Solution Manual for Elementary Differential Equations with Boundary Value Problems, 2nd Edition

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ELEMENTARYDIFFERENTIALEQUATIONS WITHBOUNDARYVALUEPROBLEMSSECONDEDITIONWerner KohlerVirginia TechLee JohnsonVirginia Tech’SSOLUTIONSMANUALJEREMYBOURDONMcMaster UniversityTERRIBOURDONVirginia Tech

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CONTENTSChapter1:Introduction to1Differential EquationsChapter2:First Order5Differential EquationsChapter3:Second and Higher Order51Linear Differential EquationsChapter4:First Order Linear Systems110Chapter5:Laplace Transforms172Chapter6:Nonlinear Systems212Chapter7:Numerical Methods248Chapter8:Series Solutions of Linear266Differential EquationsChapter9:Second Order Partial Differential298Equations and Fourier SeriesChapter 10:First Order Partial Differential Equations329and the Method of CharacteristicsChapter 11:Linear Two-Point Boundary339Value Problems

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Chapter 1Introduction to Differential EquationsSection 1.21.This D.E. is of order two because the highest derivative in the equation is′′y.2.Order is 1.3.This D.E. is of order one because the highest derivative in the equation is′y. (Note:()′≠′′′yy3)4.Order is 3.5.Differentiating gives us′ =ykekt. Substitution yieldskeektkt+=20. Therefore,k= −2.6.′′ −=⇒−=⇒= ±yyk eekktkt0012.7.Differentiating gives us′ = −yktekt222sincos. Substitution yields−+=222022ktetektktsinsincoscos. Therefore,sincos22102tekkt−+() =, and solving forkgivesusk=12.8.ykeyykekettt=′ +=−+=−−−,,.00kcan be any real number.9 (a).yCet=2. Differentiating gives us′ =⋅=yCettyt222. Therefore,′ −=yty20forany value ofC.9 (b).Substituting into the differential equation yieldsyCeCe( )112==. Using the initialcondition,yCe( )12==. Solving forC, we findCe=−21.10.′′′ =′′ =+=++=+++yytcytc tcytc tc tc22321121231223.,,.Order = 33 arbitrary constants11 (a).yCtCt=+1222sincos. Differentiating gives us′ =−yCtCt222212cossinand′′ = −−= −+= −yCtCtCtCty424242241212sincos(sincos). Therefore,′′ += −+=yyyy4440and thusy tCtCt( )sincos=+1222is a solution of theD.E.′′ +=yy40.11 (b).yCCCπ4121103( ) =+==( )( )and′( ) =−= −= −⇒=yCCCCπ412222021221( )( ).12.yeykyekeketttt=′ += −+=−=−−−−2822404444.()∴===∴==.( ).,.kyyky4024200

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2 • Chapter 1 Introduction to Differential Equations13.yct=−1. Differentiating gives us′ = −−yct2. Thus′ += −+=−=−−−yyctc tcc t2222220().Solving this forc, we find thatccc c210−=−=(). Therefore,c=0 1,.14.yett= −+−sin′ +==yyg tyy( ),( ).00′ =+−yettcos′ +=+−+=∴=+= −=−−yyetetgg tttyyttcossin( )cossin ,( )01015.ytr=. Differentiating gives us′ =−yrtr1and′′ =−−yr rtr()12.Thust ytyyr rtrttr rrtrrrr2221221220′′ −′ +=−−+=−−+=()[ ()]. Solving this forr, wefind thatr rrrrrr()()()−−+=−+=−−=122322102. Therefore,r=1 2,.16.yc ec eyc ec eyc ec eytttttt=+′ =−′′ =+=−−−12221222122222444.,∴′′ −=.yy4017.From (16),yC eC ett=+−1222, which we differentiate to get′ =−−yC eC ett221222. Using theinitial conditions,y( )02=and′=y( )00, we have two equations containingC1andC2:CC122+=and22012CC−=. Solving these simultaneous equations gives usCC121==. Thus,the solution to the initial value problem isyeettt=+=−2222cosh().18.yccccccy tet( ),,( )01 222101212122=+=−=∴===.19.From (16),y tC eC ett( )=+−1222. Using the initial conditiony( )03=, we find thatCC123+=.From the initial conditionlim( )ty t→∞=0and the equation fory t( )given to us in (16), we canconclude thatC10=(ifC10≠, thenlimt→∞= ±∞). Therefore,C23=andy tet( )=−32.20.ccy tccy tett1221210001010+==⇒=∴==→−∞lim( )( )and.21.From the graph, we can see that′ = −y1and thaty( )11=. Thusmy=′ −= − −= −1112andyy011==( ).22.′ =⇒=+= −=∴=ymtym tcytt..210020From the graph,only atAlsoFrom the graph.( ).cymm= −= −∴ −=−⇒=110 512211.23.We know that this is a freefall problem, so we can begin with the generic equation for freefallsituations:y tg tv ty( )= −++2200. The object is released from rest, sov00=. The impact timecorresponds to the time at whichy=0, so we are left with the following equation for theimpact timet:0220= −+g ty. Solving this fortyieldstyg=20. For the velocity at the time ofimpact:vygtvgtgy=′ = −+= −= −002.

