CHAPTER 1.

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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

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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! &

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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 ba b,

Based on the direction field, the amount of drug in the bloodstream approaches theequilibrium level of "#&!71 A3>238 + 0/A29?

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7 71 @.@.> ##

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 coefficient

a 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.

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All solutions approach the equilibrium solution C > ! a b23.

All solutions appear to from the sinusoid ,diverge C > =38> "a b $# % 1which 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.

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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.

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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+

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.CC & .>

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 #>

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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

.CC + .>

"

"

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

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+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 #! !"

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written as , with solution The object is released with an.@.> 1 @ > 1> @ a b !initialvelocity .@!

a b, 2. Suppose that the object is released from a height of above the ground. Usingunitsthefact that , in which is the of the object, we obtain@ .B.> B downward displacementthedifferential equation for the displacement as With the origin placed at.B.> 1> @ !the point of release, direct integration results in . Based on theB > 1> # @ >a b # !chosencoordinate system, the object reaches the ground when . Let be the timeB > 2 > Xa bthat it takes the object to reach the ground. Then . Using the1X # @ X 2# !quadraticformula to solve for ,X

X @ @ #121! !

The answer corresponds to the time it takes for the object to fall to the ground.positiveThenegative answer represents a previous instant at which the object could have beenlaunchedupward , only to ultimately fall downward with speed ,a bwith the same impact speed @!from a height of above the ground.2 units

a b a b a b- > X @ > + . The impact speed is calculated by substituting into in part That is,@ X @ #12a b ! .10 , . The general solution of the differential equation is Given thata b a b+ U > - / b U ! "!! - "!!a b , the value of the constant is given by . Hence the amount ofmgthorium-234 present at any time is given by . Furthermore, based on theU > "!! /a b hypothesis, setting results in Solving for the rate constant, we> " )#!% "!! / ! 7solution,we have . Taking the natural logarithm of both sides, it follows that!&U U /! !

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12. The differential equation governing the amount of radium-226 is ,.U.> U ! / a b a b half-life , the decay rate is given by . The7 "'#! < 68 # "'#!years per yeara bamount of radium-226, after years, is therefore Let be> U > U ! / Xa b a b !!!!%#()'>the time that it takes the isotope to decay to of its original amount. Then setting$%> X ,and , we obtain Solving for the decayU X U ! U ! U ! / a b a b a b a b$ $% % !!!!%#()'Xtime, it follows that or !!!!%#()' X 68 $% X '(#$'a b years .13. The solution of the differential equation, with , isU ! !a bU > GZ " / a b a b GV>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

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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

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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

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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 wC > & > ' > 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

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& > "! > 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 /

% > >

BBB % > # 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

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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 ! #

#) )

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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+

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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 bin which is an arbitrary constant. As becomes large, all solutions converge to the- >function C > $=38 #> # "a b a b5 .a b+

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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+

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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+

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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.

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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 "# ) '&

CHAPTER 2.

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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 +

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a b, .

a b- " B " . Evidently, the solution is defined for 22. The differential equation can be written as Integrating botha b$C % .C $B .B # #sides, we obtain Imposing the initial condition, the specific solutionC %C B - $ $is Referring back to the differential equation, we find that asC %C B " C p_$ $ wC p# $ B "#(' "&*) The respective values of the abscissas are ,

Hence the solution is valid for "#(' B "&*)

24. Write the differential equation as Integrating both sides,a b a b$ #C .C # / .B Bwe obtain Based on the specified initial condition, the solution$C C #B / - # Bcan be written as , it follows that$C C #B / " # B Completing the squareC B $# #B / "$% #B / "$% !a b B B The solution is defined if ,that is, . In that interval, , for It can "& B # C ! B 68 # a bapproximately wbe verified that . In fact, on the interval of definition. HenceC 68 # ! C B !ww wwa b a bthe solution attains a global maximum at B 68 #

26. The differential equation can be written as Integratinga b" C# ".C # " B .B a bboth sides of the equation, we obtain Imposing the given++8C #B B - #initialcondition, the specific solution is Therefore,++8C #B B C B >+8 # a b a b#B B#Observe that the solution is defined as long as It is easy to # #B B # 1 1#see that Furthermore, for and Hence#B B " #B B # B #' !' # # 1the solution is valid on the interval Referring back to the differential #' B !'

