soln manual for diffeq final

Chapter1Section1.11.42024y(t)4 2 2 4tFory> 3/2,theslopesarenegative,and,thereforethesolutionsdecrease. Fory< 3/2,theslopesarepositive,and,therefore,thesolutionsincrease. Asaresult,y 3/2ast 2.42024y(t)4 2 2 4tFory> 3/2,theslopesarepositive,and,thereforethesolutionsincrease. Fory< 3/2,theslopesarenegative,and,therefore,thesolutionsdecrease. Asaresult,ydivergesfrom3/2ast 3.142024y(t)4 2 2 4tFory> 3/2,theslopesarepositive,and,thereforethesolutionsincrease. Fory< 3/2,theslopesarenegative,and,therefore,thesolutionsdecrease. Asaresult, ydivergesfrom3/2ast 4.42024y(t)4 2 2 4tFory> 1/2,theslopesarenegative,and,thereforethesolutionsdecrease. Fory< 1/2,theslopesarepositive, and, therefore, thesolutionsincrease. Asaresult, y 1/2ast 5.42024y(t)4 2 2 4t2Fory> 1/2,theslopesarepositive,and,thereforethesolutionsincrease. Fory< 1/2,theslopesarenegative,and,therefore,thesolutionsdecrease. Asaresult, ydivergesfrom1/2ast 6.42024y(t)4 2 2 4tFory> 2, theslopesarepositive, and, thereforethesolutionsincrease. Fory< 2, theslopesarenegative, and, therefore, thesolutionsdecrease. Asaresult, ydivergesfrom 2ast 7. Forthesolutionstosatisfyy 3ast , weneedy

3andy

>0fory< 3. Theequationy

= 3 ysatisestheseconditions.8. Forthesolutionstosatisfyy 2/3ast ,weneedy

< 0fory> 2/3andy

> 0fory< 2/3. Theequationy

= 2 3ysatisestheseconditions.9. Forthesolutionstosatisfyydivergesfrom2, weneedy

>0fory>2andy

0fory> 1/3andy

< 0fory< 1/3. Theequationy

= 3y 1satisestheseconditions.11.42024y(t)4 2 2 4t3y= 0andy= 4areequilibriumsolutions;y 4ifinitialvalueispositive;ydivergesfrom0ifinitialvalueisnegative.12.420246y(t)4 2 2 4ty=0andy=5areequilibriumsolutions; ydivergesfrom5ifinitialvalueisgreaterthan5;y 0ifinitialvalueislessthan5.13.42024y(t)4 2 2 4ty=0isequilibriumsolution; y 0ifinitial valueisnegative; ydivergesfrom0ifinitialvalueispositive.14.442024y(t)4 2 2 4ty=0andy=2areequilibriumsolutions; ydivergesfrom0if initial valueisnegative;y 2ifinitialvalueisbetween0and2;ydivergesfrom2ifinitialvalueisgreaterthan2.15. (j)16. (c)17. (g)18. (b)19. (h)20. (e)21.(a) Let q(t) denotetheamount of chemical inthepondat timet. Thechemical q willbemeasuredingramsandthetimet will bemeasuredinhours. Therateatwhichthechemical isenteringthepondisgivenby300gallons/hour .01grams/gal =300 102. Therateat whichthechemical leaves thepondis givenby300gallons/hourq/1, 000, 000grams/gal =300 q106. Therefore, thedierential equationisgivenbydq/dt = 300(102q106).(b) Ast , 102 q1060. Therefore, q 104g. Thelimitingamountdoesnotdependontheamountthatwaspresentinitially.22. ThesurfaceareaofasphericalraindropofradiusrisgivenbyS=4r2. Thevolumeof aspherical raindropisgivenbyV =4r3/3. Therefore, weseethatthesurfaceareaS=cV2/3for someconstant c. If theraindropevaporates at arateproportional toitssurfacearea,thendV/dt = kV2/3forsomek > 0.23. Thedierencebetweenthetemperatureoftheobjectandtheambienttemperatureisu 70. Sincethedierenceisdecreasingifu>70(andincreasingifu 1250,thenq

> 0. Ifq< 1250,thenq

> 0. Therefore,q 1250.25.(a) Followingthediscussioninthetext,theequationisgivenbymv

= mg kv2.(b) Afteralongtime,v

0. Therefore,mg kv20,orv

mg/k(c) Weneedtosolvetheequation

.025 9.8/k=35. Solvingthisequation, weseethatk = 0.0002kg/m26.42024y(t)4 2 2 4tyisasymptotictot 3ast 27.42024y(t)4 2 2 4ty 0ast 28.642024y(t)4 2 2 4ty , 0,or dependingontheinitialvalueofy29.42024y(t)4 2 2 4ty or dependingwhethertheinitialvalueliesaboveorbelowtheliney= t/2.30.42024y(t)4 2 2 4ty or oryoscillatesdependingwhethertheinitial valueofyliesaboveorbelowthesinusoidalcurve.31.742024y(t)4 2 2 4ty orisasymptoticto2t 1dependingontheinitialvalueofy32.42024y(t)4 2 2 4ty 0andthenfailstoexistaftersometf 033.42024y(t)4 2 2 4ty or dependingontheinitialvalueofySection1.21.8Chapter2Section2.11.(a)42024y(t)4 2 2 4 6 8 10t(b) Allsolutionsseemtoconvergetoanincreasingfunctionast .(c) Theintegratingfactoris(t) = e3t. Thene3ty

+ 3e3ty= e3t(t + e2t) =(e3ty)

= te3t+ et=e3ty=

(te3t+ et) dt =13te3t19e3t+ et+ c=y=t319+ e2t+ ce3t.Weconcludethatyisasymptotictot/3 1/9ast .2.(a)42024y(t)1 0.5 0.5 1 1.5 2t1(b) All slopeseventuallybecomepositive, soall solutionswill eventuallyincreasewithoutbound.(c) Theintegratingfactoris(t) = e2t. Thene2ty

2e2ty= e2t(t2e2t) =(e2ty)

= t2=e2ty=

t2dt =t33+ c=y=t33 e2t+ ce2t.Weconcludethatyincreasesexponentiallyast .3.(a)42024y(t)1 0.5 0.5 1 1.5 2t(b) Allsolutionsappeartoconvergetothefunctiony(t) = 1.(c) Theintegratingfactoris(t) = et. Therefore,ety

+ ety= t + et=(ety)

= t + et=ety=

(t + et) dt =t22+ et+ c=y=t22 et+ 1 + cet.Therefore,weconcludethaty 1ast .4.(a)242024y(t)1 1 2 3 4 5t(b) Thesolutionseventuallybecomeoscillatory.(c) Theintegratingfactoris(t) = t. Therefore,ty

+ y= 3t cos(2t) =(ty)

= 3t cos(2t)=ty=

3t cos(2t) dt =34 cos(2t) +32t sin(2t) + c=y= +3 cos 2t4t+3 sin 2t2+ct.Weconcludethatyisasymptoticto(3 sin 2t)/2ast .5.(a)42024y(t)1 0.5 0.5 1 1.5 2t(b) All slopes eventually become positive so all solutions eventually increase without bound.(c) Theintegratingfactoris(t) = e2t. Therefore,e2ty

2e2ty= 3et=(e2ty)

= 3et=e2ty=

3etdt = 3et+ c=y= 3et+ ce2t.Weconcludethatyincreasesexponentiallyast .36.(a)42024y(t)4 2 2 4t(b) Fort > 0,allsolutionsseemtoeventuallyconvergetothefunctiony= 0.(c) Theintegratingfactoris(t) = t2. Therefore,t2y

+ 2ty= t sin(t) =(t2y)

= t sin(t)=t2y=

t sin(t) dt = sin(t) t cos(t) + c=y=sin t t cos t + ct2.Weconcludethaty 0ast .7.(a)42024y(t)4 2 2 4t(b) Fort > 0,allsolutionsseemtoeventuallyconvergetothefunctiony= 0.4(c) Theintegratingfactoris(t)=et2. Therefore, usingthetechniquesshownabove, weseethaty(t) = t2et2+ cet2. Weconcludethaty 0ast .8.(a)42024y(t)4 2 2 4t(b) Fort > 0,allsolutionsseemtoeventuallyconvergetothefunctiony= 0.(c) Theintegratingfactoris(t) = (1 + t2)2. Then(1 + t2)2y

+ 4t(1 + t2)y=11 + t2=((1 + t2)2y) =

11 + t2 dt=y= (tan1(t) + c)/(1 + t2)2.Weconcludethaty 0ast .9.(a)42024y(t)4 2 2 4t5(b) All slopes eventuallybecome positive. Therefore, all solutions will increase withoutbound.(c) Theintegratingfactoris(t) = et/2. Therefore,2et/2y

+ et/2y= 3tet/2=2et/2y=

3tet/2dt = 6tet/212et/2+ c=y= 3t 6 + cet/2.Weconcludethaty 3t 6ast .10.(a)42024y(t)1 1 2 3 4 5t(b) For y> 0, the slopes are all positive, and, therefore, the corresponding solutions increasewithout bound. For y< 0 almost all solutions have negative slope and therefore decreasewithoutbound.(c) By dividing the equation by t, we see that the integrating factor is (t) = 1/t. Therefore,y

/t y/t2= tet=(y/t)

= tet=yt=

tetdt = tetet+ c=y= t2ettet+ ct.Weconcludethaty ifc > 0,y ifc < 0andy 0ifc = 0.11.(a)642024y(t)2 1 1 2t(b) Thesolutionappearstobeoscillatory.(c) Theintegratingfactoris(t) = et. Therefore,ety

+ ety= 5etsin(2t) =(ety)

= 5etsin(2t)=ety=

5etsin(2t) dt = 2etcos(2t) + etsin(2t) + c =y= 2 cos(2t) + sin(2t) + cet.Weconcludethaty sin(2t) 2 cos(2t)ast .12.(a)42024y(t)2 1 1 2t(b) Allslopesareeventuallypositive. Therefore,allsolutionsincreasewithoutbound.(c) Theintegratingfactoris(t) = et/. Therefore,2et/2y

+ et/2y= 3t2et/2=(2et/2y)

= 3t2et/2=2et/2y=

3t2et/2dt = 6t2et/224tet/2+ 48et/2+ c=y= 3t212t + 24 + cet/2.Weconcludethatyisasymptoticto3t212t + 24ast .713. Theintegratingfactoris(t) = et. Therefore,(ety)

= 2tet=y= et

2tetdt = 2te2t2e2t+ cet.The initial condition y(0) = 1 implies 2 +c = 1. Therefore, c = 3 and y= 3et+2(t 1)e2t14. Theintegratingfactoris(t) = e2t. Therefore,(e2ty)

= t =y= e2t

t dt =t22 e2t+ ce2t.The initial conditiony(1) =0implies e2t/2+ce2t=0. Therefore, c =1/2, andy= (t21)e2t/215. Dividingtheequationbyt,weseethattheintegratingfactoris(t) = t2. Therefore,(t2y)

= t3t2+ t =y= t2

(t3t2+ t) dt =

t24 t3+12+ct2

.Theinitialconditiony(1) = 1/2impliesc = 1/12,andy= (3t44t3+ 6t2+ 1)/12t2.16. Theintegratingfactoris(t) = t2. Therefore,(t2y)

= cos(t) =y= t2

cos(t) dt = t2(sin(t) + c).Theinitialconditiony() = 0impliesc = 0andy= (sin t)/t217. Theintegratingfactoris(t) = e2t. Therefore,(e2ty)

= 1 =y= e2t

1 dt = e2t(t + c).Theinitialconditiony(0) = 2impliesc = 2andy= (t + 2)e2t.18. Afterdividingbyt,weseethattheintegratingfactoris(t) = t2. Therefore,(t2y)

= 1 =y= t2

t sin(t) dt = t2(sin(t) t cos(t) + c).The initial condition y(/2) = 1 implies c = (2/4)1 and y= t2[(2/4)1t cos t+sin t].19. Afterdividingbyt3,weseethattheintegratingfactoris(t) = t4. Therefore,(t4y)

= tet=y= t4

tetdt = t4(tetet+ c).Theinitialconditiony(1) = 0impliesc = 0andy= (1 + t)et/t4, t = 020. Afterdividingbyt,weseethattheintegratingfactoris(t) = tet. Therefore,(tety)

= tet=y= t1et

tetdt = t1et(tetet+ c) = t1(t 1 + cet).Theinitialconditiony(ln 2) = 1impliesc = 2andy= (t 1 + 2et)/t, t = 021.8(a)42024y(t)4 2 2 4tThesolutionsappeartodivergefromanoscillatorysolution. Itappearsthata0 1.Fora> 1, thesolutionsincreasewithoutbound. Fora< 1, thesolutionsdecreasewithoutbound.(b) Theintegratingfactoris(t) = et/2. Fromthis,weconcludethatthegeneralsolutionisy(t) = (8 sin(t) 4 cos(t))/5 + cet/2. Thesolutionwillbesinusoidalaslongasc = 0.The initial condition for the sinusoidal behavior is y(0) = (8 sin(0) 4 cos(0))/5 = 4/5.Therefore,a0= 4/5.(c) yoscillatesfora = a022.(a)42024y(t)4 2 2 4tAll solutions eventually increase or decrease without bound. The value a0appears to beapproximatelya0= 3.(b) Theintegratingfactoris(t) = et/2,andthegeneralsolutionisy(t) = 3et/3+ cet/2.The initial condition y(0) = a implies y= 3et/3+(a +3)et/2. The solution will behavelike(a + 3)et/2. Therefore,a0= 3.9(c) y fora = a023.(a)42024y(t)1 1 2 3 4 5tSolutionseventuallyincreaseordecreasewithoutbound,dependingontheinitialvaluea0. Itappearsthata0 1/8.(b) Dividing the equation by 3, we see that the integrating factor is (t) = e2t/3. Therefore,the solution is y= [(2 +a(3 +4))e2t/32et/2]/(3 +4). The solution will eventuallybehavelike(2 + a(3 + 4))e2t/3/(3 + 4). Therefore,a0= 2/(3 + 4).(c) y 0fora = a024.(a)42024y(t)1 2 3 4 5tIt appears that a0 .4. Ast 0, solutionsincreasewithout boundif y >a0anddecreasewithoutboundify< a0.10(b) The integrating factor is (t) = tet. The general solution is y= tet+cet/t. The initialconditiony(1) = aimpliesy= tet+ (ea 1)et/t. Ast 0,thesolutionwillbehavelike(ea 1)et/t. Fromthis,weseethata0= 1/e.(c) y 0ast 0fora = a025.(a)42024y(t)5 4 3 2 1tItappearsthata0 .4. Thatis, ast 0, fory(/2)>a0, solutionswill increasewithoutbound,whilesolutionswilldecreasewithoutboundfory(/2) < a0.(b) After dividingbyt, weseethat theintegratingfactor is t2, andthesolutionis y =cos t/t2+ 2a/4t2. Sincelimt0 cos(t)=1, solutionswill increasewithoutboundifa > 4/2anddecreasewithoutboundifa < 4/2. Therefore,a0= 4/2.(c) Fora0= 4/2,y= (1 cos(t))/t21/2ast 0.26.(a)42024y(t)0.5 1 1.5 2 2.5 3t11It appears that a0 2. For y(1) > a0, the solution will increase without bound as t 0,whilethesolutionwilldecreasewithoutboundify(t) < a0.(b) Afterdividingbysin(t),weseethattheintegratingfactoris(t) = sin(t). Asaresult,weseethatthesolutionisgivenbyy=(et+ c) sin(t). Applyingourinitial condition,weseethat our solutionis y =(et e + a sin 1)/ sin t. Thesolutionwill increaseif1 e + a sin 1>0anddecreaseif 1 e + a sin 1121516.(a) Rewriting as 4y3dy= x(x2+1)dx, then integrating both sides, we have y4= (x2+1)2/4+C. Theinitialcondition,y(0) = 1/2impliesC= 0. Therefore,y=