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Chapter 1 Introduction to Differential Equations • 324.′′ =′ =+==⇒=+xaxatvvxxat,0002020 .88811811 6423522=⇒=== =aafttxft( )/ sec .,.AtSection 1.31 (a).The equation is autonomous because′ydepends only ony.1 (b).Setting′ =y0, we have01= −+y. Solving this foryyields the equilibrium solution:y=1.2 (a).not autonomous2 (b).no equilibrium solutions, isoclines aret=constant.3 (a).The equation is autonomous because′ydepends only ony.3 (b).Setting′ =y0, we have 0=siny. Solving this foryyields the equilibrium solutions:yn= ±π.4 (a).autonomous4 (b).y yy(),,−==100 1.5 (a).The equation is autonomous because′ydoes not depend explicitly ont.5 (b).There are no equilibrium solutions because there are no points at which′ =y0.6 (a).not autonomous6 (b).y=0is equilibrium solution, isoclines are hyperbolas.7 (a).c= −1: Settingc= −1gives us−+= −y11which, solved fory, readsy=2. This is the isoclineforc= −1.c=0: Settingc=0gives us−+=y10which, solved fory, readsy=1. This is the isoclineforc=0.c=1: Settingc=1 gives us−+=y11 which, solved fory, readsy=0, the isocline forc=1.8 (a).−+= −⇒=+ytyt11−+=⇒=ytyt0−+=⇒=−ytyt119 (a).c= −1: Settingc= −1gives usyt221−= −which can be simplified toty221−=(ahyperbola). This is the isocline forc= −1.c=0: Settingc=0gives usyt220−=which can be simplified toyt= ±. This is the isoclineforc=0.c=1: Settingc=1 gives usyt221−=(a hyperbola). This is the isocline forc=1.

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4 • Chapter 1 Introduction to Differential Equations10.ffyyy( )( )()0202==′ =−′ ><<′ <− ∞ <<<< ∞yyyyy002002,forforand.11.One example that would fit these criteria is′ = −−yy()12. For this autonomous D.E.,′ =y0aty=1and′ <y0for−∞ <<y1and1<< ∞y.12.′ =y1.13.One example that would fit these criteria is′ =yysin()2π. For this autonomous D.E.,′ =y0atyn=2 .14.c.15.f.16.a.17.b.18.d.19.e.

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Chapter 2First Order Differential EquationsSection 2.11.This equation is linear because it can be written in the form′ +=yp t yg t( )( ). It isnonhomogeneous because when it is put in this form,g t( )≠0.2.nonlinear3.This equation is nonlinear because it cannot be written in the form′ +=yp t yg t( )( ).4.nonlinear5.This equation is nonlinear because it cannot be written in the form′ +=yp t yg t( )( ).6.linear, homogeneous7.This equation is nonlinear because it cannot be written in the form′ +=yp t yg t( )( ).8.nonlinear9.This equation is linear because it can be written in the form′ +=yp t yg t( )( ). It isnonhomogeneous because when it is put in this form,g t( )≠0.10.linear, homogeneous11 (a). Theorem 2.1 guarantees a unique solution for the interval(,)−∞ ∞, sincett21+andsin( )tareboth continuous for alltand−2is on this interval.11 (b). Theorem 2.1 guarantees a unique solution for the interval(,)−∞ ∞, sincett21+andsin( )tareboth continuous for alltand0is on this interval.11 (c). Theorem 2.1 guarantees a unique solution for the interval(,)−∞ ∞, sincett21+andsin( )tareboth continuous for alltandπis on this interval.12 (a).2<< ∞t12 (b).−<<22t12 (c).−<<22t12 (d).−∞ << −t2

