STUDENT SOLUTIONS MANUAL FOR ELEMENTARY DIFFERENTIAL .

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STUDENT SOLUTIONS MANUAL FORELEMENTARYDIFFERENTIAL EQUATIONSANDELEMENTARYDIFFERENTIAL EQUATIONSWITH BOUNDARY VALUEPROBLEMSWilliam F. TrenchAndrew G. Cowles Distinguished Professor EmeritusDepartment of MathematicsTrinity UniversitySan Antonio, Texas, USAwtrench@trinity.eduThis book has been judged to meet the evaluation criteria set by the Editorial Board of the American Institute of Mathematics in connection withthe Institute’s Open Textbook Initiative. It may be copied, modified, redistributed, translated, and built upon subject to the Creative CommonsAttribution-NonCommercial-ShareAlike 3.0 Unported License.

This book was published previously by Brooks/Cole Thomson LearningReproduction is permitted for any valid noncommercial educational, mathematical, or scientific purpose.However, charges for profit beyond reasonable printing costs are prohibited.

TO BEVERLYContentsChapter 1 Introduction11.2 First Order Equations1Chapter 2 First Order Equations52.12.22.32.42.52.6Linear First Order EquationsSeparable EquationsExistence and Uniqueness of Solutions of Nonlinear EquationsTransformation of Nonlinear Equations into Separable EquationsExact EquationsIntegrating Factors5811131721Chapter 3 Numerical Methods253.1 Euler’s Method3.2 The Improved Euler Method and Related Methods2529

ii Contents3.3 The Runge-Kutta Method34Chapter 4 Applications of First Order Equations394.1 Growth and Decay4.2 Cooling and Mixing4.3 Elementary Mechanics4.4 Autonomous Second Order Equations4.5 Applications to CurvesChapter 5 Linear Second Order 647579Homogeneous Linear EquationsConstant Coefficient Homogeneous EquationsNonhomgeneous Linear EquationsThe Method of Undetermined Coefficients IThe Method of Undetermined Coefficients IIReduction of OrderVariation of ParametersChapter 6 Applcations of Linear Second Order Equations856.16.26.36.485878990Spring Problems ISpring Problems IIThe RLC CircuitMotion Under a Central ForceChapter 7 Series Solutions of Linear Second Order 8Review of Power SeriesSeries Solutions Near an Ordinary Point ISeries Solutions Near an Ordinary Point IIRegular Singular Points; Euler EquationsThe Method of Frobenius IThe Method of Frobenius IIThe Method of Frobenius IIIChapter 8 Laplace Transforms1258.1 Introduction to the Laplace Transform8.2 The Inverse Laplace Transform8.3 Solution of Initial Value Problems8.4 The Unit Step Function8.5 Constant Coefficient Equations with Piecewise Continuous ForcingFunctions8.6 Convolution125127134140143152

Contents iii8.7 Constant Cofficient Equations with Impulses55Chapter 9 Linear Higher Order Equations1599.19.29.39.4159171175181Introduction to Linear Higher Order EquationsHigher Order Constant Coefficient Homogeneous EquationsUndetermined Coefficients for Higher Order EquationsVariation of Parameters for Higher Order EquationsChapter 10 Linear Systems of Differential 94201245218Introduction to Systems of Differential EquationsLinear Systems of Differential EquationsBasic Theory of Homogeneous Linear SystemsConstant Coefficient Homogeneous Systems IConstant Coefficient Homogeneous Systems IIConstant Coefficient Homogeneous Systems IIVariation of Parameters for Nonhomogeneous Linear SystemsChapter22111.1 Eigenvalue Problems for y00 C y D 011.2 Fourier Expansions I11.3 Fourier Expansions II221223229Chapter 12 Fourier Solutions of Partial Differential Equations23912.112.212.312.4239247260270The Heat EquationThe Wave EquationLaplace’s Equation in Rectangular CoordinatesLaplace’s Equation in Polar CoordinatesChapter 13 Boundary Value Problems for Second Order Ordinary Differential Equations13.1 Two-Point Boundary Value Problems13.2 Sturm-Liouville Problems273273279