(x2+ 1)/2.(b)42024y4 2 2 4x(c) < x < 17.(a) Rewritingas(2y 5)dy= (3x2ex)dx,thenintegratingbothsides,wehavey25y=x3ex+C. The initial condition, y(0) = 1 implies C= 3. Completing the square, wehave(y 5/2)2= x3ex+ 13/4. Therefore,y= 5/2

x3ex+ 13/4.(b)2042024y2 1 1 2 3 4 5x(c) 1.4445 < x < 4.6297approximately18.(a) Rewriting as (3 +4y)dy= (exex)dx, then integrating both sides, we have 3y +2y2=(ex+ex) +C. The initial condition,y(0) = 1 implies C= 7. Completing the square,wehave(y + 3/4)2= (ex+ ex)/2 + 65/16. Therefore,y= 34+1465 8ex8ex.(b)10.50.511.52y2 1 1 2x(c) |x| < 2.0794approximately19.(a) Rewriting as cos(3y)dy= sin(2x)dx, then integrating both sides, we have sin(3y)/3 =cos(2x)/2 + C. Theinitial condition, y(/2)=/3impliesC=1/2. Therefore, y=[ arcsin(3 cos2x)]/3.(b)2110.50.511.52y2 1 1 2x(c) |x /2| < 0.6155approximately20.(a) Rewritingasy2dy= arcsin(x)/1 x2dx,thenintegratingbothsides,wehavey3/3 =(arcsin(x))2/2 + C. Theinitial condition, y(0) =1/impliesC=0. Therefore, y=

32(arcsin x)2

1/3.(b)10.50.511.52y1 0.8 0.6 0.4 0.2 0.2 0.4 0.6 0.8 1x(c) 1 < x < 121. Rewritingtheequationas(3y2 6y)dy=(1 + 3x2)dxandintegratingbothsides, wehavey3 3y2=x + x3+ C. Theinitial condition, y(0)=1impliesc= 2. Therefore,y33y2xx3+2 = 0. When 3y26y= 0, the integral curve will have a vertical tangent.Inparticular,wheny= 0, 2. Fromoursolution,weseethaty= 0impliesx = 1andy= 2impliesx = 1. Therefore,thesolutionisdenedfor 1 < x < 1.22. Rewritingtheequationas (3y2 4)dy =3x2dxandintegratingbothsides, wehavey34y= x3+C. The initial condition y(1) = 0 implies C= 1. Therefore, y34yx3= 1.When3y2 4=0, theintegral curvewill haveavertical tangent. Inparticular, wheny= 2/3. Atthesevaluesfory, wehavex= 1.276, 1.598. Therefore, thesolutionisdenedfor 1.276 < x < 1.5982223. Rewritingthe equationas y2dy =(2+x)dxandintegratingbothsides, we havey1=2x+x2/2+C. The initial conditiony(0) =1 implies C =1. Therefore,y= 1/(x2/2 + 2x 1). Tondwherethefunctionattainsitminimumvalue, welookwherey

=0. Weseethaty

=0impliesy=0orx= 2. But, asseenbythesolutionformula, yisneverzero. Further, itcanbeveriedthaty

(2)>0, and, therefore, thefunctionattainsaminimumatx = 2.24. Rewritingtheequationas(3 + 2y)dy= (2 ex)dxandintegratingbothsides,wehave3y + y2=2x ex+ C. Bytheinitialconditiony(0)=0,wehaveC=1. Completingthesquare, it follows that y= 3/2+

2x ex+ 13/4. The solution is dened if 2xex+13/4 0, thatis, 1.5 x 2(approximately). Inthatinterval, y=0forx=ln 2. Itcanbeveried that y

(ln 2) < 0, and, therefore, the function attains its maximum value at x = ln 2.25. Rewritingtheequationas(3 + 2y)dy= 2 cos(2x)dxandintegratingbothsides,wehave3y +y2= sin(2x) +C. By the initial condition y(0) = 1, we have C= 2. Completing thesquare, it follows that y= 3/2+

sin(2x) + 1/4. The solution is dened for sin(2x)+1/4 0. Thatis, 0.126 x 1.44. Tondwherethesolutionattainsitsmaximumvalue, weneedtocheckwherey

=0. Weseethaty

=0when2 cos(2x) =0. Intheinterval ofdenitionabove,thatoccurswhen2x = /2,orx = /4.26. Rewritingthisequationas(1 + y2)1dy=2(1 + x)dxandintegratingbothsides, wehavetan1(y)=2x + x2+ C. TheinitialconditionimpliesC=0. Therefore,thesolutionis y= tan(x2+2x). The solution is dened as long as /2 < 2x +x2< /2. We note that2x +x2 1. Further, 2x +x2= /2 for x = 2.6 and 0.6. Therefore, the solution is validintheinterval 2.6 0 for all t or y(t) < 0 for all t. Therefore, if y0> 0 and |y/(y 4)| = Ce2t2/3,wemusthavey 4. Ontheotherhand, if y0 1 (since the term t2/(1 +t3) has a singularity at t = 1.Therefore,wecanconcludethatoursolutionwillexistfor[e3y20/21]1/3< t < .17.5242024y(t)2 1 1 2tIfy0> 0,theny 3. Ify0= 0,theny= 0. Ify0< 0,theny .18.42246810y(t)2 1 1 2tIfy0 0,theny 0. Ify0< 0,theny .19.5051015y(t)2 1 1 2tIfy0> 9,theny . Ify0< 9,theny 0.20.5342024y(t)2 2 4 6 8 10tIfy0< yc 0.019,theny . Otherwise,yisasymptotictot 1.21.(a) We know that the family of solutions given by equation (19) are solutions of this initial-valueproblem. Wewanttodetermineif oneof thesepassesthroughthepoint(1, 1).Thatis,wewanttondt0> 0suchthatify= [23(t t0)]3/2,then(t, y) = (1, 1). Thatis,weneedtondt0> 0suchthat1 =23(1 t0). But,thesolutionofthisequationist0= 1/2.(b) From the analysis in part (a), we nd a solution passing through (2, 1) by setting t0= 1/2.(c) Sinceweneedy0= [23(2 t0)]3/2,wemusthave |y0| [43]3/2.22.(a) First,itisclearthaty1(2) = 1 = y2(2). Further,y

1= 1 = t + [(t 2)2]1/22= t + (t2+ 4(1 t))1/22andy

2= t2= t + (t2t2)1/22.Thefunctiony1isasolutionfort 2. Thefunctiony2isasolutionforallt.(b) Theorem2.4.2requiresthatfandf/ybecontinuousinarectangleaboutthepoint(t0, y0) =(2, 1). Sincef isnotcontinuousif t 1.33. Thesolutionofy

+ 2y=0withy(0)=1isgivenbyy(t)=e2tfor0 t 1. Theny(1)=e2. Then, fort>1, thesolutionoftheequationy

+ y=0isy=cet. Sincewewanty(1) = e2,weneedce1= e2. Therefore,c = e1. Therefore,y(t) = e1et= e1tfort > 1.34.(a) Multiplyingtheequationbyett0p(s) ds,wehave

ett0p(s) dsy

= ett0p(s) dsg(t).Integratingwehaveett0p(s) dsy(t) = y0 +

tt0est0p(r) drg(s) ds,whichimpliesy(t) = y0ett0p(s) ds+

tt0etsp(r) drg(s) ds.(b) Assumep(t) p0> 0forallt t0and |g(t)| Mforallt t0. Therefore,

tt0p(s) ds

tt0p0 ds = p0(t t0)whichimpliesett0p(s) ds ett0p0 ds= ep0(tt0) 1fort t0.Also,

tt0etsp(r) drg(s) ds

tt0etsp(r) dr|g(s)| ds

tt0ep0(ts)M ds Mep0(ts)p0

tt0= M

1p0ep0(tt0)p0

Mp057(c) Let p(t) = 2t +1 1 for all t 0 and let g(t) = et2. Therefore, |g(t)| 1 for all t 0.Bytheanswertopart(a),y(t) = et0(2s+1) ds+

t0ets(2r+1) dres2ds= e(t2+t)+ et2t

t0esds= et2.Weseethatysatisesthepropertythatyisboundedforalltimet 0.Section2.51.0123456f0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2yThe only equilibrium point is y= 0. Since f

(0) = a > 0, the equilibrium point is unstable.00.511.522.53y(t)2 1 1 2t2.58Section2.61. HereM(x, y) = 2x + 3andN(x, y) = 2y 2. SinceMy= Nx= 0,theequationisexact.Since x= M= 2x +3, to solve for , we integrate Mwith respect to x. We conclude that=x2+ 3x + h(y). Theny=h

(y)=N=2y 2impliesh(y)=y2 2y. Therefore,(x, y) = x2+ 3x +y22y= c.420246810y10 8 6 4 2 2 4x2. HereM(x, y)=2x + 4yandN(x, y)=2x 2y. SinceMy =Nx, theequationisnotexact.3. HereM(x, y)=3x2 2xy + 2andN(x, y)=6y2 x2+ 3. SinceMy= 2x=Nx, theequation is exact. Since x= M= 3x22xy+2, to solve for , we integrate Mwith respectto x. We conclude that = x3x2y +2x+h(y). Then y= x2+h

(y) = N= 6y2x2+3implies h

(y) = 6y2+3. Therefore, h(y) = 2y3+3y and (x, y) = x3x2y+2x+2y3+3y= c.42024y4 2 2 4x4. HereM(x, y)=2xy2+ 2yandN(x, y)=2x2y + 2x. SinceMy=4xy + 2=Nx, theequation is exact. Since x= M= 2xy2+2y, to solve for , we integrate Mwith respect tox. We conclude that = x2y2+2xy +h(y). Then y= 2x2y +2x +h

(y) = N= 2x2y +2ximpliesh

(y) = 0. Therefore,h(y) = Cand(x, y) = x2y2+ 2xy= c.7642024y4 2 2 4x5. Here M(x, y) = ax+byand N(x, y) = bx+cy. Since My= b = Nx, the equation is exact.Since x= M= ax+by, to solve for , we integrate Mwith respect to x. We conclude that= ax2/2 +bxy +h(y). Then y= bx+h

(y) = N= bx+cyimplies h

(y) = cy. Therefore,h(y) = cy2/2and(x, y) = ax2/2 + bxy + cy2/2 = c.42024y4 2 2 4x6. HereM=ax byandN=bx cy. SinceMy= bandNx=b, theequationisnotexact.7. HereM(x, y) = exsin y 2y sin xandN(x, y) = excos y + 2 cos x. SinceMy= excos y sin x=Nx, theequationisexact. Sincex=M=exsin y 2y sin x, tosolvefor, weintegrateMwithrespect tox. Weconcludethat =exsin y+ 2y cos x + h(y). Theny= excos y +2 cos x+h

(y) = N= excos y +2 cos x implies h

(y) = 0. Therefore, h(y) = Cand(x, y) = exsin y + 2y cos x = c.7742024y4 2 2 4x8. HereM=exsin y+ 3yandN= 3x + exsin y. Therefore, My=excos y+ 3andNx= 3 + exsin y. SinceMy = Nx,therefore,theequationisnotexact.9. HereM(x, y)=yexycos(2x) 2exysin(2x) + 2xandN(x, y)=xexycos(2x) 3. SinceMy= exycos(2x)+xyexycos(2x)2xexysin(2x) = Nx, the equation is exact. Since x= M=yexycos(2x)2exysin(2x)+2x, to solve for , we integrate Mwith respect to x. We concludethat=exycos(2x) + x2+ h(y). Theny=xexycos(2x) + h

(y)=N=xexycos(2x) 3impliesh

(y) = 3. Therefore,h(y) = 3yand(x, y) = exycos(2x) + x23y= c.2112y2 1 1 2x10. HereM(x, y) = y/x + 6xandN(x, y) = ln(x) 2. SinceMy= 1/x = Nx,theequationisexact. Sincex=M=y/x + 6x, tosolvefor, weintegrateMwithrespecttox. Weconcludethat=y ln(x) + 3x2+ h(y). Theny=ln(x) + h

(y)=N=ln(x) 2impliesh

(y) = 2. Therefore,h(y) = 2yand(x, y) = y ln(x) + 3x22y= c.7821012y0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2x11. HereM(x, y) =x ln(y) + xyandN(x, y) =y ln(x) + xy. SinceMy=x/y+ xandNx= y/x +y,weconcludethattheequationisnotexact.12. Here M(x, y) = x/(x2+y2)3/2and N(x, y) = y/(x2+y2)3/2. Since My= Nx, the equationisexact. Sincex= M= x/(x2+ y2)3/2,tosolvefor,weintegrateMwithrespecttox.Weconcludethat= 1/(x2+ y2)1/2+ h(y). Theny=y/(x2+ y2)3/2+ h

