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solution_manual_to_accompany_boyce_elementary_differential_equations_8e_xdownx.com_/math219/219hw2.pdfMath 219, Homework 2Due date: 23.11.2005, Wednesday
This homework concerns two (fictitious) design problems about the solar car MES-e of the METU Robotics Society, which won the Formula-G trophy in September2005. Just for the purposes of this homework, assume that they want to modify thecar, and they are asking for your help on two issues.
1. The first problem is about the shock absorbing system of the car. We may modelthe shock absorber as a single linear spring. This question concerns how to adjustthe damping coefficient in order to meet certain requirements.
(a) It is known that when the pilot, weighing 80kg, gets into the car seat, the shockabsorber is compressed by 5cm. From this data, compute the spring constant k (inkg/sec2).
(b) The car (without the pilot) weighs 240kg. Write a differential equation whichgoverns the vertical motion of the car (this could for instance describe the vertical
displacement when the car goes over a speed bump). (Hint: Check Damped freevibrations from Boyce,Di Prima, section 3.8).
(c) It is required that, when the car goes over a speed bump of 5cm high, thevertical displacement x(t) should approach the equilibrium point 0 in a way that fort 1sec, |x(t)| 1cm. For several values of the damping coefficient and for thisinitial condition, sketch the graph of the solution curve for 0 t 2sec. Finally,decide which values of the damping coefficient are allowable in order to meet thisrequirement.
2. The second problem is about the power supply of the motor. The panels convertsolar energy to electrical energy and store it in the accumulator. Assume that theaccumulator provides a voltage of E(t) = 48 cos(t) Volts to the system. Thefrequency is adjusted by the gas pedal. The circuit can be modeled as a seriesL C circuit. Take L = 0.05Henry and C = 106Farad.(a) Write a differential equation for the current I(t) through the voltage source(Hint: Check Electric Circuits from Boyce,Di Prima, section 3.8. We are assumingR = 0).
(b) Solve this differential equation for I(t) in terms of using the method of unde-termined coefficients.
(c) Currents over 20 Amperes may harm the accumulator. Using your result frompart (b), find out which frequencies (in Hertz) should not be allowed. For severalvalues of graph I(t) for 0 t 0.2sec using ODE Architect, and denote whichof these are potentially harmful and which are not.
solution_manual_to_accompany_boyce_elementary_differential_equations_8e_xdownx.com_/math219/219HW3.pdfMath 219, Homework 3Due date: 9.12.2005, Friday
1. Consider the initial value problem
d2x
dt2+
dx
dt+ x = u4(t), y(0) = y
(0) = 0
(a) Solve this initial value problem using the Laplace transform.
(b) Use ODE Architect to solve the equation, and graph the solution. Also graphdx
dtwith respect to t (You can use the function Step(t, 4) to create a unit step functionwith discontinuity at t = 4).
(c) Discuss how the graphs agree with the solutions in (a): in particular determine
(if any) all the points where x(t) anddx
dtare discontinuous, behavior of these two
functions for t , their maxima and minima.
2. Write each of the following systems of differential equations in matrix form, findthe eigenvalues and eigenvectors of the coefficient matrices, and using these, findall solutions of each system. Also, graph the phase portraits (x y graph) usingODE Architect. Please use a scale which includes the point (0, 0), and graph severalsolutions in order to clearly observe the behavior around (0, 0). Also, place arrowson the solution curves which indicate the direction of increasing t, and make surethat solution curves along the eigenvector directions are graphed if there are anyreal eigenvectors.
(a)
dx
dt= 2x y
dy
dt= 3x + 3y
(b)
dx
dt= x + y
dy
dt= 3x 4y
(c)
dx
dt= 2x + 3y
dy
dt= 5x + 5y
(d)
dx
dt= 4x + 3y
dy
dt= 3x + 2y
(e)
dx
dt= x 3y
dy
dt= 2x + y
solution_manual_to_accompany_boyce_elementary_differential_equations_8e_xdownx.com_/math219/219hw4.pdfMath 219, Homework 4Due date: 30.12.2005, Wednesday
Suppose that K > 0, and f(t) is defined as
f(t) =
{1 if 4n t < 4n + 10 otherwise
where n runs through the set of integers.
(a) Determine the Fourier series for f(t).
(b) Consider the differential equation
d2x
dt2+ x = f(t).
By using ODE Architect, solve (and graph the solutions of) this equation for asmany values of K as possible between 0.5 and 10 for 0 t 100.(You can enterf(t) in ODE Architect using the built in command SqWave(t, L, K) and then setL = 4 K, and assign K definite values on the lines below). Record the maximumvalues of x(t) for each of these Ks, and plot a K vs. max(x(t)) graph by hand.
(c) Which values of K result in a resonance in the system? (Hint: you should find 5such values). Can you relate these values to the Fourier series terms? Please printthe resonance graphs.
(d) What do the Fourier coefficients correspond to on the resonance graphs?
We wish you a happy new year and success on your finals!
solution_manual_to_accompany_boyce_elementary_differential_equations_8e_xdownx.com_/math219/ch01.pdf CHAPTER 1.
________________________________________________________________________ page 1
Chapter One
Section 1.1
1.
For , the slopes are C "& negative, and hence the solutions decrease. For , theC "&slopes are , and hence the solutions increase. The equilibrium solution appears topositivebe , to which all other solutions converge.C > "&a b3.
For , the slopes are C "& :9=3tive, and hence the solutions increase. For C "&, the slopes are , and hence the solutions decrease. All solutions appear tonegativediverge away from the equilibrium solution .C > "&a b5.
For , the slopes are C "# :9=3tive, and hence the solutions increase. ForC "# , the slopes are , and hence the solutions decrease. All solutionsnegativediverge away from
CHAPTER 1.
________________________________________________________________________ page 2
the equilibrium solution .C > "#a b6.
For , the slopes are C # :9=3tive, and hence the solutions increase. For ,C #the slopes are , and hence the solutions decrease. All solutions diverge awaynegativefromthe equilibrium solution .C > #a b8. For solutions to approach the equilibrium solution , we must haveall C > #$a bC ! C #$ C ! C #$w w for , and for . The required rates are satisfied by thedifferential equation .C # $Cw
9. For solutions than to diverge from , must be an other increasingC > # C # C >a b a bfunction for , and a function for . The simplest differentialC # C #decreasingequationwhose solutions satisfy these criteria is .C C #w
10. For solutions than to diverge from , we must have other C > "$ C "$ C !a b wfor , and for . The required rates are satisfied by the differentialC "$ C ! C "$wequation .C $C "w
12.
Note that for and . The two equilibrium solutions are andC ! C ! C & C > !w a bC > & C ! C &a b . Based on the direction field, for ; thus solutions with initialwvalues than diverge from the solution . For , the slopes aregreater & C > & ! C &a bnegative between, and hence solutions with initial values and all decrease toward the! &
CHAPTER 1.
