Solution Manual For Differential Equations And Linear Algebra, 3rd Edition

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1Solutions to Section 1.1True-False Review:1.FALSE.A derivative must involvesomederivative of the functiony=f(x), not necessarily the firstderivative.2.TRUE.The initial conditions accompanying a differential equation consist of the values ofy, y′, . . .att= 0.3. TRUE.If we define positive velocity to be oriented downward, thendvdt=g,wheregis the acceleration due to gravity.4.TRUE.We can justify this mathematically by starting froma(t) =g, and integrating twice to getv(t) =gt+c, and thens(t) = 12gt2+ct+d, which is a quadratic equation.5. FALSE.The restoring force is directed in the directionoppositeto the displacement from the equilibriumposition.6.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.7.FALSE.The temperature of the object is given byT(t) =Tm+ce−kt, whereTmis the temperatureof the medium, andcandkare constants.Sincee−kt6= 0, we see thatT(t)6=Tmfor all timest.Thetemperature of the objectapproachesthe temperature of the surrounding medium, but never equals it.8. TRUE.Since the temperature of the coffee is falling, the temperaturedifferencebetween the coffee andthe room is higher initially, during the first hour, than it is later, when the temperature of the coffee hasalready decreased.9. FALSE.The slopes of the two curves arenegativereciprocals of each other.10. TRUE.If the original family of parallel lines have slopeskfork6= 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.11. FALSE.The family of orthogonal trajectories for a family of circles centered at the origin is the familyof lines passing through the origin.Problems:1.Starting from the differential equationd2ydt2=g, wheregis the acceleration of gravity andyis the unknownposition function, we integrate twice to obtain the general equations for the velocity and the position of theobject:dydt=gt+c1andy(t) =gt22+c1t+c2,wherec1, c2are constants of integration. Now we impose the initial conditions:y(0) = 0 implies thatc2= 0,

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2anddydt(0) = 0 implies thatc1= 0.Hence, the solution to the initial-value problem isy(t) =gt22.The object hits the ground at the timet0for whichy(t0) = 100. Hence 100 =gt202, so thatt0=√200g≈4.52s, where we have takeng= 9.8 ms−2.2.Starting from the differential equationd2ydt2=g, wheregis the acceleration of gravity andyis the unknownposition function, we integrate twice to obtain the general equations for the velocity and the position of theball, 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 positivedirection downward), we know thaty(0) = 0. Since the object 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= 0and 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.3.Starting from the differential equationd2ydt2=g, wheregis the acceleration of gravity andyis the unknownposition function, we integrate twice to obtain the general equations for the velocity and the position of therocket, 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.

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3(b)The time that the rocket reaches its maximum height ist=−cg≈ − −42.029.81= 4.28 s.4.Starting from the differential equationd2ydt2=g, wheregis the acceleration of gravity andyis the unknownposition function, we integrate twice to obtain the general equations for the velocity and the position of therocket, respectively:dydt=gt+candy(t) = 12gt2+ct+d,wherec, dare constants of integration.Settingy= 0 to be at the level of the platform (with positivedirection 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 .Solving 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.5.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,wheregis the acceleration of gravity andyis the unknown position function. We integrate this differentialequation twice to obtain the general equations for the velocity and the position of the object:dydt=gt+c1andy(t) =gt22+c1t+c2.Now we impose the initial conditions: sincey(0) = 0, we havec2= 0. Moreover, sincedydt(0) =−2, we havec1=−2. Hence the solution to the initial-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−2.6.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.