(y)=N=y/(x2+ y2)3/2impliesh

(y)=0. Therefore, h(y)=0and(x, y)= 1/(x2+ y2)1/2=cwhichimpliesthat(x, y) = (x2+y2) = c.2112y2 1 1 2x13. HereM(x, y) =2x y andN(x, y) =2y x. Therefore, My=Nx= 1whichimplies that the equationis exact. IntegratingMwithrespect tox, we conclude that=x2 xy + h(y). Theny= x + h

(y)=N=2y ximpliesh

(y)=2y. Therefore,h(y) = y2and we conclude that = x2xy +y2= C. The initial condition y(1) = 3 impliesc = 7. Therefore,x2xy +y2= 7. Solvingfory,weconcludethaty=12

x +28 3x2

.Therefore,thesolutionisvalidfor3x2 28.14. HereM(x, y)=9x2+ y 1andN(x, y)= 4y + x. Therefore, My=Nx=1whichimplies that the equationis exact. IntegratingMwithrespect tox, we conclude that=3x3+ xy x + h(y). Theny=x + h

(y) =N= 4y+ ximplies h

(y) = 4y.Therefore, h(y)= 2y2andweconcludethat=3x3+ xy x 2y2=C. Theinitialconditiony(1)=0impliesc=2. Therefore, 3x3+ xy x 2y2=2. Solvingfory, weconcludethaty= [x (24x3+x28x 16)1/2]/4. Thesolutionisvalidforx > 0.9846.7915. HereM(x, y)=xy2+ bx2yandN(x, y)=x3+ x2y. Therefore, My=2xy + bx2andNx=3x2+ 2xy. Inorderfortheequationtobeexact, weneedb=c. Takingthisvaluefor b, weintegratingMwithrespect tox. Weconcludethat =x2y2/2 + x3y+ h(y).Theny=x2y + x3+ h

(y)=N=x3+ x2yimpliesh

(y)=0. Therefore, h(y)=Cand(x, y) = x2y2/2 + x3y= C. Thatis,thesolutionisgivenimplicitlyasx2y2/2 + x3y= c.16. Here M(x, y) =ye2xy+xandN(x, y) =bxe2xy. ThenMy=e2xy+2xye2xyandNx=be2xy+ 2bxye2xy. Theequationwill beexactaslongasb=1. IntegratingMwithrespect to x, we conclude that = e2xy/2+x2/2+h(y). Then y= xe2xy+h

(y) = N= xe2xyimpliesh

(y) = 0. Therefore,h(y) = 0andweconcludethatthesolutionisgivenimplicitlybytheequatione2xy+ x2= C.17. Wenoticethat(x, y) = f(x) + g(y). Therefore,x= f

(x)andy= g

(y). Thatis,x= M(x, y0) y= N(x0, y).Furthermore,xy= Myandyx= Nx. Basedonthehypothesis,xy= yxandMy= Nx.18. Wenoticethat(M(x))y= 0 = (N(y))x. Therefore,theequationisexact.19. HereM(x, y)=x2y3andN(x, y)=x + xy2. Therefore,My=3x2y2andNx=1 + y2.Weseethattheequationisnotexact. Now, multiplyingtheequationby(x, y)=1/xy3,theequationbecomesxdx + (1 + y2)/y3dy= 0.NowweseethatforthisequationM=xandN=(1 + y2)/y3. Therefore, My=0=Nx.IntegratingMwithrespecttox,weseethat= x2/2 + h(y). Further,y= h

(y) = N=(1+y2)/y3= 1/y3+1/y. Therefore, h(y) = 1/2y2+ln(y) and we conclude that the solutionoftheequationisgivenimplicitlybyx21/y2+ 2 ln(y) = C.0.10.150.20.250.3y2 1 0 1 2x20. Multiplyingtheequationby(x, y) = yex,theequationbecomes(exsin y 2y sin x)dx + (excos y + 2 cos x)dy= 0.Now we see that for this equation M= exsin y2y sin x and N= excos y+2 cos x. Therefore,My=excos y 2 sin x=Nx. IntegratingMwithrespecttox, weseethat=exsin y +2y cos x + h(y). Further,y= excos y + 2 cos x + h

(y) = N= excos y + 2 cos x. Therefore,h(y) = 0andweconcludethatthesolutionoftheequationisgivenimplicitlybyexsin y +2y cos x = C.802112y2 1 1 2x21. Multiplyingtheequationby(x, y) = y,theequationbecomesy2dx + (2xy y2ey)dy= 0.NowweseethatforthisequationM= y2andN= 2xy y2ey. Therefore,My= 2y= Nx.IntegratingMwithrespecttox,weseethat= xy2+ h(y). Further,y= 2xy + h

(y) =N= 2xy y2ey. Therefore,h

(y) = y2eywhichimpliesthath(y) = ey(y22y + 2),andwe conclude that the solution of the equation is given implicitly by xy2ey(y22y+2) = C.2112y2 1 1 2x22. Multiplyingtheequationby(x, y) = xex,theequationbecomes(x2+ 2x)exsin ydx +x2excos ydy= 0.Now we see that for this equation My= (x2+2x)excos y= Nx. Integrating Mwith respecttox, weseethat=x2exsin y + h(y). Further, y=x2excos y + h

(y)=N=x2excos y.Therefore, h

(y)=0whichimpliesthatthesolutionof theequationisgivenimplicitlybyx2exsin y= C.8121012y1.2 1.4 1.6 1.8 2x23. Suppose is an integrating factor which will make the equation exact. Then multiplyingtheequationby,wehaveMdx + Ndy= 0.Thenweneed(M)y=(N)x. Thatis, weneedyM+ My=xN+ Nx. Thenwerewrite the equation as (NxMy) = yMxN. Suppose does not depend on x. Thenx= 0. Therefore,(NxMy) = yM. Usingtheassumptionthat(NxMy)/M= Q(y),wecanndanintegratingfactorbychoosingwhichsatisesy/=Q. Weconcludethat(y) = exp

Q(y) dyisanintegratingfactorofthedierentialequation.24. Suppose is an integrating factor which will make the equation exact. Then multiplyingtheequationby,wehaveMdx + Ndy= 0.Thenweneed(M)y=(N)x. Thatis, weneedyM+ My=xN+ Nx. Thenwerewritetheequationas (Nx My) =yM xN. Bythegivenassumption, weneedtosatisfyR(xM yN)=yM xN. Thisequationissatisedif y=(x)Randx= (y)R. Consider = (xy). Thenx=

yandy=

xwhere = d/dzforz= xy.Therefore,weneedtochoosetosatisfy

= R. Thisequationisseparablewithsolution = exp(

R(z) dz).25. Since(MyNx)/N= 3isafunctionofxonly,weknowthat = e3xisanintegratingfactorforthisequation. Multiplyingtheequationby,wehavee3x(3x2y + 2xy + y3)dx + e3x(x2+ y2)dy= 0.Then My= e3x(3x2+2x +3y2) = Nx. Therefore, this new equation is exact. Integrating Mwithrespecttox,weconcludethat= (x2y + y3/3)e3x+ h(y). Theny= (x2+ y2)e3x+h

(y)=N=e3x(x2+ y2). Therefore, h

(y)=0andweconcludethatthesolutionisgivenimplicitlyby(3x2y + y3)e3x= c.8221012y1.5 1.6 1.7 1.8 1.9 2x26. Since (MyNx)/N= 1 is a function of x only, we know that = exis an integratingfactorforthisequation. Multiplyingtheequationby,wehave(exexyex)dx + exdy= 0.ThenMy= ex= Nx. Therefore,thisnewequationisexact. IntegratingMwithrespecttox, weconcludethat= ex ex+ yex+ h(y). Theny=ex+ h

(y)=N=ex.Therefore, h

(y)=0andweconcludethatthesolutionisgivenimplicitlyby ex ex+yex= c.2112y2 1.5 1 0.5 0.5 1 1.5x27. Since(Nx My)/M=1/yisafunctionofyonly,weknowthat(y)=e1/y dy=yisanintegratingfactorforthisequation. Multiplyingtheequationby,wehaveydx + (x y sin y)dy= 0.Then for this equation,My= 1 = Nx. Therefore, this new equation is exact. Integrating Mwithrespecttox,weconcludethat= xy + h(y). Theny= x + h

(y) = N= x y sin y.Therefore, h

(y) = y sin ywhich implies that h(y) = siny +y cos y, and we conclude thatthesolutionisgivenimplicitlybyxy sin y + y cos y= C.832112y2 1 1 2x28. Since (NxMy)/M= (2y1)/y is a function of y only, we know that (y) = e21/y dy=e2y/yisanintegratingfactorforthisequation. Multiplyingtheequationby,wehavee2ydx + (2xe2y1/y)dy= 0.Then for this equation, My= Nx. Therefore, this new equation is exact. Integrating Mwithrespect to x, we conclude that = xe2y+h(y). Then y= 2xe2y+h

(y) = N= 2xe2y1/y.Therefore, h

(y) = 1/y which implies that h(y) = ln(y), and we conclude that the solutionisgivenimplicitlybyxe2yln(y) = C.11.21.41.61.82y2 1 0 1 2x29. Since(Nx My)/M=cot(y)isafunctionofyonly,weknowthat(y)=ecot(y) dy=sin(y)isanintegratingfactorforthisequation. Multiplyingtheequationby,wehaveexsin ydx + (excos y + 2y)dy= 0.Thenforthisequation, My=Nx. Therefore, thisnewequationisexact. IntegratingMwithrespecttox,weconcludethat= exsin y + h(y). Theny= excos y + h

(y) = N=excos y + 2y. Therefore,h

(y) = 2ywhichimpliesthath(y) = y2,andweconcludethatthesolutionisgivenimplicitlybyexsiny + y2= C.842112y2 1 1 2x30. Since(NxMy)/M= 2/yisafunctionofyonly,weknowthat(y) = e2/y dy= y2isanintegratingfactorforthisequation. Multiplyingtheequationby,wehave(4x3+ 3y)dx + (3x + 4y3)dy= 0.Then for this equation, My= Nx. Therefore, this new equation is exact. Integrating Mwithrespect tox,weconcludethat = x4+3xy +h(y). Theny= 3x +h

(y) = N= 3x +4y3.Therefore, h

(y)=4y3whichimpliesthath(y)=y4, andweconcludethatthesolutionisgivenimplicitlybyx4+ 3xy +y4= C.2112y2 1 1 2x31. Since(Nx My)/(xM yN)=1/xyisafunctionof xyonly, weknowthat(xy)=e1/xy dy=xyisanintegratingfactorforthisequation. Multiplyingtheequationby,wehave(3x2y + 6x)dx + (x3+ 3y2)dy= 0.Then for this equation, My= Nx. Therefore, this new equation is exact. Integrating Mwithrespect to x, we conclude that = x3y +3x2+h(y). Then y= x3+h

(y) = N= x3+3y2.Therefore, h

(y)=3y2whichimpliesthath(y)=y3, andweconcludethatthesolutionisgivenimplicitlybyx3y + 3x2+y3= C.852112y2 1 1 2x32. Usingtheintegratingfactor = [xy(2x +y)]1,thisequationcanberewrittenas

2x+22x +y

dx +

1y+12x +y

dy= 0.IntegratingMwithrespecttox, weseethat=2 ln |x| + ln |2x + y| + h(y). Theny=(2x + y)1+ h

(y) =N=(2x + y)1+ 1/y. Therefore, h

(y) =1/ywhichimpliesthath(y)=ln |y|. Therefore, =2 ln |x| + ln |2x + y| + ln |y| =C. Applyingtheexponentialfunction,weconcludethatthesolutionisgivenimplicitlybe2x3y +x2y2= C.Section2.71. TheEulerformulaisyn+1= yn + h(3 + tnyn)inwhichtn= t0 + nh. Sincet0= 0,wehaveyn+1= yn(1 h) + 3h +nh2.(a) Forh = 0.05,theEulerapproximationsforynatn = 2, 4, 6, 8aregivenby1.1975, 1.38549, 1.56491, 1.73658(b) Forh = 0.025,theEulerapproximationsforynatn = 4, 8, 12, 16aregivenby1.19631, 1.38335, 1.56200, 1.733082. TheEulerformulaisyn+1= yn + h(5tn3yn)inwhichtn= t0 + nh. Sincet0= 0,wehaveyn+1= yn + 5nh23hynwithy0= 2.(a) Forh = 0.05,theEulerapproximationsforynatn = 2, 4, 6, 8aregivenby1.59980, 1.29288, 1.07242, 0.930175.(b) Forh = 0.025,theEulerapproximationsforynatn = 4, 8, 12, 16aregivenby1.61124, 1.31361, 1.10012, 0.9625523. TheEulerformulaisyn+1=yn + h(2yn 3tn)inwhichtn=t0 + nh. Sincet0=0, wehaveyn+1= yn(1 + 2h) 3nh2.86Chapter3Section3.11.(a)A =

2 33 1

=A1=111

1 33 2

.Therefore,x = A1b =111

1 33 2

75

=

21

.Thatis,x1= 2, x2= 1.(b) Thelinesareintersecting,asshownbelow.1050510x22 1 1 2 3 4 5x12.(a)A =

1 22 3

=A1=17

3 22 1


3 22 1

106

=

62

.Thatis,x1= 6,x2= 2.(b) Therefore,thelinesareintersecting,asshownbelow.14202x22 4 6 8x13.(a)A =

1 32 1

=A1=17

1 32 1


1 32 1

00

=

00

.Thatis,x1= 0,x2= 0.(b) Therefore,thelinesareintersecting,asshownbelow.202468x21 1 2 3 4x14.(a)A =

1 22 4

det A = 0. Therefore, A is singular which implies either there is no solution or an innitenumberofsolutions. Multiplyingtherstequationby2,wehave 2x1 + 4x2= 8. Oursecondequationis2x1 4x2= 6. Addingthetwoequations, wehave0=2whichcannotoccur. Therefore,wehavenosolution.2(b) Therefore,thelinesareparallel,asshownbelow.1.522.533.54x21 0 1 2 3 4x15.(a)A =

