−π) = −1.

Solution.

(a) The characteristic equation 2r2 − 2r + 1 = 0 has roots r1 = 12(1− i) and



109

r2 = 12(1 + i).

(b) y(t) = e

t2

(c1 cos t

2 + c2 sin t

2).(c) The initial conditions and y (t) = 1

2et2 cos t

2(c1 + c2) + 12e

t2 sin t

2(−c1 + c2)lead to the equations −e−

π2 c2 = 1 and c2 − c1 = 2e

π2 . Solving, we find

c1 = −3eπ2 and c2 = −e

π2 . Hence, the unique solution to the initial value

problem is

y(t) = −e12(t+π)(3cos

t

2 + sin

t

2)

Problem 14.6

y + 4y + 5y = 0, y(π

2) =

1

2, y(

π

2) =

−2.

Solution.

(a) The characteristic equation r2 + 4r + 5 = 0 has roots r1 = −2 − i andr2 = −2 + i.(b) y(t) = e−2t(c1 cos t + c2 sin t).(c) The initial conditions and y (t) = e−2t cos t(c2 − 2c1)− e−2t sin t(c1 + 2c2)lead to the equations e−πc2 = 1

2 and c1 + 2c2 = 2eπ. Solving, we find c1 = eπ

and c2 = 12eπ. Hence, the unique solution to the initial value problem is

y(t) = e

π−2t

(cos t +

1

2 sin t)

Problem 14.7

y + 4π2y = 0, y(1) = 2, y(1) = 1.

Solution.

(a) The characteristic equation r2 + 4π2 = 0 has roots r1 = −2πi and r2 =2πi.(b) y(t) = c1 cos 2πt + c2 sin 2πt.

(c) The initial conditions and y(t) = 2πc2 cos 2πt − 2πc1 sin 2πt lead to theequations c1 = 2 and 2πc2 = 1. Solving, we find c1 = 2 and c2 = (2π)−1.Hence, the unique solution to the initial value problem is

y(t) = 2cos 2πt + (2π)−1 sin2πt




Problem 14.8

9y + π2y = 0, y(3) = 2, y(3) = −π.

Solution.

(a) The characteristic equation 9r2 + π2 = 0 has roots r1 = −π3 i and r2 = π

3 i.(b) y(t) = c1 cos π

3t + c2 sin π

3t.

(c) The initial conditions and y (t) = π3 c2 cos π

3 t− π3 c1 sin π

3 t lead to the equa-tions −c1 = 2 and −π

3c2 = −π. Solving, we find c1 = −2 and c2 = 3. Hence,

the unique solution to the initial value problem is

y(t) = −2cos π

3t + 3 sin

π

3t

In Problems 14.9 - 14.10, the function y(t) is a solution of the initial valueproblem y + ay + by = 0, y(t0) = y0, y(t0) = y0, where the point t0 isspecified. Determine the constants a, b, y0, and y

0.

Problem 14.9

y(t) = 2 sin 2t + cos 2t, t0 = π

4.

Solution.

The roots of the characteristic equation are r1,2 =

±2i so that the character-

istic equation is r2 + 4 = 0. Hence, the corresponding differential equationis y + 4y = 0. From this, we find a = 0 and b = 4. Now, y0 = y( π

4) =

2sin π2 + cos π

2 = 2. Finally, y 0 = y ( π

4 ) = 4 cos π2 − 2sin π

2 = −2

Problem 14.10

y(t) = et−π6 cos2t − et−π

6 sin2t, t0 = π

6.

Solution.

The roots of the characteristic equation are r1,2 = 1 ± 2i so that the char-

acteristic equation is r2

− 2r + 5 = 0. Hence, the corresponding differ-ential equation is y − 2y + 5y = 0. From this, we find a = −2 andb = 5. Now, y0 = y( π

6 ) = cos π3 − sin π

3 = 12 −

√ 32 . Finally, y0 = y( π

6 ) =

cos π3 − sin π

3 − 2cos π

3 − 2sin π

3 = −1

2 − 3

√ 3

2



111

In Problems 14.11 - 14.13, rewrite the function y(t) in the form y(t) =

Ke

αt

cos(βt − δ ), where 0 ≤ δ < 2π. Use this representation to sketch agraph of the given function, on a domain sufficiently large to display itsmain features.

Problem 14.11

y(t) = sin t + cos t.

Solution.

We have c1 = 1 and c2 = 1 so that K =√

12 + 12 =√

2. Moreover, cos δ =c1K

=√ 22

and sin δ =√ 22

. Thus, δ = π4

and

y(t) =√

2cos

t− π4

.

The graph of y(t) is given below.

Problem 14.12

y(t) = et cos t +√

3et sin t.

Solution.

We have c1 = 1 and c2 = √ 3 so that K = √ 1 + 3 = 2. Moreover, cos δ =c1K = 1

2 and sin δ =√ 32 . Thus, δ = π

3 and

y(t) = 2et cos

t − π

3

.





Problem 14.13

y(t) = e−2t cos2t− e−2t sin2t.

Solution.

We have c1 = 1 and c2 = −1 so that K =√

1 + 1 =√

2. Moreover, cos δ =c1K

= 1√ 2

and sin δ = − 1√ 2

. Thus, δ = 7π4

and

y(t) =√

2e−2t cos

2t − 7π

4

.


Problem 14.14

Consider the differential equation y +ay+9y = 0, where a is a real number.



113

Suppose that we know the Wronskian of a fundamental set of solutions of

this differential equation is constant: W (t) = 1 for all real numbers t. Findthe general solution of this differential equation.

Solution.

First we need to find a. Since W (t) = −aW (t), a = 0 so that y + 9y = 0.The characteristic equation is r2 + 9 = 0 and has complex roots r1,2 = ±3i.Thus, the general solution is given by

y(t) = c1 cos 3t + c2 sin 3t

Problem 14.15

Rewrite 2 cos 7t − 11sin7t in phase-angle form. Give the exact function (soyour answer will involve the inverse tangent function.)

Solution.

We have c1 = 2 and c2 = −11 so that K =√

4 + 121 =√

125 = 5√

5.Furthermore, tan δ = − 11

2 so that δ = − arctan112

. Hence,

y(t) = 5√

5cos

7t + arctan

11

2

Problem 14.16

Find a homogeneous linear ordinary differential equation whose general so-lution is y(t) = c1e2t cos (3t) + c2e2t sin (3t).

Solution.The roots to the characteristic equation are r1,2 = 2 ± 3i so that the char-acteristic equation is r2 − 4r + 13 = 0 and the corresponding differentialequation is

y − 4y + 13y = 0

Problem 14.17

Rewrite y(t) = 5e(5−2i)t − 3e(5+2i)t, without complex exponents, using sinesand cosines. What ODE of the form ay + by + cy = 0, has y as a solution?

Solution.

Using Euler’s formula, we have e(5−2i)t = e5t(cos 2t

−i sin2t) and e(5+2i)t =

e5t(cos 2t + i sin2t). Thus, y(t) = 2e5t cos2t − 8ie5t sin2t. The characteristicroots are r1,2 = 5± 2i so that the characteristic equation is r2− 10r + 29 andthe corresponding differential equation is

y − 10y + 29y = 0




Problem 14.18

Consider the function y(t) = 3 cos 2t − 4sin2t. Find a second order linearIVP that y satisfies.

Solution.

The roots to the characteristic equation are r1,2 = ±2i so that the charac-teristic equation is r2 + 4 = 0 and the corresponding differential equationis

y + 4y = 0, y(0) = 3, y(0) = −8

Problem 14.19

An equation of the form

t2y + αty + βy = 0, t > 0

where α and β are real constants is called an Euler equation. Show thatthe substitution x = ln t transforms Euler equation into an equation withconstant coefficients. Hint: dy

dt = dydx

dxdt .

Solution.

Since x = ln t, we find dxdt = 1

t . But dydt = dy

dxdxdt = 1

tdydx . Moreover, d2y

dt2 =

− 1t2

dydx + 1

t2d2ydx2 = 1

t2 d2ydx2 − dy

dx. Hence,

0 =t2y + αty + βy

=t2

1

t2

d2y

dx2 − dy

dx

+ αt

1

t

dy

dx

+ βy

=d2y

dx2 + (α − 1)

dy

dx + βy

Problem 14.20

Use the result of the previous problem to solve the differential equation t2y+ty + y = 0.

Solution.

Here we have α = β = 1 so that

d2y

dx2 + y = 0



115

The characteristic equation is r2 + 1 = 0 with complex roots r1,2 = ±i. The

general solution is y(x) = c1 cos x + c2 sin x

ory(t) = c1 cos (ln t) + c2 sin (ln t)






15 Series Solutions of

Differential Equations

Problem 15.1

Identify the singular and ordinary points of the differential equation

y + ty + (t2 + 2)y = 0.

Solution.

Since p(t) = t and q (t) = t2 + 2 are polynomials, they are analytic functionseverywhere. Thus, every real number is an ordinary point

Problem 15.2


(t2 − 1)y + ty + 1t

y = 0.

Solution.

We have p(t) = tt2−1 and q (t) = 1

t(t2−1) . Thus, the points 1, −1, and 0 aresingular points of the equation, any other real number is an ordinary pointof the equation

Problem 15.3


(t2

− 4)y + 3ty + y = 0.

Solution.

We have p(t) = 3tt2−4

and q (t) = 1t2−4

. Thus, the singular points are t = ±2and any other point is an ordinary point

117



118 15 SERIES SOLUTIONS OF DIFFERENTIAL EQUATIONS

Problem 15.4

The point t0 = 0 is an ordinary point of the differential equation(t2 − 4)y + 3ty + y = 0.

Determine a value of R as stated in Theorem 18.1.

Solution.

The point t0 = 0 is an ordinary point. By the previous exercise, both p(t)and q (t) are analytic in −2 < t < 2 so that a choice of R is R = 2

Problem 15.5

Find the series solution near the ordinary point 0 to the differential equation

y + ty + (t2 + 2)y = 0.

Solution.

There are no singular points. Thus, the series solution converges for everyvalue of t. The solution is of the form

y(t) =∞

n=0

antn.

Differentiating this series twice we find

y(t) = ∞

n=1 nantn−1 and y (t) = ∞

n=2 n(n− 1)antn−2.