________________________________________________________________________ page 3
solution . For , the slopes are all ; thus solutions with initialC > ! C !a b positivevaluesless than approach the solution .! C > !a b14.
Observe that for and . The two equilibrium solutions are C ! C ! C # C > !w a band . Based on the direction field, for ; thus solutions with initialC > # C ! C #a b wvalues than diverge from . For , the slopes are alsogreater # C > # ! C #a bpositive between, and hence solutions with initial values and all increase toward the! #solutionC > # C !a b . For , the slopes are all ; thus solutions with initialnegativevalues than diverge from the solution .less ! C > !a b16. Let be the total amount of the drug in the patient's body ata b a b a b+ Q > in milligramsanygiven time . The drug is administered into the body at a rate of > 2 &!! !%Q 712< a b
a b,
Based on the direction field, the amount of drug in the bloodstream approaches theequilibrium level of "#&!71 A3>238 + 0/A29?# #
or equivalently,
.@
.> 7 1 @
# #
a b, ! After a long time, Hence the object attains a given by.@.> terminal velocity@
71_ #
a b- @ 71 !!%!) 51=/- Using the relation , the required is # #_# drag coefficienta b.
19.
All solutions appear to approach a linear asymptote . It is easy toa bA3>2 =69:/ /;?+6 >9 "verify that is a solution.C > > $a b20.
CHAPTER 1.
________________________________________________________________________ page 5
All solutions approach the equilibrium solution C > ! a b23.
All solutions appear to from the sinusoid ,diverge C > =38> "a b $# % 1
which is also a solution corresponding to the initial value .C ! a b25.
All solutions appear to converge to . First, the rate of change is small. TheC > !a bslopeseventually increase very rapidly in .magnitude
26.
CHAPTER 1.
________________________________________________________________________ page 6
The direction field is rather complicated. Nevertheless, the collection of points at whichthe slope field is , is given by the implicit equation The graph ofzero C 'C #> $ #these points is shown below:
The of these curves are at , . It follows that for solutions withy-intercepts C ! 'initial values , all solutions increase without bound. For solutions with initialC 'values in the range , the slopes remain , andC ' ! C ' and negativehencethese solutions decrease without bound. Solutions with initial conditions in the range ' C ! initially increase. Once the solutions reach the critical value, given bythe equation , the slopes become negative and negative. TheseC 'C #>$ # remainsolutions eventually decrease without bound.
CHAPTER 1.
________________________________________________________________________ page 7
Section 1.2
1 The differential equation can be rewritten asa b+.C
& C .>
Integrating both sides of this equation results in , or equivalently, 68 & C > -k k "& C - / C ! C> . Applying the initial condition results in the specification ofa b !the constant as . Hence the solution is - & C C > & C & / ! !a b a b >
All solutions appear to converge to the equilibrium solution C > & a b1 Rewrite the differential equation asa b-
.C
"! #C .>
Integrating both sides of this equation results in , or 68 "! #C > -"# "k kequivalently,& C - / C ! C#> . Applying the initial condition results in the specification ofa b !the constant as . Hence the solution is - & C C > & C & / ! !a b a b #>
All solutions appear to converge to the equilibrium solution , but at a rateC > &a b fasterthan in Problem 1a
2 The differential equation can be rewritten asa b+
CHAPTER 1.
________________________________________________________________________ page 8
.C
C & .>
Integrating both sides of this equation results in , or equivalently,68 C & > -k k "C & - / C ! C> . Applying the initial condition results in the specification ofa b !the constant as . Hence the solution is - C & C > & C & / ! !a b a b >
All solutions appear to diverge from the equilibrium solution .C > &a b2 Rewrite the differential equation asa b,
.C
#C & .>
Integrating both sides of this equation results in , or equivalently,"# "68 #C & > -k k#C & - / C ! C#> . Applying the initial condition results in the specification ofa b !the constant as . Hence the solution is - #C & C > #& C #& / ! !a b a b #>
All solutions appear to diverge from the equilibrium solution .C > #&a b2 . The differential equation can be rewritten asa b-
.C
#C "! .>
Integrating both sides of this equation results in , or equivalently,"# "68 #C "! > -k kC & - / C ! C#> . Applying the initial condition results in the specification ofa b !the constant as . Hence the solution is - C & C > & C & / ! !a b a b #>
CHAPTER 1.
________________________________________________________________________ page 9
All solutions appear to diverge from the equilibrium solution .C > &a b3 . Rewrite the differential equation asa b+
.C
, +C .> ,
which is valid for . Integrating both sides results in , orC , + 68 , +C > - ""+ k kequivalently, . Hence the general solution is , +C - / C > , - / + +> +>a b a bNote that if , then , and is an equilibrium solution.C ,+ .C.> ! C > ,+a ba b,
As increases, the equilibrium solution gets closer to , from above.a b a b3 + C > !Furthermore, the convergence rate of all solutions, that is, , also increases.+ As increases, then the equilibrium solution a b33 , C > ,+a b also becomes larger. Inthis case, the convergence rate remains the same. If and both increase , a b a b333 + , but constant,+ then the equilibrium solutionC > ,+a b remains the same, but the convergence rate of all solutions increases.5 . Consider the simpler equation . As in the previous solutions, re-a b+ .C .> +C" "write the equation as
.C
C + .>
"
"
Integrating both sides results in C > - / "a b +>a b a b a b, C > C > 5 Now set , and substitute into the original differential equation. We"find that
CHAPTER 1.
________________________________________________________________________ page 10
+C ! + C 5 ," "a b .That is, , and hence . +5 , ! 5 ,+a b a b- C > - / ,+ . The general solution of the differential equation is This is+>exactly the form given by Eq. in the text. Invoking an initial condition ,a b a b"( C ! C!the solution may also be expressed as C > ,+ C ,+ / a b a b! +>6 . The general solution is , that is, .a b a b a b a b+ : > *!! - / : > *!! : *!! /> ># #!With , the specific solution becomes . This solution is a: )&! : > *!! &!/! #a b >decreasing exponential, and hence the time of extinction is equal to the number ofmonthsit takes, say , for the population to reach . Solving , we find that> *!! &!/ !0 #zero >0> # 68 *!!&! &()0 a b .monthsa b a b a b, : > *!! : *!! / The solution, , is a exponential as long as! #> decreasing: *!! *!! : *!! / !! !
#. Hence has only root, given bya b >0 one> # 68
*!!
*!! :0
!
a b a b- ,. The answer in part is a general equation relating time of extinction to the valueofthe initial population. Setting , the equation may be written as> "#0 months
*!!
*!! : /
!