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4We integrate the differential equation twice to obtain the velocity and position functions, respectively:dydt=gt+c1andy(t) =gt22+c1t+c2.Now we impose the initial conditions. Sincey(0) = 0, we havec2= 0. Moreover, sincedydt(0) =v0, we havec1=v0. Hence the solution to the initial-value problem isy(t) =gt22+v0t. We are given thaty(t0) =h.Consequently,h=gt20+v0t0. Solving forv0yieldsv0= 2h−gt202t0.7.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). SinceA6= 0 andω6= 0 and|φ|< π, we haveφ= 0. It follows thata=A.8.Taking derivatives ofy(t) =c1cos (ωt) +c2sin (ωt), we obtaindydt=−c1ωsin (ωt) +c2ωcos (ωt)andd2ydt2=−c1ω2cos (ωt)−c2ω2sin (ωt) =−ω2[c1cos (ωt) +c2cos (ωt)] =−ω2y.Consequently,d2ydt2+ω2y= 0.To determine the amplitude of the motion we write the solution to thedifferential 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.

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59.We compute the first three derivatives ofy(t) = lnt:dydt= 1t ,d2ydt2=−1t2,d3ydt3= 2t3.Therefore,2(dydt)3= 2t3=d3ydt3,as required.10.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.11.We compute the first two derivatives ofy(x) =exsinx:dydx=ex(sinx+ cosx)andd2ydx2= 2excosx.Then2ycotx−d2ydx2= 2(exsinx) cotx−2excosx= 0,as required.12.Starting from (T−Tm)−1dTdt=−k, we obtainddt(ln|T−Tm|) =−k.The preceding equation canbe integrated directly to yield ln|T−Tm|=−kt+c1.Exponentiating both sides of this equation gives|T−Tm|=e−kt+c1, which can be written asT−Tm=ce−kt,wherec=±ec1. Rearranging this, we conclude thatT(t) =Tm+ce−kt.13.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.14.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-wayfrom 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.

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615.Applying implicit differentiation to the given family of curvesx2+ 4y2=cwith respect toxgives2x+ 8y dydx= 0.Therefore,dydx=−x4y .Therefore, the collection of orthogonal trajectories satisfies:dydx= 4yx=⇒1ydydx= 4x=⇒ddx(ln|y|) = 4x=⇒ln|y|= 4 ln|x|+c1=⇒y=kx4,wherek=±ec1.1.5-0.5-0.41.0xy(x)-0.80.40.50.8-1.0-1.5Figure 1: Figure for Problem 1516.Differentiation of the given family of curvesy=cxwith respect toxgivesdydx=−cx2=−1x·cx=−yx .Therefore, the collection of orthogonal trajectories satisfies:dydx=xy=⇒y dydx=x=⇒ddx(12y2)=x=⇒12y2= 12x2+c1=⇒y2−x2=c2,wherec2= 2c1.-22-4y(x)24-4x-24Figure 2: Figure for Problem 16

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717.Solving the equationy=cx2forcgivesc=yx2. Hence, differentiation leads todydx= 2cx= 2(yx2)x= 2yx .Therefore, the collection of orthogonal trajectories satisfies:dydx=−x2y=⇒2y dydx=−x=⇒ddx(y2) =−x=⇒y2=−12x2+c1=⇒2y2+x2=c2,wherec2= 2c1.x1y(x)1.622.01.20.40.8-1.2-2-0.4-2.0-1.6-1-0.8Figure 3: Figure for Problem 1718.Solving the equationy=cx4forcgivesc=yx4. Hence,dydx= 4cx3= 4(yx4)x3= 4yx .Therefore, the collection of orthogonal trajectories satisfies:dydx=−x4y=⇒4y dydx=−x=⇒ddx(2y2) =−x=⇒2y2=−12x2+c1=⇒4y2+x2=c2,wherec2= 2c1.1.5-0.41.0y(x)x-0.80.40.8-1.0-1.5Figure 4: Figure for Problem 1819.Implicit differentiation of the given family of curvesy2= 2x+cwith respect toxgives 2ydydx= 2. Thatis,dydx= 1y. Therefore, the collection of orthogonal trajectories satisfies:dydx=−y=⇒y−1dydx=−1=⇒ddx(ln|y|) =−1=⇒ln|y|=−x+c1=⇒y=ke−x,