Substituting these series in the given equation we find

∞n=2

n(n− 1)antn−2 + t

∞n=1

nantn−1 + t2∞

n=0

antn + 2

∞n=0

antn =0

∞n=2

n(n− 1)antn−2 +∞

n=1

nantn +∞

n=0

antn+2 +∞

n=0

2antn =0

∞

n=0

(n + 2)(n + 1)an+2tn +∞

n=1

nantn +∞

n=2

an−2tn +

∞

n=0

2antn =0.

We shift the index of summation in the first series by 2, replacing n withn + 2 and using the initial value n = 0. We shift the index of summation inthe third series by −2, replacing n by n−2 and using the initial value n = 2.



119

Since we want to express everything in only one summation sign, we have to

start the summation at n = 2 in every series

∞n=0

(n + 2)(n + 1)an+2tn +∞

n=1

nantn +∞

n=2

an−2tn +∞

n=0

2antn =

2a2 + 6a3t +∞

n=2

(n + 2)(n + 1)an+2tn + a1t +∞

n=2

nantn+

∞n=2

an−2tn + 2a0 + 2a1t +∞

n=2

2antn =0

or

(2a0+2a2)+(6a3+3a1)t+∞

n=2

[(n + 2)(n + 1)an+2 + (n + 2)an + an−2] tn = 0.

Equating all the coefficients to zero we find

2a0 + 2a2 =0

3a1 + 6a3 =0

(n + 1)(n + 2)an+2 + (n + 2)an + an−2 =0.

Thus,

a2

=−

a0

a3 = − a1

2

an+2 = − (n + 2)an + an−2

(n + 1)(n + 2) , n ≥ 2.




This last condition is called a recurrence formula, we can express each

an+2 in terms of a previous coefficient an. From the above, we can see that

a2 = − a0

a3 = − a1

2

a4 =1

4a0

a5 = 3

40a1

a6 = − 1

60a0

a7 = − 11680

a1.

Notice that each even coefficient is expressed in terms of a0 and each oddcoefficient is expressed in terms of a1. Then, the general solution is:

y(t) = a0

1 − t2 +

1

4t4 − 1

60t6 + · · ·

+a1

t − 1

2t3 +

3

40t5 − 1

1680t7 + · · ·

Problem 15.6


y − ty − t2y = 0.

Solution.


y(t) =∞

n=0

antn.


y(t) = ∞


n=2 n(n− 1)antn−2.



121


∞n=2

n(n− 1)antn−2 − t∞

n=1

nantn−1 − t2∞

n=0

antn =0

∞n=2

n(n− 1)antn−2 −∞

n=1

nantn −∞

n=0

antn+2 =0

∞n=0

(n + 2)(n + 1)an+2tn −∞

n=1

nantn −∞

n=2

an−2tn =0.

We shift the index of summation in the third series by −2, replacing n by

n− 2 and using the initial value n = 2.Since we want to express everything in only one summation sign, we have tostart the summation at n = 2 in every series

∞n=0

(n + 2)(n + 1)an+2tn −∞

n=1

nantn −∞

n=2

an−2tn =

2a2 + 6a3t +∞

n=2

(n + 2)(n + 1)an+2tn − a1t −∞

n=2

nantn −∞

n=2

an−2tn =

2a2 + (6a3

−a1)t +

∞

n=2

[(n + 1)(n + 2)an+2

−nan

−an−2] tn =0.


2a2 =0

6a3 − a1 =0

(n + 1)(n + 2)an+2 − nan − an−2 =0.

Thus,

a2 =0a3 =

a1

6

an+2 = nan + an−2

(n + 2)(n + 1), n ≥ 2.






a2 =0

a3 =a1

6

a4 = 1

12a0

a5 = 3

40a1

a6 = 1

90a0

a7 = 131008

a1.


y(t) = a0

1 +

1

12t4 +

1

90t6 + · · ·

+ a1

t +

1

6t3 +

3

40t5 +

13

1008t7 + · · ·

Problem 15.7


(t2 + 1)y − y + y = 0.

Solution.


y(t) =∞

n=0

antn.


y(t) = ∞


n=2 n(n− 1)antn−2.



123


t2∞

n=2

n(n− 1)antn−2 +∞

n=2

n(n− 1)antn−2 − ∞n=1

nantn−1 +∞

n=0

antn =0

∞n=2

n(n− 1)antn +∞

n=2

n(n− 1)antn−2 −∞

n=1

nantn−1 +∞

n=0

antn =0

∞n=2

n(n− 1)antn +∞

n=0

(n + 2)(n + 1)an+2tn −∞

n=0

(n + 1)an+1tn +∞

n=0

antn =0.

Since we want to express everything in only one summation sign, we have tostart the summation at n = 2 in every series

∞n=2

n(n− 1)antn +∞

n=0

(n + 2)(n + 1)an+2tn −∞

n=0

(n + 1)an+1tn +∞

n=0

antn =

∞n=2

n(n− 1)antn + 2a2 + 6a3t +∞

n=2

(n + 2)(n + 1)an+2tn−

a1 − 2a2t −∞

n=2

(n + 1)an+1tn + a0 + a1t +∞

n=2

antn =0.

This can be written in the form

(2a2−a1+a0)+(6a3−2a2+a1)t+

∞n=2

(n + 2)(n + 1)an+2 + (n2

− n + 1)an − (n + 1)an+1 tn

= 0.


2a2 − a1 + a0 =0

6a3 − 2a2 + a1 =0

(n + 1)(n + 2)an+2 + (n2 − n + 1)an − (n + 1)an+1 =0.

Thus,

a2 =a1 − a0

2a3 =

2a2 − a1

6 = −a0

6

an+2 =(n + 1)an+1 − (n2 − n + 1)an

(n + 1)(n + 2) , n ≥ 2.






a2 =a1 − a0

2

a3 = − a0

6

a4 = 1

24(2a0 − 3a1)

a5 = 1

40(3a0 − a1)

a6 = 1

90a0

a7 = 1720

(36a1 − 17a0)

The general solution is:

y(t) = a0

1 − 1

2t2 − 1

6t3 +

1

12t4 +

3

40t5 + · · ·

+a1

t +

1

2t2 − 1

8t4 − 1

40t5 + · · ·

Problem 15.8

Write the following as a single power series:

n=0

nan(t

−2)n+1 +n=2

n2an(t

−2)n.

Solution.

We haven=0

nan(t − 2)n+1 +n=2

n2an(t − 2)n =n=1

(n− 1)an−1(t − 2)n +n=2

n2an(t − 2)n

=n=2

(n− 1)an−1(t − 2)n +n=2

n2an(t − 2)n

=

n=2

[(n− 1)an−1 + n2an](t − 2)n

Problem 15.9


y + ty + y = 0.



125

Solution.

We seek solution of the form

y(t) =∞

n=0

antn.


y(t) = ∞

n=1 nantn−1 and y(t) = ∞

n=2 n(n− 1)antn−2.


∞n=2

n(n− 1)antn−2 + t∞

n=1

nantn−1 +∞

n=0

antn =0

∞n=2

n(n − 1)antn−2 +∞

n=1

nantn +∞

n=0

antn =0

∞n=0

(n + 1)(n + 2)an+2tn +∞

n=1

nantn +∞

n=0

antn =0.

Since we want to express everything in only one summation sign, we have tostart the summation at n = 1 in every series

(2a2 + a0) +∞

n=1

[(n + 2)(n + 1)an+2 + (n + 1)an] tn = 0.


2a2 + a0 =0

(n + 1)(n + 2)an+2 + (n + 1)an =0.

Thus,

a2 = − a0

2

an+2 = − an

n + 2, n ≥ 1.





an+2 in terms of a previous coefficient an. From the above, we can see thata2 = − a0

2

a3 = − a1

3

a4 =a0

8

a5 =a1

15

a6 = − a0

48

a7 =−

a1

105.


y(t) = a0

1 − 1

2t2 +

1

8t4 − 1

48t6 + · · ·

+a1

t − 1

3t3 +

1

15t5 − 1

105t7 + · · ·

Problem 15.10


(1− t)2y − 2y = 0.

Solution.

We seek solution of the form

y(t) =∞

n=0

antn.


y(t) = ∞


n=2 n(n− 1)antn−2.


(1 − 2t + t2)∞

n=2

n(n− 1)antn−2 − 2∞

n=0

antn =0

∞n=2

n(n− 1)antn − 2 ∞n=2

n(n− 1)antn−1 + ∞n=2

n(n− 1)antn−2 − 2 ∞n=0

antn =0

∞n=2

n(n− 1)antn − 2∞

n=1

n(n + 1)an+1tn + sum∞n=0(n + 1)(n + 2)an+2tn − 2

∞n=0

antn =0



127

Since we want to express everything in only one summation sign, we have to

start the summation at n = 2 in every series

(2a2−2a0)+(6a3−2a1−4a2)t+∞

n=2

(n + 2)(n + 1)an+2 − 2n(n + 1)an+1 + (n2 − n− 2)an

tn = 0


2a2 − 2a0 =0

6a3 − 2a1 − 4a2 =0

(n + 2)(n + 1)an+2 − 2n(n + 1)an+1 + (n2 − n− 2)an =0.

Thus,

a2 =a0

a3 =a1 + 2a0

3

an+2 =2nan+1 − (n− 2)an

n + 2 , n ≥ 2.

From the above, we can see that

a2 =a0

a3 = a1 + 2a0

3a4 =a3

a5 =a3

a6 =a3

a7 =a3.

The general solution is:

y(t) = a01 + t2 + 2

3

t3 + 2

3

t4 + 2

3

t5 +

· · ·+a1t + 1

3

t3 + 1

3

t4 + 1

3

t5 +

· · ·






16 The Structure of the

General Solution of Linear

Nonhomogeneous Equations

In Problems 16.1- 16.6, answer the following three questions.(a) Verify that the given function, y p(t), is a particular solution of the differ-ential equation.(b) Determine the general solution,yh, of the homogeneous equation.(c) Find the general solution to the differential equation and impose theinitial conditions to obtain the unique solution of the initial value problem.

Problem 16.1

y − 2y − 3y = e2t, y(0) = 1, y(0) = 0, y p(t) = −13

e2t.

Solution.

(a) y p = −23

e2t, y p = −43

e2t.

y p − 2y p − 3y p = − 4

3e2t +

4

3e2t + e2t

=e2t.

(b) The associated characteristic equation r2− 2r− 3 = 0 has roots r1 = −1and r2 = 3. Hence, the general solution to the homogeneous differentialequation is

yh(t) = c1e−t + c2e3t.