',
which has solution . Since is the initial population, the appropriate: )*(('*" :! !answer is .: )*)! mice
7 . The general solution is . Based on the discussion in the text, time isa b a b+ : > : / >! measured in . Assuming , the hypothesis can be expressed asmonths month days" $!: / #: < 68 #! !
"As , the exponential term vanishes, and hence the limiting value is .> p_ U GZP
14 . The rate of the chemical is . At anya b a b+ !!" $!!accumulation grams per hourgiven time , the of the chemical in the pond is > U > "!concentration grams per gallona b '.Consequently, the chemical the pond at a rate of .leaves grams per houra b a b$ "! U >%Hence, the rate of change of the chemical is given by
.U
.> $ !!!!$U >a b gm/hr .
Since the pond is initially free of the chemical, .U ! !a ba b, . The differential equation can be rewritten as
.U
"!!!! U !!!!$ .>
Integrating both sides of the equation results in . 68 "!!!! U !!!!$> Gk kTakingthe natural logarithm of both sides gives . Since , the"!!!! U - / U ! !!!!!$> a bvalue of the constant is . Hence the amount of chemical in the pond at any- "!!!!timeis . Note that . SettingU > "!!!! " / "a b a b!!!!$> grams year hours )('!> )('! U )('! *#((((, the amount of chemical present after is ,one year gramsa bthat is, .*#(((( kilograms
a b- . With the rate now equal to , the governing equation becomesaccumulation zero.U.> !!!!$U >a b Resetting the time variable, we now assign the newgm/hr .initial value as .U ! *#((((a b gramsa b a b a b. - U > *#(((( / . The solution of the differential equation in Part is !!!!$>Hence, one year the source is removed, the amount of chemical in the pond isafterU )('! '(!"a b .grams
CHAPTER 1.
________________________________________________________________________ page 13
a b/ >. Letting be the amount of time after the source is removed, we obtain the equation"! *#(((( / !!!!$ > !!!!$> Taking the natural logarithm of both sides, 68 "!*#(((( > ## ((' #'a b or .hours yearsa b0
15 . It is assumed that dye is no longer entering the pool. In fact, the rate at which thea b+dye the pool is leaves kg/min gm per hour#!! ; > '!!!! #!! '!"!!! ; > '!c d a bc da b a b.Hence the equation that governs the amount of dye in the pool is
.;
.> !# ; a bgm/hr .
The initial amount of dye in the pool is .; ! &!!!a b gramsa b, . The solution of the governing differential equation, with the specified initial value,is ; > &!!! / a b !# >a b- > %. The amount of dye in the pool after four hours is obtained by setting . That is,; % &!!! / ##%''% '! !!!a b !) . Since size of the pool is , thegrams gallonsconcentration grams/gallon of the dye is .!!$(%
a b. X. Let be the time that it takes to reduce the concentration level of the dye to!!# " #!! . At that time, the amount of dye in the pool is . Usinggrams/gallon gramsthe answer in part , we have . Taking the natural logarithm ofa b, &!!! / "#!!!# Xboth sides of the equation results in the required time .X ("% hours
a b/ !# #!!"!!!. Note that . Consider the differential equation.; "!!! ; .
Here the parameter corresponds to the , measured in .< flow rate gallons per minuteUsing the same initial value, the solution is given by In order; > &!!! / a b < >"!!!to determine the appropriate flow rate, set and . (Recall that of> % ; "#!! "#!! gm
CHAPTER 1.
________________________________________________________________________ page 14
dye has a concentration of ). We obtain the equation !!# "#!! &!!! / gm/gal < #&!Taking the natural logarithm of both sides of the equation results in the required flow rate< $&( .gallons per minute
CHAPTER 1.
________________________________________________________________________ page 15
Section 1.3
1. The differential equation is second order, since the highest derivative in the equationis of order . The equation is , since the left hand side is a linear function of two linear Candits derivatives.
3. The differential equation is , since the highest derivative of the function fourth order Cis of order . The equation is also , since the terms containing the dependentfour linearvariable is linear in and its derivatives.C
4. The differential equation is , since the only derivative is of order . Thefirst order onedependent variable is , hence the equation is .squared nonlinear
5. The differential equation is . Furthermore, the equation is ,second order nonlinearsince the dependent variable is an argument of the , which is a linearC sine function notfunction.
7. . Hence C > / C > C > / C C ! " "" " "a b a b a b> w ww > wwAlso, and . Thus C > -9=2 > C > =382 > C > -9=2 > C C ! # " # #a b a b a bw ww ww #9. . Substituting into the differential equation, we haveC > $> > C > $ #>a b a b# w> $ #> $> > $> #> $> > >a b a b# # # # . Hence the given function is a solution.10. and Clearly, isC > >$ C > "$ C > C > C > ! C >" "" " " "a b a b a b a b a b a bw ww www wwwwa solution. Likewise, , ,C > / >$ C > / "$ C > /# # #a b a b a b> w > ww >C > / C > /www > wwww ># #a b a b, . Substituting into the left hand side of the equation, wefind that . Hence both/ % / $ / >$ / %/ $/ > >> > > > > >a b a bfunctions are solutions of the differential equation.
11. and . Substituting into the leftC > > C > > # C > > %" "# "# $#" "a b a b a bw wwhand side of the equation, we have
#> > % $> > # > > # $ > # >
!
# $# "# "# "# "# "#
Likewise, and . Substituting into the leftC > > C > > C > # ># " # $# #a b a b a bw wwhand side of the differential equation, we have #> # > $> > > % > #a b a b$ # " " $ > > !" " . Hence both functions are solutions of the differential equation.
12. and . Substituting into the left handC > > C > #> C > ' >" # " "$ %a b a b a bw wwside of the differential equation, we have > ' > &> #> % > ' > #a b a b% $ # # "! > % > ! C > > 68 > C > > #> 68 ># # #
$ $#. Likewise, anda b a b2 w
C > & > ' > 68 >ww#% %a b . Substituting into the left hand side of the equation, we have
> & > ' > 68 > &> > #> 68 > % > 68 > & > ' > 68 > #a b a b a b% % $ $ 2 2 2
CHAPTER 1.
________________________________________________________________________ page 16
& > "! > 68 > % > 68 > ! 2 2 2 Hence both functions are solutions of thedifferential equation.
13. andC > -9= > 68 -9= > > =38 > C > =38 > 68 -9= > > -9= >a b a b a b a bwC > -9= > 68 -9= > > =38 > =/- >wwa b a b . Substituting into the left hand side of thedifferential equation, we have a b a ba b -9= > 68 -9= > > =38 > =/- > -9= > 68 -9= > > =38 > -9= > 68 -9= > > =38 > =/- > -9= > 68 -9= > > =38 > =/- >a b a b .Hence the function is a solution of the differential equation.C >a b15. Let . Then , and substitution into the differential equationC > / C > < /a b a b ww # results in . Since , we obtain the algebraic equation < / # / ! / ! < # !# #
The roots of this equation are < 3 # "# 17. and . Substituting into the differentialC > / C > < / C > < /a b a b a b w ww # equation, we have . Since , we obtain the algebraic< / > C > < > C > < < " >a b a b a b a b< w < ww %> < > % > !# < < %< > % > ! > !a b < <
` ?