CHAPTER 1Introduction1.2 BASIC CONCEPTS1.2.2. (a) If y D ce 2x , then y 0 D 2ce 2x D 2y.x2c2xc2x 2 cx2c0(b) If y DC , then y 0 D,soxyCyDCC D x2.23x3x3x3x(c) If122y D C ce x ; then y 0 D 2xce x2and 12222C ce x D 2xce x C x C 2cxe x D x:y 0 C 2xy D 2xce x C 2x2(d) IfyD1 C ce1 cex 2 2x 2 2theny0.1Dcex 2 2/. cxe.1x 2 2/cxe.1 C cex 2 2 /2x 2 2/cxe22cxe x 2.1 ce x2 2 /2Dandy21DD1 C ce1ce.1 C cex 2 2x 2 2x 2 2 21/.1 cex.1 ce 2 2/22D!24ce x 2;.1 ce x2 2/21x 2 2 2/x 2 2

2Chapter 1 Basic Conceptsso4cx C 4cxD 0:.1 ce x2 2 /2 3 3 3 xxxC c , then y 0 D x 2 sec2C c D x 2 1 C tan2D x 2 .1 C y 2 /.(e) If y D tan33 x3Ccy D .c1 C c2x/e C sin x C x 2 ; then(f) Ify 0 D .c1 C 2c2x/e x C cos x C 2x;2y 0 C x.y 2y02y 0 C y1/ DD .c1 C 3c2x/e x sin x C 2;andy 00D c1 e x .1 2 C 1/ C c2 xe x .3 4 C 1/sin x 2 cos x C sin x C 2 4x C x 2D2 cos x C x 2 4x C 2:224and y 00 D c1 e x C 3 , so .1 x/y 00 C xy 0 y D(g) If y D c1e x C c2x C , then y 0 D c1 e x C c22xxx4.1 x/ 224.1 x x 2 /c1 .1 x C x 1/ C c2 .x x/ CDx3x xx3c1 sin x C c2 cos xcc2 sin x c1 sin x C c2 cos x1 cos x0(h) If y DC 4x C 8 then y DC 4 and1 21 2xx2x 3 2 c1 sin x C c2 cos x c1 sin x c2 cos x 3 c1 sin x C c2 cos x12 000200y DC, so x y Cxy C xyD44x 1 2x 3 2x 5 2 3c1x 3 2 sin x x 1 2 cos x C x 1 2 sin x C x 1 2 cos x4 1 1 231 1 23 2xsin x C x sin xxsin x C c2x 3 2 cos x C x 1 2 sin x C x 1 2 cos x24 41 1 21 1 211 23 22x sin xxcos x C x cos xxcos x C 4x C x.4x C 8/ D 4x 3 C 8x 2 C2443x 2.R1.2.4. (a) If y 0 D xe x , then y D xe x C e x dx C c D .1 x/e x C c, and y.0/ D 1 ) 1 D 1 C c,so c D 0 and y D .1 x/e x . r 1 202(b) If y D x sin x , then y Dcos x C c; yD 1 ) 1 D 0 C c, so c D 1 and221yD1cos x 2 .21 dsin x(c) Write y 0 D tan x DD.cos x/. Integrating this yields y D ln j cos xj C c;cos xcos x dx ppy. 4/pD 3 ) 3 Dp ln .cos. 4// C c, or 3 D ln 2 C c, so c D 3 ln 2, so y D ln.j cos xj/ C3 ln 2 D 3 ln. 2j cos xj/.x53237(d) If y 00 D x 4 , then y 0 DC c1 ; y 0 .2/ D 1 )C c1 D 1 ) c1 D, so y 0 D5515x537x6376447. Therefore, y D.x 2/ C c2; y.2/ D 1 )C c2 D 1 ) c2 D, so51530153015647 37xyD.x 2/ C.15530ZRxe 2x1xe 2xe 2xxe 2xe 2x(e) (A) xe 2x dx De 2x dx D. Therefore, y 0 DC c1 ;222424155xe 2xe 2x5xe 2xy 0 .0/ D 1 )C c1 D ) c1 D , so y 0 DC ; Using (A) again, y D4442444e 2xe 2x5xe 2xe 2x5129C x C c2 DC x C c2 ; y.0/ D 7 )C c2 D 7 ) c2 D, so884444442x2xxee529yDC xC .4 R444RR(B) x cos x dx D x sin xR (f) (A) x sin x dx D x cos x00 C cos x dx D x cos x C sin x andsin x dx D x sin x C cos x. If y D x sin x, then (A) implies that y 0 D x cos x sin x C c1 ; y 0 .0/ D3 ) c D 3, so y 0 D x cos x sin x 3. Now (B) implies that y D x sin x Ccos x Ccos x 3x Cc2 Dx sin x C 2 cos x 3x C c2 ; y.0/ D 1 ) 2 C c2 D 1 ) c2 D 1, so y D x sin x C 2 cos x 3x 1.