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6 • Chapter 2 First Order Differential Equations13 (a). For this equation,p t( )is continuous for allt≠−22,andg t( )is continuous for allt≠3.Therefore, Theorem 2.1 guarantees a unique solution for( ,)3∞, the largest interval thatincludest=5.13 (b). For this equation,p t( )is continuous for allt≠−22,andg t( )is continuous for allt≠3.Therefore, Theorem 2.1 guarantees a unique solution for(, )−2 2, the largest interval thatincludest= −32.13 (c). For this equation,p t( )is continuous for allt≠−22,andg t( )is continuous for allt≠3.Therefore, Theorem 2.1 guarantees a unique solution for(, )−2 2, the largest interval thatincludest=0.13 (d). For this equation,p t( )is continuous for allt≠−22,andg t( )is continuous for allt≠3.Therefore, Theorem 2.1 guarantees a unique solution for(,)−∞ −2, the largest interval thatincludest= −5.13 (e). For this equation,p t( )is continuous for allt≠−22,andg t( )is continuous for allt≠3.Therefore, Theorem 2.1 guarantees a unique solution for(, )−2 2, the largest interval thatincludest=32.14.lnlntttttt+−=−−+11222undefined att=0 2,.14 (a).2<< ∞t.14 (b). 02<<t.14 (c).−∞ <<t0.14 (d).−∞ <<t0.15.y tet( )=32. Differentiating gives us′ ==yettyt3222(). Substituting these values into the givenequation yields20typ t y+=( ). Solving this forp t( ), we find thatp tt( )= −2. Puttingt=0 intothe equation forygives usy03=.16(a).yCtr=′ =−yCrtr1260tyy′ −=∴−=−=⇒−=⇒−=⇒=()()262602602603CrtCtrCtryrrrrryCCCCr()()()−=−=⇒≠∴−=⇒= −2280281316 (b).−∞ <<t0sincep tt( )= −316 (c).y ttt( ),= −− ∞ << ∞3.

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Chapter 2 First Order Differential Equations • 717.y t( )=0satisfies all of these conditions.Section 2.21 (a).First, we will integratep t( )=3to findP tt( )=3. The general solution, then,isy tCeCeP tt( )( )==−−3.1 (b).yC( )03== −. Therefore, the solution to the initial value problem isyet= −−33.2 (a).′ −=yy120, (),eyyCett−′ ==220.2 (b).yCeCe(),−===−1221212y tet( )()=+21 23 (a).We can rewrite this equation into the conventional form:′ −=yty20. Then we will integratep tt( )= −2to findP tt( )= −2. The general solution, then, isy tCeCeP tt( )( )==−2.3 (b).yCe( )13==. Solving forCyieldsCe=−31. Therefore, the solution to the initial valueproblem isy teeett( )()==−−331122.4 (a).tyyyt y′ −=⇒′ −=4040.−= −= −∴=∫44144t dttttlnln()μ140454tytyty′ −=′ =−()yCt=4.4 (b).yCy tt( )( )114==∴=.5 (a).For this D.E.,p t( )= −3. Integrating gives usP tt( )= −3. An integrating factor is,then,μ( )tet=−3. Multiplying the D.E. byμ( )t, we obtaineyeyeyetttt−−−−′ −=′ =333336().Integrating both sides yieldseyeCtt−−= −+332. Solving forygives usyCet= −+23.5 (b).yC( )012== −+. Solving forCyieldsC=3, and thus our final solution isyet= −+233.6 (a).′ −==yyeyt2033,( ).()eyeeyeCyeCetttttt−−′ =⇒=+⇒=+22326 (b).yCCyeett( ),0132232=+=⇒==+.7 (a).Putting this D.E. in the conventional form, we have′ +=yyet3212. For this D.E.,p t( )=32 .Integrating gives usP tt( )=32 . An integrating factor is, then,μ( )tet=32. Multiplying the D.E.byμ( )t, we obtaineyeyeyetttt323232523212′ +=′ =(). Integrating both sides yieldseyeCtt325215=+.Solving forygives usyeCett=+−1532.