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equation, the solution is at Since on the entire intervalstationary B " C B !wwa bofdefinition, the solution attains a global minimum at B "

28 . Write the differential equation as Integratinga b a b a b+ C % C .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-

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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 $! >a b increasemosquito population is The population by . Hence the T > T / / " days a b a b! #!!!! T / T ( #T #T T / " ! " ! ! !a b a b (< Based on the data, set , that is, The growthrate is determined as Therefore the population,< 68 # ( !**!# a b per dayincluding the by birds, is predation T > # "! / #!" **( / " a b a b& !**> !**> #!" **($ "*(($ / !**>

16 . The is a b a b c d+ C > /B: #"! >"! #-9=>"! #*'$# doubling-time 7a b a b a b, .C.> C"! C > C ! / The differential equation is , with solution The>"!doubling-time is given by 7 "!68 # '*$"& 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@ .@.B 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< / ?

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Section 2.4

2. Considering the roots of the coefficient of the leading term, the ODE has uniquesolutions on intervals not containing or . Since , the initial value problem! % # ! %a bhas a unique solution on the interval a b! % 3. The function is discontinuous at of . Since , the>+8 > odd multiples 1 1 1# # #

$1initial value problem has a unique solution on the interval 1 1# #$ 5. and . These functions are discontinuous at: > #> 1 > $> a b a ba b a b% > % ># ##B # # # . The initial value problem has a unique solution on the interval a b6. The function is defined and continuous on the interval . Therefore the68 > ! _a binitial value problem has a unique solution on the interval .a b! _7. The function is continuous everywhere on the plane, along the straight0 > Ca b exceptline The partial derivative has the C #>& `0`C (> #> &Ca b# sameregion of continuity.

9. The function is discontinuous along the coordinate axes, and on the hyperbola0 > Ca b> C "# # . Furthermore,

`0 " C 68 >C`C C " > C # " > Ca b k ka b# # # # #

has the points of discontinuity.same

10. is continuous everywhere on the plane. The partial derivative is also0 > C `0`Ca bcontinuous everywhere.

12. The function is discontinuous along the lines and . The0 > C > 5 C "a b 1partial derivative has the region of continuity.`0`C -9> > " Ca b a b# same 14. The equation is separable, with Integrating both sides, the solution.CC #> .> #is given by For , solutions exist as long as .C > C " C > C ! > "Ca b ! ! ! !# #For , solutions are defined for .C ! >! all

15. The equation is separable, with Integrating both sides and invoking.CC .> $the initial condition, Solutions exist as long as ,C > C #C > " #C > " !a b ! ! !that is, . If , solutions exist for . If , then the#C > " C ! > "#C C !! ! ! !solution exists for all . If , solutions exist for .C > ! > C ! > "#Ca b ! !16. The function is discontinuous along the straight lines and .0 > C > " C !a bThe partial derivative is discontinuous along the same lines. The equation is`0`C

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separable, with Integrating and invoking the initial condition, theC .C > .> # a b" >$solution is Solutions exist as long asC > 68 " > C a b k k #$ $ #! "#

#$ 68 " > C !

$ #! ,that is, For all it can be verified that yields a validC 68 " > C C !! ! !# $#$ k ksolution, even though Theorem does not guarantee one , solutions exists as long as#%# k k a b" > /B: $C # > "$ #! From above, we must have . Hence the inequalitymay be written as It follows that the solutions are valid for> /B: $C # " $ #a b!c da b/B: $C # " > _! "$# .17.

18.

