Solution Manual for Differential Equations and Linear Algebra, 4th Edition

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SOLUTIONSMANUALDIFFERENTIALEQUATIONS ANDLINEARALGEBRAFOURTHEDITIONStephen W. GoodeandScott A. AnninCalifornia State University, Fullerton

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ContentsChapter 11Chapter 2123Chapter 3207Chapter 4256Chapter 5348Chapter 6390Chapter 7442Chapter 8534Chapter 9631Chapter 10762Chapter 11861Appendix A979Appendix B982Appendix C987

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1Chapter 1 SolutionsSolutions to Section 1.1True-False Review:(a): FALSE.A derivative must involvesomederivative of the functiony=f(x), not necessarily the firstderivative.(b):FALSE.The order of a differential equation is the order of thehighest, not the lowest, derivativeappearing in the differential equation.(c): FALSE.This differential equation has order two, since the highest order derivative that appears in theequation is the second order expressiony′′.(d):FALSE.The carrying capacity refers to the maximum population size that the environment cansupport in the long run; it is not related to the initial population in any way.(e): TRUE.The valuey(0) is called an initial condition to the differential equation fory(t).(f ):TRUE.According to Newton’s Law of Cooling, the rate of cooling is proportional to thedifferencebetween the object’s temperature and the medium’s temperature.Since that difference is greater for theobject at 100◦Fthan the object at 90◦F, the object whose temperature is 100◦Fhas a greater rate ofcooling.(g): FALSE.The temperature of the object is given byT(t) =Tm+ce−kt, whereTmis the temperatureof the medium, andcandkare constants.Sincee−kt= 0, we see thatT(t)=Tmfor all timest.Thetemperature of the objectapproachesthe temperature of the surrounding medium, but never equals it.(h):TRUE.Since the temperature of the coffee is falling, the temperaturedifferencebetween the coffeeand the room is higher initially, during the first hour, than it is later, when the temperature of the coffeehas already decreased.(i): FALSE.The slopes of the two curves arenegativereciprocals of each other.(j): TRUE.If the original family of parallel lines have slopeskfork= 0, then the family of orthogonal tra-jectories are parallel lines with slope−1k. If the original family of parallel lines are vertical (resp. horizontal),then the family of orthogonal trajectories are horizontal (resp. vertical) parallel lines.(k): FALSE.The family of orthogonal trajectories for a family of circles centered at the origin is the familyof lines passing through the origin.(l): TRUE.Ifv(t) denotes the velocity of the object at timetanda(t) denotes the velocity of the objectat timet, then we havea(t) =v′(t), which is a differential equation for the unknown functionv(t).(m): FALSE.The restoring force is directed in the directionoppositeto the displacement from the equi-librium position.(n): TRUE.The allometric relationshipB=B0m3/4, whereB0is a constant, relates the metabolic rateand total body mass for any species.Problems:1.The order is 2.2.The order is 1.

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23.The order is 3.4.The order is 2.5.We compute the first three derivatives ofy(t) = lnt:dydt= 1t ,d2ydt2=−1t2,d3ydt3= 2t3.Therefore,2(dydt)3= 2t3=d3ydt3,as required.6.We compute the first two derivatives ofy(x) =x/(x+ 1):dydx=1(x+ 1)2andd2ydx2=−2(x+ 1)3.Theny+d2ydx2=xx+ 1−2(x+ 1)3=x3+ 2x2+x−2(x+ 1)3= (x+ 1) + (x3+ 2x2−3)(x+ 1)3=1(x+ 1)2+x3+ 2x2−3(1 +x)3,as required.7.We compute the first two derivatives ofy(x) =exsinx:dydx=ex(sinx+ cosx)andd2ydx2= 2excosx.Then2ycotx−d2ydx2= 2(exsinx) cotx−2excosx= 0,as required.8.(T−Tm)−1dTdt=−k=⇒ddt(ln|T−Tm|) =−k. The preceding equation can be integrated directly toyield ln|T−Tm|=−kt+c1. Exponentiating both sides of this equation gives|T−Tm|=e−kt+c1, whichcan be written asT−Tm=ce−kt,wherec=±ec1. Rearranging yieldsT(t) =Tm+ce−kt.9.After 4 p.m.In the first two hours after noon, the water temperature increased from 50◦F to 55◦F, an increase of five degrees.Because the temperature of the water has grown closer to the ambient airtemperature, the temperature difference|T−Tm|is smaller, and thus, the rate of change of the temperatureof the water grows smaller, according to Newton’s Law of Cooling. Thus, it will take longer for the watertemperature to increase another five degrees. Therefore, the water temperature will reach 60◦F more thantwo hours later than 2 p.m., or after 4 p.m.10.The object temperature cools a total of 40◦F during the 40 minutes, but according to Newton’s Law ofCooling, it cools faster in the beginning (since|T−Tm|is greater at first). Thus, the object cooled half-way