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8 • Chapter 2 First Order Differential Equations7 (b).yC( )0015==+. Solving forCyieldsC= −15, and thus our final solution isyeett=−−151532.8 (a).′ +=+=∴′ =+−yyetye yetttt1220222cos(),()()cosπe yetCyetCetttt=++⇒=++−−sinsin212.8 (b).yCeCeyetett();sin()ππππ21012222=+=⇒= −=+−−−−−.9 (a).Putting this D.E. in the conventional form, we have′ += −yt ytcos( )cos( )232. For thisD.E.,p tt( )cos( )=2. Integrating gives usP tt( )sin( )=2. An integrating factor is, then,μ( )sin( )tet=2. Multiplying the D.E. byμ( )t, we obtaineyt eyeyt ettttsin( )sin( )sin( )sin( )cos( )()cos( )2222232′ +=′ = −. Integrating both sides yieldseyeCttsin( )sin( )223= −+. Solving forygives usyCet= −+−32sin( ).9 (b).yC( )043= −= −+. Solving forCyieldsC= −1, and thus our final solution isyet= −−−32sin( ).10 (a).′ +=++−=′ =++−yyetyeeyeteettttt211222,(), ()yeeteeeCyetCettttttt22222121412214=+−++⇒=+++−−.10 (b).yeCeeCe()−=−++=⇒=−112141422∴=+++−−+()yetett2141421.11.We can rewrite this equation into the conventional form:′ +=yt y40. Then we will integratep tt( )=4to findP ttt( )lnln==44. The general solution, then, isy tCeCeCeCtP ttt( )( )lnln====−−−−444.12.μ =−∴=−−exp(cos )( )(cos )tty tCett.13.First, we will integratep tt( )cos()= −22to findP tt( )sin()= −2. The general solution, then,isy tCeCeP tt( )( )sin()==−2.14.(() )ty210+′ =yCt=+21.

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Chapter 2 First Order Differential Equations • 915.We can rewrite this equation into the conventional form:′ −+=yty3102(). Then we willintegratep tt( )()= −+312to findP ttt( )= −−33 . The general solution, then,isy tCeCeP ttt( )( )==−+33.16.′ +=∴= −−−−∫yeyedtettt0()−′ =−eyet0yCeet=−.17.For this D.E.,p t( )=2. Integrating gives usP tt( )=2. An integrating factor is, then,μ( )tet=2.Multiplying the D.E. byμ( )t, we obtaineyeyeyetttt22222′ +=′ =(). Integrating both sidesyieldseyeCtt2212=+. Therefore, the general solution isy tCet( )=+−122.18.′ +=⇒′ =⇒=+⇒=+−−−yyeeyeeyeCyeCettttttt2222().19.For this D.E.,p t( )=2. Integrating gives usP tt( )=2. An integrating factor is, then,μ( )tet=2.Multiplying the D.E. byμ( )t, we obtaineyeyeyttt22221′ +=′ =(). Integrating both sidesyieldseytCt2=+. Therefore, the general solution isy tteCett( )=+−−22.20.′ +=⇒′ =⇒=+⇒=+−ytyteyteeyeCyCettttt2121222222().21.Putting this equation into the conventional form gives us′ +=yt yt2. For this D.E.,p tt( )=2 .Integrating gives usP tt( )ln=2. An integrating factor is, then,μ( )lntett==22. Multiplying theD.E. byμ( )t, we obtaint ytyt yt2232′ +=′ =(). Integrating both sides yieldst ytC2414=+.Therefore, the general solution isy ttCt( )=+−1422.22.()(),ln()tytyttyttytett22222424242442+′ +=+⇒′ ++===++μ∴+′ =+=+⇒+=++(() )()()tytttttyttC222422534444543yCttt=+++5354324().23.For this D.E.,p t( )=1. Integrating gives usP tt( )=. An integrating factor is, then,μ( )tet=.Multiplying the D.E. byμ( )t, we obtaine ye ye ytetttt′ +=′ =(). Integrating both sidesyieldse yteeCttt=−+. Therefore, the general solution isy ttCet( )=−+−1.24.′ +=⇒′ =yyteyettt23322cos()cosuet=2dvtdt=cos 3duedtt=22vt=133sinetdtetetdtttt222333233cossinsin=−∫∫