2 31 2

=A1=17

2 31 2


2 31 2

45

=

12

.Thatis,x1= 1,x2= 2.(b) Therefore,thelinesareintersecting,asshownbelow.43210x23 2 1 1 2 3x16.(a)A =

3 26 4

det(A) = 0. Therefore,therearenosolutionsoraninnitenumberofsolutions. Multi-plying the rst equation by 2, we see the equations represent the same line. Therefore,there are aninnite number of solutions. Inparticular, anypair (x1, x2) suchthatx2= 3x1/2willsatisfythesystemofequations.3(b) Therefore,thelinesarecoincident,asshownbelow.42024x23 2 1 1 2 3x17.(a)A =

2 34 6

det(A)=0, therefore, therearenosolutionsoraninnitenumberof solutions. Mul-tiplyingtherstequationby 2, weseethetwoequationsareequationsforthesameline. Therefore, thereareaninnitenumberof solutions. Inparticular, anynumbers(x1, x2)suchthatx2= (2x16)/3.(b) Therefore,thelinesarecoincident,asshownbelow.2101x21 1 2 3 4 5x18.(a)A =

4 14 3

=A1=116

3 14 4


3 14 4

012

=

3/43

.Thatis,x1= 3/4,x2= 3.4(b) Therefore,thelinesareintersecting,asshownbelow.86420246x210.50.5 1 1.5 2x19.(a)A =

1 44 1

=A1= 115

1 44 1

.Therefore,x = A1b = 115

1 44 1

1010

=

22


1 11 2

=A1=13

2 11 1

.5Therefore,x = A1b =13

2 11 1

14

=

21


4 32 5

=A1=114

5 32 4


5 32 4

00

=

00

.Thatis,x1= 0,x2= 0.(b) Therefore,thelinesareintersecting,asshownbelow.4224x23 2 1 1 2 3x112.6(a)A =

2 54 10

det(A) = 0. Therefore,therearenosolutionsoraninnitemany. Multiplyingtherstequation by 2, we see the two equations are equations of the same line. Therefore, thereareaninnitenumberof solutions. Inparticular, anypairof theform(x1, x2)wherex2= 2x1/5willsatisfythesystemofequations.(b) Therefore,thelinesarecoincident,asshownbelow.10.500.51x23 2 1 1 2 3x113. det(A I) = 2 2 = 0 = = 2, 1. First,1= 2impliesA 1I=

1 22 4

.Therefore,x1=

21

isaneigenvectorfor1.Second,2= 1impliesA 2I=

4 22 1

.Therefore,x2=

12

isaneigenvectorfor2.14. det(A I) = 22 + 5 = 0 = = 1 + 2i, 1 2i. First,1= 1 + 2iimpliesA 1I=

2 2i 24 2 2i

.7Therefore,x1=

11 i

isaneigenvectorfor1.Second, 2= 12i is the complex conjugate of 1implies that x2 x1is an eigenvectorfor2. Inparticular,x2=

11 + i

isaneigenvectorfor2.15. det(A I) = 22 + 1 = 0 = = 1. Now, = 1impliesA 1I=

2 41 2

.Therefore,x1=

21

isaneigenvectorfor.16. det(A I) = 2+ 3 + 2 = 0 = = 1, 2. First,1= 1impliesA 1I=

2 23 3

.Therefore,x1=

11


3 23 2

.Therefore,x2=

23

isaneigenvectorfor2.17. det(A I) = 2+ 2 + 5 = 0 = = 1 + 2i, 1 2i. First,1= 1 + 2iimpliesA 1I=

2i 41 2i

.Therefore,x1=

2i1

isaneigenvectorfor1.8Second,since2= 1 2iisthecomplexconjugateof1,x2 x1=

2i1

isaneigenvectorfor2.18. det(A I) = 2 + 1/4 = 0 = = 1/2. Now,1= 1/2impliesA 1I=

3/4 3/43/4 3/4

.Therefore,x1=

11

isaneigenvectorfor.19. det(A I) = 2+ 2 + 1 = 0 = = 1. Now, = 1impliesA 1I=

1/2 11/4 1/2

.Therefore,x1=

11/2

isaneigenvectorfor.20. det(A I) = 21 = 0 = = 1, 1. First,1= 1impliesA 1I=

1 13 3

.Therefore,x1=

11


3 13 1

.Therefore,x2=

13

isaneigenvectorfor2.21. det(A I) = 2+ 1 = 0 = = i, i. First,1= iimpliesA 1I=

2 i 51 2 i

.9Therefore,x1=

2 + i1

isaneigenvectorfor1.Second,since2isthecomplexconjugateof1,x2 x1=

2 i1

isaneigenvectorfor2.22. det(A I) = 27 = 0 = = 0, 7. First,1= 0impliesA 1I=

6 32 1

.Therefore,x1=

12


1 32 6

.Therefore,x2=

31

isaneigenvectorfor2.23. det(A I) = 2+ 6 = 0 = = 2, 3. First,1= 2impliesA 1I=

1 14 4

.Therefore,x1=

11


4 14 1

.Therefore,x2=

14

isaneigenvectorfor2.1024. det(A I) = 2 + 5/2 = 0 = = (1 3i)/2. First,1= (1 + 3i)/2impliesA 1I=

32 32i 529532 32i

.Therefore,x1=

135 35i

isaneigenvectorfor1.Second,since2isthecomplexconjugateof1,thenx2 x1=

135+35i

isaneigenvectorfor2.25. det(A I) = 2+ + 1/42 = 0 = = 1/2. Now, = 2impliesA 1I=

52525252

.Therefore,x1=

11

isaneigenvectorfor.26. det(A I) = 2+ 2 + 2 = 0 = = 1 i. First,1= 1 + iimpliesA 1I=

2 i 15 2 i

.Therefore,x1=

12 i

isaneigenvectorfor1.Second,2isthecomplexconjugateof1impliesx2 x1=

12 + i

isaneigenvectorfor2.27. det(A I) = 2+ 2 = 0 = = 0, 2. First,1= 0impliesA 1I=

143943

.Therefore,x1=

431

11isaneigenvectorfor1.Second,2= 2impliesA 2I=

343941

.Therefore,x2=

491

isaneigenvectorfor2.28. det(A I) = 2+ 4 + 3 = 0 = = 1, 3. First,1= 1impliesA 1I=

1 11 1

.Therefore,x1=

11


1 11 1

.Therefore,x2=

11

isaneigenvectorfor2.29. det(A I) = 2+ 9 = 0 = = 3i. First,1= 3iimpliesA 1I=

1 3i 25 1 3i

.Therefore,x1=

11+3i2

isaneigenvectorfor1.Second,since2isthecomplexconjugateof1,x2 x1=

113i2

isaneigenvectorfor2.30. det(A I) = 2+ 4 + 4 = 0 = = 2. Now, = 2impliesA I=

1 122 1

.12Therefore,x1=

12

isaneigenvectorfor.31. det(A I) = 2104 + 1 = 0 = = 2, 1/2. First,1= 2impliesA 1I=

34343434

.Therefore,x1=

11

isaneigenvectorfor1.Second,2= 1/2impliesA 2I=

34343434

.Therefore,x2=

11

isaneigenvectorfor2.32. det(A I) = 23 +94= 0 = = 3/2. Now, = 3/2impliesA I=

12121212

.Therefore,x1=

11

isaneigenvectorfor.33.(a) det(A I) = 2+ 6 = 0 = = (1 25 + 4)/2.(b) If25 + 4>0, thentherewillbetworealeigenvalues. If25 + 4=0, thentherewillbeonerealeigenvalue. If25 + 4 < 0,thentherewillbetwoeigenvalueswithnon-zeroimaginarypart(inparticular,theywillbeacomplexconjugatepair).34.(a) det(A I) = 25 + 6 + 4 = 0 = = (5 1 16)/2.(b) If 1 16>0thentherewill betworeal eigenvalues. If 1=16, thentherewill beonereal eigenvalue. If 1 160forall since(2)2 4(25)0,thentherewillbetworealeigenvalues. Now1 22>0ifandonlyif1/2 < < 1/2. If = 1/2, then there will be one real eigenvalue. If > 1/2or < 1/2, then there will be a complex conjugate pair of eigenvalues with non-zeroimaginarypart.37. LetA =

a11a12a21a22

.LetA1=

b cd e

.NowA1A = Iimplies

b cd e

a11a12a21a22

=

1 00 1

.Therefore,wehavethefollowingsystemofequationsba11 + ca21= 1ba12 + ca22= 0da11 + ea21= 0da12 + ea22= 1.Multiplying the rst equation by a12, the second equation by a11and adding them impliesc(a21a12a11a22= a12. Therefore,c =1a21a12a11a22a12=1det(A)a12.Pluggingthisvalueforcintotherstequation,weareabletosolveforb. Inparticular,ba11a12det(A)a21= 1 =ba11=a11a22det(A).Therefore,b =a22det(A).14Similarly, multiplyingthethirdequationbya12, thefourthequationby a11andaddingthem,weseethate(a11a22a21a12) = a11. Therefore,e =a11det(A).Pluggingthisvalueforeintothethirdequation,weseethatd = a21det(A).38. Suppose is an eigenvalue for A. Then is a solution of det(AI) = 0. In particular, will satisfy (a11)(a22)a12a21= 0. Therefore, 2(a11+a22)+a11a22a12a21= 0.Now, = 0 is an eigenvalue if and only if is a solution of this equation, which is true if andonlyifa11a22a12a21= 0. Inotherwords, = 0isaneigenvalueifandonlyifdet(A) = 0.Section3.21. Thesystemisautonomous,nonhomogeneous.2. Thesystemisnonautonomous,nonhomogeneous.3. Thesystemisnonautonomous,homogeneous.4. Thesystemisautonomous,nonhomogeneous.5. Thesystemisautonomous,homogeneous.6. Thesystemisnonautonomous,homogeneous.7. Thesystemisnonautonomous,nonhomogeneous.8. Thesystemisautonomous,homogeneous.9.(a) Thesystemx

= x + y + 1 = 0y

= x + y 3 = 0canberewritteninmatrixformas

1 11 1

xy

=

1 3

.NowA =

1 11 1

=A1=12

1 11 1

.Therefore,

xy

= A1

13

=12

1 11 1

13

=

21

.Therefore,theequilibriumsolutionis

21

.15(b)2101234y1 1 2 3 4 5x(c) Solutionsinthevicinityofthecriticalpoint,tendawayfromthecriticalpoint.10.(a) Thesystemx

= x 4y 4 = 0y

= x y 6 = 0canberewritteninmatrixformas

1 41 1

xy

=

4 6

.NowA =

1 41 1

=A1=15

1 41 1

.Therefore,

xy

= A1

46

=15

1 41 1

46

=

42


42

.(b)1664202y2 4 6 8x(c) Solutionsinthevicinityofthecriticalpoint,spiralintothecriticalpoint.11.(a) Thesystemx

= 0.25x 0.75y + 8 = 0y

= 0.5x + y 11.5 = 0canberewritteninmatrixformas

0.25 0.750.5 1

xy

=

8 11.5

.NowA =

0.25 0.750.5 1

=A1= 8

1 3/41/2 1/4

.Therefore,

xy

= A1

811.5

= 8

1 3/41/2 1/4

811.5

=

42


59

.(b)6789101112y2 3 4 5 6 7 8x17(c) Solutionsinthevicinityofthecriticalpoint,tendawayfromthecriticalpoint.12.(a) Thesystemx

= 2x + y 11 = 0y

= 5x + 4y 35 = 0canberewritteninmatrixformas

2 15 4

xy

=

11 35

.NowA =

2 15 4

=A1= 13

4 15 2

.Therefore,

xy

= A1

1135

= 13

4 15 2

1135

=

35


35

.(b)34567y5 4 3 2 1x(c) Solutionsinthevicinityofthecriticalpoint,tendawayfromthecriticalpoint.13.18(a) Thesystemx

= x + y 3 = 0y

= x + y + 1 = 0canberewritteninmatrixformas

1 11 1

xy

=

3 1

.NowA =

1 11 1

=A1=12

1 11 1

.Therefore,

xy

= A1

31

=12

1 11 1

31

=

21


21

.(b)2101234y1 1 2 3 4 5x(c) Solutionsinthevicinityofthecriticalpoint,spiralawayfromthecriticalpoint.14.(a) Thesystemx

= 5x + 4y 35 = 0y

= 2x + y 11 = 0canberewritteninmatrixformas

5 42 1

xy

=

35 11

.19NowA =

5 42 1

=A1=13

1 42 5

.Therefore,

xy

= A1

3511

=13

1 42 5

3511

=

35


35

.(b)34567y5 4 3 2 1x(c) Solutionsinthevicinityofthecriticalpoint,tendtowardsthecriticalpoint.15. Letx1= uandx2= u

. Thenx

1= x2andx

2= u

= 2u 0.5u= 2x10.5x2.Therefore,weobtainthesystemofequationsx

1= x2x

2= 2x10.5x2.16. Letx1= uandx2= u

. Thenx

1= x2andx

2= u

= 0.5u

8u + 6 sin(2t)= 0.5x28x1 + 6 sin(2t).20Therefore,weobtainthesystemofequationsx

1= x2x

2= 8x10.5x2 + 6 sin(2t).17. Firstdividetheequationbyt2. Wearriveattheequationu

= 1t u

1 14t2

u.Nowletx1= uandx2= u

. Thenx

1= x2andx

2= u

= 1t u

1 14t2

u= 1t x2

1 14t2

x1.Therefore,weobtainthesystemofequationsx

1= x2x

2=

1 14t2

x11t x2.18. First,dividetheequationbyt2. Wearriveattheequationu

+3t u

+5t2u = 1 +4t2 .Nowletx1= uandx2= u

. Thenx

1= x2andx

2= u

= 3t u

5t2 u + 1 +4t2= 3t x25t2x1 + 1 +4t2.Therefore,weobtainthesystemofequationsx

1= x2x

2= 5t2x13t x2 + 1 +4t2.19. Letx1= uandx2= u

. Thenx

1= x2andx

2= u

= 0.25u

4u + 2 cos 3t= 4x10.25x2 + 2 cos 3t.21Nowu(0)=1impliesx1(0)=1andu

(0)= 2impliesx2(0)= 2. Therefore, weobtainthesystemofequationsx

1= x2x

2= 4x10.25x2 + 2 cos 3twithinitialconditionsx1(0) = 1x2(0) = 2.20. First,dividetheequationbyt. Wearriveattheequationu