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3from 70◦F to 30◦F in less than half the total cooling time. Therefore, it took less than 20 minutes for theobject to reach 50◦F.11.The given family of curves satisfies:x2+ 9y2=c=⇒2x+ 18y dydx= 0 =⇒dydx=−x9y.Orthogonal trajectories satisfy:dydx= 9yx=⇒1ydydx= 9x=⇒ddx(ln|y|) = 9x=⇒ln|y|= 9 ln|x|+c1=⇒y=kx9,wherek=±ec1.1.5-0.5-0.41.0xy(x)-0.80.40.50.8-1.0-1.5Figure 0.0.1: Figure for Problem 1112.Given family of curves satisfies:y=cx2=⇒c=yx2. Hence,dydx= 2cx=c(yx2)x= 2yx .Orthogonal trajectories satisfy:dydx=−x2y=⇒2y dydx=−x=⇒ddx(y2) =−x=⇒y2=−12x2+c1=⇒2y2+x2=c2,wherec2= 2c1.13.Given a family of curves satisfies:y=cx=⇒x dydx+y= 0 =⇒dydx=−yx.Orthogonal trajectories satisfy:dydx=xy=⇒y dydx=x=⇒ddx(12y2)=x=⇒12y2= 12x2+c1=⇒y2−x2=c2,wherec2= 2c1.14.The given family of curves satisfies:y=cx5=⇒c=yx5. Hence,dydx= 5cx4= 5(yx5)x4= 5yx .

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4x1y(x)1.622.01.20.40.8-1.2-2-0.4-2.0-1.6-1-0.8Figure 0.0.2: Figure for Problem 12-22-4y(x)24-4x-24Figure 0.0.3: Figure for Problem 13Orthogonal trajectories satisfy:dydx=−x5y=⇒5y dydx=−x=⇒ddx(52y2)=−x=⇒52y2=−12x2+c1=⇒5y2+x2=c2,wherec2= 2c1.

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51.5-0.41.0y(x)x-0.80.40.8-1.0-1.5Figure 0.0.4: Figure for Problem 1415.Given family of curves satisfies:y=cex=⇒dydx=cex=y. Orthogonal trajectories satisfy:dydx=−1y=⇒y dydx=−1 =⇒ddx(12y2)=−1 =⇒12y2=−x+c1=⇒y2=−2x+c2.21y(x)1x-1-2-1Figure 0.0.5: Figure for Problem 1516.Given family of curves satisfies:y2= 2x+c=⇒dydx= 1y.Orthogonal trajectories satisfy:dydx=−y=⇒y−1dydx=−1 =⇒ddx(ln|y|) =−1 =⇒ln|y|=−x+c1=⇒y=c2e−x.

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6113-3-22-4-12-143y(x)x4Figure 0.0.6: Figure for Problem 1617.y=cxm=⇒dydx=cmxm−1, butc=yxmsodydx=myx. Orthogonal trajectories satisfy:dydx=−xmy=⇒y dydx=−xm=⇒ddx(12y2)=−xm=⇒12y2=−12m x2+c1=⇒y2=−1m x2+c2.18.y=mx+c=⇒dydx=m.Orthogonal trajectories satisfy:dydx=−1m=⇒y=−1m x+c1.19.y2=mx+c=⇒2y dydx=m=⇒dydx=m2y.Orthogonal trajectories satisfy:dydx=−2ym=⇒y−1dydx=−2m=⇒ddx(ln|y|) =−2m=⇒ln|y|=−2m x+c1=⇒y=c2e−2xm.20.y2+mx2=c=⇒2y dydx+ 2mx= 0 =⇒dydx=−mxy.Orthogonal trajectories satisfy:dydx=ymx=⇒y−1dydx=1mx=⇒ddx(ln|y|) =1mx=⇒mln|y|= ln|x|+c1=⇒ym=c2x.21.The given family of curves satisfies:x2+y2= 2cx=⇒c=x2+y22x. Hence,2x+ 2y dydx= 2c=x2+y2x.