Based on the direction field, and the differential equation, for , the slopesC !!eventually become negative, and hence solutions tend to . For , solutions_ C !!increase without bound if Otherwise, the slopes become negative, and> ! ! eventuallysolutions tend to . Furthermore, is an . Note that slopeszero equilibrium solutionC !!are along the curves and .zero C ! >C $

19.

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For initial conditions satisfying , the respective solutions all tend to .a b> C >C $! ! zeroSolutions with initial conditions the hyperbola eventually tend toabove or below >C $_ C !. Also, is an .! equilibrium solution

20.

Solutions with all tend to . Solutions with initial conditions to the> ! _ > C! ! !a bright of the parabola asymptotically approach the parabola as . Integral> " C >p_#curves with initial conditions the parabola and also approach the curve.above a bC !!The slopes for solutions with initial conditions the parabola and are allbelow a bC !!negative. These solutions tend to _

21. Define , in which is the Heaviside step function.C > > - ? > - ? >- $#a b a b a b a b#$Note that and C - C ! ! C - $# "- - - #$a b a b a b a b a b+ - " $# Let #$a b a b, - # $# Let #$a b a b a b a b a b a b- C # # C > # ! - # C # ! Observe that , for , and that for! - -$# $## #$ $- # - ! C # # # So for any , -a b c d26 Recalling Eq. in Section ,a b a b+ $& #"

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C = 1 = .= " -> >. ..a b a b( a b a bIt is evident that and .C > C > = 1 = .=" #a b a b a b a b'" "> >. .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 bSetting ,C #C!

7 68 " "$ " #$< #$ " "$ a bc da ba bc da bThat 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 ?

CHAPTER 2.

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proportion of carriers . Therefore the proportion of the population that escapesvanishesthe epidemic is the proportion of left at that time, susceptibles B /B: C ! !a b! "25 . Note that , and . So ifa b a b c d a b a ba b+ 0 B B V V +B 0 B V V $+B- -# w #a b a bV V ! B ! 0 ! !- , the only equilibrium point is . , and hence the solution w9a b> ! is .asymptotically stablea b a b a b, V V ! B ! V V + . If , there are equilibrium points ,- -three Now , and . Hence the solution is ,0 ! ! 0 V V + ! !w wa b a b - 9 unstableand the solutions are .9 V V +a b- asymptotically stablea b- .

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Section 2.6

1. and . Since , the equation isQ B C #B $ R B C #C # Q R !a b a b C Bexact. Integrating with respect to , while holding constant, yields Q B C

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R C B with respect to , while holding constant, results in

CHAPTER 2.

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28. The equation is not exact, since R Q #C " R Q Q B C B C. However, a b #C " C C Ca b a b is a function of alone. Hence there exists , which is a solution. .

of the differential equation . The latter equation is . .w # "Ca b separable, with. # "C C /B: #C 68 C / C. . .. One solution is . Now rewrite thea b a b #Cgiven ODE as . This equation is , and it is easy to/ .B #B / "C .C !#C #Ca b exactsee that

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Section 2.7

2 . The Euler formula is .a b a b a b+ C C 2 #C " " #2 C 28" 8 8 8a b. The differential equation is linear, with solution C > " / # a b a b#>4 . a b+ The Euler formula is .C " #2 C $2 -9= >8" 8 8a ba b a b a b. C > '-9= > $=38 > ' / &. The exact solution is .#>5.

All solutions seem to converge to .9a b> #&*6.

Solutions with positive initial conditions seem to converge to a specific function. On theother hand, solutions with coefficients decrease without bound. is annegative 9a b> !equilibrium solution.

7.

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All solutions seem to converge to a specific function.

8.

Solutions with initial conditions to the of the curve seem to diverge. On'left' > !"C#the other hand, solutions to the of the curve seem to converge to . Also, 'right' zero 9a b>is an equilibrium solution.

9.

All solutions seem to diverge.

10.