(c) The general solution to the differential equation is y(t) = c1e−t + c2e3t −13e2t. The derivative of this function is given by y (t) = −c1e−t + 3c2e3t− 2

3e2t.

129



13016 THE STRUCTURE OF THE GENERAL SOLUTION OF LINEAR NONHOMO

The condition y(0) = 1 leads to c1 + c2 = 43 . The condition y (0) = 0 leads to

−c1 + 3c2 =

2

3 . Solving for c1 and c2, we find c1 =

5

6 and c2 =

1

2 . The uniquesolution is given by

y(t) = 5

6e−t +

1

2e3t − 1

3e2t

Problem 16.2

y − y − 2y = 10, y(−1) = 0, y(−1) = 1, y p(t) = −5.

Solution.

(a) y p = y p = 0.

y p −

y p −

2y p = 0−

0−

2(−

5) = 10.

(b) The associated characteristic equation r2−r−2 = 0 has roots r1 = −1 andr2 = 2. Hence, the general solution to the homogeneous differential equationis

yh(t) = c1e−t + c2e2t.

(c) The general solution to the differential equation is y(t) = c1e−t + c2e2t −5. The derivative of this function is given by y(t) = −c1e−t + 2c2e2t. Thecondition y(−1) = 0 leads to c1e + c2e−2 = 5. The condition y (−1) = 1 leadsto −c1e + 2c2e−2 = 1. Solving for c1 and c2, we find c1 = 3

e and c2 = 2e2. The

unique solution is given by

y(t) = 3e−(t+1) + 2e2t+2 − 5

Problem 16.3

y + y = 2e−t, y(0) = 2, y(0) = 2, y p(t) = −2te−t.

Solution.

(a) y p = −2e−t + 2te−t, y p = 4e−t − 2te−t.

y p + y p =4e−t − 2te−t − 2e−t + 2te−t

=2e−t.

(b) The associated characteristic equation r2 + r = 0 has roots r1 = 0 andr2 = −1. Hence, the general solution to the homogeneous differential equationis

yh(t) = c1 + c2e−t.



131

(c) The general solution to the differential equation is y(t) = c1+c2e−t−2te−t.

The derivative of this function is given by y (t) = −c2e−t

− 2e−t

+ 2te−t

. Thecondition y(0) = 2 leads to c1 + c2 = 2. The condition y(0) = 2 leads to−c2 − 2 = 2. Solving for c1 and c2, we find c1 = 6 and c2 = −4. The uniquesolution is given by

y(t) = 6 − 4e−t − 2te−t

Problem 16.4

y + 4y = 10et−π, y(π) = 2, y(π) = 0, y p(t) = 2et−π.

Solution.

(a) y p = y p = 2e

t

−π

.

y p + 4y p =2et−π + 8et−π

=10et−π.

(b) The associated characteristic equation r2 +4 = 0 has roots r1 = −2i andr2 = 2i. Hence, the general solution to the homogeneous differential equationis

yh(t) = c1 cos 2t + c2 sin 2t.

(c) The general solution to the differential equation is y(t) = c1 cos 2t +c2 sin 2t+2et−π. The derivative of this function is given by y(t) =

−2c1 sin 2t+

2c2 cos 2t + 2et−π. The condition y(π) = 2 leads to c1 + 2 = 2. The conditiony(π) = 0 leads to 2c2 + 2 = 0. Solving for c1 and c2, we find c1 = 0 andc2 = −1. The unique solution is given by

y(t) = − sin2t + 2et−π

Problem 16.5

y − 2y + 2y = 5 sin t, y(π

2) = 1, y(

π

2) = 0, y p(t) = 2 cos t + sin t.

Solution.

(a) y p = −2sin t + cos t, y p = −2cos t − sin t.

y p − 2y p + 2y p = − 2cos t− sin t + 4 sin t − 2cos t + 4 cos t + 2 sin t

=5sin t.




(b) The associated characteristic equation r2−2r +2 = 0 has roots r1 = 1− i

and r2 = 1 + i. Hence, the general solution to the homogeneous differentialequation isyh(t) = et(c1 cos t + c2 sin t).

(c) The general solution to the differential equation is y(t) = et(c1 cos t +c2 sin t) + 2 cos t + sin t. The derivative of this function is given by y(t) =et cos t(c1 + c2) + et sin t(−c1 + c2) − 2sin t + cos t. The condition y( π

2) = 1

leads to eπ2 c2 + 1 = 1. The condition y( π

2) = 0 leads to −e

π2 c1 − 2 = 0.

Solving for c1 and c2 we find c1 = −2e−π2 and c2 = 0. The unique solution is

given byy(t) = −2et−π

2 cos t + 2 cos t + sin t

Problem 16.6

y−2y + y = t2 +4+2si n t, y(0) = 1, y(0) = 3, y p(t) = t2 + 4t +10+cos t.

Solution.

(a) y p = 2t + 4 − sin t, y p = 2− cos t.

y p − 2y p + y p =2 − cos t − 4t − 8 + 2 sin t + t2 + 4t + 10 + cos t

=t2 + 4 + 2 sin t.

(b) The associated characteristic equation r2 − 2r + 1 = 0 has roots r1 =r2 = 1. Hence, the general solution to the homogeneous differential equation

isyh(t) = c1et + c2tet.

(c) The general solution to the differential equation is y(t) = c1et + c2tet +t2 + 4t + 10 + cos t. The derivative of this function is given by y (t) = c1et +c2et + c2tet + 2t + 4 − sin t. The condition y(0) = 1 leads to c1 + 10 + 1 = 1.The condition y(0) = 3 leads to c1 + c2 + 4 = 3. Solving for c1 and c2, wefind c1 = −10 and c2 = 9. The unique solution is given by

y(t) = −10et + 9tet + t2 + 4t + 10 + cos t

The functions u1, u2, and u3 are solutions to the following differential equa-

tionsu + p(t)u + q (t)u =2et + 1

u + p(t)u + q (t)u =2e−t − t − 1

u + p(t)u + q (t)u =3t.



133

In Problems 16.7 - 16.8, use the functions u1, u2 and u3 to construct a par-

ticular solution of the differential equation.

Problem 16.7

u + p(t)u + q (t)u = et + 2t + 1

2.

Solution.

The right-hand side of the given equation can be written as et + 2t + 12 =

12

(2et +1)+ 23

(3t) so that by Theorem 16.2, the function u(t) = 12

u1(t)+ 23

u3(t)is the required particular solution

Problem 16.8

u + p(t)u + q (t)u = et + e−t

2 .

Solution.

The right-hand side of the given equation can be written as et+e−t

2 = 14(2et +

1) + 14

(2e−t − t − 1) + 112

(3t) so that by Theorem 16.2, the function u(t) =14u1(t) + 1

4u2(t) + 112u3(t) is the required particular solution

In Problems 16.9 - 16.11, determine the function g(t).

Problem 16.9

y − 2y = g(t), y p(t) = 3t +√

t, t > 0.

Solution.

We have y p = 3 + 1

2√

t and y p = − 1

4t32

. Thus,

g(t) =y p − 2y p

=−

1

4t−

32

−6−

t−12

Problem 16.10

y + y = g(t), y p(t) = ln (1 + t), t > −1.




Solution.

We have y p =

1

1+t and y p = −(1 + t)−2

. Thus,g(t) =y p + y p

= − 1

(1 + t)2 +

1

1 + t

Problem 16.11

y + 2y + y = g(t), y p(t) = t − 2.

Solution.

We have y p = 1 and y p = 0. Thus,

g(t) =y p + 2y p + y p

=0 + 2 + t − 2

=t

In Problems 16.12 - 16.13, the general solution of the nonhomogeneous dif-ferential equation y + αy + βy = g(t) is given, where c1 and c2 are arbitraryconstants. Determine the constants α and β and the function g(t).

Problem 16.12

y(t) = c1et + c2tet + t2et.

Solution.

From the given general solution we see that the roots of the characteristicequation are r1 = r2 = 1. Thus, the characteristic equation is (r−1)(r−1) =r2−2r +1 = 0. The associated differential equation is y−2y+y = 0. Hence,α = −2 and β = 1. Now,

g(t) =y p − 2y p + y p

=2et + 4tet + t2et − 4tet − 2t2et + t2et

=2et

Problem 16.13

y(t) = c1 sin 2t + c2 cos 2t− 1 + sin t.



135

Solution.

From the given general solution we see that the roots of the characteristicequation are r1 = −2i and r2 = 2i. Thus, the characteristic equation is(r−2i)(r+2i) = r2+4 = 0. The associated differential equation is y+4y = 0.Hence, α = 0 and β = 4. Now,

g(t) =y p + 4y p

=− sin t + 4 sin t− 4

=3sin t − 4

Problem 16.14

Given that the function et

5 satisfies the differential equation y + 4y = et,

write a general solution of the differential equation y + 4y = et.

Solution.

First, we find yh. The characteristic equation r2 + 4 = 0 has the rootsr1,2 = ±2i. Thus, yh(t) = c1 cos 2t + c2 sin 2t. The general solution to thenonhomogeneous equation is

y(t) = c1 cos 2t + c2 sin 2t + et

5

Problem 16.15

Find the general solution to the differential equation

y(4) + 9y = 24 + 108t2

given a particular solution y p(t) = cos 3t + sin 3t + t4.

Solution.

Let z = y. Then the given equation reduces to a second order differentialequation

z + 9z = 24 + 108t2.

The characteristic equation is r2 + 9 = 0 so that the roots are r1,2 = ±3i.Thus, z h(t) = c1 cos 3t + c2 sin 3t. Integrating this function twice we find

yh(t) = c1 cos 3t + c2 sin 3t + c3t + c4. Hence, the general solution to the givendifferential equation is

y(t) = c1 cos 3t + c2 sin 3t + c3t + c4 + t4






17 The Method of Variation of

Parameters

Problem 17.1

Solve y + y = sec t by variation of parameters.

Solution.

The characteristic equation r2 + 1 = 0 has roots r = ±i and

yh(t) = c1 cos t + c2 sin t.

Also, y1(t) = cos t and y2(t) = sin t so that W (t) = cos2 t + sin2 t = 1. Now,

u1

=− sin t sec tdt = d(cos t)

cos t = ln

|cos t

|and

u2 =

cos t sec tdt =

dt = t.

Hence, the particular solution is given by

y p(t) = ln | cos t| cos t + t sin t

and the general solution is

y(t) = c1 cos t + c2 sin t + ln | cos t| cos t + t sin t

Problem 17.2

Solve y − y = et by variation of parameters.