+1tu

+u = 0.Nowletx1= uandx2= u

. Thenx

1= x2andx

2= 1tu

u= 1tx2x1.Nowu(1) = 1 =x1(1) = u(1) = 1andu

(1) = 0 =x2(1) = u

(1) = 0. Therefore,weobtainthesystemofequationsx

1= x2x

2= x11tx2withinitialconditionsx1(1) = 1x2(1) = 0.21.(a) Taking a clockwise loop around each of the paths, it is easy to see that voltage drops aregivenbyv1v2= 0andv2v3= 0.(b) Consider the right node. The current in is given by i1+i2. The current leaving the nodeis i3. Therefore, thecurrentpassingthroughthenodeis(i1 + i2) (i3). BasedonKirchosrstlaw,i1 +i2 +i3= 0.(c) Inthecapacitor,Cv

1= i1. Intheresistor,v2= Ri2. Intheinductor,Li

3= v3.(d) Basedonpart(a),v3= v2= v1. Basedonparts(b)and(c),Cv

1 +1Rv2 +i3= 0.ItfollowsthatCv

1= 1Rv1i3and Li

3= v1.2222. Leti1, i2, i3andi4bethecurrentthroughthe1ohmresistor, 2ohmresistor, inductorandcapacitor, respectively. Assignv1, v2, v3andv4astherespectivevoltagedrops. BasedonKirchossecondlaw,thenetvoltagedropsaroundeachloopsatisfyv1 + v3 + v4= 0, v1 + v3 + v2= 0 and v4v2= 0.ApplyingKirchosrstlawtotheupperrightnode,wehavei1i3= 0.Likewise,intheremainingnodes,wehavei2 + i4i1= 0 and i3i4i2= 0.Combiningtheaboveequations,wehavev4v2= 0, v1 + v3 + v4= 0 and i2 + i4i3= 0.Usingthecurrent-voltagerelations,wehavev1= R1i1, v2= R2i2, Li

3= v3, Cv

4= i4.Combiningtheseequations,wehaveR1i3 + Li

3 + v4= 0 and Cv

4= i31R2v4.Nowseti3= iandv4= v,toobtainthesystemofequationsLi

= R1i v and Cv

= i 1R2v.Finally,usingthefactthatR1= 1,R2= 2,L = 1andC= 1/2,wehavei

= i v and v

= 2i v,asclaimed.23. Leti1, i2, i3andi4bethecurrentthroughtheresistors,inductorandcapacitor,respec-tively. Assignv1, v2, v3andv4astherespectivevoltagedrops. BasedonKirchossecondlaw,thenetvoltagedropsaroundeachloopsatisfyv1 + v3 + v4= 0, v1 + v3 + v2= 0 and v4v2= 0.ApplyingKirchosrstlawtotheupperrightnode,wehavei3(i2 + i4) = 0.Likewise,intheremainingnodes,wehavei1i3= 0 and i2 + i4i1= 0.23Combiningtheaboveequations,wehavev4v2= 0, v1 + v3 + v4= 0 and i2 + i4i3= 0.Usingthecurrent-voltagerelations,wehavev1= R1i1, v2= R2i2, Li

3= v3, Cv

4= i4.Combiningtheseequations,wehaveR1i3 + Li

3 + v4= 0 and Cv

4= i31R2v4.Nowseti3= iandv4= v,toobtainthesystemofequationsLi

= R1i v and Cv

= i 1R2v.24.(a) Let Q1(t) and Q2(t) be the amount of salt in the respective tanks at time t. The volumeof eachtankremainsconstant. Basedonconservationof mass, therateof increaseofsaltisgivenbyrateofincrease = ratein rateout.ForTank1,therateofsaltowinginisrin= 3q1 +Q2100. TherateofowoutofTank1isrout=Q115 . Therefore,dQ1dt= 3q1 +Q2100 Q115.Similarly,forTank2,dQ2dt= q2 +Q130 3Q2100 .TheinitialconditionsareQ1(0) = Q01andQ2(0) = Q02.(b) The equilibrium values are obtained by solving Q1/15+Q2/100+3q1= 0 and Q1/303Q2/100 + q2= 0. ItssolutionisgivenbyQE1= 54q1 + 6q2andQE2= 60q1 + 40q2.(c) Wewouldneedtobeabletosolvethesystem54q1 + 6q2= 6060q1 + 40q2= 50.The solution of this system is q1= 7/6 and q2= 1/2 which is not a physically realisticsolution. Therefore,itisnotpossibletohavethosevaluesasanequilibriumstate.(d) Wecanwriteq2= 9q1 +QE16q2= 32q1 +QE24024whicharetheequationsoftwolinesintheq1q2plane. Theinterceptsoftherstlineare (QE1 /54, 0) and(0, QE1 /6). The intercepts of the secondline are (QE2 /60, 0) and(0, QE2 /40). Therefore, thesystemwill haveauniquesolutionintherstquadrantaslongasQE1 /54 QE2 /60orQE2 /40 QE1 /6. Thatis,10/9 QE2 /QE1 20/3.Section3.31. WelookforeigenvaluesandeigenvectorsofA =

3 22 2

.We see that det(AI) = 2 2 = ( 2)( +1). Therefore, the eigenvalues are givenby = 2, 1.First,1= 2impliesA 1I=

1 22 4

.Therefore,v1=

21

isaneigenvectorassociatedwith1andx1(t) = e2t

21

isonesolutionofthesystem.Second,2= 1impliesA 2I=

4 22 1

.Therefore,v2=

12

isaneigenvectorassociatedwith2andx2(t) = et

12

isasecondsolutionofthesystem.Therefore,thegeneralsolutionisgivenbyx(t) = c1e2t

21

+ c2et

12

.2521012x22 1 1 2x1If the initial condition is a multiple of

1 2

t, then the solution will tend to the origin alongthe eigenvector

1 2

t. Otherwise, the solution will grow, following the eigenvector

2 1

t.2. WelookforeigenvaluesandeigenvectorsofA =

1 23 4

.We see that det(AI) = 2+3+2 = (+2)(+1). Therefore, the eigenvalues are givenby = 1, 2.First,1= 1impliesA 1I=

2 23 3

.Therefore,v1=

11


11


3 23 2

.Therefore,v2=

23


23

isasecondsolutionofthesystem.26Therefore,thegeneralsolutionisgivenbyx(t) = c1et

11

+ c2e2t

23

.21012x22 1 1 2x1If theinitial conditionis amultipleof

2 3

t, thenthesolutionwill tendtotheoriginalongtheeigenvector

2 3

t. Otherwise, thesolutionwill tendtotheoriginfollowingtheeigenvector

1 1


2 13 2

.Weseethatdet(AI) = 21 = ( 1)( + 1). Therefore,theeigenvaluesaregivenby = 1, 1.First,1= 1impliesA 1I=

1 13 3

.Therefore,v1=

11


11


3 13 1

.Therefore,v2=

13

27isaneigenvectorassociatedwith2andx2(t) = et

13

isasecondsolutionofthesystem.Therefore,thegeneralsolutionisgivenbyx(t) = c1et

11

+ c2et

13

.3210123x23 2 1 1 2 3x1If the initial condition is a multiple of

1 3


1 3


1 1


1 14 2

.We see that det(AI) = 2+ 6 = ( +3)( 2). Therefore, the eigenvalues are givenby = 2, 3.First,1= 2impliesA 1I=

1 14 4

.Therefore,v1=

11


11


4 14 1

.28Therefore,v2=

14


14


11

+ c2e3t

14

.42024x24 2 2 4x1If theinitial conditionisamultipleof

1 4


1 4

t. Otherwise,thesolutionwillgrow,followingtheeigenvector

1 1


4 38 6

.Weseethatdet(A I)=2+ 2=( + 2). Therefore, theeigenvaluesaregivenby = 0, 2.First,1= 0impliesA 1I=

4 38 6

.Therefore,v1=

34

isaneigenvectorassociatedwith1andx1(t) =

34

29isonesolutionofthesystem.Second,2= 2impliesA 2I=

6 38 4

.Therefore,v2=

12


12

isasecondsolutionofthesystem.Therefore,thegeneralsolutionisgivenbyx(t) = c1

34

+ c2e2t

12

.42024x24 2 2 4x1Thebehaviorofthesolutionsast issimilartothebehaviorofsolutionsinExample5ofthetext.6. WelookforeigenvaluesandeigenvectorsofA =

2 11 2

.We see that det(AI) = 2+4+3 = (+1)(+3). Therefore, the eigenvalues are givenby = 1, 3.First,1= 1impliesA 1I=

1 11 1

.Therefore,v1=

11


11


1 11 1

.Therefore,v2=

11


11


11

+ c2e3t

11


1 1

t, thenthesolutionwill tendtotheoriginalong the eigenvector

1 1

t. Otherwise, the solution will tend to the origin, following theeigenvector

1 1


5/4 3/43/4 5/4

.Weseethatdet(A I) = 252 + 1 = ( 2)( 1/2). Therefore,theeigenvaluesaregivenby = 2, 1/2.31First,1= 2impliesA 1I=

3/4 3/43/4 3/4

.Therefore,v1=

11


11

isonesolutionofthesystem.Second,2= 1/2impliesA 2I=

3/4 3/43/4 3/4

.Therefore,v2=

11

isaneigenvectorassociatedwith2andx2(t) = et/2

11


11

+ c2et2

11


1 1

t, then the solution will grow following the theeigenvector

1 1


1 1


3/4 7/41/4 5/4

.32Weseethatdet(A I) = 212 12= ( 1)( + 1/2). Therefore,theeigenvaluesaregivenby = 1, 1/2.First,1= 1impliesA 1I=

7/4 7/41/4 1/4

.Therefore,v1=

11


11


1/4 7/41/4 7/4

.Therefore,v2=

71


71


11

+ c2et2

71


7 1


7 1


1 1


1/4 3/41/2 1

.Weseethatdet(A I)=234 +18=

12

14

. Therefore, theeigenvaluesaregivenby = 1/2, 1/4.First,1= 1/2impliesA 1I=

3/4 3/41/2 1/2

.Therefore,v1=

11


11


1/2 3/41/2 3/4

.Therefore,v2=

32


32

isasecondsolutionofthesystem.Therefore,thegeneralsolutionisgivenbyx(t) = c1et/2

11

+ c2et/4

32


3 2

t, then the solution will stay on the eigenvector

3 2


1 1


5 13 1

.We see that det(AI) = 26+8 = (4)(2). Therefore, the eigenvalues are givenby = 4, 2.First,1= 4impliesA 1I=

1 13 3

.Therefore,v1=

11


11


3 13 1

.Therefore,v2=

13


13

isasecondsolutionofthesystem.35Therefore,thegeneralsolutionisgivenbyx(t) = c1e4t

11

+ c2e2t

13

.21012x22 1 1 2x1Iftheinitial conditionisamultipleof

1 3

t, thenthesolutionwill grow, stayingontheeigenvector

1 3


1 1


2 15 4

.We see that det(AI) = 223 = (3)(+1). Therefore, the eigenvalues are givenby = 3, 1.First,1= 3impliesA 1I=

5 15 1

.Therefore,v1=

15


15


1 15 5

.Therefore,v2=

11


11


15

+ c2et

11


1 1


1 1


1 5


3 61 2

.Weseethat det(A I) =2 =( 1). Therefore, theeigenvaluesaregivenby = 0, 1.First,1= 0impliesA 1I=

3 61 2

.Therefore,v1=

21

isaneigenvectorassociatedwith1andx1(t) =

21


2 61 3

.37Therefore,v2=

31


31

isasecondsolutionofthesystem.Therefore,thegeneralsolutionisgivenbyx(t) = c1

21

+ c2et

31

.21012x23 2 1 1 2 3x1Theentirelinespannedbytheeigenvector

2 1

tconsistsof equilibriumpoints. Thedirectioneldchangesacrossthelinex1 + 2x2= 0.13.A =

1 23 4

impliesdet(A I) = 2+ 3 + 2 = ( + 2)( + 1). Therefore,theeigenvaluesare = 2and = 1.First,1= 2impliesA 1I=

3 23 2

.Therefore,v1=

23

isaneigenvectorfor1. Therefore,x1(t) = e2t

23

isonesolution.38Second,2= 1impliesA 2I=

2 23 3

.Therefore,v2=

11

isaneigenvectorfor2. Therefore,x2(t) = et

11

isasecondsolution.Therefore,thegeneralsolutionisx(t) = c1e2t

23

+ c2et

11

.Theinitialconditionx(0) =

31

timplies

2 13 1

c1c2

=

31

.Therefore,

c1c2

=

1 13 2

31

=

27

.Therefore,thesolutionisx(t) = 2e2t

23

+ 7et

11

.00.511.522.53x11 2 3 4 5t0.511.52x20 1 2 3 4 5tBothcomponentstendtozeroast .14.A =

2 13 2

39impliesdet(A I)=2 1=( + 1)( 1). Therefore, theeigenvaluesare=1and = 1.First,1= 1impliesA 1I=

1 13 3

.Therefore,v1=

11


11

isonesolution.Second,2= 1impliesA 2I=

3 13 1

.Therefore,v2=

13


13

isasecondsolution.Therefore,thegeneralsolutionisx(t) = c1et

11

+ c2et

13


25

timplies

1 11 3

c1c2

=

25

.Therefore,

c1c2

=12

3 11 1

25

=

1/23/2

.Therefore,thesolutionisx(t) =12et

11

+32et

13

.4010203040506070x10 1 2 3 4 5t10203040506070x20 1 2 3 4 5tBothcomponentsx1andx2tendto+ ast .15.A =

5 13 1

impliesdet(A I)=2 6 + 8=( 4)( 2). Therefore, theeigenvaluesare=2and = 4.First,1= 2impliesA 1I=