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Solutions with initial conditions increase without bound. Solutions withpositivenegativeinitial conditions decrease without bound. Note that is an equilibrium solution.9a b> !11. The Euler formula is . The initial value is .C C $2 C &2 C #8" 8 8 !12. The iteration formula is C " $2 C 2 > C > C ! !& 8" 8 8 ! !8a b a b a b# . 14. The iteration formula is C " 2 > C 2 C "! > C ! " 8" 8 8 ! !8a b a b a b$17. The Euler formula is

C C 2 C #> C$ >8" 88 8 8

8#

a b#.

The initial point is a b a b> C " # ! !18 . See Problem 8.a b+19a b+

a b, C C 2 C 2 >. . The critical value of appearsThe iteration formula is 8" 8 8 8# # !to be near . For , the iterations diverge.! !! ! ! !')"& C

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20 . The ODE is a b+ linear, with general solution . Invoking the specifiedC > > - /a b >initial condition, , we have Hence ThusC > C C > - / - C > / a b a b! ! ! ! ! ! > >! !the solution is given by .9a b a b> C > / >! ! > >!a b, . The Euler formula is C " 2 C 2 2 > 5 8 "8" 8 8a b . Now set .a b a b a b a b- C " 2 C 2 2 > " 2 C > > 2 >. We have . Rearranging" ! ! ! " ! !the terms, . Now suppose that ,C " 2 C > > C " 2 C > >" ! ! " 5 ! ! 55a ba b a b a bfor some . Then . Substituting for , we find that5 " C " 2 C 2 2 > C5" 5 5 5a bC " 2 C > " 2 > 2 2 > " 2 C > > 25" ! ! 5 5 ! ! 55" 5"a b a b a b a b a b .Noting that , the result is verified.> > 55" 5

a b a b. 2 > > 8 > >. Substituting , with ,! 8C " C > >> >88 ! !

!8 a b .

Taking the limit of both sides, as , and using the fact that ,8p_ " +8 /lim8_

8 +a bpointwise convergence is proved.

21. The exact solution is . The Euler formula is 9a b> /> C " 2 C8" 8a b . It is easyto see that Given , set . Taking the limit,C " 2 C " 2 > ! 2 >88 !8 8a b a bwe find that lim lim

8_ 8_8 >C " >8 / 8 a b

23. The exact solution is . The Euler formula is 9a b> ># /#> C " #2 C 8" 8a b 2# 2 > C " C " #2 2# " #2 > # 8 ! " ". Since , It is easy toa b a bshow by mathematical induction, that . For , set C " #2 > # > ! 2 >88 88a band thus . Taking the limit, we find that > > C " #>8 ># 8 8 8lim lim8_ 8_c da b / >##> . Hence pointwise convergence is proved.

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Section 2.8

2. Let and . It follows that .D C $ > " .D. .D.> .>. .D.>7 7 7a ba bFurthermore, . Hence . The new initial.D.> .C.> " C .D. " D $$ $7 a bcondition is D ! ! a b73. The approximating functions are defined recursively by .9 98" 8!

>a b c d' a b> # = " .=Setting , . Continuing, , ,9 9 9 9! " # $a b a b a b a b> ! > #> > #> #> > > #> #># $ #%$9%a b> > > #> #> # %$ $% $ # , . Given convergence, set

9 9 9 9a b a b c d" a b a b"

> > > >

#> >+5 x

" 5" 5

5

5"

_

5#

_5 .

Comparing coefficients, , , . It follows that ,+ $x %$ + %x #$ + )$ % $+ "'% ,and so on. We find that in general, that . Hence+ #5 5

9a b "> >5 x / "

5"

_5

#>

#5

.

5. The approximating functions are defined recursively by

9 98" 8!

>a b c d( a b> = # = .= .Setting , . Continuing, ,9 9 9 9! " # $a b a b a b a b> ! > > # > > # > "# > > # # # $ # > "# > *' > > # > "# > *' > *'! $ % # $ % &, , . Given convergence, set9%a b

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9 9 9 9a b a b c d" a b a b"

> > > >

> # >+5 x

" 5" 5

5

5"

_

# 5

5$

_.