137



138 17 THE METHOD OF VARIATION OF PARAMETERS

Solution.


2

− 1 = 0 for y − y = 0 has roots r = ±1. Thehomogeneous solution is

yh(t) = c1et + c2e−t.

The homogeneous solution yh(t) = c1et + c2e−t found above implies y1 =et, y2 = e−t is a suitable independent pair of solutions. Their Wronskian isW = −2. The variation of parameters formula applies:

y p(t) = et

e−t

2 etdt− e−t

et

2 etdt.

Integration, followed by setting all constants of integration to zero, givesy p(t) = tet

2 − et

4 .The general solution is

y(t) = c1et + c2e−t + 1

2tet

Problem 17.3

Solve the following 2nd order equation using the variation of parametersmethod:

y + 4y = t2 + 8cos 2t.

Solution.The characterisitc equation r2 + 4 = 0 has roots r = ±2i so that yh(t) =c1 cos 2t + c2 sin 2t. Hence, y1(t) = cos 2t, y2(t) = sin 2t, and W (t) = 2. Thus,

y p = − cos2t

sin2t(t2 + 8cos 2t)

2 dt + sin 2t

cos2t(t2 + 8cos 2t)

2 dt

= − cos2t(1

4t sin2t +

1

8 cos 2t − 1

4t2 cos2t − cos2 2t)

+sin2t(1

4t cos2t − 1

8 sin 2t +

1

4t2 sin2t + 2t +

1

2 sin 4t)

=

− 1

8

+ 1

4

t2 + cos2 2t cos2t + 2t sin2t + 1

2

sin 4t sin2t.

The general solution is

y(t) = c1 cos 2t + c2 sin 2t − 1

8 +

1

4t2 + 2t sin2t



139

Problem 17.4

Find a particular solution by the variation of parameters to the equationy + 2y + y = e−t ln t.

Solution.

The characteristic equation

r2 + 2r + 1 = 0

has roots r1 = r2 = −1, so the fundamental solutions of the reduced equationare

y1(t) = e−t, y2(t) = te−t.

Compute the Wronskian.

W (t) = e−t te−t

−e−t e−t − te−t

=e−t(e−t − te−t) + e−t · te−t

=e−2t − te−2t + te−2t

=e−2t.

Compute u1(t).

u1(t) =−

y2(t)g(t)

W (t) dt

=− te−t

·e−t ln t

e−2t dt

=−

t ln tdt = −t2

2 ln t +

t2

2 · 1

tdt

=− t2

2 ln t +

t2

4 .

Compute u2(t).

u2(t) =

y1(t)g(t)

W (t) dt

= e−t

·e−t ln t

e−2t dt

=

ln tdt = t ln t−

t · 1

tdt

=t ln t − t.




Note. We used integration by parts to compute the integrals

t ln tdt and

ln tdt.The particular solution to our complete equation is

y p(t) =u1(t)y1(t) + u2(t)y2(t)

=

−t2

2 ln t +

t2

4

e−t + (t ln t − t)te−t

=t2

2 ln te−t − 3t2

4 e−t

=(t2

2 ln t − 3t2

4 )e−t

Problem 17.5

Solve the following initial value problem by using variation of parameters:

y + 2y − 3y = tet, y(0) = − 1

64, y(0) =

59

64.

Solution.

From the characteristic equation, we obtain y1(t) = et, y2(t) = e−3t andW (t) = −4e−2t. Integration then yields

u1(t) = −

e−3ttet

−4e−2tdt =

t2

8

u2(t) = ettet

−4e−2t dt = − 1

16te4t

+

e4t

64 .

Thus. y p(t) = et

64(8t2 − 4t + 1) and the general solution is

y(t) = c1et + c2e−3t + t2

8 et − 1

16tet.

Initial conditions:

y(0) =c1 + c2 = − 1

64

y(0) =c1 − 3c2 − 4

64 =

59

64.

These are satisfied by c1 = 1564 and c2 = − 1

4 . Finally the solution to the initialvalue problem is

y = et

64(8t2 − 4t + 15) − 1

4e−3t



141

Problem 17.6

(a) Verify that {e

√ t

, e−

√ t

} is a fundamental set for the equation4ty + 2y − y = 0

on the interval (0,∞). You may assume that the given functions are solutionsto the equation.(b) Use the method of variation of parameters to find one solution to theequation

4ty + 2y − y = 4√

te√

t.

Solution.

(a) Usually the first thing to do would be to check that y1(t) = e√

t and

y2(t) = e−√ t really are solutions to the equation. However, the question saysthat this can be assumed and so we move on to the next step, which is tocheck that the Wronskian of the two solutions is non-zero on (0 ,∞). We have

y1 = 12√

te√

t and y 2 = − 12√

te−

√ t

and so

W (t) = y1y2 − y1y2 = − 1

2√

t− 1

2√

t= − 1√

t.

This is indeed non-zero and so {e√

t, e−√

t} is a fundamental set for the ho-mogeneous equation.

(b) The variation of parameters formula says that

y = −y1

y2g

W (t)dt + y2

y1g

W (t)dt

is a solution to the nonhomogeneous equation in the form y + py + qy = g.To get the right g, we have to divide the equation through by 4t and sog = 1√

te√

t. Thus

y = − e√

t

e−

√ t( 1√

t)e√

t

−1/√

tdt + e−

√ t

e

√ t( 1√

t)e√

t

−1/√

tdt

=e√

t

dt − e−√

t

e2√

tdt

=te√

t − e−√

t

e2

√ tdt.




To evaluate the integral, we substitute u = 2√

t so that dt = 12udu. We get

e2

√ tdt =

12

ueudu =

12

(u − 1)eu = (√

t − 1/2)e2√

t.

Thusy = (t −

√ t + 1/2)e

√ t

is one solution to the equation. You might notice that the 1/2 can be dropped

(because (1/2)e√

t is a solution to the homogeneous equation) so that

y = (t−√

t)e√

t

would also work

Problem 17.7

Use the method of variation of parameters to find the general solution to theequation

y + y = sin t.

Solution.

The characteristic equation r2 + 1 = 0 has roots r = ±i so that the solutionto the homogeneous equation is yh(t) = c1 cos t + c2 sin t. The Wronskian

is W (cos t, sin t) = 1. Now u1(t) = − sin2 t = cos(2t)−12 . Hence, u1(t) =

12(

12 sin(2t) − t). Similarly, u2(t) = sin t cos t. Hence, u2(t) =

12 sin

2

t. Soy p(t) = −1

2t cos t + 12 sin t. The general solution is given by

y(t) = c1 cos t + c2 sin t− 1

2t cos t

Problem 17.8

Consider the differential equation

t2y + 3ty − 3y = 0, t > 0.

(a) Determine r so that y = tr is a solution.

(b) Use (a) to find a fundamental set of solutions.(c) Use the method of variation of parameters for finding a particular solutionto

t2y + 3ty − 3y = 1

t3, t > 0.



143

Solution.

(a) Inserting y, y, and y into the equation we find r

2

+ 2r − 3 = 0. Solvingfor r to obtain r1 = 1 and r2 = −3.(b) Let y1(t) = t and y2(t) = t−3. Since

W (t) =

t t−3

1 −3t−4

= −4t−3

{y1, y2} is a fundamental set of solutions for t > 0.(c) Recall that the variation of parameters formula states that if y1 and y2form a fundamental solution set for y + p(t)y + q (t)y = 0, then y p(t) =u1(t)y1(t) + u2(t)y2(t) is a particular solution to the equation y + p(t)y +q (t)y = g(t), where

u1(t) = − t−3t−5

−4t−3dt = − 1

16t−4

u2(t) =

t · t−5

−4t−3dt = −1

4 ln t.

Thus,

y p(t) = − 1

16t−3 − 1

4t−3 ln t

Problem 17.9

Use the method of variation of parameters to find the general solution to thedifferential equations

y + y = sin2 t.

Solution.

The characterisitc equation r2 + 1 = 0 has roots r = ±i so that y1(t) =cos t, y2(t) = sin t, and W (t) = 1. Hence,

u1(t) =−

sin t sin2 tdt =

(1− cos2 t)d(cos t) = cos t − 1

3 cos3 t

u2(t) =

cos t sin2 tdt =

1

3 sin3 t.

Thus,

y p(t) = cos2 t − 13

cos4 t + 13

sin4 t

and

y(t) = c1 cos t + c2 sin t + cos2 t− 1

3 cos4 t +

1

3 sin4 t






18 The Laplace Transform:

Basic Definitions and Results

Problem 18.1

Determine whether the integral ∞0

11+t2

dt converges. If the integral con-verges, give its value.

Solution.

We have ∞0

1

1 + t2dt = lim

A→∞

A

0

1

1 + t2dt = lim

A→∞[arctan t]A

0

= limA→∞

arctan A = π

2.

So the integral is convergent

Problem 18.2

Determine whether the integral ∞0

t1+t2 dt converges. If the integral con-

verges, give its value.

Solution.

We have

∞

0

t

1 + t2dt =

1

2 limA→∞

A

0

2t

1 + t2dt =

1

2 limA→∞ ln (1 + t2)A

0

=1

2 limA→∞

ln (1 + A2) = ∞.

Hence, the integral is divergent

145



14618 THE LAPLACE TRANSFORM: BASIC DEFINITIONS AND RESULTS

Problem 18.3

Determine whether the integral ∞0 e−t

cos(e−t

)dt converges. If the integralconverges, give its value.

Solution.

Using the substitution u = e−t we find

∞0

e−t cos(e−t)dt = limA→∞

e−A

1

− cos udu

= limA→∞

[− sin u]e−A

1 = limA→∞

[sin1− sin(e−A)]

=sin1.

Hence, the integral is convergent

Problem 18.4

Using the definition, find L[e3t], if it exists. If the Laplace transform existsthen find the domain of F (s).

Solution.

We have

L[e3t] = limA→∞

A

0

e3te−stdt = limA→∞

A

0

et(3−s)dt

= limA→∞

et(3−s)

3 − s

A

0

= limA→∞

eA(3−s)

3 − s − 1

3 − s

=

1

s−

3, s > 3

Problem 18.5

Using the definition, find L[t− 5], if it exists. If the Laplace transform existsthen find the domain of F (s).



147

Solution.