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7Therefore,2y dydx=x2+y2x−2x=y2−x2x,so thatdydx=y2−x22xy.Orthogonal trajectories satisfy:dydx=−2xyy2−x2=2xyx2−y2.22.u=x2+ 2y2=⇒0 = 2x+ 4y dydx=⇒dydx=−x2y.Orthogonal trajectories satisfy:dydx= 2yx=⇒y−1dydx= 2x=⇒ddx(ln|y|) = 2x=⇒ln|y|= 2 ln|x|+c1=⇒y=c2x2.x1y(x)1.622.01.20.40.8-1.2-2-0.4-2.0-1.6-1-0.8Figure 0.0.7: Figure for Problem 2223.m1= tan (a1) = tan (a2−a) =tan (a2)−tan (a)1 + tan (a2) tan (a) =m2−tan (a)1 +m2tan (a) .24.d2ydt2=g=⇒dydt=gt+c1=⇒y(t) =gt22+c1t+c2.Now impose the initial conditions.y(0) = 0 =⇒c2= 0.dydt(0) =⇒c1= 0.Hence, the solution to the initial-value problem is:y(t) =gt22.The object hits theground at time,t0, wheny(t0) = 100. Hence 100 =gt202, so thatt0=√200g≈4.52 s, where we have takeng= 9.8 ms−2.

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825.Fromd2ydt2=g, we integrate twice to obtain the general equations for the velocity and the position ofthe ball, respectively:dydt=gt+candy(t) = 12gt2+ct+d, wherec, dare constants of integration. Settingy= 0 to be at the top of the boy’s head (and positive direction downward), we know thaty(0) = 0. Since theobject hits the ground 8 seconds later, we have thaty(8) = 5 (since the ground lies at the positiony= 5).From the values ofy(0) andy(8), we find thatd= 0 and 5 = 32g+ 8c. Therefore,c= 5−32g8.(a).The ball reaches its maximum height at the moment wheny′(t) = 0. That is,gt+c= 0. Therefore,t=−cg= 32g−58g≈3.98 s.(b).To find the maximum height of the tennis ball, we computey(3.98)≈ −253.51 feet.So the ball is 253.51 feetabovethe top of the boy’s head, which is 258.51 feet above the ground.26.Fromd2ydt2=g, we integrate twice to obtain the general equations for the velocity and the position ofthe rocket, respectively:dydt=gt+candy(t) = 12gt2+ct+d, wherec, dare constants of integration. Settingy= 0 to be at ground level, we know thaty(0) = 0. Thus,d= 0.(a).The rocket reaches maximum height at the moment wheny′(t) = 0. That is,gt+c= 0. Therefore, thetime that the rocket achieves its maximum height ist=−cg. At this time,y(t) =−90 (the negative signaccounts for the fact that the positive direction is chosen to be downward). Hence,−90 =y(−cg)= 12g(−cg)2+c(−cg)=c22g−c2g=−c22g .Solving this forc, we find thatc=±√180g. However, sincecrepresents the initial velocity of the rocket,and the initial velocity is negative (relative to the fact that the positive direction is downward), we choosec=−√180g≈ −42.02 ms−1, and thus the initial speed at which the rocket must be launched for optimalviewing is approximately 42.02 ms−1.(b).The time that the rocket reaches its maximum height ist=−cg≈ − −42.029.81= 4.28 s.27.Fromd2ydt2=g, we integrate twice to obtain the general equations for the velocity and the position ofthe rocket, respectively:dydt=gt+candy(t) = 12gt2+ct+d, wherec, dare constants of integration. Settingy= 0 to be at the level of the platform (with positive direction downward), we know thaty(0) = 0. Thus,d= 0.(a).The rocket reaches maximum height at the moment wheny′(t) = 0. That is,gt+c= 0. Therefore, thetime that the rocket achieves its maximum height ist=−cg. At this time,y(t) =−85 (this is 85 m abovethe platform, or 90 m above the ground). Hence,−85 =y(−cg)= 12g(−cg)2+c(−cg)=c22g−c2g=−c22g .