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8113-3-22-4-12-143y(x)x4Figure 5: Figure for Problem 19wherek=±ec1.20.Differentiating the given family of curvesy=cexwith respect toxgivesdydx=cex=y. Therefore, thecollection of orthogonal trajectories satisfies:dydx=−1y=⇒y dydx=−1=⇒ddx(12y2)=−1=⇒12y2=−x+c1=⇒y2=−2x+c2,wherec2= 2c1.21y(x)1x-1-2-1Figure 6: Figure for Problem 2021.Differentiating the given family of curvesy=mx+cwith respect toxgivesdydx=m. Therefore, thecollection of orthogonal trajectories satisfies:dydx=−1m=⇒y=−1m x+c1.22.Differentiating the given family of curvesy=cxmwith respect toxgivesdydx=cmxm−1. Sincec=yxm,we havedydx=myx. Therefore, the collection of orthogonal trajectories satisfies:dydx=−xmy=⇒y dydx=−xm=⇒ddx(12y2)=−xm=⇒12y2=−12m x2+c1=⇒y2=−1m x2+c2,

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9wherec2= 2c1.23.Implicit differentiation of the given family of curvesy2+mx2=cwith respect toxgives 2y dydx+2mx= 0.That is,dydx=−mxy. Therefore, the collection of orthogonal trajectories satisfies:dydx=ymx=⇒y−1dydx=1mx=⇒ddx(ln|y|) =1mx=⇒mln|y|= ln|x|+c1=⇒ym=c2x,wherec2=±ec1.24.Implicit differentiation of the given family of curvesy2=mx+cwith respect toxgives 2y dydx=m.That is,dydx=m2y. Therefore, the collection of orthogonal trajectories satisfies:dydx=−2ym=⇒y−1dydx=−2m=⇒ddx(ln|y|) =−2m=⇒ln|y|=−2m x+c1=⇒y=c2e−2xm,wherec2=±ec1.25.Consider the coordinate curveu=x2+ 2y2(i.e.uis constant). Differentiating implicitly with respect toxgives 0 = 2x+ 4y dydx. Therefore,dydx=−x2y. Therefore, the collection of orthogonal trajectories satisfies:dydx= 2yx=⇒y−1dydx= 2x=⇒ddx(ln|y|) = 2x=⇒ln|y|= 2 ln|x|+c1=⇒y=c2x2,wherec2=±ec1.x1y(x)1.622.01.20.40.8-1.2-2-0.4-2.0-1.6-1-0.8Figure 7: Figure for Problem 2526.We havem1= tan (a1) = tan (a2−a) =tan (a2)−tan (a)1 + tan (a2) tan (a) =m2−tan (a)1 +m2tan (a) .

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10Solutions to Section 1.2True-False Review:1.FALSE.The order of a differential equation is the order of thehighestderivative appearing in thedifferential equation.2. TRUE.This is condition 1 in Definition 1.2.11.3. TRUE.This is the content of Theorem 1.2.15.4.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 Definition1.2.11 for the general solution.5.FALSE.There are solutions toy′′+y= 0 that do not have the formc1cosx+ 5c1sinx, such asy(x) = cosx+ sinx.Therefore,c1cosx+ 5c1sinxdoes not meet the second requirement set form inDefinition 1.2.11 for the general solution.6.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:1.2, nonlinear.2.3, linear.3.2, nonlinear.4.2, nonlinear.5.4, linear.6.3, nonlinear.7.We can quickly compute the first two derivatives ofy(x):y′(x) = (c1+2c2)excos 2x+(−2c1+c2)exsin 2xandy′′(x) = (−3c1+4c2)excos 2x+(−4c1−3c2)exsinx.Then we havey′′−2y′+ 5y= [(−3c1+ 4c2)excos 2x+ (−4c1−3c2)exsinx]−2 [(c1+ 2c2)excos 2x+ (−2c1+c2)exsin 2x]+5(c1excos 2x+c2exsin 2x),which cancels to 0, as required. This solution is valid for allx∈R.8.We can quickly compute the first two derivatives ofy(x):y′(x) =c1ex−2c2e−2xandy′′(x) =c1ex+ 4c2e−2x.Then we havey′′+y′−2y= (c1ex+ 4c2e−2x) + (c1ex−2c2e−2x)−2(c1ex+c2e−2x) = 0.Thusy(x) =c1ex+c2e−2xis a solution of the given differential equation for allx∈R.