Using integration by parts we find

L[t− 5] = limA→∞

A

0

(t − 5)e−stdt = limA→∞

−(t − 5)e−st

s

A

0

+ 1

s

A

0

e−stdt

= limA→∞

−(A− 5)e−sA + 5

s −

e−st

s2

A

0

= 1

s2 − 5

s, s > 0

Problem 18.6

Using the definition, find L[e(t−1)2 ], if it exists. If the Laplace transform

exists then find the domain of F (s).Solution.

We have ∞0

e(t−1)2e−stdt =

∞0

e(t−1)2−stdt.

Since limt→∞(t − 1)2 − st = limt→∞ t2

1− (2+s)t + 1

t2

= ∞, for any fixed s

we can choose a positive C such that (t− 1)2− st ≥ 0 for t ≥ C. In this case,e(t−1)2−st ≥ 1 and this implies that

∞0 e(t−1)2−stdt ≥ ∞

C dt. The integral onthe right is divergent so that the integral on the left is also divergent by thecomparison theorem of improper integrals. Hence, f (t) = e(t−1)2 does not

have a Laplace transform

Problem 18.7

Using the definition, find L[(t − 2)2], if it exists. If the Laplace transformexists then find the domain of F (s).

Solution.

We have

L[(t − 2)2] = limT →∞

T

0

(t − 2)2e−stdt.

Using integration by parts with u = e−st and v = (t − 2)2 we find

T

0

(t − 2)2e−stdt = − (t − 2)2e−st

s

T

0

+ 2s

T

0

(t − 2)e−stdt

=4

s − (T − 2)2e−sT

s +

2

s

T

0

(t− 2)e−stdt.




Thus,

limT →∞ T

0(t − 2)2e−stdt =

4

s + 2

s limT →∞ T

0(t − 2)e−stdt.

Using by parts with u = e−st and v = t − 2 we find T

0

(t − 2)e−stdt =

−(t − 2)e−st

s − 1

s2e−st

T

0

.

Letting T → ∞ in the above expression we find

limT →∞

T

0

(t − 2)e−stdt = −2

s +

1

s2, s > 0.

Hence,

F (s) = 4

s +

2

s

−2

s +

1

s2

=

4

s − 4

s2 +

2

s3, s > 0

Problem 18.8

Using the definition, find L[f (t)], if it exists. If the Laplace transform existsthen find the domain of F (s).

f (t) =

0, 0 ≤ t < 1t− 1, t ≥ 1.

Solution.We have

L[f (t)] = limT →∞

T

1

(t − 1)e−stdt.

Using integration by parts with u = e−st and v = t − 1 we find

limT →∞

T

1

(t − 1)e−stdt = limT →∞

−(t − 1)e−st

s − 1

s2e−st

T

1

= e−s

s2 , s > 0

Problem 18.9

Using the definition, find L[f (t)], if it exists. If the Laplace transform exists

then find the domain of F (s).

f (t) =

0, 0 ≤ t < 1t− 1, 1 ≤ t < 2

0, t ≥ 2.



149

Solution.

We have

L[f (t)] =

21

(t−1)e−stdt =

−(t− 1)e−st

s − 1

s2e−st

21

= −e−2s

s +

1

s2(e−s−e−2s), s = 0

Problem 18.10

Let n be a positive integer. Using integration by parts establish the reductionformula

tne−stdt = − tne−st

s +

n

s

tn−1e−stdt, s > 0.

Solution.

Let u = e−st and v = tn. Then u = −e−sts and v = ntn−1. Hence,

tne−stdt = −tne−st

s +

n

s

tn−1e−stdt, s > 0

Problem 18.11

For s > 0 and n a positive integer evaluate the limits

(a) limt→0 tne−st (b) limt→∞ tne−st.

Solution.

(a) limt→0 tne−st = limt→0tn

est = 01 = 0.

(b) Using L’Hopital’s rule repeatedly we find

limt→∞

tne−st = · · · = limt→∞

n!

snest = 0

Problem 18.12

(a) Use the previous two problems to derive the reduction formula for theLaplace transform of f (t) = tn,

L[t

n

] =

n

sL[t

n

−1

], s > 0.

(b) Calculate L[tk], for k = 1, 2, 3, 4, 5.(c) Formulate a conjecture as to the Laplace transform of f (t), tn with n apositive integer.




Solution.

(a) Using the two previous problems we find

L[tn] = limT →∞

T

0

tne−stdt = limT →∞

−

tne−st

s

T

0

+ n

s

T

0

tn−1e−stdt

=n

s limT →∞

T

0

tn−1e−stdt = n

sL[tn−1], s > 0.

(b) We have

L[t] = 1

s2

L[t2

] =

2

sL[t] =

2

s3

L[t3] =3

sL[t2] =

6

s4

L[t4] =4

sL[t3] =

24

s5

L[t5] =5

sL[t4] =

120

s5 .

(c) By induction, one can easily show that for n = 0, 1, 2, · · ·

L[tn] = n!

sn+1, s > 0

From a table of integrals, eαu sin βudu =eαu α sin βu − β cos βu

α2 + β 2 eαu cos βudu =eαu α cos βu + β sin βu

α2 + β 2

Problem 18.13

Use the above integrals to find the Laplace transform of f (t) = cos ωt, if itexists. If the Laplace transform exists, give the domain of F (s).

Solution.We have

L[cos ωt] = limT →∞

−

e−st

−s cos ωt + ω sin ωt

s2 + ω2

T

0

=

s

s2 + ω2, s > 0



151

Problem 18.14

Use the above integrals to find the Laplace transform of f (t) = sin ωt, if itexists. If the Laplace transform exists, give the domain of F (s).

Solution.

We have

L[sin ωt] = limT →∞

−

e−st

−s sin ωt + ω cos ωt

s2 + ω2

T

0

=

ω

s2 + ω2, s > 0

Problem 18.15

Use the above integrals to find the Laplace transform of f (t) = cos ω(t− 2),if it exists. If the Laplace transform exists, give the domain of F (s).

Solution.

Using a trigonometric identity we can write f (t) = cos ω(t − 2) = cos ωt cos2ω+sin ωt sin2ω. Thus, using the previous two problems we find

L[cos ω(t − 2)] = s cos2ω + ω sin2ω

s2 + ω2 , s > 0

Problem 18.16

Use the above integrals to find the Laplace transform of f (t) = e3t sin t, if itexists. If the Laplace transform exists, give the domain of F (s).

Solution.

We have

L[e3t sin t] = limT →∞

T

0

e−(s−3)t sin tdt

= limT →∞

−

e−(s−3)t

(s− 3) sin t + cos t

(s− 3)2 + 1

T

0

= 1

(s− 3)2 + 1, s > 3

Problem 18.17Consider the function f (t) = tan t.(a) Is f (t) continuous on 0 ≤ t < ∞, discontinuous but piecewise continuouson 0 ≤ t < ∞, or neither?(b) Are there fixed numbers a and M such that |f (t)| ≤ Meat for 0 ≤ t < ∞?




Solution.

(a) Since f (t) = tan t =

sin t

cos t and this function is discontinuous at t = (2n +1) π2 . Since this function has vertical asymptotes there it is not piecewise

continuous.(b) The graph of the function does not show that it can be bounded byexponential functions. Hence, no such numbers a and M

Problem 18.18

Consider the function f (t) = et2

e2t+1 .(a) Is f (t) continuous on 0 ≤ t < ∞, discontinuous but piecewise continuouson 0 ≤ t < ∞, or neither?(b) Are there fixed numbers a and M such that |f (t)| ≤ Meat for 0 ≤ t < ∞?

Solution.

(a) Since et2 and e2t +1 are continuous everywhere, f (t) = et2

e2t+1 is continuous

on 0 ≤ t < ∞.(b) Since e2t + 1 ≤ e2t + e2t = 2e2t, f (t) ≥ 1

2et2e−2t = 1

2et2−2t. But for t ≥ 4

we have t2−2t > t2

2 . Hence, f (t) > 1

2et2

2 . So f (t) is not of exponential order

Problem 18.19

Find L−1 3s−2

.

Solution.

Since L 1s−a

= 1s−a , s > a we find

L−1

3

s− 2

= 3L−1

1

s− 2

= 3e2t, t ≥ 0

Problem 18.20

Find L−1− 2

s2 + 1s+1

.

Solution.

Since L[t] = 1s2

, s > 0 and L 1s−a

= 1s−a

, s > a we find

L−1− 2

s2 +

1s + 1

= − 2L−1

1s2

+ L−1

1s + 1

= − 2t + e−t, t ≥ 0



19 Further Studies of Laplace

Transform

Problem 19.1

Use Table L to find L[2et + 5].

Solution.

L[2et + 5] = 2L[et] + 5L[1] = 2

s− 1 +

5

s, s > 1

Problem 19.2

Use Table L to find L[e3t−3h(t − 1)].

Solution.

L[e3t−3h(t − 1)] = L[e3(t−1)h(t − 1)] = e−sL[e3t] = e−s

s− 3 , s > 3

Problem 19.3

Use Table L to find L[sin2 ωt].

Solution.

L[sin2 ωt] = L[1− cos2ωt

2 ] =

1

2(L[1]−L[cos 2ωt]) =

1

2

1

s − s

s2 + 4ω2

, s > 0

Problem 19.4

Use Table

L to find

L[sin3t cos3t].

Solution.

L[sin3t cos3t] = L

sin6t

2

=

1

2L[sin6t] =

3

s2 + 36, s > 0

153



154 19 FURTHER STUDIES OF LAPLACE TRANSFORM

Problem 19.5

Use Table L to find L[e

2t

cos3t].Solution.

L[e2t cos3t] = s − 2

(s − 2)2 + 9, s > 2

Problem 19.6

Use Table L to find L[e4t(t2 + 3t + 5)].

Solution.

L[e4t(t2+3t+5)] = L[e4tt2]+3L[e4tt]+5L[e4t] = 2(s − 4)3 + 3(s− 4)2+ 5s− 4 , s > 4

Problem 19.7

Use Table L to find L−1[ 10s2+25

+ 4s−3

].

Solution.

L−1[ 10

s2 + 25+

4

s− 3] = 2L−1[

5

s2 + 25]+4L−1[

1

s− 3] = 2sin 5t+4e3t, t ≥ 0

Problem 19.8

Use Table L to find L−1

[ 5(s−3)4 ].

Solution.

L−1[ 5

(s − 3)4] =

5

6L−1[

3!

(s − 3)4] =

5

6e3tt3, t ≥ 0

Problem 19.9

Use Table L to find L−1[ e−2s

s−9 ].