Section 1.2 Basic Concepts3RR(g) If y 000 D x 2 e x , then y 00 D x 2 e x dx D x 2 e x 2 xe x dx D x 2 eRx 2xe x C 2e x C c1 ;y 00 .0/ D 3 ) 2 C c1RD 3 ) c1 D 1, so (A) y 00 D .x 2 2x C 2/e x C 1. Since .x 2 2x C 2/e x dx D.x 2 2x C 2/e x.2x 2/e x dx D .x 2 2x C 2/e x .2x 2/e x C 2e x D .x 2 4x C 6/e x ,0x0(A) implies that y D .x 2 4x C 6/eR 2C x C c2 ; yx .0/ D 22 ) 6 C c2xD R 2 ) c2 xD 8, so (B)02xy D .x 4x C 6/e C x 8; Since .x 4x C 6/e dx D .x 4x C 6/e.2x 4/e dx D .x 2x28x Cc3;4x C6/e x .2x 4/e x C2e x D .x 2 6x C12/e x , (B) implies that y D .x 2 6x C12/e x C22xy.0/ D 1 ) 12 C c3 D 1 ) c3 D 11, so y D .x 2 6x C 12/e x C8x 11.2cos 2x17C c1 ; y 00 .0/ D 3 )C c1 D 3 ) c1 D ,(h) If y 000 D 2 C sin 2x, then y 00 D 2x222cos 2x7sin 2x700020so y D 2xC . Then y D xC x C c2 ; y .0/ D 6 ) c2 D 6, so224237xcos 2x 7 21702 sin 2xC x 6. Then y DCC x 6xCc3 ; y.0/ D 1 ) Cc3 D 1 ) c3 D ,y Dx4238488x3cos 2x7 27so y DCC x6x C .3848(i) If y 000 D 2x C 1, then y 00 D x 2 C x C c1; y 00 .2/ D 7 ) 6 C c1 D 7 ) c1 D 1; so y 00 D x 2 C x C 1.x3 x21426x3 x2Then y 0 DCC .x 2/ C c2; y 0 .2/ D 4 )C c2 D 4 ) c2 D, so y 0 DCC32333243xx1268526.x 2/. Then y DCC .x 2/2.x 2/ C c3 ; y.2/ D 1 ) C c3 D 1 ) c3 D,3126233343x1265xso y DCC .x 2/2.x 2/.1262331.2.6. (a) If y D x 2 .1 C ln x/, then y.e/ D e 2 .1 C ln e/ D 2e 2; y 0 D 2x.1 C ln x/ C x D 3x C 2x ln x,so y 0 .e/ D 3e C 2e ln e D 5e; (A) y 00 D 3 C 2 C 2 ln x D 5 C 2 ln x. Now, 3xy 0 4y D 3x.3x C2x ln x/ 4x 2.1 C ln x/ D 5x 2 C 2x 2 ln x D x 2 y 00 , from (A).x211225(b) If y DC x 1, then y.1/ D C 1 1 D ; y 0 D x C 1, so y 0 .1/ D C 1 D ; (A)33 3333 2x22200202y D . Now xxy C y C 1 D xxxC1 CC x 1 C 1 D x 2 D x 2 y 00 , from (A).3333(c) If y D .1 C x 2 / 1 2 , then y.0/ D .1 C 02 / 1 2 D 1; y 0 D x.1 C x 2 / 3 2 , so y 0 .0/ D 0; (A)00y D .2x 2 1/.1Cx 2 / 5 2 . Now, .x 2 1/y x.x 2 C1/y 0 D .x 2 1/.1Cx 2 / 1 2 x.x 2 C1/. x/.1C.x 2 1/y x.x 2 C 1/y 0x 2 / 3 2 D .2x 2 1/.1 C x 2 / 1 2 D y 00 .1 C x 2 /2 from (A), so y 00 D.x 2 C 1/2x21 41x.x 2/. 1 2/. 3 2/(d) If y D, then y.1 2/ DD ; y0 D, so y 0 .1 2/ DD 3;1 x1 1 22.1 x/2.1 1 2/22x2xx 2 .x 2/x20(A) y 00 D.Now,(B)xCyDxCDand(C)xyyDD.1 x/31 x1 x.1 x/21 x2330xxx 002.x C y/.xyy/. From (B) and (C), .x C y/.xy 0 y/ DDy , so y 00 D.233.1 x/.1 x/2x1.2.8. (a) y D .x c/a is defined and x c D y 1 a on .c; 1/; moreover, y 0 D a.x a 1a y 1 aD ay .a 1/ a .(b) if a 1 or a 0, then y 0 is a solution of (B) on . 1; 1/.c/a1D1.2.10. (a) Since y 0 D c we must show that the right side of (B) reduces to c for all values of x in some