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119.The derivative ofy(x) =1x+ 4 isy′(x) =−1(x+ 4)2=−y2. Thusy(x) =1x+ 4 is a solution of the givendifferential equation forx∈(−∞,−4) orx∈(−4,∞).10.The derivative ofy(x) =c1√xisy′(x) =c12√x=y2x.Thusy(x) =c1√xis a solution of the givendifferential equation for allx >0.11.We can quickly compute the first two derivatives ofy(x) =c1e−xsin (2x):y′(x) = 2c1e−xcos (2x)−c1e−xsin (2x)andy′′(x) =−3c1e−xsin (2x)−4c1e−xcos (2x).Therefore, we havey′′+ 2y′+ 5y=−3c1e−xsin (2x)−4c1e−xcos (2x) + 2[2c1e−xcos (2x)−c1e−xsin (2x)] + 5[c1e−xsin (2x)] = 0,which shows thaty(x) =c1e−xsin (2x) is a solution to the given differential equation for allx∈R.12.We can quickly compute the first two derivatives ofy(x) =c1cosh (3x) +c2sinh (3x):y′(x) = 3c1sinh (3x) + 3c2cosh (3x)andy′′(x) = 9c1cosh (3x) + 9c2sinh (3x).Therefore, we havey′′−9y= [9c1cosh (3x) + 9c2sinh (3x)]−9[c1cosh (3x) +c2sinh (3x)] = 0,which shows thaty(x) =c1cosh (3x) +c2sinh (3x) is a solution to the given differential equation for allx∈R.13.We can quickly compute the first two derivatives ofy(x) =c1x3+c2x:y′(x) =−3c1x4−c2x2andy′′(x) = 12c1x5+ 2c2x3.Therefore, we havex2y′′+ 5xy′+ 3y=x2(12c1x5+ 2c2x3)+ 5x(−3c1x4−c2x2)+ 3(c1x3+c2x)= 0,which shows thaty(x) =c1x3+c2xis a solution to the given differential equation for allx∈(−∞,0) orx∈(0,∞).14.We can quickly compute the first two derivatives ofy(x) =c1√x+ 3x2:y′(x) =c12√x+ 6xandy′′(x) =−c14√x3+ 6.Therefore, we have2x2y′′−xy′+y= 2x2(−c14√x3+ 6)−x(c12√x+ 6x)+ (c1√x+ 3x2) = 9x2,which shows thaty(x) =c1√x+ 3x2is a solution to the given differential equation for allx >0.

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1215.We can quickly compute the first two derivatives ofy(x) =c1x2+c2x3−x2sinx:y′(x) = 2c1x+ 3c2x2−x2cosx−2xsinxandy′′(x) = 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.16.We can quickly compute the first two derivatives ofy(x) =c1eax+c2ebx:y′(x) =ac1eax+bc2ebxandy′′(x) =a2c1eax+b2c2ebx.Substituting these results into 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.17.We can quickly compute the first two derivatives ofy(x) =eax(c1+c2x):y′(x) =eax(c2) +aeax(c1+c2x) =eax(c2+ac1+ac2x)andy′′(x) =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 equation for allx∈R.18.We can quickly compute the first two derivatives ofy(x) =eax(c1cosbx+c2sinbx):y′(x) =eax(−bc1sinbx+bc2cosbx) +aeax(c1cosbx+c2sinbx)=eax[(bc2+ac1) cosbx+ (ac2−bc1) sinbx],y′′(x) =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].