Solution.

L−1[ e−2s

s− 9] = e9(t−2)h(t − 2) =

0, 0 ≤ t < 2e9(t−2), t ≥ 2

Problem 19.10

Use Table L to find L−1[ e−3s(2s+7)s2+16 ].



155

Solution.

We have

L−1[e−3s(2s + 7)

s2 + 16 ] =2L−1[

e−3ss

s2 + 16] +

7

4L−1[

e−3s4

s2 + 16]

=2cos4(t − 3)h(t − 3) + 7

4 sin 4(t − 3)h(t − 3), t ≥ 0

Problem 19.11

Graph the function f (t) = h(t − 1) + h(t − 3) for t ≥ 0, where h(t) is theHeaviside step function, and use Table L to find L[f (t)].

Solution.Note that

f (t) =

0, 0 ≤ t < 11, 1 ≤ t < 32, t ≥ 3.

The graph of f (t) is shown below. Using Table L we find

L[f (t)] = L[h(t − 1)] + L[h(t− 3)] = e−s

s +

e−3s

s , s > 0

Problem 19.12

Graph the function f (t) = t[h(t − 1) − h(t − 3)] for t ≥ 0, where h(t) is theHeaviside step function, and use Table L to find L[f (t)].

Solution.Note that

f (t) =

0, 0 ≤ t < 1t, 1 ≤ t < 30, t ≥ 3.





L[f (t)] =L[(t − 1)h(t − 1) + h(t − 1) − (t− 3)h(t − 3) − 3h(t − 3)]

=L[(t − 1)h(t − 1)] + L[h(t− 1)] − L[(t − 3)h(t− 3)]− 3L[h(t − 3)]

=e−s

s2 +

e−s

s − e−3s

s2 − 3e−3s

s , s > 1

Problem 19.13

Graph the function f (t) = 3[h(t − 1) − h(t − 4)] for t ≥ 0, where h(t) is theHeaviside step function, and use Table L to find L[f (t)].

Solution.

Note that

f (t) =

0, 0 ≤ t < 13, 1 ≤ t < 40, t

≥4.


L[f (t)] = 3L[h(t − 1)]− 3L[h(t − 4)] = 3e−s

s − 3e−4s

s , s > 0

Problem 19.14

Graph the function f (t) = |2− t|[h(t− 1)− h(t− 3)] for t ≥ 0, where h(t) isthe Heaviside step function, and use Table L to find L[f (t)].



157

Solution.

Note that

f (t) =

0, 0 ≤ t < 1|2 − t|, 1 ≤ t < 3

0, t ≥ 3.


L[f (t)] =(2 − t)h(t− 1) + 2(t − 2)h(t− 2) − (t − 2)h(t − 3)

=L[−(t − 1)h(t − 1) + h(t − 1) + 2(t − 2)h(t − 2) − (t− 3)h(t − 3) − h(t − 3)]

= − L[(t − 1)h(t − 1)] + L[h(t− 1)] + 2L[(t − 2)h(t − 2)]

−L[(t− 3)h(t − 3)]− L[h(t − 3)]= − e−s

s2 +

e−s

s +

2e−2s

s2 − e−3s

s2 − e−3s

s , s > 0

Problem 19.15

Graph the function f (t) = h(2− t) for t ≥ 0, where h(t) is the Heaviside stepfunction, and use Table L to find L[f (t)].

Solution.

Note that

f (t) =

1, 0 ≤ t ≤ 20, t > 2.

The graph of f (t) is shown below. From this graph we see that f (t) =h(t) − h(t− 2). Using Table L we find

L[f (t)] = L[h(t)]− L[h(t − 2)] = 1 − e−2s

s , s > 0




Problem 19.16

Graph the function f (t) = h(t − 1) + h(4 − t) for t ≥ 0, where h(t) is theHeaviside step function, and use Table L to find L[f (t)].

Solution.

Note that

f (t) =

1, 0 ≤ t < 12, 1 ≤ t ≤ 41, t > 4.


L[f (t)] = L[h(t−1)]+L[h(4−t)] = e−s

s +

40

e−stdt = 1 + e−s − e−4s

s , s > 0

Problem 19.17

The graph of f (t) is given below. Represent f (t) as a combination of Heav-iside step functions, and use Table L to calculate the Laplace transform of f (t).



159

Solution.

From the graph we see thatf (t) = (t − 2)h(t − 2) − (t − 3)h(t− 3) − h(t − 4).

Thus,

L[f (t)] = L[(t−2)h(t−2)]−L[(t−3)h(t−3)]−L[h(t−4)] = e−2s − e−3s

s2 −e−4s

s , s > 0

Problem 19.18

The graph of f (t) is given below. Represent f (t) as a combination of Heav-iside step functions, and use Table L to calculate the Laplace transform of

f (t).

Solution.

From the graph we see that

f (t) = h(t − 1) + h(t − 2) − 2h(t− 3).

Thus,

L[f (t)] = L[h(t−1)]−2L[h(t−3)]+L[h(t−2)] = e−s − 2e−3s + e−2s

s , s > 0

Problem 19.19

Use Laplace transform technique to solve the initial value problem

y + 4y = g(t), y(0) = 2

where

g(t) =

0, 0 ≤ t < 112, 1 ≤ t < 30, t ≥ 3




Solution.

Note first that g(t) = 12[h(t− 1) − h(t − 3)] so that

L[g(t)] = 12L[h(t− 1)]− 12L[h(t − 3)] = 12(e−s − e−3s)

s , s > 0.

Now taking the Laplace transform of the DE and using linearity we find

L[y] + 4L[y] = L[g(t)].

But L[y] = sL[y]− y(0) = sL[y] − 2. Letting L[y] = Y (s) we obtain

sY (s)

−2 + 4Y (s) = 12

e−s − e−3s

s

.

Solving for Y (s) we find

Y (s) = 2

s + 4 + 12

e−s − e−3s

s(s + 4) .

But

L−1

2

s + 4

= 2e−4t

and

L−1 12e−s − e−3s

s(s + 4)

=3L−1 (e−s − e−3s)

1s − 1

s + 4

=3L−1

e−s

s

− 3L−1

e−3s

s

− 3L−1

e−s

s + 4

+ 3L−1

e−3s

s + 4

=3h(t − 1) − 3h(t − 3) − 3e−4(t−1)h(t − 1) + 3e−4(t−3)h(t − 3).

Hence,

y(t) = 2e−4t+3[h(t−1)−h(t−3)]−3[e−4(t−1)h(t−1)−e−4(t−3)h(t−3)], t ≥ 0



20 The Laplace Transform and

the Method of Partial Fractions

In Problems 20.1 - 20.4, give the form of the partial fraction expansion forF (s). You need not evaluate the constants in the expansion. However, if thedenominator has an irreducible quadratic expression then use the completingthe square process to write it as the sum/difference of two squares.

Problem 20.1

F (s) = s3 + 3s + 1

(s − 1)3(s− 2)2.

Solution.

F (s) = A1

(s− 1)3 +

A2

(s − 1)2 +

A3

s− 1 +

B1

(s− 2)2 +

B2

s− 2

Problem 20.2

F (s) = s2 + 5s− 3

(s2 + 16)(s− 2).

Solution.

F (s) = A1s + A2

s2 + 16 +

B1

s−

2

Problem 20.3

F (s) = s3 − 1

(s2 + 1)2(s + 4)2.

161



16220 THE LAPLACE TRANSFORM AND THE METHOD OF PARTIAL FRACTIO

Solution.

F (s) = A1s + A2

(s2 + 1)2 + A3s + A4

s2 + 1 + B1

(s + 4)2 + B2

s + 4

Problem 20.4

F (s) = s4 + 5s2 + 2s− 9

(s2 + 8s + 17)(s− 2)2.

Solution.

F (s) = A1

(s− 2)2 +

A2

s− 2 +

B1s + B2

s2 + 8s + 17

Problem 20.5

Find L−1

1(s+1)3

.

Solution.

Using Table L we find L−1

1(s+1)3

= 1

2e−tt2, t ≥ 0

Problem 20.6

Find L−1

2s−3

s2−3s+2 .

Solution.

We factor the denominator and split the rational function into partial frac-tions:

2s− 3

(s− 1)(s− 2) =

A

s− 1 +

B

s− 2.

Multiplying both sides by (s− 1)(s− 2) and simplifying to obtain

2s− 3 = A(s− 2) + B(s − 1)

= (A + B)s

−2A

−B.

Equating coefficients of like powers of s we obtain the system A + B = 2−2A − B = −3.



163

Solving this system by elimination we find A = 1 and B = 1. Now finding

the inverse Laplace transform to obtain

L−1

2s− 3

(s− 1)(s− 2)

= L−1

1

s− 1

+ L−1

1

s− 2

= et + e2t, t ≥ 0

Problem 20.7

Find L−14s2+s+1

s3+s

.

Solution.


4s2 + s + 1s(s2 + 1)

= As

+ B s + C s2 + 1

.

Multiplying both sides by s(s2 + 1) and simplifying to obtain

4s2 + s + 1 = A(s2 + 1) + (Bs + C )s

= (A + B)s2 + Cs + A.

Equating coefficients of like powers of s we obtain A + B = 4, C = 1, A = 1.Thus, B = 3. Now finding the inverse Laplace transform to obtain

L−1 4s2 + s + 1

s(s2 + 1) = L−1 1

s+3L−1 s

s2 + 1+L−1 1

s2 + 1 = 1+3 cos t+sin t, t ≥ 0.

Problem 20.8

Find L−1

s2+6s+8s4+8s2+16

.

Solution.


s2 + 6s + 8

(s2 + 4)2 =

B1s + C 1s2 + 4

+ B2s + C 2(s2 + 4)2

.

Multiplying both sides by (s2 + 4)2 and simplifying to obtain

s2 + 6s + 8 = (B1s + C 1)(s2 + 4) + B2s + C 2

= B1s3 + C 1s2 + (4B1 + B2)s + 4C 1 + C 2.




Equating coefficients of like powers of s we obtain B1 = 0, C 1 = 1, B2 = 6,

and C 2 = 4. Now finding the inverse Laplace transform to obtain

L−1

s2 + 6s + 8

(s2 + 4)2

= L−1

1

s2 + 4

+ 6L−1

s

(s2 + 4)2

+ 4L−1

1

(s2 + 4)2

= 1

2 sin 2t + 6

t

4 sin 2t

+ 4

1

16[sin2t − 2t cos2t]

=

3

2t sin2t +

3

4 sin 2t − 1

2t cos2t, t ≥ 0

Problem 20.9

Use Laplace transform to solve the initial value problem

y + 2y = 26sin 3t, y(0) = 3.