Comparing coefficients, , , , . We+ $x ""# + %x "*' + &x "*'! $ % &find that , , , In general, . Hence+ "# + "% + ") + #$ % & 5 5"

9a b a b"> >#5 x % / #> %

5#

_5

>

5#

# .

6. The approximating functions are defined recursively by

9 98" 8!

>a b c d( a b> = " = .= .Setting , , , , 9 9 9 9 9! " $ %a b a b a b a b a b> ! > > > # > > > ' > > > #% > # $ %2 > > "#! & , . Given convergence, set

9 9 9 9a b a b c d" a b a b

> > > >

> > # > # > ' > ' > #% > ! !

" 5" 55"

_

# # $ $ %

.

Note that the terms can be rearranged, as long as the series converges uniformly.

CHAPTER 2.

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8 . a b+ The approximating functions are defined recursively by9 98" 8

!

>a b a b( > = = = .=# .Set . The iterates are given by , ,9 9 9! "a b a b a b> ! > > # > > # > "!# # &29 9$ %a b a b> > # > "! > )! > > # > "! > )! > ))! # & ) # & ) "", , .Upon inspection, it becomes apparent that

98$ 8"

$ 5"

a b c d" c d> > " > ># # & # & ) # & ) # $

> # & ) # $

#$ '

#

5"

8

a b a ba b a b>

8 ">

5 "

a b, .

The iterates appear to be converging.

9 . The approximating functions are defined recursively bya b+9 98"

!

>

8a b a b( > = = .=# # .Set . The first three iterates are given by , ,9 9 9! "a b a b a b> ! > > $ > > $ > '$$ $ (29$a b> > $ > '$ #> #!(* > &*&$&$ ( "" "& .

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a b, .

The iterates appear to be converging.

10 . The approximating functions are defined recursively bya b+9 98"

!

>

8a b a b( > " = .=$ .Set . The first three iterates are given by , ,9 9 9! "a b a b a b> ! > > > > > %2 %9$a b> > > % $> #) $> "'! > )$$% ( "! "$ .a b, .

The approximations appear to be diverging.

12 . The approximating functions are defined recursively bya b+9 98" !

>

8a b ( a ba b> .=$= %= ## = "# .

Note that . For computational purposes, replace the" #C # C Sa b !"#5!

'5 C(

above iteration formula by

CHAPTER 2.

________________________________________________________________________ page 71

9 98"!

>

8a b a b( "> $= %= # = .="# # 55!'

.

Set . The first four approximations are given by ,9 9! "a b a b> ! > > > > ## $92a b> > > # > ' > % > & > #% # $ % & ' ,9$a b> > > # > "# $> #! %> %& # % & ' ,9%a b> > > # > ) (> '! > "& # % & 'a b, .

The approximations appear to be converging to the exact solution,

9a b > " " #> #> > # $13. Note that and , . Let . Then .9 9 98 8 8a b a b a b a b! ! " " a 8 " + ! " + +8Clearly, . Hence the assertion is true.lim

8_+ !8

14 . , . Let . Then a b a b a b+ ! ! a 8 " + ! " + #8+ / #8+/ 9 98 8 8+ 8+# #Using l'Hospital's rule, . Hence lim lim lim

D_ D_ 8_+D +D#+D/ "D/ ! + ! # # 98a b

a b ' , #8B / .B / " / . Therefore,!" !"8B 8B 8# #lim lim8_ 8_! !

" "( (a b a b9 98 8B .B B .B .15. Let be fixed, such that . Without loss of generality, assume that> > C > C Ha b a b" #C C 0 C" # . Since is differentiable with respect to , the mean value theorem asserts thatb C C 0 > C 0 > C 0 > C C such that Taking the absolute0 0a b a b a b a ba b" # " # " #Cvalue of both sides, Since, by assumption,k k k k k ka b a b a b0 > C 0 > C 0 > C C " # " #C 0`0`C H 0 is continuous in , attains a C maximum on any closed and bounded subset of H.