4Chapter 1 Basic Conceptsinterval. If y D c 2 C cx C 2c C 1,x 2 C 4x C 4ypD x 2 C 4x C 4c 2 C 4cx C 8c C 4D x 2 C 4.1 C c/x C 4.c 2 C 2c C 1/D x 2 C 4.1 C c/ C 2.c C 1/2 D .x C 2c C 2/2 :x 2 C 4x C 4y D x C 2c C 2 and the right side of (B) reduces to c if x 2c 2.x.x C 4/xC2(b) If y1 D, then y10 Dand x 2 C 4x C 4y D 0 for all x. Therefore, y1 satisfies42(A) on . 1; 1/.Therefore,

CHAPTER 2First Order Equations2.1 LINEAR FIRST ORDER EQUATIONSy02.1.2.D 3x 2; j ln jyj D x 3 C k; y D cey2.1.4.y0Dy2.1.6.y0DyyDe3;xln jyj D 3 ln jxj C k D1CxDx1xln jxj3 C k;1; j ln jyj Dxln jxj.ln x/2 2yDy0Dyy D 3x k .2.1.10.y10Dy1k; j ln jyj Dx3; ln jy1 j De 3x1C c; y D C ce33y10Dy13xln jxjc.x3x C k; y Dln j sin xj C k Dk ln jxj C k1 D ln jx3x; y1 D e3xce x; y.1/ D 1 ) c D e;xkln jx sin xj C k; y Dj C k1 ; y D cjxj; y D ue3x; u0 e3xk;c;x sin xy.1/ D 3 ) c D 3;D 1; u0 D e 3x ; u D.2x; ln jy1 j D x 2 ; y1 D e 2 xx22uDC c; y D e xCc .22x2y10Dy1; y D ue11u; ln jy1 j D ln jxj; y1 D ; y D ;xxx3x 27 ln jxj 3xcu D 7 ln jxj CC c; y DCC .2x2x2.1.16.1y0Dcot x; j ln jyj Dyx y. 2/ D 2 ) c D ; y D.x sin x2.1.14. y D ce.x 1/2.1.8.2.1.12.x35x2; u0 ex2D xex2; u0 D x;u077D 2 C 3; u0 DC 3x;xxx