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9Solving this forc, we find thatc=±√170g. However, sincecrepresents the initial velocity of the rocket,and the initial velocity is negative (relative to the fact that the positive direction is downward), we choosec=−√170g≈ −40.84 ms−1, and thus the initial speed at which the rocket must be launched for optimalviewing is approximately 40.84 ms−1.(b).The time that the rocket reaches its maximum height ist=−cg≈ − −40.849.81= 4.16 s.28.Ify(t) denotes the displacement of the object from its initial position at timet, the motion of the objectcan be described by the initial-value problemd2ydt2=g,y(0) = 0,dydt(0) =−2.We first integrate this differential equation:d2ydt2=g=⇒dydt=gt+c1=⇒y(t) =gt22+c1t+c2.Nowimpose the initial conditions.y(0) = 0 =⇒c2= 0.dydt(0) =−2 =⇒c1=−2.Hence the solution to theinitial-value problem isy(t) =gt22−2t. We are given thaty(10) =h. Consequently,h=g(10)22−2·10 =⇒h= 10(5g−2)≈470 m where we have takeng= 9.8 ms−2.29.Ify(t) denotes the displacement of the object from its initial position at timet, the motion of the objectcan be described by the initial-value problemd2ydt2=g,y(0) = 0,dydt(0) =v0.We first integrate the differential equation:d2ydt2=g=⇒dydt=gt+c1=⇒y(t) =gt22+c1t+c2. Now imposethe initial conditions.y(0) = 0 =⇒c2= 0.dydt(0) =v0=⇒c1=v0. Hence the solution to the initial-valueproblem isy(t) =gt22+v0t. We are given thaty(t0) =h. Consequently,h=gt20+v0t0. Solving forv0yieldsv0= 2h−gt202t0.30.Fromy(t) =Acos (ωt−φ), we obtaindydt=−Aωsin (ωt−φ)andd2ydt2=−Aω2cos (ωt−φ).Hence,d2ydt2+ω2y=−Aω2cos (ωt−φ) +Aω2cos (ωt−φ) = 0.Substitutingy(0) =a, we obtaina=Acos(−φ) =Acos(φ).Also, fromdydt(0) = 0, we obtain 0 =−Aωsin(−φ) =Aωsin(φ). SinceA= 0 andω= 0 and|φ|< π, we haveφ= 0. It follows thata=A.31.y(t) =c1cos (ωt) +c2sin (ωt) =⇒dydt=−c1ωsin (ωt) +c2ωcos (ωt) =⇒d2ydt2=−c1ω2cos (ωt)−c2ω2sin (ωt) =−ω2[c1cos (ωt) +c2cos (ωt)] =−ω2y.Consequently,d2ydt2+ω2y= 0.To determine the

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10amplitude of the motion we write the solution to the differential equation in the equivalent form:y(t) =√c21+c22[c1√c21+c22cos (ωt) +c2√c21+c22sin (ωt)].We can now define an angleφbycosφ=c1√c21+c22andsinφ=c2√c21+c22.Then the expression for the solution to the differential equation isy(t) =√c21+c22[cos (ωt) cosφ+ sin (ωt) sinφ] =√c21+c22cos (ωt+φ).Consequently the motion corresponds to an oscillation with amplitudeA=√c21+c22.32.In this problem we havem0= 3g,M= 2700g,a= 1.5. Substituting these values into Equation (1.1.26)yieldsm(t) = 2700{1−[1−(1900)1/4]e−1.5t/(4(2700)1/4)}4.Therefore the mass of the heron after 30 days ism(30) = 2700{1−[1−(1900)1/4]e−45/(4(2700)1/4)}4≈1271.18 g.33.In this problem we havem0= 8g,M= 280g,a= 0.25. Substituting these values into Equation (1.1.26)yieldsm(t) = 280{1−[1−(135)1/4]e−t/(16(280)1/4)}4.We need to find the time,twhen the mass of the rat reaches 75% of its fully grown size. Therefore we needto findtsuch that75100·280 = 280{1−[1−(135)1/4]e−t/(16(280)1/4)}4.Solving algebraically fortyieldst= 16·(280)1/4·ln[1−(1/35)1/41−(75/100))1/4]≈140 days.Solutions to Section 1.2True-False Review:(a): TRUE.This is condition 1 in Definition 1.2.8.