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13Substituting 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.19.Fromy(x) =erx, we obtainy′(x) =rerxandy′′(x) =r2erx. Substituting these results into the givendifferential equation yieldserx(r2+ 2r−3) = 0, so thatrmust satisfyr2+ 2r−3 = 0, or (r+ 3)(r−1) = 0.Consequentlyr=−3 andr= 1 are the only values ofrfor whichy(x) =erxis a solution to the givendifferential equation. The corresponding solutions arey(x) =e−3xandy(x) =ex.20.Fromy(x) =erx, we obtainy′(x) =rerxandy′′(x) =r2erx. Substituting these results into the givendifferential equation yieldserx(r2−8r+ 16) = 0, so thatrmust satisfyr2−8r+ 16 = 0, or (r−4)2= 0.Consequently the only value ofrfor whichy(x) =erxis a solution to the differential equation isr= 4. Thecorresponding solution isy(x) =e4x.21.Fromy(x) =xr, we obtainy′(x) =rxr−1andy′′(x) =r(r−1)xr−2. Substituting these results into thegiven differential equation yieldsxr[r(r−1) +r−1] = 0, so thatrmust satisfyr2−1 = 0. Consequentlyr=−1 andr= 1 are the only values ofrfor whichy(x) =xris a solution to the given differential equation.The corresponding solutions arey(x) =x−1andy(x) =x.22.Fromy(x) =xr, we obtainy′(x) =rxr−1andy′′(x) =r(r−1)xr−2.Substituting these results intothe given differential equation yieldsxr[r(r−1) + 5r+ 4] = 0, so thatrmust satisfyr2+ 4r+ 4 = 0, orequivalently (r+ 2)2= 0. Consequentlyr=−2 is the only value ofrfor whichy(x) =xris a solution tothe given differential equation. The corresponding solution isy(x) =x−2.23.Fromy(x) =12x(5x2−3) =12(5x3−3x), we obtainy′(x) =12(15x2−3) andy′′(x) = 15x. Substitutingthese results into the Legendre equation 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= 3.24.Fromy(x) =a0+a1x+a2x2, we obtainy′(x) =a1+ 2a2xandy′′(x) = 4a2. Substituting these resultsinto the given differential equation yields (1−x2)(2a2)−x(a1+ 2a2x) + 4(a0+a1x+a2x2) = 0. That is,3a1x+2a2+4a0= 0. For this equation to hold for allxwe require 3a1= 0, and 2a2+4a0= 0. Consequently,a1= 0 anda2=−2a0.The corresponding solution to the differential equation isy(x) =a0(1−2x2).Imposing the normalization conditiony(1) = 1 requires thata0=−1. Hence, the required solution to thedifferential equation isy(x) = 2x2−1.25.Differentiatingxsiny−ex=cimplicitly with respect toxyieldsxcosy dydx+ siny−ex= 0.Thus,dydx=ex−sinyxcosy.