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Hence k k k ka b a b0 > C 0 > C O C C " # " #16. For a interval of , , . Since satisfies asufficiently small > > > H 09 98" 8a b a bLipschitz condition, Herek k k ka b a b a b a ba b a b0 > > 0 > > O > > 9 9 9 98 8" 8 8"O 7+B 0k kC .17 . Hence , ina b a b a b k k k k k k' ' 'a b a b+ > 0 = ! .= > 0 = ! .= Q.= Q >9 9" "! ! !> k k k k> >which is the maximum value of on .Q 0 > C Hk ka ba b a b a b c d' a b a ba b, > > 0 = = 0 = ! .=. By definition, . Taking the absolute9 9 9# " "!>value of both sides, . Based on thek k k ka b a b c d' a b a ba b9 9 9# " "!>> > 0 = = 0 = ! .=k kresults in Problems 16 and 17, .k k k k k ka b a b a b' '9 9 9# " "! !> >> > O = ! .= OQ = .=k k k kEvaluating the last integral, we obtain .k k k ka b a b9 9# "> > QO > ##a b- . Suppose that

k ka b a b k k9 93 3" "> > QO >3 x3 3for some . By definition, .3 " > > 0 > = 0 = = .=9 9 9 93" 3 3 3"!

>a b a b c d' a b a ba b a bIt follows that

k k k ka b a b a b a b( a b a b( k ka b a b( k kk ka b a b

9 9 9 9

9 9

3" 3 3 3"!

!3 3"

!

"

" "

> > 0 = = 0 = = .=

O = = .=

O .=QO =3 x

QO > QO 23 " x 3 " x

k kk kk k

>

>

> 3 3

3 3 33

Hence, by mathematical induction, the assertion is true.

18 . Use the triangle inequality, .a b k k k k k k+ + , + ,a b k k k k k k a ba b a b a b, > 2 > Q2 > > QO 2 8 x. For , , and . Hence9 9 9" 8 8" 8" 8

k ka b ""a b

98"

> Q O 23 x

Q O2O 3 x

3"

8 3 3

3"

8 3

CHAPTER 2.

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a b a b - , / " . The sequence of partial sums in converges to By the QO O2 comparisontest bounded, the sums in also converge. Furthermore, the sequence , anda b+ k ka b98 > is hence has a convergent subsequence. Finally, since individual terms of the series musttend to zero, it follows that the sequence k k k ka b a b a b9 9 98 8" 8> > p ! >, and is convergent.19 . Let and Then by ofa b a b a b a b a b' 'a b a b+ > 0 = = .= > 0 = = .= 9 9 < > linearitythe integral, 9 < 9 > 0 = = 0 = = .= . It follows that 9 < 9 a b- . We know that satisfies a Lipschitz condition,0 k k k ka b a b0 > C 0 > C O C C" # " # ,based on in . Therefore,k k`0`C O H

k k k ka b a b a b a b( a b a b( k ka b a b

9 < 9 0 = = 0 = = .=

O = = .=

!

>

!

>

.

CHAPTER 2.

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Section 2.9

1. Writing the equation for each , , , 8 ! C !* C C !* C C !* C" ! # " $ #and so on, it is apparent that . The terms constitute an C !* C8 !8a b alternatingseries zero, which converge to , regardless of .C!

3. Write the equation for each , , , , .8 ! C $ C C %# C C &$ C " ! # " $ # Upon substitution, we find that , , C % $ # C C & % $ $ # C # " $ ! a b a b a bIt can be proved by mathematical induction, that

C C" 8 # x# 8x

8 " 8 # C"#

8 a b a ba b

!

! .

This sequence is divergent, except for .C !!

4. Writing the equation for each , , , , ,8 ! C C C C C C C C" ! # " $ # % $and so on, it can be shown that

C C 8 %5 8 %5 " C 8 %5 # 8 %5 $8!

! , for or , for or

The sequence is convergent .only for C !!