`> + =38 B =38 +>
#
##
#
## #
"
"
- - -
- - -
It is easy to see that . Likewise, given , we have+ ? B > =38 B +># #` ? ` ?`B `># #
# #" " a b a b
` ?
`B =38 B +>
` ?
`> + =38 B +>
#
#
#
##
#
#
a ba b
Clearly, is also a solution of the partial differential equation.? B >#a b28. Given the function , the partial derivatives are? B > > /a b 1 B % ># #!
? > / > B /
# > % >
? > /
#>
B /
% > >
BB
B % > # B % >
# % #
>
B % >
#
# B % >
# #
1 1
! !
1 1
!
# # # #
# # # #
! !
! !
It follows that .!# ? ? BB ># >B /
% > >
1 !!# # B % ># #
# #
!
Hence is a solution of the partial differential equation.? B >a b
CHAPTER 1.
________________________________________________________________________ page 18
29 .a b+
a b, The path of the particle is a circle, therefore are intrinsic to thepolar coordinatesproblem. The variable is radial distance and the angle is measured from the vertical.< )Newton's Second Law states that In the direction, the equation of!F a 7 tangentialmotion may be expressed as , in which the , that is,!J 7+) ) tangential accelerationthe linear acceleration the path is is in the directionalong positive+ P. .> +) )# #)of increasing . Since the only force acting in the tangential direction is the component) of weight, the equation of motion is
71 =38 7P .
.>)
)#
#
Note that the equation of motion in the radial direction will include the tension in therod .
a b- . Rearranging the terms results in the differential equation. 1
.> P =38 !
#
#
))
solution_manual_to_accompany_boyce_elementary_differential_equations_8e_xdownx.com_/math219/ch02.pdf CHAPTER 2.
________________________________________________________________________ page 18
Chapter Two
Section 2.1
1a b+
a b, Based on the direction field, all solutions seem to converge to a specific increasingfunction.
a b a b a b- > / C > >$ "* / - / The integrating factor is , and hence . $> #> $>It follows that all solutions converge to the function C > >$ "* "a b2a b+
a b, . All slopes eventually become positive, hence all solutions will increase withoutbound.
a b a b a b- > / C > > / $ - / The integrating factor is , and hence It is. #> $ #> #>evident that all solutions increase at an exponential rate.
3a b+
CHAPTER 2.
________________________________________________________________________ page 19
a b a b, C > " . All solutions seem to converge to the function !a b a b a b- > / C > > / # " - / The integrating factor is , and hence It is. #> # > >clear that all solutions converge to the specific solution .C > "!a b4 .a b+
a b, . Based on the direction field, the solutions eventually become oscillatory.a b a b- > > The integrating factor is , and hence the general solution is.
C > =38 #> $-9= #> $ -
%> # >a b a ba b
in which is an arbitrary constant. As becomes large, all solutions converge to the- >function C > $=38 #> # "a b a b5 .a b+
CHAPTER 2.
________________________________________________________________________ page 20
a b, . All slopes eventually become positive, hence all solutions will increase withoutbound.
a b a b a b'- > /B: #.> / The integrating factor is The differential equation. #>canbe written as , that is, Integration of both/ C #/ C $/ / C $/ #> w #> > #> >wa bsides of the equation results in the general solution It follows thatC > $/ - / a b > #>all solutions will increase exponentially.
6a b+
a b a b, C > ! All solutions seem to converge to the function !a b a b- > > The integrating factor is , and hence the general solution is. #
C > -9= > =38 #> -
> > >a b a b a b
# #
in which is an arbitrary constant. As becomes large, all solutions converge to the- >function C > ! !a b7 .a b+
CHAPTER 2.
________________________________________________________________________ page 21
a b a b, C > ! All solutions seem to converge to the function !a b a b a b a b- > /B: > C > > / - / The integrating factor is , and hence It is. # # > ># #clear that all solutions converge to the function .C > !!a b8a b+
a b a b, C > ! All solutions seem to converge to the function !a b a b a b c da b- > C > >+8 > G Since , the general solution is . a b a b" > " ># ## #"It follows that all solutions converge to the function .C > !!a b9a b+
CHAPTER 2.
________________________________________________________________________ page 22
a b, . All slopes eventually become positive, hence all solutions will increase withoutbound.
a b a b '- > /B: .> / The integrating factor is . The differential equation can. "# >#be written as , that is, Integration/ C / C# $> / # $> / #># w ># ># ># / C#># wof both sides of the equation results in the general solution AllC > $> ' - / a b >#solutions approach the specific solution C > $> ' !a b10 .a b+
a b, C !. For , the slopes are all positive, and hence the corresponding solutionsincreasewithout bound. For , almost all solutions have negative slopes, and hence solutionsC !tend to decrease without bound.
a b- > First divide both sides of the equation by . From the resulting , thestandard formintegrating factor is . The differential equation can be.a b '> /B: .> "> ">written as , that is, Integration leads to the generalC > C> > / C> > / w # > >wa bsolution For , solutions C > >/ - > - !a b > diverge, as implied by the directionfield. For the case , the specific solution is - ! C > >/a b >, which evidentlyapproaches as .zero > p_
11 .a b+
a b, The solutions appear to be oscillatory.
CHAPTER 2.