3 13 1

.Therefore,v1=

13


13


1 13 3

.Therefore,v2=

11


11


13

+ c2e4t

11

.41Theinitialconditionx(0) =

21

timplies

1 31 1

c1c2

=

21

.Therefore,

c1c2

=

3/27/2

.Therefore,thesolutionisx(t) = 32e2t

13

+72e4t

11

.020406080100120140160180x10.2 0.4 0.6 0.8 1t020406080100120140160x20.2 0.4 0.6 0.8 1tBothcomponentstendto+ ast .16.A =

2 15 4

impliesdet(A I)=2 2 3=( + 1)( 3). Therefore, theeigenvaluesare=3and = 1.First,1= 3impliesA 1I=

5 15 1

.Therefore,v1=

15


15


1 15 5

.42Therefore,v2=

11


11


15

+ c2et

11


13

timplies

c1c2

=

1/21/2

.Therefore,thesolutionisx(t) =12e3t

15

+12et

11

.246810x10 0.2 0.4 0.6 0.8 1t1020304050x20 0.2 0.4 0.6 0.8 1tBothcomponentstendto+ ast .17.(a)4321012x22 1 1 2x1(b)1012345x21 1 2 3 4 5x1(c) Thegeneralsolutionisgivenbyx(t) = c1et

12

+ c2e2t

12


23

timplies

c1c2

=

1/47/4

.Therefore,thesolutionpassingthroughtheinitialpoint(2, 3)isgivenbyx(t) = 14et

12

+74e2t

12

.Thecomponentplotsareshownbelow.4400.511.522.531 2 3 4 5t18.(a)21012x22 1 1 2x1(b)42024x21 1 2 3 4 5x1(c) Thegeneralsolutionisgivenbyx(t) = c1et

12

+ c2e2t

12


23

timplies

c1c2

=

1/47/4


12

+74e2t

12

.Thecomponentplotsareshownbelow.32101230.2 0.4 0.6 0.811.2 1.4 1.6 1.82t19.(a)21012x22 1 1 2x1(b)461012345x21 1 2 3 4 5x1(c) Thegeneralsolutionisgivenbyx(t) = c1et

12

+ c2e2t

12


23

timplies

c1c2

=

1/47/4


12

+74e2t

12

.Thecomponentplotsareshownbelow.0204060801001201401601800.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2t20.(a)4721012x22 1 1 2x1(b)1012345x21 1 2 3 4 5x1(c) Thegeneralsolutionisgivenbyx(t) = c1et

12

+ c2e2t

12


23

timplies

c1c2

=

7/41/4

.Therefore,thesolutionpassingthroughtheinitialpoint(2, 3)isgivenbyx(t) =74et

12

+14e2t

12

.Thecomponentplotsareshownbelow.4805101520250.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2t21.(a)21012x22 1 1 2x1(b)1012345x21 1 2 3 4 5x1(c) Thegeneralsolutionisgivenbyx(t) = c1e0.5t

14

+ c2e0.5t

41


23

timplies

1 44 1

c1c2

=

23

.Therefore,

c1c2

= 115

1 44 1

23

=

2/31/3

.Therefore,thesolutionpassingthroughtheinitialpoint(2, 3)isgivenbyx(t) =23e0.5t

14

+13e0.5t

41

.Thecomponentplotsareshownbelow.510152025300 1 2 3 4 5t22.(a)21012x22 1 1 2x1(b)501012345x21 1 2 3 4 5x1(c) Thegeneralsolutionisgivenbyx(t) = c1e0.5t

21

+ c2e0.8t

12


23

timplies

2 11 2

c1c2

=

23

.Therefore,

c1c2

=15

2 11 2

23

=

7/54/5


21

+45e0.8t

12

.Thecomponentplotsareshownbelow.0.511.522.530 1 2 3 4 5t23.51(a)21012x22 1 1 2x1(b)1012345x21 1 2 3 4 5x1(c) Thegeneralsolutionisgivenbyx(t) = c1e0.3t

12

+ c2e0.6t

13


23

timplies

1 12 3

c1c2

=

23

.Therefore,

c1c2

=15

3 12 1

23

=

3/51


21

+45e0.8t

12

.Thecomponentplotsareshownbelow.5210203040500 1 2 3 4 5t24.(a)21012x22 1 1 2x1(b)1012345x21 1 2 3 4 5x1(c) Thegeneralsolutionisgivenbyx(t) = c1e1.5t

12

+ c2et

31


23

timplies

1 32 1

c1c2

=

23

.Therefore,

c1c2

= 17

1 32 1

23

=

11

.Therefore,thesolutionpassingthroughtheinitialpoint(2, 3)isgivenbyx(t) = e1.5t

12

+ et

31

.Thecomponentplotsareshownbelow.100001000200030001 2 3 4 5t25. ThegeneralsolutioninExample3isx(t) = c1e0.25t

12

+ c2e0.05t

32

.Theinitialconditionx1(0) = 13,x2(0) = 10impliesthatc1= 7andc2= 2. Therefore,x1(t) = 7e0.25t+ 6e0.05tx2(t) = 14e0.25t+ 4e0.05t.The rst component x1is monotone decreasing. By solving the equation 7e0.25t+6e0.05t=0.5, weseethatx1(T)=0.5whenT=49.7. Therefore, 0 x1(t) 0.5forall t 49.7.Thesecondcomponent,x2ismonotonedecreasingafterreachingitsmaximum. Bysolvingtheequation 14e0.25t+ 4e0.05t=0.5, wecanseethat0 x2(t) 0.5forall t 41.57.Therefore, we conclude that the component functions will satisfy the specied bounds for allt 49.7.26.54(a) For = 0.5, the characteristic equation is 22+4+1 = 0. Therefore, the eigenvalues are = 1 1/2. The eigenvector corresponding to1= 1 +1/2 is v1=

2 1

t.Theeigenvectorcorrespondingto2= 1 1/2isv2=2 1

t. Therefore, thegeneralsolutionisx(t) = c1e(1+1/2)t

21

+ c2e(11/2)t21

.Sincebotheigenvaluesarenegative,theequilibriumpointisastablenode.(b) For = 2,the characteristic equation is 2+2 1 = 0. Therefore,the eigenvalues are = 1 2. The eigenvector corresponding to 1= 1 +2 is v1=

1 2

t. Theeigenvectorcorrespondingto2= 1 2isv2=

12

t. Therefore, thegeneralsolutionisx(t) = c1e(1+2)t

12

+ c2e(12)t

12

.Sincetheeigenvalueshaveoppositesign,theequilibriumpointisasaddlepoint.(c) For general , the characteristic equation is 2+2+1 = 0. The eigenvalues are givenby = 1 . For0.5 2,botheigenvaluesarerealandclearly 1 < 0.Therefore, wejustneedtodeterminewhether 1 +=0. Weseethisoccurswhen = 1. Atthatvaluetheequilibriumpointswitchesfromasaddlepoint(for > 1)toastablenode(for < 1).27.(a) Forthegivendata,thesystemcanbewrittenasddt

iv

=

1/2 1/23/2 5/2

iv

.The characteristic equation for this system is 2+3+2 = 0. Therefore, the eigenvaluesaregivenby= 1, 2. For= 1, acorrespondingeigenvectorisgivenbyv1=

1 1

t. For= 2, acorrespondingeigenvectorisgivenbyv2=

1 3

t. Therefore,thegeneralsolutionis

iv

= c1et

11

+ c2e2t

13

.(b) Since both eigenvalues are real and negative, the equilibrium point (0, 0) is a stable node.Therefore,i(t) 0andv(t) 0ast .28.(a) Forthegeneralequation(i),thecharacteristicequationisgivenby2+

L + CR1R2LCR2

+R1 + R2LCR2= 0.55The eigenvalues are real and distinct provided that the discriminant is positive. That is,

L +CR1R2LCR2

24

R1 +R2LCR2

> 0,whichsimpliestothecondition

1CR2R1L

24LC> 0.(b) SincethesumoftherootsofthecharacteristicequationisL +CR1R2LCR2< 0andtheproductoftherootsisR1 +R2LCR2> 0,itfollowsthatbothrootsarenegative. Therefore,i = 0, v= 0isastablenode.29. u = x xcimpliesu

= x

x

c= Ax +b= A(u +xc) +b= Au +Ax

c +b= Au +A(A1b) +b= Au b +b= Au.Therefore,u

= Auasclaimed.Section3.41.A =

3 24 1

impliesdet(A I) = 22 + 5.Therefore,theeigenvaluesaregivenby = 1 2i. Now,1= 1 + 2iimpliesA 1I=

2 2i 24 2 2i

.Therefore,v1=

11 i

56Chapter4Section4.11. Thedierentialequationislinear.2. Thedierentialequationisnonlinear.3. Thedierentialequationislinear.4. Thedierentialequationislinear.5. Thedierentialequationisnonlinear.6. Wewill usetheequationmg kL=0. If themassis8lb, thenmg=8. If themassstretchesthespring6inches, thenL=6. Therefore, 8 6k=0whichmeansk=4/3lb/inch.7. Again, weusetheequationmg kL=0. Herethemassis10kg. Theforceduetogravityisg=9.8m/s2. Therefore, mg=98Newtons. Themassstretchesthespring.7meters. Therefore,k = 98/.7 = 140N/m.8. Thespringconstantisk= 2/(1/2) = 4lb/ft. Themassm = 2/32 = 1/16lbs2/ft. Theequationofmotionis116y

+ 4y= 0ory

+ 64y= 0withinitialconditionsy(0) = 1/4ft,y

(0) = 0ft/sec.9. Thespringconstantisk=.98/.05=19.6N/m. Themassm=.1kg. Theequationofmotionis.1y

+ 19.6y= 0ory

+ 196y= 0withinitialconditionsy(0) = 0m,y

(0) = .1m/sec.10. Thespringconstant is k=3/(1/4) =12lb/ft. Themass m=3/32lbs2/ft. Theequationofmotionis332y

+ 12y= 0ory

+ 128y= 0withinitialconditionsy(0) = 1/12ft,y

(0) = 2ft/sec.11. TheinductanceL=1henry. TheresistanceR=0. ThecapacitanceC=0.25 106farads. Therefore,theequationforchargeqisq

+ (4 106)q= 0withinitialconditionsq(0) = 106coulombs,q

(0) = 0coulombs/sec.112. Thespringconstantisk= .196/.05 = 3.92N/m. Themassm = .02kg. Thedampingconstantis= 400dyne-sec/cm=.4N-sec/cm. Therefore,theequationofmotionis.02y

+ .4y

+ 3.92y= 0ory

+ 20y

+ 196y= 0withinitialconditionsy(0) = .02m,y

(0) = 0m/sec.13. Thespringconstant is k=16/(1/4) =64lb/ft. Themass m=1/2lbs2/ft. Thedampingcoecientis= 2lb-sec/ft. Therefore,theequationofmotionis12y

+ 2y

+ 64y= 0ory

+ 4y

+ 128y= 0withinitialconditionsy(0) = 0ft,y

(0) = 1/4ft/sec.14. The spring constant is k = 3/.1 = 30 N/m. The mass m = 2 kg. The damping coecientis= 3/5N-sec/m. Therefore,theequationofmotionis2y

+35y

+ 30y= 0ory

+ .3y

+ 15y= 0withinitialconditionsy(0) = .05m,y

(0) = .1m/sec.15. TheinductanceL=0.2henry. TheresistanceR=3 102ohms. ThecapacitanceC= 105farads. Therefore,theequationforchargeqis0.2q

+ 300q

+ 105q= 0orq

+ 1500q

+ 500, 000q= 0withinitialconditionsq(0) = 106coulombs,q

(0) = 0coulombs/sec.16. WeknowthenetforceactingonthemasssatisesFnet= mx

(t)wherex

istheaccelerationofthemass. Here,weassumethespringispullingthemasstotheleft. Wewillconsiderthattobethenegativedirection. TheforceduetothespringisgivenbyFs= k(L + x).Theforceduetoairresistance, Fd, isactinginthedirectionoppositetothedirectionofmotionofthemass,andisproportionaltothevelocityofthemass. Therefore,Fd= x

(t)2where 0. Inparticular,ifthemassismovingtotheleft,thenthevelocityx

(t)0inwhichcase, Fd 0. Finally, weareassumingthatanexternal forceF(t)isactingonthemass. Puttingtogetherthesefactors,wehaveFnet= Fs +Fd +F(t)=mx

(t) = k(L +x) x

(t) +F(t).From equation (13), we know that mg= kL. Since the mass lies on a frictionless, horizontalsurface,theforceduetogravityiszero. Therefore, kL=0,meaningL=0. Inparticular,thespringisnotstretchedbythemasswhenthespringisinequilibrium. Therefore, weconcludethatmx

(t) + x

(t) +kx(t) = F(t),asinequation(18). Thederivationdiersheresinceitisassumedthereisnoforceduetogravityactingonthespring.17.(a) WeknowthatthenetforceisgivenbyFnet= mx

(t). Here,weareassumingtherearenodampingof external forcespresent. Also, weareassumingthereisnoforceduetogravity. Therefore,Fnet= Fs=mx

= kx x3=mx

+kx + x3= 0.(b) Ifweassumethatthemaximumdisplacementissmall, thenx30, inwhichcasethelinearizedequationismx

+kx = 0.18. WeknowthenetforceactingonthemassisgivenbyFnet=mx

. Theforceexertedonthemassbyspring1isgivenbyFs1= k1x. Theforceexertedonthemassbyspring2isgivenbyFs2= k2x. Also, weareassumingairresistanceisactingonthemasswithcoecient. Theairresistanceisproportionaltothevelocityx

(t)ofthemass. Therefore,wehaveFnet= Fs1 +Fs2 +Fd=mx

= k1x k2x x

=mx

+ x

+ (k1 +k2)x = 0.19. Letu(t)bethedepthoftheblockintothewater. LetmbethemassoftheblockandFB(t) be the upward buoyant force exerted by the uid. The equation of motion is given bymu

(t) = FB(t).3Sincethedensityof thecubicblockis, themassm= l3. Sincethebuoyantforceisequaltotheweightofthedisplaceduid,weseethatFB= 0 u(t) l2 g. Therefore, l3u