6Chapter 2 First Order Equations2y10Dy121e xue x2x; ln jy1 j D ln jxj x 2 ; y1 D; yD;xxx x4x3c2C c; y D e xC.u0 D x 3 ; u D44x2.1.18.2u0 e xD x2exx2;y10D tan x; ln jy1 j D ln j cos xj; y1 D cos x; y D u cos x; u0 cos x D cos x; u0 D 1;y1u D x C c; y D .x C c/ cos x.ˇˇˇ .x 2/5 ˇy104x 351ˇˇ;DD; ln jy1 j D 5 ln jx 2j ln jx 1j D ln ˇ2.1.22.y1.x 2/.x 1/x 2 x 1x 1 ˇ.x 2/5u.x 2/5u0 .x 2/5.x 2/2111; yD;D; u0 D; uDCy1 D3x 1x 1x 1x 1.x 2/2 .x 2/21 .x 2/3.x 2/5c; y DCc.2 .x 1/.x 1/2.1.20.2.1.24.y10Dy1u D xe x3; ln jy1 j D 3 ln jxj D ln jxj 3;xexexce x C c; y D 2C 3.3xxxy1 D1u; y D 3;3xxu0exD; u0 D xe x ;x3x2y10Dy14x1; ln jy1 j D 2 ln.1 C x 2 / D ln.1 C x 2 / 2 ; y1 D; y D1 C x2.1 C x 2 /20u22x C cu;D; u0 D 2; u D 2x C c; y D; y.0/ D 1 )222222.1 C x /.1 C x /.1 C x /.1 C x 2 /22x C 1c D 1; y D.1 C x 2 /22.1.26.2.1.28.y10Dy12.1.30.y10Dy1u0D cos x; u0 Dsin xsin xsin2 x1sin x cos x; u DC c; y DC c csc x; y. 2/ D 1 ) c D 21 ; y D .sin x C csc x/.222cot x; ln jy1 j D3; ln jy1 j Dln j sin xj; y1 D3 ln jx1u; y D;sin xsin x1j D ln jx1j3; y1 D11/3x 1.xu01sin x10DC; u DC sin x; u D ln jx 1j.x 1/3.x 1/4.x 1/3x 1ln jx 1j cos x C cln jx 1j cos x; y.0/ D 1 ) c D 0; y D.x 1/3.x 1/32.1.32.uD; ln jy1 j D3 ln jx1j D ln jx1j3; y1 D1u1/3.x;cos x C c; y Dy102D; ln jy1 j D 2 ln jxj D ln.x 2 /; y1 D x 2 ; y D ux 2 ; u0 x 2 Dy1xln jxj C c; y D x 2 .c ln jxj/; y.1/ D 1 ) c D 1; y D x 2 .1 ln x/.y103Dy1xu01 C .xD.x 1/3.x2.1.34.; y Dx; u0 D; y Du1;x;1.x.x 1/31/ sec2 x1ln jx 1j C tan x C c; u0 DCsec2 x; u D ln jx 1jCtan xCc; y D;1/4x 1.x 1/3ln jx 1j C tan x C 1y.0/ D 1 ) c D 1; y D.x 1/31/3