________________________________________________________________________ page 23
a b a b a b a b a b- > / C > =38 #> # -9= #> - / The integrating factor is , and hence . > >It is evident that all solutions converge to the specific solution C > =38 #> #!a b a b-9= #>a b .12a b+
a b, . All solutions eventually have positive slopes, and hence increase without bound.a b a b- > / The integrating factor is . The differential equation can be. #>written as , that is, Integration of both/ C / C# $> # / C# $> #># w ># # ># #w sides of the equation results in the general solution ItC > $> "#> #% - / a b # >#follows that all solutions converge to the specific solution .C > $> "#> #%!a b #14. The integrating factor is . After multiplying both sides by , the. .a b a b> / >#>equation can be written as Integrating both sides of the equation results / C > 2> win the general solution Invoking the specified condition, weC > > / # - / a b # #> #>require that . Hence , and the solution to the initial value/ # - / ! - "## #problem is C > > " / # a b a b# #>16. The integrating factor is . Multiplying both sides by ,. .a b a b '> /B: .> > >#> #the equation can be written as Integrating both sides of the equationa b a b> C -9= > # wresults in the general solution Substituting and settingC > =38 > > - > > a b a b # # 1the value equal to zero gives . Hence the specific solution is - ! C >a b =38 > > a b #17. The integrating factor is , and the differential equation can be written as.a b> /#> Integrating, we obtain Invoking the specified initial a b/ C " / C > > - 2 2> >wcondition results in the solution C > > # / a b a b #>19. After writing the equation in standard orm0 , we find that the integrating factor is. .a b a b '> /B: .> > >%> % . Multiplying both sides by , the equation can be written as a b a b> C > / > C > > " / - % > % >w Integrating both sides results in Letting> " and setting the value equal to zero gives Hence the specific solution of- ! the initial value problem is C > > > / a b $ % >21 .a b+
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The solutions appear to diverge from an apparent oscillatory solution. From thedirectionfield, the critical value of the initial condition seems to be . For , the+ " + "!solutions increase without bound. For , solutions decrease without bound.+ "
a b, The integrating factor is . The general solution of the differential.a b> / #>equation is . The solution is sinusoidal as longC > )=38 > %-9= > & - /a b a ba b a b >#as . The - ! initial value of this sinusoidal solution is+ ! a ba b a b)=38 ! %-9= ! & %& a b a b- , See part .22a b+
All solutions appear to eventually initially increase without bound. The solutions increaseor decrease, depending on the initial value . The critical value seems to be + + " !
a b, The integrating factor is , and the general solution of the differential.a b> / #>equation is Invoking the initial condition , theC > $/ - / C ! +a b a b>$ >#solutionmay also be expressed as Differentiating, follows thatC > $/ + $ / a b a b>$ >#C ! " + $ # + " # wa b a b a b The critical value is evidently + " !
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a b- + ". For , the solution is ! C > $/ # / >a b a b>$ >#, which for large isdominated by the term containing / >#
is .C > )=38 > %-9= > & - /a b a ba b a b >#23a b+
As , solutions increase without bound if , and solutions decrease> p ! C " + %a bwithout bound if C " + % a ba b a b ', > /B: .> > / . The integrating factor is The general solution of the. >"> >differential equation is . Invoking the specified value ,C > > / - / > C " +a b a b> >we have . That is, . Hence the solution can also be expressed as" - + / - + / "C > > / + / " / >a b a b> > . For small values of , the second term is dominant.>Setting , critical value of the parameter is + / " ! + "/ !
a b- + "/ + "/. For , solutions increase without bound. For , solutions decreasewithout bound. When , the solution is + "/ C > > / ! >p !a b >, which approaches as .
24 .a b+
As , solutions increase without bound if , and solutions decrease> p ! C " + %a bwithout bound if C " + % a b
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a b a b a b a b, C + C > + -9= > >. Given the initial condition, , the solution is # %1 1#Since , solutions increase without bound if , and solutionslim
>
#
!-9= > " + %1
decrease without bound if Hence the critical value is+ % 1#+ % !%)%(! 1
# .
a b a b a b a b- + % C > " -9= > > C > "# For , the solution is , and . Hence the1#>lim
!
solution is bounded.
25. The integrating factor is Therefore general solution is.a b '> /B: .> / "# >#C > %-9= > )=38 > & - / a b c da b a b #> Invoking the initial condition, the specificsolution is . Differentiating, it follows thatC > %-9= > )=38 > * / &a b c da b a b >#
C > %=38 > )-9= > %& / &
C > %-9= > )=38 > ##& / &
w >
ww >
a b a b a b a b a b a b #
#
Setting , the first solution is , which gives the location of the C > ! > "$'%$wa b " firststationary point. Since . TheC > !wwa b" , the first stationary point in a local maximumcoordinates of the point are .a b"$'%$ )#!!)26. The integrating factor is , and the differential equation.a b '> /B: .> /#$ ># $canbe written as The general solution is a b a b/ C / > / # C > # $ # $ # $> > >w #" '>) #" '>) #")- / C > C /# $ # $!
> >. Imposing the initial condition, we have .a b a bSince the solution is smooth, the desired intersection will be a point of tangency. Takingthe derivative, Setting , the solutionC > $% #C #"% / $ C > !w > wa b a b a b! # $is Substituting into the solution, the respective at the> 68 #" )C * " !$# c da b valuestationary point is . Setting this result equal to C > 68 $ 68 #" )Ca b a b" !$ * *# % ) zero,we obtain the required initial value C #" * / ) "'%$ ! %$a b27. The integrating factor is , and the differential equation can be written as.a b> />4a b a b/ C $ / # / -9= #> > > >w 4 4 4 The general solution is
C > "# )-9= #> '%=38 #> '& - / a b c da b a b > 4Invoking the initial condition, , the specific solution isC ! !a b
C > "# )-9= #> '%=38 #> ()) / '& a b a b a b > 4As , the exponential term will decay, and the solution will oscillate about an> p_averagevalue amplitude of , with an of "# ) '&
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29. The integrating factor is , and the differential equation can be written.a b> /$ #>as The general solution is a b a b/ C $> / # / C > #> %$ % / $ # $ # #> > > >w - / C > #> %$ % / C "'$ / $ # $ #!
> > > Imposing the initial condition, a b a bAs , the term containing will the solution. Its > p_ /$ #> dominate sign will determinethe divergence properties. Hence the critical value of the initial condition isC ! "'$The corresponding solution, , will also decrease withoutC > #> %$ % /a b >bound.