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11(b): TRUE.This is the content of Theorem 1.2.12.(c):FALSE.There are solutions toy′′+y= 0 that do not have the formc1cosx+ 5c2cosx, such asy(x) = sinx.Therefore,c1cosx+ 5c2cosxdoes not meet the second requirement set forth in Definition1.2.8 for the general solution.(d):FALSE.There are solutions toy′′+y= 0 that do not have the formc1cosx+ 5c1sinx, suchasy(x) = cosx+ sinx.Therefore,c1cosx+ 5c1sinxdoes not meet the second requirement set form inDefinition 1.2.8 for the general solution.(e): TRUE.Since the right-hand side of the differential equation is a function ofxonly, we can integrateboth sidesntimes to obtain the formula for the solutiony(x).Problems:1.Linear.2.Non-linear, because of they2expression on the right side of the equation.3.Non-linear, because of the termyy′′on tthe left side of the equation.4.Non-linear, because of the expression tanyappearing on the left side of the equation.5.Linear.6.Non-linear, because of the expression1y′on the left side of the equation.7.y(x) =c1e−5x+c2e5x=⇒y′=−5c1e−5x+ 5c2e5x=⇒y′′= 25c1e−5x+ 25c2e5x=⇒y′′−25y=(25c1e−5x+ 25c2e5x)−25(c1e−5x+c2e5x) = 0.Thusy(x) =c1e−5x+c2e5xis a solution of the givendifferential equation for allx∈R.8.y(x) =c1cos 2x+c2sin 2x=⇒y′=−2c1sin 2x+ 2c2cos 2x=⇒y′′=−4c1cos 2x−4c2sin 2x=⇒y′′+ 4y= (−4c1cos 2x−4c2sin 2x) + 4(c1cos 2x+c2sin 2x) = 0.Thusy(x) =c1cos 2x+c2sin 2xis asolution of the given differential equation for allx∈R.9.y(x) =c1ex+c2e−2x=⇒y′=c1ex−2c2e−2x=⇒y′′=c1ex+ 4c2e−2x=⇒y′′+y′−2y= (c1ex+4c2e−2x) + (c1ex−2c2e−2x)−2(c1ex+c2e−2x) = 0. Thusy(x) =c1ex+c2e−2xis a solution of the givendifferential equation for allx∈R.10.y(x) =1x+ 4 =⇒y′=−1(x+ 4)2=−y2.Thusy(x) =1x+ 4 is a solution of the given differentialequation forx∈(−∞,−4) orx∈(−4,∞).11.y(x) =c1√x=⇒y′=c12√x=y2x. Thusy(x) =c1√xis a solution of the given differential equation forallx∈ {x:x >0}.12.y(x) =c1e−xsin (2x) =⇒y′= 2c1e−xcos (2x)−c1e−xsin (2x) =⇒y′′=−3c1e−xsin (2x)−4c1e−xcos (2x) =⇒y′′+ 2y′+ 5y=−3c1e−xsin (2x)−4c1e−xcos (2x) + 2[2c1e−xcos (2x)−c1e−xsin (2x)] + 5[c1e−xsin (2x)] = 0.Thusy(x) =c1e−xsin (2x) is a solution to the given differential equation for allx∈R.13.y(x) =c1cosh (3x) +c2sinh (3x) =⇒y′= 3c1sinh (3x) + 3c2cosh (3x) =⇒y′′= 9c1cosh (3x) +