= 0 u(t) l2 g,orlu

+ 0gu = 0.20.(a)10.80.60.40.200.20.40.60.81x15 10 15 20t10.80.60.40.200.20.40.60.81x25 10 15 20t(b)21012x22 1 1 2x1(c) Thecriticalpointisstable.21.(a)410.80.60.40.200.20.40.60.81x15 10 15 20t42024x25 10 15 20t(b)86422468x22 1 1 2x1(c) Thecriticalpointisstable.22.(a)10.80.60.40.200.20.40.60.81x12 2 4 6 8 10t21012x22 2 4 6 8 10t(b)520246x22 1 1 2x1(c) Thecriticalpointisasymptoticallystable.23.(a)10.80.60.40.200.20.40.60.81x12 2 4 6 8 10t10.80.60.40.200.20.40.60.81x22 2 4 6 8 10t(b)202468x20.5 1 1.5 2x1(c) Thecriticalpointisasymptoticallystable.624.(a)6040200204060x12 2 4 6 8 10t1005050100150200x22 2 4 6 8 10t(b)300200100100200300400x2100 100 200x1(c) Thecriticalpointisunstable.25.(a)0.40.200.20.4x1, x35 10 15 20t7(b)0.80.60.40.200.20.40.60.8x1, x35 10 15 20t26.(a) Ify1= cos(rt),theny

1= r sin(rt)andy

1= r2cos(rt). Therefore,inorderfory1tobeasolutionofourequation,weneedmr2cos(rt)k cos(rt) = 0or r =

k/m. Similarly, for r =

k/m, y2=sin(rt) will beasolutionof thisequation. Ify= c1y1 + c2y2,theny

= c1y

1+ c2y

2impliesmy

+ ky= m(c1y

1+ c2y

2) + k(c1y1 + c2y2) = c1(my

1+ ky1) + c2(my

2+ ky2) = 0.Therefore,y= c1y1 + c2y2willbeasolutionify1andy2aresolutions.(b) By part (a), y1(t) = cos(rt) and y2(t) = sin(rt) will be solutions as long as r =

k/m.Further,anysolutionoftheformy= c1 cos(rt) + c2 sin(rt)willbeasolution.(i) Inthiscase,r= 1. Notethatwecantaker= 1asr= 1willjustchangethecoecientsc1andc2above. Nowifr=1, theny(t)=c1 cos(t) + c2 sin(t). Theny(0)=c1=1andy

(t)= sin(t) + c2 cos(t)impliesy

(0)=c2= 1. Therefore,oursolutionisy(t) = cos(t) sin(t).(ii) Inthiscase, r = 4. Asabove, wewill taker =4. Theny(t) =c1 cos(4t) +c2 sin(4t). Therefore, y(0)=c1=1andy

(t)= 4 sin(4t) + 4c2 cos(4t)impliesy

(0) = 4c2= 1 =c2= 1/4. Therefore,y(t) = cos(4t) 14 sin(4t).8(c)10.500.512 4 6 8 10tThe soft spring has a higher amplitude and a longer period. The sti spring has a smalleramplitudeandashorterperiod.Section4.21. WritetheIVPasy

+3ty

= 1.Since the function p(t) = 3/t is continuous for all t > 0 and t0= 1 > 0, the IVP is guaranteedtohaveauniquesolutionforallt > 0.2. WritetheIVPasy

3tt 1y

+4t 1y=sin tt 1.Sincethecoecientfunctionsarecontinuousforall t 0. First,welookforeigenvaluesofA. WehaveA I=

1c/a b/a

.Therefore,det(A I) = 2+ba +ca. Therefore,det(A I) = 0 =a2+ b + c = 0.Therefore,1,2= b b24ac2a.Here we are assuming that b24ac > 0. Therefore, the two eigenvalues are real and distinct.Nowwelookforeigenvectors. Fori,i = 1, 2,wehaveA iI=

i1c/a b/a i

.Since iis an eigenvalue, the matrix AiIis singular. Further, vi=

x1ix2i

twill be aneigenvectoraslongas ix1i + x2i= 0. Inparticular,weseethatvi=

1i

satisesthenecessaryequation. Weconcludethatx1(t) = e1t

11

andx2(t) = e2t

12

22are two linearly independent solutions. Consequently, the general solution of equation (1) isgivenbytherstcomponentofc1x1 + c2x2.18. Asinthesolutiontoexercise17above,wenotethattheeigenvaluesaregivenby1,2= b b24ac2a.In this case, we are assuming that b24ac = 0. Therefore, we have one repeated eigenvalue= b/2a. Asinthesolutiontoexercise17above, anassociatedeigenvectorisgivenbyv =

1 1

t. Therefore,x1(t) = e1t

11

isonesolution. Inordertolookforanothersolution, weneedtolookforwsatisfyingtheequation(A I)w = vfor = b/2aandv =

1 1

t. Bysubstitutingthesevaluesinforandv,ourequationreducesto

b/2a 1c/a b/2a

w1w2

=

1b/2a

.Weseethatw =

01

isasolutionofthisequation. Therefore,x2(t) =

t

11

+

01

e1tisanothersolutionofouroriginalequation. Therefore,x1(t) = e1t

11

andx2(t) = e2t

t1 +1t

form a fundamental set of solutions of Equation (2). Further, the general solution of Equation(1)correspondstotherstcomponentofc1x1 + c2x2.19. The characteristic equation is given by 2+ 2 = 0. Therefore, the two distinct rootsare = 2, 1. Therefore,thegeneralsolutionisgivenbyy(t) = c1e2t+ c2et.Therefore,y

(t) = 2c1e2t+ c2et.23Nowusingtheinitialconditions,weneedc1 + c2= 12c1 + c2= 1.The solution of this system of equations is c1= 0 and c2= 1. Therefore, the specic solutionisy(t) = et.2468101214161820y0 0.5 1 1.5 2 2.5 3tThesolutiony ast .20. Thecharacteristicequationis givenby92 12 + 4=0. Therefore, thereis onerepeatedroot, = 2/3. Therefore,thegeneralsolutionisgivenbyy(t) = c1e2t/3+ c2te2t/3.Therefore,y

(t) =2c13e2t/3+ c2

1 +2t3

e2t/3.Nowusingtheinitialconditions,weneedc1= 22c13+ c2= 1.Thesolutionof thissystemof equationsisc1=2andc2= 7/3. Therefore, thespecicsolutionisy(t) = 2e2t/373te2t/3.3020100y0.5 1 1.5 2 2.5 3t24Thesolutiony ast .21. The characteristic equation is given by 2+4+3 = 0. Therefore, the two distinct rootsare = 1, 3. Therefore,thegeneralsolutionisgivenbyy(t) = c1et+ c2e3t.Therefore,y

(t) = c1et3c2e3t.Nowusingtheinitialconditions,weneedc1 + c2= 2c13c2= 1.Thesolutionofthissystemofequationsisc1= 5/2andc2= 1/2. Therefore,thespecicsolutionisy(t) =52et12e3t.0.20.40.60.811.21.41.61.82y0 0.5 1 1.5 2 2.5 3tThesolutiony 0ast .22. Thecharacteristicequationisgivenby62 5 + 1=0. Therefore, thetwodistinctrealrootsare = 1/3, 1/2. Therefore,thegeneralsolutionisgivenbyy(t) = c1et/3+ c2et/2.Therefore,y

(t) =c13 et/3+c22 et/2.Nowusingtheinitialconditions,weneedc1 + c2= 4c13+c22= 0.Thesolutionof thissystemof equationsisc1=12andc2= 8. Therefore, thespecicsolutionisy(t) = 12et/38et/2.2532101234y0.5 1 1.5 2 2.5 3tThesolutiony ast .23. The characteristic equation is given by 26+9 = 0. Therefore, there is one repeatedroot, = 3. Therefore,thegeneralsolutionisgivenbyy(t) = c1e3t+ c2te3t.Therefore,y

(t) = 3c1e3t+ c2(1 + 3t)e3t.Nowusingtheinitialconditions,weneedc1= 03c1 + c2= 2.The solution of this system of equations is c1= 0 and c2= 2. Therefore, the specic solutionisy(t) = 2te3t.02004006008001000120014001600y0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2tThesolutiony ast .24. Thecharacteristicequationisgivenby2+ 3=0. Therefore, thetwodistinctrealrootsare = 0, 3. Therefore,thegeneralsolutionisgivenbyy(t) = c1 + c2e3t.Therefore,y

(t) = 3c2e3t.26Nowusingtheinitialconditions,weneedc1 + c2= 23c2= 3.Thesolutionof thissystemof equationsisc1= 1andc2= 1. Therefore, thespecicsolutionisy(t) = 1 e3t.21.81.61.41.21y0 0.5 1 1.5 2 2.5 3tThesolutiony 1ast .25. The characteristic equation is given by 2+4+4 = 0. Therefore, there is one repeatedroot, = 2. Therefore,thegeneralsolutionisgivenbyy(t) = c1e2t+ c2te2t.Therefore,y

(t) = 2c1e2t+ c2(1 2t)e2t.Nowusingtheinitialconditions,weneedc1e2c2e2= 22c1e2+ 3c2e2= 1.Thesolutionofthissystemofequationsisc1= 7e2andc2= 5e2. Therefore,thespecicsolutionisy(t) = 7e2(1+t)+ 5te2(1+t).0.511.52y1 1 2 3t27Thesolutiony 0ast .26. The characteristic equation is given by2+5 +3 = 0. Therefore,the two distinct realrootsare = (5 13)/2. Therefore,thegeneralsolutionisgivenbyy(t) = c1e(5+13)t/2+ c2e(513)t/2.Therefore,y

(t) =(5 +13)c12e(5+13)t/2+(5 13)c22e(513)t/2.Nowusingtheinitialconditions,weneedc1 + c2= 1(5 +13)c12+(5 13)c22= 0.Thesolutionof thissystemof equationsisc1=(1 + 5/13)/2andc2=(1 5/13)/2.Therefore,thespecicsolutionisy(t) =1 + 5/132e(5+13)t/2+1 5/132e(513)t/2.0.20.40.60.81y0 0.5 1 1.5 2 2.5 3tThesolutiony 0ast .27. The characteristic equation is given by 22+ 4 = 0. Therefore,the two distinct realrootsare = (1 33)/4. Therefore,thegeneralsolutionisgivenbyy(t) = c1e(1+33)t/4+ c2e(133)t/4.Therefore,y

(t) =(1 +33)c14e(1+33)t/4+(1 33)c24e(133)t/4.Nowusingtheinitialconditions,weneedc1 + c2= 0(1 +33)c14+(1 33)c24= 1.28Thesolutionof thissystemof equationsisc1=2/33andc2= 2/33. Therefore, thespecicsolutionisy(t) = 233e(1+33)t/4233e(133)t/4.24681012y0 0.5 1 1.5 2 2.5 3tThesolutiony ast .28. The characteristic equation is given by 2+8 9 = 0. Therefore,the two distinct realrootsare = 9, 1. Therefore,thegeneralsolutionisgivenbyy(t) = c1e9t+ c2et.Therefore,y

(t) = 9c1e9t+ c2et.Nowusingtheinitialconditions,weneedc1e9+ c2e = 19c1e9+ c2e = 0.The solution of this system of equations is c1=110e9and c2=910e1. Therefore, the specicsolutionisy(t) =110e99t+910e1+t.2468y0.5 1 1.5 2 2.5 3t29Thesolutiony ast .29. Thecharacteristicequationisgivenby42 1=0. Therefore, thetwodistinctrealrootsare = 1/2, 1/2. Therefore,thegeneralsolutionisgivenbyy(t) = c1et/2+ c2et/2.Therefore,y

(t) =c12 et/2c22 et/2.Nowusingtheinitialconditions,weneedc1e1+ c2e = 1c12 e1c22 e = 1.The solution of this system of equations is c1= e/2 and c2=32e1. Therefore, the specicsolutionisy(t) = 12e1+t/2+32e1t/2.65432101y2 1 1 2 3tThesolutiony ast .30. Weneedtondacharacteristicequationofdegreetwowithroots=2, 3. Wetakep() = ( 2)( + 3) = 2+ 6. Therefore,thedierentialequationisy

+ y

6y= 0.31. Weneedtondacharacteristicequationofdegreetwowithonerepeatedroot = 2.We take p() = (+2)2= 2+4+4. Therefore, the dierential equation is y

+4y

+4y= 0.32. Thecharacteristicequationis2(2 1) +( 1) = 0. Solvingthisequation,weseethattherootsare = , 1. Therefore,thegeneralsolutionisy(t) = c1et+ c2e(1)t.Inorderforthesolutiontotendtozero, weneed, 1 0. Thiswilloccurexactlywhen > 1.3033. Thecharacteristicequationis2+ (3 ) 2( 1)=0. Solvingthisequation,weseethattherootsare = 1, 2. Therefore,thegeneralsolutionisy(t) = c1e(1)t+ c2e2t.In order for the solution to tend to zero, we need 1 < 0. Therefore, the solutions will alltend to zero as long as < 1. Due the term c2e2t, we can never guarantee that all solutionswillbecomeunboundedast .34. Supposetherootsaredistinct,1, 2. Thenthesolutionisy(t) = c1e1t+ c2e2t.Solvingtheequationy(t) =0, weseethatwemusthavec1e1t= c2e2twhichimpliese(12)t= c2/c1. First, inorder toguaranteeanysolutionof this equation, wewouldneedc2/c1< 0. Then,applyingthenaturallogarithmfunctiontotheequation,weseethatt = ln(c2/c1)/(12).Iftherootsarenotdistinct,thenthesolutionisgivenbyy(t) = c1et+ c2tet.Therefore, y(t)=0implies(c1+ c2t)et=0. Sinceet=0, wemusthavec1+ c2t =0.Therefore,thesolutionwillbezeroonlywhent = c1/c2.35.(a) Ifyisaconstantsolution,thentheequationreducestocy= d,whichimpliesy= d/c.(b) Ifyeisanequilibriumsolution,thenye= d/c. Therefore,Y= y d/csatisesaY + bY + cY= ay

+ by

+ c(y d/c) = 0sinceyisasolutionofay

+ by

+ cy= d.36. Therootsofthecharacteristicequationare = b +b24ac2a.Therefore,therootswillbe(a) real,dierentandnegativeif(1)b24ac > 0and(2)b >b24ac.(b) realwithoppositesignsif(1)b24ac > 0and(2)b 0and(2) b >b24ac.3137. Lety2(t) = t2v(t). Sinceysatisesthedierentialequation,wehavet2(t2v