Section 2.1 Linear First Order Equations7y102xD 2; ln jy1 j D ln jx 2 1j; y1 D x 2 1; y D u.x 2 1/; u0 .x 2 1/ D x;y1x1 x112202; u D ln jx1j C c; y D .x1/ln jx1j C c ; y.0/ D 4 ) c D 4;u D 2x1 22 1y D .x 2 1/ln jx 2 1j 4 .22.1.36.y022222.1.38. 1 D 2x; ln jy1 j D x 2 ; y1 D e x ; y D ue x ; u0 e x D x 2 ; u0 D x 2 e x ; u Dy1 Z xZ xZ x22222cCt 2 e t dt; y D e x c Ct 2 e t dt ; y.0/ D 3 ) c D 3; y D e x 3 Ct 2 e t dt .000y10e x tan xtan xD 1; ln jy1 j D x; y1 D e x ; y D ue x ; u0 e x D; u0 D;yZ1xx Z xZ xxtan ttan ttan tuDcCdt ; y D e x c Cdt ; y.1/ D 0 ) c D 0; y D e xdt.ttt1112.1.40.y10Dy121e xue xu0 e xex; ln jy1 j D xln jxj; y1 D; y D;D;x Zxxx Z xxxx e222u0 D e x e x ; u D c Ce t e t dt; y DcCe t e t dt ; y.1/ D 2 ) c D 2e;x1 Z x11.x 1/xt t2yD2eCee e dt .x12.1.42.12.1.44. (b) Eqn. (A) is equivalent to2Dxy01x.B/y1021D ; ln jy1 j D 2 ln jxj; y1 D x 2 ; y D ux 2 ; u0 x 2 D;y1xx111u0 D; uDC c, so y D C cx 2 is the general solution of (A) on . 1; 0/ and .0; 1/.32x2x2(c) From the proof of (b), any solution of (A) must be of the form8̂1 C c1x 2 ; x 0;2yD.C/:̂ 1 C c2x 2 ; x 0;2on . 1; 0/ and .0; 1/. Herefor x 0, and any function of the form (C) satisfies (A) for x 0. To complete the proof we must showthat any function of the form (C) is differentiable and satisfies (A) at x D 0. By definition,y 0 .0/ D limx!0if the limit exists. Buty.x/y.x/x1 2xy.0/y.x/ 1 2D limx!00xD c1 x; x 0c2 x; x 0;so y 0 .0/ D 0. Since 0y 0 .0/ 2y.0/ D 0 0 2.1 2/ D 1, any function of the form (C) satisfies (A) atx D 0.(d) From (b) any solution y of (A) on . 1; 1/ is of the form (C), so y.0/ D 1 2.

8Chapter 2 First Order Equations1 2and c2 arbitrary is a solutionx02of the initial value problem on . 1; 1/. Since these functions are all identical on .0; 1/, this does notcontradict Theorem 2.1.1, which implies that (B) (so (A)) has exactly one solution on .0; 1/ such thaty.x0 / D y0 . A similar argument applies if x0 0.(e) If x0 0, then every function of the form (C) with c1 Dy02.1.46. (a) Let y D c1 y1 C c2 y2 . Theny 0 C p.x/yD.c1 y1 C c2 y2 /0 C p.x/.c1 y1 C c2 y2 /Dc1y10 C c2y20 C c1p.x/y1 C c2 p.x/y2Dc1.y10 C p.x/y1 / C c2 .y2 C p.x/y2 / D c1f1 .x/ C c2f2 .x/:(b) Let f1 D f2 D f and c1 D c2 D 1.(c) Let f1 D f , f2 D 0, and c1 D c2 D 1.3x0 3x2.1.48. (a) If D tan y, then 0 D .sec2 y/y 0 , so 0 3 D 1; 1 D e 3x ; D ue ; u e D 1;3xe11u0 D e 3x ; u DC c; D C ce 3x D tan y; y D tan 1C ce 3x .33311u u01222(b) If D e y , then 0 D 2yy 0 e y , so 0 C D 2 ; 1 D 2 ; D 2 ; 2 D 2 ; u0 D 1;xxxxxx 1 2c1c1y2u D x C c; D C 2 D e ; y D lnC 2.xxxx0y2y021(c) Rewrite the equation as C ln y D 4x. If D ln y, then 0 D , so 0 C D 4x; 1 D 2 ;yxyxx cc u u020342 D 2 ; 2 D 4x; u D 4x ; u D x C c; D x C 2 D ln y; y D exp x C 2 .xxxx1y0131u u0300(d) If D, then D,so C D; D; D;D;11Cy.1 C y/2xx2xx xx233 ln jxj C c1xu0 D; u D 3 ln jxj c; DD; y D 1C.xx1Cy3 ln jxj C c2.2 SEPARABLE EQUATIONSBy2.2.2. inspection, y k (k Dinteger) is a constant solution. Separate variables to find others:cos y 0y D sin x; ln.j sin yj/ D cos x C c.sin y ln y.ln y/22.2.4. y 0 is a constant solution. Separate variables to find others:y0 D x2;Dy2x3C c.312.2.6. y 1 and y 1 are constant solutions. For others, separate variables: .y 2 1/ 3 2 yy 0 D 2 ;x 211 C cxxx21 221 22.y1/Dc D; .y1/D; .y1/ D;xx1 C cx1 C cx 2 2 !1 2xx2y D1C; y D 1C.1 C cx1 C cx