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129c2sinh (3x) =⇒y′′−9y= [9c1cosh (3x) + 9c2sinh (3x)]−9[c1cosh (3x) +c2sinh (3x)] = 0. Thusy(x) =c1cosh (3x) +c2sinh (3x) is a solution to the given differential equation for allx∈R.14.y(x) =c1x3+c2x=⇒y′=−3c1x4−c2x2=⇒y′′= 12c1x5+ 2c2x3=⇒x2y′′+ 5xy′+ 3y=x2(12c1x5+ 2c2x3)+5x(−3c1x4−c2x2)+ 3(c1x3+c2x)= 0. Thusy(x) =c1x3+c2xis a solution to the given differential equationfor allx∈(−∞,0) orx∈(0,∞).15.y(x) =c1x2lnx=⇒y′=c1(2xlnx+x) =⇒y′′=c1(2 lnx+ 3) =⇒x2y′′−3xy′+ 4y=x2·c1(2 lnx+3)−3x·c1(2xlnx+x) + 4c1x2lnx=c1x2[(2 lnx+ 3)−3(2xlnx+ 1) + 4 lnx] = 0. Thusy(x) =c1x2lnxis a solution of the given differential equation for allx >0.16.y(x) =c1x2cos(3 lnx) =⇒y′=c1[2xcos(3 lnx)−3xsin(3 lnx)] =⇒y′′=c1[−7 cos(3 lnx)−6 sin(3 lnx)] =⇒x2y′′−3xy′+13y=x2·c1[−7 cos(3 lnx)−9 sin(3 lnx)]−3x·c1[2xcos(3 lnx)−3xsin(3 lnx)]+13c1x2cos(3 lnx) =c1x2{[−7 cos(3 lnx)−9 sin(3 lnx)]−3[2 cos(3 lnx)−3 sin(3 lnx)] + 13 cos(3 lnx)}= 0. Thusy(x) =c1x2cos(3 lnx)is a solution of the given differential equation for allx >0.17.y(x) =c1√x+ 3x2=⇒y′=c12√x+ 6x=⇒y′′=−c14√x3+ 6 =⇒2x2y′′−xy′+y= 2x2(−c14√x3+ 6)−x(c12√x+ 6x)+(c1√x+3x2) = 9x2. Thusy(x) =c1√x+3x2is a solution to the given differential equationfor allx∈ {x:x >0}.18.y(x) =c1x2+c2x3−x2sinx=⇒y′= 2c1x+ 3c2x2−x2cosx−2xsinx=⇒y′′= 2c1+ 6c2x+x2sinx−2xcosx−2xcos−2 sinx. Substituting these results into the given differential equation yieldsx2y′′−4xy′+ 6y=x2(2c1+ 6c2x+x2sinx−4xcosx−2 sinx)−4x(2c1x+ 3c2x2−x2cosx−2xsinx)+ 6(c1x2+c2x3−x2sinx)= 2c1x2+ 6c2x3+x4sinx−4x3cosx−2x2sinx−8c1x2−12c2x3+ 4x3cosx+ 8x2sinx+ 6c1x2+ 6c2x3−6x2sinx=x4sinx.Hence,y(x) =c1x2+c2x3−x2sinxis a solution to the differential equation for allx∈R.19.y(x) =c1eax+c2ebx=⇒y′=ac1eax+bc2ebx=⇒y′′=a2c1eax+b2c2ebx. Substituting these resultsinto the differential equation yieldsy′′−(a+b)y′+aby=a2c1eax+b2c2ebx−(a+b)(ac1eax+bc2ebx) +ab(c1eax+c2ebx)= (a2c1−a2c1−abc1+abc1)eax+ (b2c2−abc2−b2c2+abc2)ebx= 0.Hence,y(x) =c1eax+c2ebxis a solution to the given differential equation for allx∈R.20.y(x) =eax(c1+c2x) =⇒y′=eax(c2) +aeax(c1+c2x) =eax(c2+ac1+ac2x) =⇒y′′=eaax(ac2) +aeax(c2+ac1+ac2x) =aeax(2c2+ac1+ac2x). Substituting these into the differential equation yieldsy′′−2ay′+a2y=aeax(2c2+ac1+ac2x)−2aeax(c2+ac1+ac2x) +a2eax(c1+c2x)=aeax(2c2+ac1+ac2x−2c2−2ac1−2ac2x+ac1+ac2x)= 0.Thus,y(x) =eax(c1+c2x) is a solution to the given differential eqaution for allx∈R.