6. Writing the equation for each ,8 !

C !& C '" ! C !& C ' !& !& C ' ' !& C ' !& '# " ! !a b a b a b# C !& C ' !& !& C ' ' !& C ' " !& !&$ # " !a b a b c da b$ # C !& C "# " !&8 !8 8a b c da bwhich can be verified by mathematical induction. The sequence is convergent for all ,C!and in fact .C p"#8

7. Let be the balance at the end of the . TheC 8 C "

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C C #& #&" " 8 !3 3 38 ,

in which . Here is the annual interest rate, given as Therefore3 "

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________________________________________________________________________ page 76

a b!!)$$ #%! a b a bC "# "!!! "# "!!!!" !"!is .C "!$ '#%'#! dollars

15.

16. For example, take and :3 $& ? ""!

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19 . .a b a b a b a b a b+ $%%* $ $&%% $%%* %(#'$$ 3 3 3 3# # " $ #a b, "!! "!! "##. % diff % k k k k$ $$ %''*#%($'$%''*## a b- . Assuming , a b a b3 3 3 3 $ 3$ # % $ % $&'%$a b. "' $&'&. A period solutions appears near .3

a b a b a b/ . Note that With the assumption that , we have3 3 $ 3 3 $ $8" 8 8 8" 88"a b a b3 3 $ 3 3 !8" 8 8 8" 8" 8" C C 8 $, which is of the form , . It follows thata b a b3 3 $ 3 35 5" $ #$5 5 % for . Then3 3 3 3 3 3 3 3 3 3

3 3 3 3 3 $ $ $

3 3 3 3 3 $$

8 " # " $ # % $ 8 8"

" # " $ #" # $8

" # " $ #

%8

"

"

" "

a b a b a b a ba b a bc da b a b Hence Substitution of the appropriate values yieldslim

8_ "3 3 3 38 # $ # a b $$

lim8_

38 $&'**

CHAPTER 2.

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Miscellaneous Problems

1. Linear C -B B & # $

2. Homogeneous .++8 CB 68 B C - a b # #3. Exact . B BC $C C ! # $

4. Linear in .B C B - / C / a b C C5. Exact . B C BC B - # #

6. Linear . C B " / " "Ba b7. Let .? B B C " - / # # # C#

8. Linear . C % -9= # -9= B B a b #9. Exact . B C B C - # #

10. .. . B C B CB - a b # $ #11. Exact . B $ BC / - $ C

12. Linear . C - / / 68 " / B B Ba b13. Homogeneous . # CB 68 B - k k14. Exact/Homogeneous . B #BC #C $% # #

15. Separable . C --9=2 B# #a b16. Homogeneous . # $ ++8 #C B $ B 68 B - k k17. Linear . C - / / $B #B

18. Linear/Homogeneous . C - B B #

19. .. . B $C #BC "!B ! a b $20. Separable . / / - B C

21. Homogeneous . / 68 B - CB k k22. Separable . C $C B $B # $ $

23. Bernoulli . "C B B / .B -B ' # #B24. Separable . =38 B =38 C - #

25. Exact . B C ++8 CB - # a b26. .. . B B #B C C - a b # # #27. .. . B =38 B -9= #C =38 B - a b "# #28. Exact . #BC BC B - $ $

29. Homogeneous . +

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Chapter Three

Section 3.1

1. Let , so that and . Direct substitution into the differentialC / C < / C < / w ww equation yields . Canceling the exponential, the characteristica b< #< $ / !# equation is The roots of the equation are , . Hence the< #< $ ! < $ "#general solution is .C - / - /" #> $>

2. Let . Substitution of the assumed solution results in the characteristic equationC /< $< # ! < # "# The roots of the equation are , . Hence the generalsolution is .C - / - /" #> #>

4. Substitution of the assumed solution results in the characteristic equationC /#< $< " ! < "# "# The roots of the equation are , . Hence the generalsolution is .C - / - /" #># >

6. The characteristic equation is , with roots . Therefore the%< * ! < $##general solution is .C - / - /" #$># $>#

8. The characteristic equation is , with roots . Hence the< #< # ! <