Note on Problems 31-34 :
Let be , and consider the function , in which 1 > C > C > 1 > C > p_a b a b a b a b a bgiven " "as . Differentiating, . Letting be a > p_ C > C > 1 > +w w wa b a b a b" constant, it followsthat C > +C > C > +C > 1 > +1 > w w wa b a b a b a b a b a b" " Note that the hypothesis on thefunction will be satisfied, if . That is, HenceC > C > +C > ! C > - / " " ""a b a b a b a bw +>C > - / 1 > C +C 1 > +1 > a b a b a b a b+> w w, which is a solution of the equation For convenience, choose .+ "
31. Here , and we consider the linear equation The integrating1 > $ C C $ a b wfactor is , and the differential equation can be written as The.a b a b> / / C $/ > > >wgeneral solution is C > $ - / a b >33. Consider the linear equation The integrating1 > $ > C C " $ > a b wfactor is , and the differential equation can be written as .a b a b a b> / / C # > / > > >wThe general solution is C > $ > - / a b >34. Consider the linear equation The integrating1 > % > C C % #> > a b # w #factor is , and the equation can be written as .a b a b a b> / / C % #> > / > > # >wThe general solution is C > % > - / a b # >
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Section 2.2
2. For , the differential equation may be written as B " C .C .B c da bB " B# $Integrating both sides, with respect to the appropriate variables, we obtain the relationC # 68 - C B 68 - # " #$ $k k k k" B " B$ $ That is, a b 3. The differential equation may be written as Integrating bothC .C =38B .B #sides of the equation, with respect to the appropriate variables, we obtain the relation C -9= B - G -9= B C " G" That is, , in which is an arbitrary constant.a bSolving for the dependent variable, explicitly, .C B " G -9= Ba b a b5. Write the differential equation as , or -9= #C .C -9= B .B =/- #C .C -9= B .B# # # #Integrating both sides of the equation, with respect to the appropriate variables, we obtainthe relation >+8 #C =38 B -9= B B -
7. The differential equation may be written as Integratinga b a bC / .C B / .B C Bboth sides of the equation, with respect to the appropriate variables, we obtain therelationC # / B # / - # C # B
8. Write the differential equation as Integrating both sides of thea b" C# .C B .B #equation, we obtain the relation , that is, C C $ B $ - $C C B G$ $ $ $
9 . The differential equation is separable, with Integrationa b a b+ C .C " #B .B #yields Substituting and , we find that C B B - B ! C "' - ' " #Hence the specific solution is . The isC B B '" # explicit formC B " a b a bB B '#a b,
a b a ba b- B B ' B # B $. Note that . Hence the solution becomes # singular atB # B $ and
10 a b a b + C B #B #B % #
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10a b,
11 Rewrite the differential equation as Integrating both sidesa b+ B / .B C .C Bof the equation results in Invoking the initial condition, weB / / C # - B B #obtain Hence - "# C #/ #B / "# B B The of the solution isexplicit formC B a b #/ #B / " C ! "B B The positive sign is chosen, since a ba b,
a b- B "( B !(' The function under the radical becomes near and negative11 Write the differential equation as Integrating both sides of thea b+ < .< . # ") )equation results in the relation Imposing the condition , we < 68 - < " #" ) a bobtain . - "# The of the solution is explicit form < # " # 68 a b a b) )
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a b,
a b- ! Clearly, the solution makes sense only if Furthermore, the solution becomes)singular when , that is, 68 "# / ) ) 13 a b a b a b+ C B # 68 " B % #a b,
14 . Write the differential equation as Integrating botha b a b+ C .C B " B .B $ "##sides of the equation, with respect to the appropriate variables, we obtain the relation C # " B - - $# # # Imposing the initial condition, we obtain Hence the specific solution can be expressed as The C $ # " B # # explicitform positive of the solution is The C B " $ # " B a b # sign is chosen tosatisfy the initial condition.
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a b,
a b- The solution becomes singular when # " B $ B # . That is, at 15 a b a b + C B "# B "&% #a b,
16 . a b+ Rewrite the differential equation as Integrating both%C .C B B " .B $ #a bsidesof the equation results in Imposing the initial condition, we obtainC % - % a bB "# #- ! %C ! Hence the solution may be expressed as The forma bB "# # % explicitof the solution is The is chosen based on C B C ! a b a ba b B " # " ## sign
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a b,
a b- B The solution is valid for all .17 a b a b + C B B / "$% $ Ba b,
a b- B "%& . The solution is valid for This value is found by estimating the root of%B %/ "$ ! $ B
18 . Write the differential equation as Integrating botha b a b a b+ $ %C .C / / .B B Bsides of the equation, with respect to the appropriate variables, we obtain the relation$C #C / / - C ! "# B Ba b a b Imposing the initial condition, , we obtain- ( $C #C / / ( Thus, the solution can be expressed as Now by# B Ba bcompleting the square on the left hand side, # C $% a b# / / '&)a bB B .Hence the explicit form of the solution is C B $% '&"' -9=2 B a b
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a b,
a b k k- '& "' -9=2 B ! B #" . Note the , as long as Hence the solution is valid onthe interval . #" B #"
19 a b+ C B $ =38 $ -9= B a b a b1 "$ " #a b,
20 a b+ Rewrite the differential equation as IntegratingC .C + " > .> " " "both sides of the equation, we obtain Taking68 C 68 C % %> %68 " > - k k k k k kthe of both sides, it follows that It followsexponential k k a ba bC C % G / " > %> %that as , . That is, > p_ C C % " % C % p_ C > p % k k k k a ba b a ba b a b, C ! # G " Setting , we obtain that . Based on the initial condition, the solutionmay be expressed as Note that , for allC C % / " > C C % !a b a b a b%> %> ! C % > ! Hence for all Referring back to the differential equation, it followsthat is always . This means that the solution is . We findC w positive monotone increasingthat the root of the equation is near / " > $** > #)%% %> a b%a b a b- C > % Note the is an equilibrium solution. Examining the local direction field,
we see that if , then the corresponding solutions converge to . ReferringC ! ! C %a bback to part , we have , for Settinga b a b c d a ba b+ C C % C C % / " > C % ! ! !%%>> # C C % $/ C # C # % , we obtain Now since the function! !a b a b a b a ba b# %0 C C C % C % C %a b a b is for and , we need only solve the equationsmonotoneC C % C C % ! ! ! !# #
% %a b a b $** $/ %!" $/a b a band The respective solutionsare and C $''## C %%!%# ! !
30a b0
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31a b-
32 . Observe that Hence the differential equationa b a b + B $C #BC # # " $# B # BC C"is .homogeneous
a b a b, C B @ @ B@ B $B @ #B @. The substitution results in . Thew # # # #transformed equation is This equation is , with general@ " @ #B@ w #a b separablesolution In terms of the original dependent variable, the solution is@ " - B #B C - B # # $
a b-
33a b-
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34 Observe that Hence thea b a b + %B $C #B C # # C CB B "differential equation is .homogeneous
a b a b, C B @ @ B@ # @ # @. The substitution results in . The transformedwequation is This equation is , with general@ @ &@ % # @B w #a b separablesolution In terms of the original dependent variable, the solutiona b k k@% @" GB # $is a b k k%B C BC G#a b-
35 .a b-
36 Divide by to see that the equation is homogeneous. Substituting , wea b+ B C B@#obtain The resulting differential equation is separable.B @ " @ w #a ba b a b, " @ .@ B .B Write the equation as Integrating both sides of the equation, "#we obtain the general solution In terms of the original " " @ 68 B - a b k kdependent variable, the solution is C B G 68 B B c dk k "
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a b-
37 The differential equation can be expressed as . Hence thea b + C w " $# B # BC C"equation is homogeneous. The substitution results in .C B@ B @ " &@ #@w #a bSeparating variables, we have #@ ""&@ B# .@ .B
a b, Integrating both sides of the transformed equation yields "&68 68 B -k k" &@# k k ,that is, In terms of the original dependent variable, the general" &@ G B # &k ksolution is &C B G B # # $k ka b-
38 The differential equation can be expressed as . Hence thea b + C w $ "# B # BC C "equation is homogeneous. The substitution results in , thatC B@ B @ @ " #@w #a bis, #@ "@ " B# .@ .B
a b k k, 68 68 B - Integrating both sides of the transformed equation yields ,k k@ "#that is, In terms of the original dependent variable, the general solution@ " G B # k kis C G B B B # # #k k
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a b-
CHAPTER 2.