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1321.y(x) =eax(c1cosbx+c2sinbx) so,y′=eax(−bc1sinbx+bc2cosbx) +aeax(c1cosbx+c2sinbx)=eax[(bc2+ac1) cosbx+ (ac2−bc1) sinbx] so,y′′=eax[−b(bc2+ac1) sinbx+b(ac2+bc1) cosbx] +aeax[(bc2+ac1) cosbx+ (ac2+bc1) sinbx]=eax[(a2c1−b2c1+ 2abc2) cosbx+ (a2c2−b2c2−abc1) sinbx].Substituting these results into the differential equation yieldsy′′−2ay′+ (a2+b2)y= (eax[(a2c1−b2c1+ 2abc2) cosbx+ (a2c2−b2c2−abc1) sinbx])−2a(eax[(bc2+ac1) cosbx+ (ac2−bc1) sinbx]) + (a2+b2)(eax(c1cosbx+c2sinbx))=eax[(a2c1−b2c1+ 2abc2−2abc2−2a2c1+a2c1+b2c1) cosbx+ (a2c2−b2c2−2abc1+ 2abc1−2a2c2+a2c2+b2c2) sinbx]= 0.Thus,y(x) =eax(c1cosbx+c2sinbx) is a solution to the given differential equation for allx∈R.22.y(x) =erx=⇒y′=rerx=⇒y′′=r2erx. Substituting these results into the given differential equationyieldserx(r2−r−6) = 0, so thatrmust satisfyr2−r−6 = 0, or (r−3)(r+ 2) = 0. Consequentlyr= 3andr=−2 are the only values ofrfor whichy(x) =erxis a solution to the given differential equation. Thecorresponding solutions arey(x) =e3xandy(x) =e−2x.23.y(x) =erx=⇒y′=rerx=⇒y′′=r2erx. Substituting these results into the given differential equationyieldserx(r2+ 6r+ 9) = 0, so thatrmust satisfyr2+ 6r+ 9 = 0, or (r+ 3)2= 0. Consequentlyr=−3 isthe only value ofrfor whichy(x) =erxis a solution to the given differential equation. The correspondingsolution arey(x) =e−3x.24.y(x) =xr=⇒y′=rxr−1=⇒y′′=r(r−1)xr−2. Substitution into the given differential equation yieldsxr[r(r−1) +r−1] = 0, so thatrmust satisfyr2−1 = 0. Consequentlyr=−1 andr= 1 are the onlyvalues ofrfor whichy(x) =xris a solution to the given differential equation. The corresponding solutionsarey(x) =x−1andy(x) =x.25.y(x) =xr=⇒y′=rxr−1=⇒y′′=r(r−1)xr−2. Substitution into the given differential equation yieldsxr[r(r−1) + 5r+ 4] = 0, so thatrmust satisfyr2+ 4r+ 4 = 0, or equivalently (r+ 2)2= 0. Consequentlyr=−2 is the only value ofrfor whichy(x) =xris a solution to the given differential equation.Thecorresponding solution isy(x) =x−2.26.y(x) =12x(5x2−3) =12(5x3−3x) =⇒y′=12(15x2−3) =⇒y′′= 15x. Substitution into the Legendreequation withN= 3 yields (1−x2)y′′−2xy′+ 12y= (1−x2)(15x) +x(15x2−3) + 6x(5x2−3) = 0.Consequently the given function is a solution to the Legendre equation withN= 3.27.y(x) =a0+a1x+a2x2=⇒y′=a1+2a2x=⇒y′′= 4a2. Substitution into the given differential equationyields (1−x2)(2a2)−x(a1+2a2x)+4(a0+a1x+a2x2) = 0 =⇒3a1x+2a2+4a0= 0. For this equation to holdfor allxwe require 3a1= 0, and 2a2+ 4a0= 0. Consequentlya1= 0, anda2=−2a0. The correspondingsolution to the differential equation isy(x) =a0(1−2x2). Imposing the normalization conditiony(1) = 1requires thata0=−1. Hence, the required solution to the differential equation isy(x) = 2x2−1.28.xsiny−ex=c=⇒xcosy dydx+ siny−ex= 0 =⇒dydx=ex−sinyxcosy.

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