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1426.Differentiatingxy2+ 2y−x=cimplicitly with respect toxyields 2xy dydx+y2+ 2dydx−1 = 0. Thus,dydx=1−y22(xy+ 1) .27.Differentiatingexy−x=cimplicitly with respect toxyieldsexy[x dydx+y]−1 = 0. Therefore, we havexexydydx+yexy= 1.Hence,dydx= 1−yexyxexy.Given thaty(1) = 0, we havee0(1)−1 =c, so thatc= 0.Therefore,exy−x= 0. Rearranging this equation and taking logarithms, we conclude thaty= lnxx.28.Differentiatingey/x+xy2−x=cimplicitly with respect toxyieldsey/xx dydx−yx2+ 2xy dydx+y2−1 = 0.A short algebraic simplification yieldsdydx=x2(1−y2) +yey/xx(ey/x+ 2x2y).29.Differentiatingx2y2−sinx=cimplicitly with respect toxyields 2x2y dydx+2xy2−cosx= 0. Rearranging,we obtaindydx= cosx−2xy22x2y.Sincey(π) = 1π, we haveπ2(1π)2−sinπ=c.Therefore,c= 1.Hence,x2y2−sinx= 1 so thaty2=1 + sinxx2.Sincey(π) =1π, take the branch ofywherex >0, so thaty(x) =√1 + sinxx.30.By integratingdydx= sinxwith respect tox, we obtainy(x) =−cosx+cfor allx∈R.31.By integratingdydx=x−1/2with respect tox, we obtainy(x) = 2x1/2+cfor allx >0.32.By integratingd2ydx2=xextwice with respect tox, we obtaindydx=xex−ex+c1andy(x) =xex−2ex+c1x+c2for allx∈R.33.We consider three cases:Case 1:n=−1:In this case, two integrations with respect toxyieldsdydx= ln|x|+c1andy(x) =xln|x|+c1x+c2for allx∈(−∞,0) orx∈(0,∞).Case 2:n=−2:In this case, two integrations with respect toxyieldsdydx=−x−1+c1andy(x) =c1x+c2−ln|x|for allx∈(−∞,0) orx∈(0,∞).

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15Case 3:n6=−1,−2:In this case, two integrations with respect toxyieldsdydx=xn+1n+ 1 +c1andy(x) =xn+2(n+ 1)(n+ 2) +c1x+c2for allx∈R.34.Integratingdydx= lnxwith respect toxyieldsy(x) =xlnx−x+c. Sincey(1) = 2, we have 2 = 1(0)−1+c,so thatc= 3. Thus,y(x) =xlnx−x+ 3.35.Integratingd2ydx2= cosxtwice with respect toxyieldsdydx= sinx+c1andy(x) =−cosx+c1x+c2.Fromy′(0) = 1, we obtainc1= 1, and fromy(0) = 2, we obtainc2= 3. Thus,y(x) = 3 +x−cosx.36.Integratingd3ydx3= 6xthree times with respect tox, we obtaind2ydx2= 3x2+c1,dydx=x3+c1x+c2,y(x) = 14x4+ 12c1x2+c2x+c3.Fromy′′(0) = 4, we obtainc1= 4, fromy′(0) =−1, we obtainc2=−1, and fromy(0) = 1, we obtainc3= 1. Thus,y(x) =14x4+ 2x2−x+ 1.37.Integratingy′′(x) =xextwice with respect tox, we obtainy′(x) =xex−ex+c1andy(x) =xex−2ex+c1x+c2.Fromy′(0) = 4, we obtainc1= 5, and fromy(0) = 3, we obtainc2= 5. Thus,y(x) =xex−2ex+ 5x+ 5.38.Starting withy(x) =c1ex+c2e−x, we find thaty′(x) =c1ex−c2e−xandy′′(x) =c1ex+c2e−x. Thus,y′′−y= 0, soy(x) =c1ex+c2e−xis a solution to the differential equation on (−∞,∞). Next we establishthat every solution to the differential equation has the formc1ex+c2e−x.Suppose thaty=f(x) is anysolution to the differential equation. Then according to Theorem 1.2.15,y=f(x) is the unique solution tothe initial-value problemy′′−y= 0,y(0) =f(0),y′(0) =f′(0).However, consider the functiony(x) =f(0) +f′(0)2ex+f(0)−f′(0)2e−x.This is of the formy(x) =c1ex+c2e−x, wherec1=f(0)+f′(0)2andc2=f(0)−f′(0)2, and therefore solves thedifferential equationy′′−y= 0. Furthermore, evaluation this function atx= 0 yieldsy(0) =f(0)andy′(0) =f′(0).Consequently, this function solves the initial-value problem above.However, by assumption,y(x) =f(x)solves the same initial-value problem. Owing to the uniqueness of the solution to this initial-value problem,it follows that these two solutions are the same:f(x) =c1ex+c2e−x.

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