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Section 2.3
5 . Let be the amount of salt in the tank. Salt enters the tank of water at a rate ofa b+ U# " =38 > =38 > #U"!! " " " "% # # % It leaves the tank at a rate of 9D738 9D738Hence the differential equation governing the amount of salt at any time is
.U " "
.> # % =38 > U&!
The initial amount of salt is The governing ODE is U &! 9D ! linear, with integratingfactor Write the equation as The.a b > / / U / >&! >&! >&!w " "# % =38 >specific solution is U > #& "#&=38 > '#&-9= > '$"&! / #&!" 9D a b >&!a b,
a b- The amount of salt approaches a , which is an oscillation of amplitudesteady state"% #& 9D about a level of
6 . The equation governing the value of the investment is . The value ofa b+ .W.> < Wthe investment, at any time, is given by Setting , the requiredW > W / W X #Wa b a b! !time is X 68 # < a ba b, < ( !( X ** C !)>a bAfter years, the balances are and thirty $ $W $$( ($% W #&! &(*E F
a b, < W $! '%! / For an rate , the balances after years are andunspecified thirty E $! C"! C > C ! / The differential equation is , with solution The>"!doubling-time is given by 7 "!68 # '*$"& a ba b a b- .C.> !& =38# > C& . Consider the differential equation The equation is1separable, with Integrating both sides, with respect to the" "C &.C !" =38# > .> 1appropriate variable, we obtain Invoking the initial68 C > -9=# > "! - a b1 1 1condition, the solution is The isC > /B: " > -9=# > "! a b c da b1 1 1 doubling-time7 '$)!% , The approaches the value found in part .doubling-time a ba b. .
17 . The differential equation is , with integrating factora b a b+ .C.> < > C 5 linear. . .a b a b a b a b '> /B: < > .> C 5 > Write the equation as Integration of bothw
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sides yields the general solution . In this problem,C 5 . C ! > ' a b a b a b. 7 7 . .!theintegrating factor is .a b c da b> /B: -9= > > &
a b a b, C > ! > > The population becomes , if , for some . Referring toextinct part ,a b+we find that C > ! a b
( c da b!
>"&
-
/B: -9= & . & / C 7 7 7
It can be shown that the integral on the left hand side increases , from monotonically zeroto a limiting value of approximately . Hence extinction can happen &!)*$ only if& / C &!)*$ C !)$$$ "& - -, that is,
a b a b a b- , C > ! . Repeating the argument in part , it follows that ( c da b!
>"&
-
/B: -9= & . / C "
57 7 7
Hence extinction can happen , that is, only if / C 5 &!)*$ C %"''( 5 "& - -
a b. C 5. Evidently, is a function of the parameter .- linear19 . Let be the of carbon monoxide in the room. The rate of ofa b a b+ U > volume increaseCO CO leaves the room is The amount of at a rate ofa ba b!% !" !!!% 0> 738 $a b a b a b!" U > "#!! U > "#!!! 0> 738 Hence the total rate of change is given by$the differential equation This equation is and.U.> !!!% U > "#!!! a b linearseparable, with solution Note that U > %) %) /B: >"#!!! U !a b a b 0> 0> $ $!Hence the at any time is given by .concentration %B > U > "#!! U > "#a b a b a ba b a b a b, B > % %/B: >"#!!!. The of in the room is A levelconcentration CO %of corresponds to . Setting , the solution of the equation!!!!"# !!"# B !!"#% a b7% %/B: >"#!!! !!"# $'a b is .7 minutes
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20 The concentration is It is easy to seea b a b a b+ - > 5 T< - 5 T< / ! Zthat - >p_ 5 T< a ba b a b, - > - / X 68#Z < X 68"!Z B > $")%& %%" > #))%& /B: >%& a b a b a b, the position is given by Hencethe is .maximum height mB > %&()a b"a b a b, B > ! > &"#) Setting , the ball hits the ground at .# # seca b-
23 The differential equation for the motion is ,a b+ 7.@.> @ 71upward . #in which . This equation is , with Integrating. ""$#& .@ .> separable 7@ 71. #
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both sides and invoking the initial condition, Setting@ > %%"$$ >+8 %#& ### > a b a b@ > ! > "*"' @ >a b a b" ", the ball reaches the maximum height at . Integrating , thesecposition is given by Therefore theB > "*)(& 68 -9= !### > !%#& %)&( a b c da bmaximum height m is .B > %)&'a b"a b, 7.@.> @ 71 The differential equation for the motion is downward . #This equation is also separable, with For convenience, set at771 @. # .@ .> > !the of the trajectory. The new initial condition becomes . Integrating bothtop @ ! !a bsides and invoking the initial condition, we obtain 68 %%"$ @ %%"$ @ >##&c da b a bSolving for the velocity, Integrating , the@ > %%"$ " / " / @ >a b a b a b a b> >##& ##&position is given by To estimate theB > **#* 68 / " / ")'# a b a b > > ###& ##&duration sec of the downward motion, set , resulting in . Hence theB > ! > $#('a b# #total time sec that the ball remains in the air is .> > &"*#1 #
a b-
24 Measure the positive direction of motion . Based on Newton's a b+ #downward ndlaw,the equation of motion is given by
7 .@
.>
!(& @ 71 ! > "! "# @ 71 > "! , ,
Note that gravity acts in the direction, and the drag force is . During thepositive resistivefirst ten seconds of fall, the initial value problem is , with initial.@.> @(& $#velocity This differential equation is separable and linear, with solution@ ! ! a b fps@ > #%! " / @ "! "('( a b a b a b>(& . Hence fpsa b a b, B > !. Integrating the velocity, with , the distance fallen is given by
B > #%! > ")!! / ")!!a b >(& .Hence .B "! "!(%&a b ft
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a b- > ! For computational purposes, reset time to . For the remainder of the motion,the initial value problem is , with specified initial velocity.@.> $#@"& $#@ ! "('( @ > "& "'"( / > p_a b a b The solution is given by As ,fps $# >"&@ > p @ "& B ! "!(%&a b a bP Integrating the velocity, with , the distance fallenfpsafter the parachute is open is given by To find theB > "& > (&) / ""&!$ a b $# >"&duration of the second part of the motion, estimate the root of the transcendental equation"& X (&) / ""&!$ &!!! X #&'' $# X"& The result is sec
a b.