Solution.

Taking the Laplace of both sides to obtain

L[y] + 2L[y] = 26L[sin3t].

Using Table L the last equation reduces to

sY (s)−

y(0) + 2Y (s) = 26 3

s2 + 9 .

Solving this equation for Y (s) we find

Y (s) = 3

s + 2 +

78

(s + 2)(s2 + 9).

Using the partial fraction decomposition we can write

1

(s + 2)(s2 + 9) =

A

s + 2 +

B s + C

s2 + 9 .

Multipliying both sides by (s + 2)(s2 + 9) to obtain

1 = A(s2 + 9) + (Bs + C )(s + 2)

= (A + B)s2 + (2B + C )s + 9A + 2C.



165

Equating coefficients of like powers of s we find A + B = 0, 2B + C = 0, and

9A + 2C = 1. Solving this system we find A =

1

13 , B = −1

13 , and C =

2

13 .Thus,

Y (s) = 9

s + 2 − 6

s

s2 + 9

+ 4

3

s2 + 9

.

Finally,

y(t) = L−1[Y (s)] = 9L−1

1

s + 2

− 6L−1

s

s2 + 9

+ 4L−1

3

s2 + 9

= 9e−2t − 6cos3t + 4 sin 3t, t ≥ 0

Problem 20.10


y + 2y = 4t, y(0) = 3.

Solution.


L[y] + 2L[y] = 4L[t].


sY (s) − y(0) + 2Y (s) = 4

s2 .


Y (s) = 3

s + 2 +

4

(s + 2)s2.


1

(s + 2)s2 =

A

s + 2 +

B s + C

s2 .

Multipliying both sides by (s + 2)s2 to obtain

1 = As2 + (Bs + C )(s + 2)

= (A + B)s2 + (2B + C )s + 2C.




Equating coefficients of like powers of s we find A + B = 0, 2B + C = 0, and

2C = 1. Solving this system we find A =

1

4 , B = −1

4 , and C =

1

2 . Thus,

Y (s) = 4

s + 2 − 1

s + 2

1

s2.

Finally,

y(t) = L−1[Y (s)] = 4L−1

1

s + 2

− L−1

1

s

+ 2L−1

1

s2

= 4e−2t − 1 + 2t, t ≥ 0

Problem 20.11


y + 3y + 2y = 6e−t, y(0) = 1, y(0) = 2.

Solution.


L[y] + 3L[y] + 2L[y] = 6L[e−t].


s2

Y (s) − sy(0) − y(0) + 3(sY (s) − y(0)) + 2Y (s) = 6

s + 1 .


Y (s) = s + 5

(s + 1)(s + 2) +

6

(s + 2)(s + 1)2 =

s2 + 6s + 11

(s + 1)2(s + 2).


s2 + 6s + 11

(s + 2)(s + 1)2 =

A

s + 2 +

B

s + 1 +

C

(s + 1)2.

Multipliying both sides by (s + 2)(s + 1)2 to obtain

s2 + 6s + 11 = A(s + 1)2 + B(s + 1)(s + 2) + C (s + 2)

= (A + B)s2 + (2A + 3B + C )s + A + 2B + 2C.



167

Equating coefficients of like powers of s we find A + B = 1, 2A+ 3B + C = 6,

and A + 2B + 2C = 11. Solving this system we find A = 3, B = −2, andC = 6. Thus,

Y (s) = 3

s + 2 − 2

s + 1 +

6

(s + 1)2.

Finally,

y(t) = L−1[Y (s)] = 3L−1

1

s + 2

− 2L−1

1

s + 1

+ 6L−1

1

(s + 1)2

= 3e−2t − 2e−t + 6te−t, t ≥ 0

Problem 20.12


y + 4y = cos 2t, y(0) = 1, y(0) = 1.

Solution.


L[y] + 4L[y] = L[cos2t].


s2Y (s)

−sy(0)

−y(0) + 4Y (s) =

s

s2 + 4.


Y (s) = s + 1

s2 + 4 +

s

(s2 + 4)2.

Using Table L again we find

y(t) = L−1

s

s2 + 4

+

1

2L−1

2

s2 + 4

+ L−1

s

(s2 + 4)2

= cos2t +

1

2 sin 2t +

t

4 sin 2t, t ≥ 0

Problem 20.13


y − 2y + y = e2t, y(0) = 0, y(0) = 0.





169

Solution.

Taking the Laplace of both sides to obtainL[y] + 9L[y] = L[g(t)] = 6L[h(t) − h(t − π)].


s2Y (s) − sy(0) − y(0) + 9Y (s) = 6

s − 6e−πs

s .


Y (s) = s + 3

s2 + 9 +

6

s(s2 + 9)(1− e−πs).

Using the partial fraction decomposition, we can write

6

s(s2 + 9) =

A

s +

Bs + C

s2 + 9 .

Multipliying both sides by s(s2 + 9) to obtain

6 = A(s2 + 9) + (Bs + C )s

= (A + B)s2 + Cs + 9A.

Equating coefficients of like powers of s we find A + B = 0, C = 0, and

9A = 6. Solving this system we find A = 23 , B = −23 , and C = 0. Thus,

Y (s) = s

s2 + 9 +

3

s2 + 9 + (1− e−πs)

2

3

1

s − 2

3

s

s2 + 9

.

Finally,

y(t) = L−1[Y (s)] = cos 3t + sin 3t + 2

3(1− cos3t) − 2

3(1− cos3(t − π))h(t − π)

= cos 3t + sin 3t + 2

3(1− cos3t) − 2

3(1 + cos 3t)h(t − π), t ≥ 0

Problem 20.15Determine the constants α, β, y0, and y 0 so that Y (s) = 2s−1

s2+s+2 is the Laplace

transform of the solution to the initial value problem

y + αy + βy = 0, y(0) = y0, y(0) = y 0.




Solution.

Taking the Laplace transform of both sides we finds2Y (s) − sy0 − y0 + αsY (s) − αy0 + βY (s) = 0.

Solving for Y (s) we find

Y (s) = sy0 + (y0 + αy0)

s2 + αs + β =

2s − 1

s2 + s + 2.

By identification we find α = 1, β = 2, y0 = 2, and y 0 = −3



21 Laplace Transforms of

Periodic Functions

Problem 21.1

Find the Laplace transform of the periodic function whose graph is shown.

Solution.

The function is of period T = 2. Thus,

3

10

e−stdt +

21

e−stdt =

−3

se−st

10

−

e−st

s

21

= 1

s(3− 2e−s − e−2s).

Hence,

L[f (t)] =

3

−2e−s

−e−2s

s(1− e−2s)

Problem 21.2


171



172 21 LAPLACE TRANSFORMS OF PERIODIC FUNCTIONS

Solution.


2 10

e−st

dt + 31

e−st

dt = −2

s e−st1

0− e−st

s31

= 1

s (2− e−s

− e−3s

).

Hence,

L[f (t)] = 2 − e−s − e−3s

s(1− e−4s)

Problem 21.3


Solution.


2

1

(t − 1)e−stdt =

2

1

te−stdt−

2

1

e−stdt

=− t

se−st − e−st

s2 +

e−st

s

21

= − e−s

s2 [(1 + s)e−s − 1]



173

Hence,

L[f (t)] = e−s

s2(1− e−2s)[1− (s + 1)e−s]

Problem 21.4


Solution.


2

0

te−stdt = −1

s2(st + 1)e−st

2

0

= − 1

s2[(2s + 1)e−2s − 1].

Hence,

L[f (t)] = 1

s2(1− e−2s)[1− (2s + 1)e−2s]

Problem 21.5

State the period of the function f (t) and find its Laplace transform where

f (t) = sin t, 0 ≤ t < π

f (t + 2π) = f (t), t≥

0.0, π ≤ t < 2π

Solution.

The graph of f (t) is shown below.




The function f (t) is of period T = 2π. The Laplace transform of f (t) is

L[f (t)] = π0 sin te−st

dt1 − e−2πs

Using integration by parts twice we find sin te−stdt = − e−st

1 + s2(cos t + s sin t)

Thus,

π

0

sin te−stdt = − e−st

1 + s2(cos t + s sin t)

π

0

= e−πs

1 + s2 +

1

1 + s2

=1 + e−πs

1 + s2

Hence,

L[f (t)] = 1 + e−πs

(1 + s2)(1− e−2πs)

Problem 21.6

State the period of the function f (t) = 1− e−t

, 0 ≤ t < 2, f (t + 2) = f (t),and find its Laplace transform.

Solution.

The graph of f (t) is shown below



175

The function is periodic of period T = 2. Its Laplace transform is

L[f (t)] = 20 (1−

e−t)e−stdt

1 − e−2s .

But 20

(1−e−t)e−stdt =

e−st

−s

20

+

e−(s+1)t

s + 1

20

= 1

s(1−e−2s)− 1

s + 1(1−e−2(s+1)).

Hence,

L[f (t)] = 1

s − 1 − e−2(s+1)

(s + 1)(1 − e−2s)

Problem 21.7Using Example 21.3 find

L−1

s2 − s

s3 +

e−s

s(1 − e−s)

.

Solution.

Note first that

s2 − s

s3 +

e−s

s(1 − e−s) =

1

s −

1

s2 − se−s

s2(1− e−s)

.

Using Example 22.3, we find

g(t) = 1 − f (t)

where f (t) is the sawtooth function shown below




Problem 21.8


ay + by + cy = f (t), y(0) = y (0) = 0, t > 0.

Suppose that the transfer function of this system is given by Φ(s) = 12s2+5s+2

.(a) What are the constants a, b, and c?(b) If f (t) = e−t, determine F (s), Y (s), and y(t).

Solution.(a) Taking the Laplace transform of both sides we find as2Y (s) + bsY (s) +cY (s) = F (s) or

Φ(s) = Y (s)

F (s) =

1

as2 + bs + c =

1

2s2 + 5s + 2.

By identification we find a = 2, b = 5, and c = 2.(b) If f (t) = e−t then F (s) = L[e−t] = 1

s+1 . Thus,

Y (s) = Φ(s)F (s) = 1

(s + 1)(2s2

+ 5s + 2)

.