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10 • Chapter 2 First Order Differential Equationsuet=2dvtdt=sin 3duedtt=22vt= −133cosetdtetetdtttt222333233sincoscos= −+∫∫∴=−−+ ⇒+=+sincos()(sincos)IetetIIettttt222332333231493323∴=+(sincos)Iettt3133232∴=++⇒=++−(sincos)(sincos)eyettCyttCettt22231332331332325 (a). #225 (b). #325 (c). #126.y ty et( )=−0α41003==−−y ey eαα,Divide:412422=⇒==eααlnlnandyeeey tet034823288====∴=−αlnln()(ln).( ).27.First, we should put the equation into our conventional form:′ −=yt yα0. Integratingp tt( )= −αgives usP ttt( )lnln= −=−αα. The general solution, then,isy tCeCeCeCtP ttt( )( )lnln====−−−ααα. Using the general solution and the point( , )2 1, we cansolve forCin terms ofα:yC( )212==⋅α;C=−2α. We can then substitute this value forCinto the general solution at the point( ,)4 4:y( )//442444422==⋅=⋅=−−ααααα. Setting theexponents equal to each other yields122==α α;. Finally, solving fory0,yy02212114==⋅=−( ).28 (a). The general solution isycet=−2, so the corresponding graph is graph 2.y02( ) =.28 (b). The general solution isycet=−()sin 22, so the corresponding graph is graph 4.y03( ) =.28 (c). The general solution isycett=−0 122.sin, so the corresponding graph is graph 1.y01( ) =.28 (d). The general solution isycet=10, so the corresponding graph is graph 3.y02( ) =.29 (a).B cA cAdBdcdAdckB( ) =( ) −== −*and differentiating gives us,BA( )*0= −.

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Chapter 2 First Order Differential Equations • 1129 (b).B cA eA cAA cAekckc( )( )( )()***= −=−=−−−, and substitution gives us1. The activity does notever exceedA*becauseA cAc( )*only approaches the value ofasapproaches∞.Alternatively, the value of1−()−ekcis never higher than 1.29 (c). Substituting the condition into our given equation, we have0 951.()**AAekc=−−.Simplification gives us−= −⇒ −== −−0 0520120.ln()ln()ekckc, and solving forcyieldsln().ck0 95120=.30.′ +==yt ytt44αμ,t yt ytt yt ytCytCt43544624466′ +==′ ⇒=+⇒=+−ααα()yCCyt( ),11361360232= −=+⇒= −−≡⇒= −= −ααα.31.Multiplying both sides of the equation by the integrating factor,μ( )tet=2, wehaveeyeCetetCtttt222211=++=++−()(). Differentiating givesus()( )()()eyeetettttt222212123′ =++=+. Therefore,()(( ) )( )( )()( )eyt ytg tetg tttt222323′ =′ =⋅=+⇒=+μμandμ( )( )( )( )teeP ttp ttP t==⇒=⇒=222.32.20222tCepCep tttt+=⇒= −( ). Substituting,()()( )Cet Cetg tttt2222244+′ −+= −⇒= −.33.Multiplying both sides of the equation by the integrating factor,μ( )tt=, wehavetyt CttC=+=+−()11. Differentiating gives us()ty′ =1. Therefore,()(( ) )( )( )( )()( )tyt ytg tttg tt′ =′ =⋅==⇒=−−μμ111andμ( )( )ln( )( )tteP ttp tttP t==⇒=⇒==−11.34.()()( ),etettg ttytt−−+−′ ++−=⇒==1100.35.y teetyytt( )sin( )= −++⇒== −++= −−2021010.Ify teettt( )sin= −++−2, then′ =++−yeettt2cos.Substituting in′ +=yyg t( ),(cos )(sin )cossin( )222eeteetettg tttttt−−+++ −++=++=.36.′ ++=+==+yt ytyett(cos )cos ,( ),sin1103μ.()(cos )()sinsinsinsinsin(sin)eyt eeeyeCyCetttttttttttt+++++−+′ =+=′ ⇒=+⇒=+11.yCCyey tttt( )lim( )(sin)0132121=+=⇒=∴=+=−+→∞and.