Section 2.2 Separable Equations2.2.8. By inspection, y 0 is a constant solution. Separate variables to find others:ln jyj Dy0Dyc1ln.1 C x 2 / C k; y D p, which includes the constant solution y 0.21 C x22.2.10. .y 1/2 y 0 D 2xC3;1/3.y39x;1 C x2D x 2 C3xCc; .y 1/3 D 3x 2 C9xCc; y D 1 C 3x 2 C 9x C c/1 3 .ˇˇ 2ˇ y ˇy0110ˇˇ D x C k; y D ce x2 2 ; y.2/ DD x;y D x; ln ˇ2.2.12.y.y C 1/y yC1y C 1ˇ2yC1x 2 22ceee2222; y D .y C 1/ce x 2 ; y.1 ce x 2/ D ce x 2 ; y D1)cD;settingcD2221 ce x 2.x 2 4/ 2eyields y D.22 e .x 4/ 2 11 11 11 11132y02.2.14.D;Cy0 D;Cy0 D.y C 1/.y 1/.y 2/xC16yC1 2y 13y 2xC1yC1 y 1y 26.y C 1/.y 2/2c; ln jy C 1j 3 ln jy 1j C 2 ln jy 2j D 6 ln jx C 1j C k;D;xC1.y 1/3.x C 1/6.y C 1/.y 2/2256y.1/ D 0 ) c D 256;D.3.y 1/.x C 1/6! y01yjyjy202.2.16.D 2x;y D 2x; ln pD x 2 C k; pD ce x ;2222y.1 C y /y y C1y C1y C12ex1y1222y.0/ D 1 ) c D p ; pD p ; 2y 2 D .y 2 C1/e x ; y 2 .2 e x / D e 2x ; y D p.22x22y C12e 2 1ˇˇ ˇy 2ˇy011ˇ D x 2 C k; y 2 D ce x2 ;2.2.18.D 2x;y 0 D 2x; ln ˇˇ.y 1/.y 2/y 2 y 1y 1 ˇ!y 12x2x2x2x2eeee4 e x1 y 2.y.0/ D 3 ) c D ;D;y 2 D.y 1/; y 1D2; yD2 y 122222 e x2The interval of validity is . 1; 1/.ˇˇ ˇy 2ˇy0111 011 0ˇ D 2x C k;2.2.20.D 1;y D 1;y D 2; ln ˇˇy.y 2/2 y 2 yy 2 yy ˇy 2y 2D ce 2x ; y.0/ D 1 ) c D 1;D e 2x ; y 2 D ye 2x ; y.1 C e 2x / D 2;yy2yD. The interval of validity is . 1; 1/.1 C e 2x2.2.22. y 2 is a constant solution of the differential equation, and it satisfies the initial condition.Therefore, y 2 is a solution of the initial value problem. The interval of validity is . 1; 1/.y01D; tan 1 y D tan21Cy1 C x2tan A C tan Btan.A C B/ Dwith A D tan1 tan A tan Bc D tan k.2.2.24.1x C k; y D tan.tan1x and B D tan11x C k/. Now use the identityc to rewrite y as y DxCc, where1 cx