+ 4tv

+ 2v) 4t(t2v

+ 2tv) + 6t2v= 0.After collecting terms, we have t4v

= 0. Therefore, v(t) = c1+c2t. Thus y2(t) = c1t2+c2t3.Sincewealreadyhavethesolutiony1(t)=t2, wesetc1=0andc2=1. Therefore, wegetthesolutiony2(t) = t3.38. Lety2(t) = tv(t). Sinceysatisesthedierentialequation,wehavet2(tv

+ 2v

) + 2t(tv

+ v) 2tv= 0.After collectingterms, wehavet3v

+ 4t2v

=0. This equationis linear inv

. Solvingthis equationfor v

, wehavev

(t) =ct4, and, therefore, v(t) =c1t3+ c2. Therefore,y2(t) = c1t2+ c2t. Sincewealreadyhavethesolutiony1(t) = t,wesetc1= 1andc2= 0.Therefore,wegetthesolutiony2(t) = t2.39. Lety2(t) = t1v(t). Sinceysatisesthedierentialequation,wehavet2(t1v

2t2v

+ 2t3v) + 3t(t1v

t2v) + t1v= 0.After collectingterms, wehavetv

+ v

=0. This equationis linear inv

. Solvingthisequationforv

, wehavev

(t)=ct1, and, therefore, v(t)=c1 ln t + c2. Therefore, y2(t)=c1t1ln t + c2t1. Sincewealreadyhavethesolutiony1(t) = t1,wesetc1= 1andc2= 0.Therefore,wegetthesolutiony2(t) = t1ln t.40. Let y2(t) = tv(t). Substituting y into the dierential equation, we conclude that v

v

=0. This equation is linear in v

. We conclude that v

(t) = cetand, therefore, v(t) = c1et+c2.Therefore, y2(t)=c1tet+ c2t. Sincewealreadyhavethesolutiony1(t)=t, wesetc1=1andc2= 0. Therefore,wegetthesolutiony2(t) = tet.41. Lety2(x) = sin(x2)v(x). Substitutingyintothedierentialequation,weconcludethatv

+ [4x2cos(x2) sin(x2)]v

= 0.Thisequationislinearinv

. Weconcludethatv

(x) = cx/[sin(x2)]2and,therefore,v(x) = c1cos(x2)sin(x2)+ c2.Therefore,y2(x) = c1 cos(x2) + c2 sin(x2).Sincewealreadyhavethesolutiony1(x) = sin(x2),wetakethesolutiony2(x) = cos(x2).42. Lety2(x) = exv(x). Substitutingyintothedierentialequation,weconcludethatv

+x 2x 1v

= 0.32Thisequationislinearinv

. Anintegratingfactoris(x) =exx 1. Rewritingtheequationas

exv

x 1

= 0,we conclude that v

(x) = c(x 1)ex. Therefore, v(x) = c1xex+c2and y2(x) = c1x +c2ex.Sincewealreadyhavethesolutiony1(x) = ex,wetakethesolutiony2(x) = x.43. Let y2(x) = x1/4e2xv(x). Substituting yinto the dierential equation, we conclude that2x9/4v

+ (4x7/4+ x5/4)v


. Anintegratingfactoris(x) = exp 2x1/2+12x

dx

=x exp(4x).Rewritingtheequationasx exp(4xv

= 0,we conclude that v

(x) = c exp(4x)/x. Integrating, we have v(x) = c1 exp(4x) +c2,and,therefore,y2(x) = c1x1/4e(2x) + c2x1/4e(2x). Sincewealreadyhavethesolutiony1(x) = x1/4e2x,wetakethesolutiony2(x) = x1/4e2x.44. Lety2(x)=x1/2sin xv(x). Substitutingyintothedierential equation, weconcludethatsin xv

+ 2 cos xv


. Itssolutionisgivenbyv

(x)=c/ sin(x)2. Integratingwithrespect to x, we have v(x) = c1 cot(x)+c2. Therefore, y2(x) = c1x1/2cos(x)+c2x1/2sin(x).Since we alreadyhave the solutiony1(x) =x1/2sin(x), we take the solutiony2(x) =x1/2cos(x).45.(a) Fory1= ex,wehavey

1= y

1= ex. Therefore,xy

1 (x + N)y

1 + Ny1= xex(x + N)ex+ Nex= 0.(b) Lety2= exv. Theninorderfory2tosatisfythedierentialequationweneedxv

+ (x N)v

= 0.Solvingthisequation,wehavev

= CxNex. Therefore,v= c1

xNexdx + c2,whichimpliesy2= c1ex

xNexdx + c2ex.33Sincewealreadyhavethesolutiony1(x) = ex,wetakey2(x) = cex

xNexdx.ForN= 1,wehavey2(x) = c(1 x).ForN= 2,wehavey2(x) = c(2 2x x2).Therefore,lettingc = 1/N!,forN= 1,wehavey2(x) = 1 + xandforN= 2,wehavey2(x) = 1 + x +12x2.46. Direct substitution veries that y1(x) = ex2/2is a solution of this dierential equation.Tondanothersolution, wesety2(x)=ex2/2v(x). Then, inorderfory2tosatisfythedierentialequation,weneedvtosatisfyv

xv

= 0.Solvingthisequationforv

,weseethatthesolutionisgivenbyv

(x) = c1ex2/2.Integrating,weobtainv(x) = c1

x0eu2/2du + c2.Therefore,y2(x) = c1ex2/2

x0eu2/2du + c2ex2/2.Settingc2= 0,wearriveatthesolutiony2(x) = ex2/2

x0eu2/2du.Section4.41. Forallpartsbelow,weletz1= a1 + ib1andz2= a2 + ib2.(a)z1 + z2= a1 + ib1 + a2 + ib2= (a1 + a2) + i(b1 + b2).Therefore,z1 + z2= (a1 + a2) i(b1 + b2)= (a1ib1) + (a2ib2)= z1 + z2.34(b)z1z2= (a1 + ib1)(a2 + ib2)= (a1a2b1b2) + i(a1b2 + a2b1).Therefore,z1z2= (a1a2b1b2) i(a1b2 + a2b1)= a1a2ia1b2b1b2ia2b1= a1(a2ib2) b1i(a2ib2)= (a2ib2)(a1ib1)= z1 z2(c)z1z2=a1 + ib1a2 + ib2=a1 + ib1a2 + ib2a2ib2a2ib2=(a1a2 + b1b2) + i(a2b1a1b2)a22 + b22.Therefore,

z1z2

=(a1a2 + b1b2) + i(a1b2a2b1)a22 + b22Next,weseethatz1z2=a1ib1a2ib2=a1ib1a2ib2a2 + ib2a2 + ib2=(a1a2 + b1b2) + i(a1b2a2b1)a22 + b22.Therefore,theresultholds.(d) If z= 0, then z= 0+0i which implies |z| = 02+02= 0. Also, if |z| = 0, then a2+b2= 0whichimpliesa = 0, b = 0,and,therefore,z= 0 + 0i.(e) Asinthesolutiontopart(b),weknowthatz1z2= (a1a2b1b2) + i(a1b2 + a2b1).35Therefore,|z1z2| = (a1a2b1b2)2+ (a1b2 + a2b1)2= (a1a2)22a1a2b1b2 + (b1b2)2+ (a1b2)2+ 2a1b2a2b1 + (a2b1)2= a21a22 + b21b22 + a21b22 + a22b21.Also,|z1||z2| = |a1 + ib1||a2 + ib2|= (a21 + b21)(a22 + b22)= a21a22 + a21b22 + b21a22 + b21b22.Therefore,thedesiredresultholds.(f) Asinthesolutiontopart(c),weknowthatz1z2=(a1a2 + b1b2) + i(a2b1a1b2)a22 + b22.Therefore,

z1z2

=

(a1a2 + b1b2) + i(a2b1a1b2)a22 + b22

=

a1a2 + b1b2a22 + b22

2+

a2b1a1b2a22 + b22

2=a21a22 + 2a1a2b1b2 + b21b22 + a22b212a2b1a1b2 + a21b22(a22 + b22)2=a21a22 + b21b22 + a22b21 + a21b22(a22 + b22)=a21(a22 + b22) + b21(a22 + b22)(a22 + b22)2=a21 + b21a22 + b22= |z1||z2|.2. Bythelawofcosines,|z1 + z2|2= |z1|2+ |z2|22|z1||z2| cos =|z1 + z2|2 |z1|2+ 2|z1||z2| + |z2|2= (|z1| + |z2|)2.Takingthesquarerootofbothsidesabove,weseethat|z1 + z2| |z1| + |z2|.36Wenoticethat |z1 + z2|= |z1| + |z2| ifandonlyifcos = 1. Underthatcondition, thelawofcosinesleadsustotheequation|z1 + z2|2= |z1|2+ |z2|2+ 2|z1||z2| = (|z1| + |z2|)2=|z1 + z2| = |z1| + |z2|.Theconditionthatcos = 1means = .3.(a)z1z2= (r1 cos 1 + ir1 sin 1)(r2 cos 2 + ir2 sin 2)= r1r2 cos 1 cos 2 + ir1r2 cos 1 sin 2 + ir1r2 sin 1 cos 2r1r2 sin 1 sin 2= r1r2(cos 1 cos 2sin 1 sin 2) + ir1r2(cos 1 sin 2 + sin 1 cos 2)= |z1||z2| cos(1 + 2) + i|z1||z2| sin(1 + 2)= |z1||z2|(cos(1 + 2) + i sin(1 +2))= |z1||z2|ei(1+2).(b)4321012344 2 2 44. e1+2i= e1e2i= e(cos(2) + i sin(2)) = e cos(2) + ie sin(2).5. e23i= e2e3i= e2(cos(3) + i sin(3)) = e2cos(3) ie2sin(3).6. ei= (cos() + i sin()) = 1.7. e2(/2)i= e2e(/2)i= e2(cos(/2) + i sin(/2)) = ie2.8. 21i= e(1i)(ln 2)= eln 2ei ln 2= eln2(cos(ln 2)+i sin(ln 2)) = 2(cos(ln 2)i sin(ln 2)) =2 cos(ln 2) 2i sin(ln 2).9. 1+2i=e(1+2i) ln =eln e2i ln =eln(cos(2 ln ) + i sin(2 ln ))=1 cos(2 ln ) +i sin(2 ln ).10.37(a) Ify(0) = y0andy

(0) = y1arerealnumbers,theny(0) = c1 + c2= y0isreal,andy

(0) = c1( + i) + c2( i) = y1isreal. Weneedtoshowthatc2= c1. Letc1= a1 + ib1andc2= a2 + ib2. Weneedtoshowthata1= a2andb2= b1. Webeginbysubstitutingthecomplexexpressionsforc1andc2intotheequationfory0. Upondoingso,wehave(a1 + a2) + i(b1 + b2) = y0is real. In order for this to be real, we need the imaginary part to be zero. In particular,weneedb1 + b2= 0,orb2= b1. Thenusingtheequationfory1,wehave(a1 + ib1)( + i) + (a2 + ib2)( i) = y1isreal. Multiplyingandsimplifying,wehave(a1 b1 + a2 + b2) + i(a1 + b1 a2 + b2) = y1.Since,b2= b1,thisequationreducesto(a1 b1 + a2 b1) + i(a1 a2) = y1is real. In order for this to be real, we need the imaginary part to be zero. In particular,weneeda1= a2. Therefore,c2= c1asclaimed.(b) Inordertoshowthaty(t)isreal,wewillshowthaty(t) = y(t). Weseethaty(t) = ce(+i)t+ ce(i)t= ce(i)t+ ce(+i)t= y(t).Therefore,y(t)isreal.(c) Ifc = + i,theny(t) = ( + i)e(+i)t+ ( i)e(i)t= ( + i)et(cos(t) + i sin(t)) + ( i)et(cos(t) i sin(t))= et[cos(t) + isin(t) + i cos(t) sin(t) + cos(t) isin(t) i cos(t) sin(t)]= et[2cos(t) 2 sin(t)].Therefore,c1= 2andc2= 2inEquation10.11.38(a) Thecharacteristicequationis 2 2 + 2=0. Therefore, theroots are=1 i.Therefore,onecomplexsolutionisy(t) = e(1+i)t= et(cos(t) + i sin(t)).Consideringthe real andimaginaryparts of this function, we arrive at the generalsolutiony(t) = c1etcos(t) + c2etsin(t).(b) Fromoursolutioninpart(a),weseethaty

(t) = c1et(cos(t) sin(t)) + c2et(sin(t) + cos(t)).Therefore,x =

yy

= c1et

cos(t)cos(t) sin(t)

+ c2et

sin(t)sin(t) + cos(t)

.3210123x23 2 1 1 2 3x1(c) Weseethatthecriticalpoint(0, 0)isanunstablespiralpoint.12.(a) Thecharacteristicequationis2 2 + 6=0. Therefore, therootsare=1 i5.Therefore,onecomplexsolutionisy(t) = e(1+i5)t= et(cos(5t) + i sin(5t)).Consideringthe real andimaginaryparts of this function, we arrive at the generalsolutiony(t) = c1etcos(5t) + c2etsin(5t).(b) Fromoursolutioninpart(a),weseethaty

(t) = c1et(cos(5t) 5 sin(5t)) + c2et(sin(5t) +5 cos(5t)).Therefore,x =

yy

= c1et

cos(5t)cos(5t) 5 sin(5t)

+ c2et

sin(5t)sin(5t) +5 cos(5t)

.393210123x23 2 1 1 2 3x1(c) Weseethatthecriticalpoint(0, 0)isanunstablespiralpoint.13.(a) Thecharacteristicequationis 2+ 2 8=0. Therefore, therootsare= 4, 2.Therefore,thegeneralsolutionisy(t) = c1e4t+ c2e2t(b) Fromoursolutioninpart(a),weseethaty

(t) = 4c1e4t+ 2c2e2t.Therefore,x =

yy

= c1e4t

14

+ c2e2t

12

.3210123x23 2 1 1 2 3x1(c) Weseethatthecriticalpoint(0, 0)isanunstablesaddlepoint.14.40(a) Thecharacteristicequationis2+ 2 + 2

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