25 . Measure the positive direction of motion . The equation of motion isa b+ upwardgiven by . The initial value problem is ,7.@.> 5 @ 71 .@.> 5@7 1with . The solution is Setting@ ! @ @ > 715 @ 715 / a b a b a b! ! 5>7@ > ! > 75 68 71 5 @ 71 a b a b c da b7 7, the maximum height is reached at time !Integrating the velocity, the position of the body is
B > 71 >5 1 " / 7 7@
5 5a b # 5>7!
Hence the maximum height reached is
B B > 1 68 7@ 7 71 5 @
5 5 717 7
! !a b #
a b a b, " 68 " Recall that for , $ $ $ $ $ $" " "# $ %# $ %26 . a b a b, 5 @ 71 / 1> lim lim
5! 5!
71 5 @ 71 /5 7
> 5>7a b! 5>7!
a b - @ / ! / ! , since lim lim7! 7!
71 715 5
5>7 5>7!
28 . In terms of displacement, the differential equation is a b+ 7@ [email protected] 5 @ 71 This follows from the : . The differential equation ischain rule .@ .@ .B .@.> .B .> .> @separable, with
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B @ 68 7@ 7 1 71 5 @
5 5 71a b ##
The inverse , since both and are monotone increasing. In terms of the givenexists B @parameters, B @ "#& @ "&$" 68 !!)"' @ " a b k k
a b a b, B "! "$%& 5 !#% . The required value is .metersa b a b- + @ "! B "! In part , set and .m/s meters29 Let represent the height above the earth's surface. The equation of motion isa b+ Bgiven by , in which is the universal gravitational constant. The7 K K.@ Q7.> VBa b#symbols and are the and of the earth, respectively. By the chain rule,Q V mass radius
7@ K.@ Q7
.B V Ba b# .This equation is separable, with Integrating both sides,@ .@ KQ V B .B a b#andinvoking the initial condition , the solution is @ ! #1V @ #KQ V B a b a b # " #1V #KQV 1 KQV From elementary physics, it follows that . Therefore#
@ B #1 V V B a b a b Note that mi/hr .1 () &%& #a b , .B.> #1 V V B We now consider . This equation is also separable,with By definition of the variable , the initial condition is V B.B #1 V .> BB ! ! B > #1 V > V V a b a b Integrating both sides, we obtain $ ## $ $# #$Setting the distance , and solving for , the duration of such aB X V #%! !!! Xa bflight would be X %* hours .
32 Both equations are linear and separable. The initial conditions are a b a b+ @ ! ? -9=Eand . The two solutions are and A ! ?=38E @ > ? -9=E / A > 1< a b a b a b ? =38E 1< / a b
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a b a b, + Integrating the solutions in part , and invoking the initial conditions, thecoordinates are andB > -9=E " /a b a b?<
C > 1>< 1 ?< =38E 2< < =38E 1< / ?
>. .a b a b .a b a b a b a b ', /B: : > .> C : > : > C . By definition, . Hence " "> >w. .a b a b" "That is, C : > C !" "w a ba b a b a b a b a b a b a b a b '- C : > = 1 = .= > 1 > : > C 1 > # ! #w " "> >>. .a b a b. .That is, C : > C 1 > # #w a b a b30. Since , set . It follows that and 8 $ @ C #C # $.@ .@.> .> .> # .>
.C .C C$
Substitution into the differential equation yields , which further C CC# .>.@$ & 5 $
results in The latter differential equation is linear, and can be written as@ # @ # w & 5a b a b/ # @ > # > / -/ # > # > # >w& & &5 5 The solution is given by Converting back tothe original dependent variable, C @ "#
31. Since , set . It follows that and 8 $ @ C #C # $.@ .@.> .> .> # .>.C .C C$
The differential equation is written as , which upon -9= > X C CC# .>.@$ a b> 5 $
further substitution is This ODE is linear, with integrating@ # -9= > X @ #w a b>factor The solution is. > >a b a b a b '> /B: # -9= > X .> /B: # =38 > #X > @ > #/B: # =38 > #X > /B: # =38 #X . - /B: # =38 > #X > a b a b a b a b(> > 7 7 7 >
!
>
Converting back to the original dependent variable, C @ "#
33. The solution of the initial value problem , is C #C ! C ! " C > / " " " "w #>a b a bTherefore y On the interval the differential equation isa b a b a b" C " / "_ " #C C ! C > -/ C " C " -/ # # # #
w > ", with Therefore Equating the limitsa b a b a bC " C " - / a b a b , we require that Hence the global solution of the initial value"problem is
C > / ! > "
/ > "a b #>">, ,
Note the discontinuity of the derivative
C > #/ ! > "
/ > "a b #>"> ,,
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Section 2.5
1.
For , the only equilibrium point is . , hence the equilibriumC ! C ! 0 ! + !! wa bsolution is .9a b> ! unstable2.
The equilibrium points are and . , therefore theC +, C ! 0 +, ! wa bequilibrium solution is .9a b> +, asymptotically stable3.
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4.
The only equilibrium point is . , hence the equilibrium solutionC ! 0 ! ! wa b9a b> !is .unstable
5.
The only equilibrium point is . , hence the equilibrium solutionC ! 0 ! ! wa b9a b> !is .+=C7:>9>3-+66C stable
6.
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7 .a b,
8.
The only equilibrium point is . Note that , and that for .C " 0 " ! C ! C " w wa bAs long as , the corresponding solution is . Hence theC "! monotone decreasingequilibrium solution is .9a b> " semistable9.
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10.
The equilibrium points are , . . The equilibrium solutionC ! " 0 C " $C w #a b9a b> ! is , and the remaining two are .unstable asymptotically stable11.
12.
The equilibrium points are , . . The equilibrium solutionsC ! # 0 C )C %C w $a b9 9a b a b> # > # and are and , respectively. Theunstable asymptotically stableequilibrium solution is .9a b> ! semistable
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13.
The equilibrium points are and . . Both equilibriumC ! " 0 C #C 'C %C w # $a bsolutions are .semistable
15 . Inverting the Solution , Eq. shows as a function of the population a b a b a b+ "" "$ > Candthe carrying capacity . With ,O C O$!
> 68 " "$ " CO
< CO " "$ a bc da ba bc da b
Setting ,C #C!
7 68 " "$ " #$
< #$ " "$ a bc da ba bc da b
That is, If , .7 7 68 % < !!#& &&%&"< per year years
a b a b, "$ C O CO In Eq. , set and . As a result, we obtain! ! "X 68
" "
< " c dc d! "" !
Given , and , .! " 7 !" !* < !!#& "(&()per year years
16 .a b+
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17. Consider the change of variable Differentiating both sides with? 68 CO a brespectto , Substitution into the Gompertz equation yields , with> ? C C ?