Using partial fraction decomposition

1

(s + 1)(2s + 1)(s + 2) =

A

s + 1 +

B

2s + 1 +

C

s + 2



177

Multiplying both sides by s + 1 and setting s = −1 we find A = −1. Next,

multiplying both sides by 2s+1 and setting s = −1

2 we find B =

4

3 . Similarly,multiplying both sides by s + 2 and setting s = −2 we find C = 13 . Thus,

y(t) = − L−1

1

s + 1

+

2

3L−1

1

s + 12

+

1

3L−1

1

s + 2

= − e−t +

2

3e−

t2 +

1

3e−2t, t ≥ 0

Problem 21.9


ay + by + cy = f (t), y(0) = y (0) = 0, t > 0

Suppose that an input f (t) = t, when applied to the above system producesthe output y(t) = 2(e−t − 1) + t(e−t + 1), t ≥ 0.(a) What is the system transfer function?(b) What will be the output if the Heaviside unit step function f (t) = h(t)is applied to the system?

Solution.

(a) Since f (t) = t we find F (s) = 1s2 . Aslo, Y (s) = L[y(t)] = L[2e−t − 2 +

te−t + t] = 2s+1 − 2

s + 1

(s+1)2 + 1

s2 = 1

s2(s+1)2 . But Φ(s) = Y (s)F (s)

= 1(s+1)2 .

(b) If f (t) = h(t) then F (s) = 1

s and Y (s) = Φ(s)F (s) = 1

s(s+1)2 . Usingpartial fraction decomposition we find

1

s(s + 1)2 =

A

s +

B

s + 1 +

C

(s + 1)2

1 =A(s + 1)2 + Bs(s + 1) + Cs

1 =(A + B)s2 + (2A + B + C )s + A

Equating coefficients of like powers of s we find A = 1, B = −1, andC = −1. Therefore,

Y (s) = 1s − 1

s + 1 − 1

(s + 1)2

andy(t) = L−1[Y (s)] = 1 − e−t − te−t, t ≥ 0




Problem 21.10


y + y + y = f (t), y(0) = y (0) = 0,

where

f (t) =

1, 0 ≤ t ≤ 1f (t + 2) = f (t)

−1, 1 < t < 2

(a) Determine the system transfer function Φ(s).(b) Determine Y (s).

Solution.(a) Taking the Laplace transform of both sides we find

s2Y (s) + sY (s) + Y (s) = F (s)

so that

Φ(s) = Y (s)

F (s) =

1

s2 + s + 1.

(b) But

20

f (t)e−stdt = 10

e−stdt− 21

e−stdt

=

e−st

−s

10

−

e−st

−s

21

=1

s(1− e−s) +

1

s(e−2s − e−s)

=(1− e−s)2

s

Hence,

F (s) =

(1

−e−s)2

s(1− e−2s) =

(1

−e−s)

s(1 + e−s)

and

Y (s) = Φ(s)F (s) = (1 − e−s)

s(1 + e−s)(s2 + s + 1)



179

Problem 21.11

Consider the initial value problemy − 4y = et + t, y(0) = y (0) = y (0) = 0.

(a) Determine the system transfer function Φ(s).(b) Determine Y (s).

Solution.

(a) Taking Laplace transform of both sides we find

s3Y (s) − 4Y (s) = F (s).

Thus,

Φ(s) =

Y (s)

F (s) =

1

s3 − 4 .(b) We have

F (s) = L[et + t] = 1

s− 1 +

1

s2 =

s2 + s − 1

(s− 1)s2 .

Hence,

Y (s) = s2 + s − 1

s2(s− 1)(s3 − 4)

Problem 21.12


y + by + cy = h(t), y(0) = y0, y(0) = y 0, t > 0.

Suppose that L[y(t)] = Y (s) = s2+2s+1s3+3s2+2s . Determine the constants b, c, y0,

and y0.

Solution.

Take the Laplace transform of both sides to obtain

s2Y (s) − sy0 − y0 + bsY (s) − by0 + cY (s) = 1

s.

Solve to find

Y (s) =

s2y0 + s(y0 + by0) + 1

s3 + bs2 + cs

= s2 + 2s + 1

s3 + 3s2 + 2s.

By comparison we find b = 3, c = 2, y0 = 1, and y 0 + by0 = 2 or y 0 = −1






22 Convolution Integrals

Problem 22.1

Consider the functions f (t) = g(t) = h(t), t ≥ 0 where h(t) is the Heavisideunit step function. Compute f

∗g in two different ways.

(a) By directly evaluating the integral.(b) By computing L−1[F (s)G(s)] where F (s) = L[f (t)] and G(s) = L[g(t)].

Solution.

(a) We have

(f ∗ g)(t) =

t

0

f (t − s)g(s)ds =

t

0

h(t− s)h(s)ds =

t

0

ds = t, t ≥ 0.

(b) Since F (s) = G(s) = L[h(t)] = 1s

we have (f ∗ g)(t) = L−1[F (s)G(s)] =L−1[ 1s2 ] = t, t ≥ 0

Problem 22.2

Consider the functions f (t) = et and g(t) = e−2t, t ≥ 0. Compute f ∗ g intwo different ways.(a) By directly evaluating the integral.(b) By computing L−1[F (s)G(s)] where F (s) = L[f (t)] and G(s) = L[g(t)].

Solution.

(a) We have

(f ∗ g)(t) =

t

0

f (t− s)g(s)ds =

t

0

e(t−s)e−2sds

=et t

0

e−3sds =e(t−3s)

−3

t

0

=et − e−2t

3 , t ≥ 0.

181



182 22 CONVOLUTION INTEGRALS

(b) Since F (s) = L[et] = 1s−1 and G(s) = L[e−2t] = 1

s+2 we find (f ∗ g)(t) =

L−1

[F (s)G(s)] = L−1

[

1

(s−1)(s−2) ]. Using partial fractions decomposition wefind1

(s− 1)(s + 2) =

1

3(

1

s− 1 − 1

s + 2).

Thus,

(f ∗g)(t) = L−1[F (s)G(s)] = 1

3

L−1[

1

s− 1] − L−1[

1

s + 2]

=

et − e−2t

3 , t ≥ 0

Problem 22.3

Consider the functions f (t) = sin t and g(t) = cos t, t ≥ 0. Compute f ∗ g in

two different ways.(a) By directly evaluating the integral.(b) By computing L−1[F (s)G(s)] where F (s) = L[f (t)] and G(s) = L[g(t)].

Solution.

(a) Using the trigonometric identity 2 sin p cos q = sin ( p + q )+sin( p − q ) wefind that 2 sin (t− s)cos s = sin t + sin (t − 2s). Hence,

(f ∗ g)(t) =

t

0

f (t − s)g(s)ds =

t

0

sin(t− s)cos sds

=1

2[

t

0

sin tds + t

0

sin(t−

2s)ds]

=t sin t

2 +

1

4

t

−t

sin udu

=t sin t

2 .t ≥ 0.

(b) Since F (s) = L[sin t] = 1s2+1 and G(s) = L[cos t] = s

s2+1 we find

(f ∗ g)(t) = L−1[F (s)G(s)] = L−1[ s

(s2 + 1)2] =

t

2 sin t, t ≥ 0

Problem 22.4Compute and graph f ∗ g where f (t) = h(t) and g(t) = t[h(t) − h(t − 2)].

Solution.

Since f (t) = h(t), F (s) = 1s . Similarly, since g(t) = th(t) − th(t − 2) =



183

th(t)− (t−2)h(t−2)−2h(t−2), G(s) = 1s2− e−2s

s2 − 2e−2s

s . Thus, F (s)G(s) =1

s3 − e−2s

s3 − 2e−2s

s2 . It follows that

(f ∗ g)(t) = t2

2 − (t − 2)2

2 h(t − 2) − 2(t− 2)h(t − 2), t ≥ 0.

The graph of (f ∗ g)(t) is given below

Problem 22.5

Compute and graph f ∗ g where f (t) = h(t)− h(t− 1) and g(t) = h(t− 1)−2h(t − 2).

Solution.

Since f (t) = h(t) − h(t − 1), F (s) = 1s − e−s

s . Similarly, since g(t) = h(t −

1) − 2h(t− 2), G(s) = e−s

s − 2e−2s

s . Thus,

F (s)G(s) =e−s − 3e−2s + 2e−3s

s2

=e−s

s2 − 3

e−2s

s2 + 2

e−3s

s2

It follows that

(f ∗ g)(t) = (t− 1)h(t − 1) − 3(t − 2)h(t − 2) + 2(t − 3)h(t − 3), t ≥ 0.

The graph of (f ∗ g)(t) is given below



184 22 CONVOLUTION INTEGRALS

Problem 22.6

Compute t

∗t

∗t.

Solution.

By the convolution theorem we have L[t∗t∗ t] = (L[t])3 = 1s2

3= 1

s6. Hence,

t ∗ t ∗ t = L−1 1s6

= t5

5! = t5

120, t ≥ 0

Problem 22.7

Compute h(t) ∗ e−t ∗ e−2t.

Solution.

By the convolution theorem we have L[h(t)∗e−t∗e−2t] = L[h(t)]L[e−t]L[e−2t] =1s

· 1s+1

· 1s+2 . Using the partial fractions decomposition we can write

1

s(s + 1)(s + 2) =

1

2s − 1

s + 1 +

1

2 · 1

s + 2.

Hence,

h(t) ∗ e−t ∗ e−2t = 1

2 − e−t +

1

2e−2t, t ≥ 0

Problem 22.8

Compute t ∗ e−t ∗ et.

Solution.

By the convolution theorem we have L[t ∗ e−t ∗ et] = L[t]L[e−t]L[et] = 1s2 ·

1s+1 · 1

s−1 . Using the partial fractions decomposition we can write

1

s2(s + 1)(s− 1) = − 1

s2 +

1

2 · 1

s− 1 − 1

2 · 1

s + 1.



185

Hence,

t ∗ e−t ∗ et = −t + et

2 − e−t

2 , t ≥ 0

Problem 22.9

Suppose it is known that

n functions h(t) ∗ h(t) ∗ · · · ∗ h(t) = Ct8. Determine the con-

stants C and the positive integer n.

Solution.

We know that L[

n functions

h(t) ∗ h(t) ∗ · · · ∗ h(t)] = (L[h(t)])n = 1

sn so that L−1[ 1

sn] =

tn−1

(n−1)!

= Ct8. It follows that n = 9 and C = 18!

Problem 22.10

Use Laplace transform to solve for y(t) :

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