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12 • Chapter 2 First Order Differential Equations37.Putting this D.E. in the conventional form, we have′ +=−−yyet22. For this D.E.,p t( )=2. Anintegrating factor is, then,μ( )tet=2. Multiplying the D.E. byμ( )t, weobtaineyeyeyeettttt222222′ +=′ =−(). Integrating both sides yieldseyeeCttt22=−+.Solving forygives usyeCett=−+−−12, and with our initial condition,yC( )0211= −=−+.Solving forCyieldsC= −2 , and thus our final solution isyeett=−−−−122.Therefore,lim( )ty t→∞= −1.38.yceteyctt=+⇒=−−0.′( ) = −+−−−−yty eetettt0.′( ) =⇒ −+−()=−yye1011001∴=y00.39.The general solution of the D.E. isyCeytCt=+≠=+=−λλλλ10,,0;. Therefore, therelevant limits are:limty→∞=<does not exist forandλλ00;limtyC→∞=⋅+>010λλλ= 1 for.40.On[ , ]1 2:′ +==yt yty1311,( ). An integrating factor isμ( )tt=. Multiplying the D.E. byμ( )t, weobtain(),( )tyttytCytCtyCC′ =⇒=+⇒=+=+=⇒=−311102321. Therefore, thesolution for12≤≤tisyt=2andy( )24=.On[ , ]2 3 :′ +==yt yy1024,( ). An integrating factor isμ( )tt=. Multiplying the D.E. byμ( )t, weobtain(),( )tytyCyCtyCC′ =⇒=⇒===⇒=−022481. Therefore, the solution for23≤≤tisyt=8.41.On[ ,]0π:′ +==yt yty(sin )sin ,( )03. An integrating factor isμ( )costet=−. Multiplying the D.E. byμ( )t,we obtaineyet yeyt etttt−−−−′ +=′ =coscoscoscos(sin )()(sin ). Integrating both sidesyieldseyeCtt−−=+coscos. Solving forygives usyCet=+1cos, and with our initialcondition,yCeCe( )03121==+⇒=−. Therefore, the solution for0≤≤tπisyet=+−121cosandye()π=+−122.On[ ,]ππ2:′ += −=+−yt ytye(sin )sin ,()π122. Multiplying the D.E. byμ( )costet=−, we obtaineyet yeyt etttt−−−−′ +=′ = −coscoscoscos(sin )()(sin ).

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Solution Manual for Elementary Differential Equations with Boundary Value Problems, 2nd Edition - Page 16 preview image

Chapter 2 First Order Differential Equations • 13Integrating both sides yieldseyeCtt−−= −+coscos. Solving forygives usyCet= −+1cos, andwith our initial condition,yeCeCee()π=+= −+⇒=+−−−121222111. Therefore, the solutionforππ≤≤t2isyeett= −+++−12211coscos.42.On[ , ]0 1:′ ==yy201,( ).ytCyCC=+==⇒=2011,( ).Therefore, the solution for01≤≤tisyt=+21andy( )13=.On[ , ]1 2:′ +==yt yy1213,( ). An integrating factor isμ( )tt=. Multiplying the D.E. byμ( )t,we obtain(),( )tyttytCytCtyCC′ =⇒=+⇒=+=+=⇒=−2113221. Therefore, thesolution for12≤≤tisytt=+2.43.On[ , ]0 1:′ +−==ytyy(),( )21003. An integrating factor isμ( )tett=−2. Multiplying the D.E. byμ( )t,we obtaineyetyeytttttt222210−−−′ +−=′ =()(). Integrating both sides yieldseyCtt2−=. Solvingforygives usyCett=−2, and with our initial condition,yC( )03==. Therefore, the solutionfor01≤≤tisyett=−32andy( )13=.On[ , ]1 3:′ +=′ ==yyyy( ),( )0013. Integrating gives usyC==3. Therefore, the solution for13≤≤tisy=3andy( )33=.On[ ,]3 4:′ + −()==yyyt1033,( ). An integrating factor isμ( )lntett==−1. Multiplying the D.E. byμ( )t,we obtain11120tttyyy′ −=′ =(). Integrating both sides yields1tyC=. Solving forygivesusyCt=, and with our initial condition,yCC( )( )3331==⇒=. Therefore, the solution for34≤≤tisyt=.44.y tt Si tSi( )( )( )=−+{}13Section 2.31 (a).To begin,Q( )00=and′ =−QQ( . )( )( )0 23100 3 . Putting the second equation in the conventionalform, we have′ +=QQ0 030 6... Multiplying both sides of this equation by the integratingfactorμ( ).tet=0 03gives us()...eQett0 030 030 6′ =. Integrating both sides

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