10 Chapter 2 First Order Equations ) c D 0, so (A) cos y D sin x. To obtain2y explicity we note that sin x D cos.x C 2/, so (A) can be rewritten as cos y D cos.x C 2/. Thisequation holds if an only if one of the following conditions holds for some integer k:2.2.26. .sin y/y 0 D cos x;cos y D sin x C c; y. / D(B) y D x C C 2k I mbox.C / y D2x C 2k :2Among these choices the only way to satisfy the initial condition is to let k D 1 in (C), so y D2.2.28. Rewrite the equation as P 0 DxC3 :21 /. By inspection, P 0 and P 1 areP011constant solutions. Separate variables to find others:D a ;P 0 D a;P.P1 /P1 Pˇˇˇ P 1 ˇ1ˇ D at C k; (A) P 1 D ce t ; P .1 ce t / D 1 ; (B) P D.ln ˇˇˇPP .1 ce t /P0 1 P0From (A), P .0/ D P0 ) c D. Substituting this into (B) yields P D.P0 P0 C .1 P0/e atFrom this limt !1 P .t/ D 1 .a P .P2.2.30. If q D rS the equation for I reduces to I 0 DrI 2 , soI0DI2r;1DIrt1; soI0I0and limt !1 I.t/ D 0. If q rS , then rewrite the equation for I as I 0 D rI.I /1 C rI0 tˇˇ ˇI ˇqI011 0ˇˇDwith D S. Separating variables yieldsD r;I D r ; ln ˇrI.I /I II ˇI r t C k; (A)D ce r t ; I.1 ce r t / D ; (B) I D. From (A), I.0/ D I0 )I1 ce r tI0 I0c D. Substituting this into (B) yields I D. If q rS , then 0 andI0I0 C . I0 /e r tqlimt !1 I.t/ D D S. If q rS , then 0 and limt !1 I.t/ D 0.rI D2.2.34. The given equation is separable if f D ap, where a is a constant. In this case the equation isy 0 C p.x/y D ap.x/:.A/Let P be an antiderivative of p; that is, P 0 D p.S OLUTIONBYS EPARATIONOFVARIABLES . y 0 Dp.x/.ya/;y0yaDp.x/; ln jyaj DP .x/ C k; y a D ce P.x/ ; y D a C ce P.x/ .S OLUTION BY VARIATION OF PARAMETERS . y1 D e P.x/ is a solution of the complementaryequation, so solutions of (A) are of the form y D ue P.x/ where u0 e P.x/ D ap.x/. Hence, u0 Dap.x/e P.x/ ; u D ae P.x/ C c; y D a C ce P.x/ .2x52yD. y1 D x 2 is a solution of y 0y D 0.2xyCxxx5x3xLook for solutions of (A) of the form y D ux 2 . Then u0 x 2 DD; u0 D;2.u C 1/xuC1uC1 pp.1 C u/2x2c.u C 1/u0 D x;DC ; u D 1 x 2 C c; y D x 2 1 x 2 C c .2222.2.36. Rewrite the given equation as (A) y 0

Section 2.3 Existence and Uniqueness of Solutions of Nonlinear Equations112.2.38. y1 D e 2x is a solution of y 0 2y D 0. Look for solutions of the nonlinear equation of thexe 2xx.1 u/21form y D ue 2x . Then u0 e 2x D; u0 D; .1 u/u0 D x;D .x 2 c/;1 u 1 u22 ppu D 1 c x 2 ; y D e 2x 1 c x 2 .2.3 EXISTENCE AND UNIQUENESS OF SOLUTIONS OF NONLINEAR EQUATIONSex C y12y.e x C y/and fy .x; y/ D 2are both continuous at all .x; y/ 222x Cyx Cy.x 2 C y 2 /2.0; 0/. Hence, Theorem 2.3.1 implies that if .x0 ; y0 / .0; 0/, then the initial value problem has a aunique solution on some open interval containing x0. Theorem 2.3.1 does not apply if .x0 ; y0 / D .0; 0/.2.3.2. f .x; y/ Dx2 C y22yx2 C y2and fy .x; y/ Dare both continuous at all .x; y/ suchln xyln xyx.ln xy/2that xy 0 and xy 1. Hence

STUDENT SOLUTIONS MANUAL FOR ELEMENTARY DIFFERENTIAL EQUATIONS AND ELEMENTARY DIFFERENTIAL EQUATIONS WITH BOUNDARY VALUE PROBLEMS William F. Trench . 8.3 Solution ofInitial Value Problems 134 8.4 The Unit Step Function 140 8.5 Constant Coefficient Equationswith Piecewise Continuous Forcing Functions 143

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