STUDENT SOLUTIONS MANUAL FOR ... - Trinity …

1 STUDENT SOLUTIONS MANUAL FORELEMENTARYDIFFERENTIAL EQUATIONSANDELEMENTARYDIFFERENTIAL EQUATIONSWITH BOUNDARY VALUEPROBLEMSW illiam F. TrenchAndrew G. Cowles Distinguished Professor EmeritusDepartment of MathematicsTrinity UniversitySan Antonio, Texas, book has been judged to meet the evaluation criteria setby the Edi-torial Board of the American Institute of Mathematics in connection withthe Institute sOpen Textbook Initiative. It may be copied, modified, re-distributed, translated, and built upon subject to the Creative CommonsAttribution-NonCommercial-ShareAl ike Unported book was published previously by Brooks/Cole Thomson LearningReproduction is permitted for any valid noncommercial educational, mathematical, or scientific , charges for profit beyond reasonable printing costs are BEVERLYC ontentsChapter 1 First Order Equations1 Chapter 2 First Order Linear First Order Separable Existence and Uniqueness of SOLUTIONS of Nonlinear Transformation of Nonlinear Equations into Separable Exact Integrating Factors21 Chapter 3 Numerical Euler s The Improved Euler Method and Related The Runge-Kutta Method34 Chapter 4 Applications of First Order Growth and Cooling and Elementary Autonomous Second Order Applications to Curves46 Chapter 5 Linear Second Order Homogeneous Linear Constant Coefficient Homogeneous Nonhomgeneous Linear The Method of Undetermined Coefficients The Method of Undetermined Coefficients Reduction of Variation of Parameters79 Chapter 6 Applcations of Linear Second Order Spring Problems Spring Problems The RLC Motion Under a Central Force90 Chapter 7 Series SOLUTIONS of Linear Second

2 Order Review of Power Series SOLUTIONS Near an Ordinary Point Series SOLUTIONS Near an Ordinary Point Regular Singular Points; Euler The Method of Frobenius The Method of Frobenius The Method of Frobenius III118 Chapter 8 Laplace Introduction to the Laplace The Inverse Laplace Solution of Initial Value The Unit Step Constant Coefficient Equations with Piecewise Continuous Constant Cofficient Equations with Impulses55 Chapter 9 Linear Higher Order Introduction to Linear Higher Order Higher Order Constant Coefficient Homogeneous Undetermined Coefficients for Higher Order Variation of Parameters for Higher Order Equations181 Chapter 10 Linear Systems of Differential Introduction to Systems of Differential Linear Systems of Differential Basic Theory of Homogeneous Linear Constant Coefficient Homogeneous Systems Constant Coefficient Homogeneous Systems Constant Coefficient Homogeneous Systems Variation of Parameters for Nonhomogeneous Linear Eigenvalue Problems for y00C Fourier Expansions Fourier

3 Expansions II229 Chapter 12 Fourier SOLUTIONS of Partial Differential The Heat The Wave Laplace s Equation in Rectangular Laplace s Equation in Polar Coordinates270 Chapter 13 Boundary Value Problems for Second Order Ordinary Differential Equations Two-Point Boundary Value Sturm-Liouville Problems279 CHAPTER (a)IfyDce2x, theny0D2ce2xD2y.(b)IfyDx23 Ccx, theny0D2x3 cx2, soxy0 CyD2x23 cxCx23 CcxDx2.(c)IfyD12 Cce x2;theny0D 2xce x2andy0C2xyD 2xce x2C2x 12 Cce x2 D 2xce x2 CxC2cxe x2Dx:(d)IfyD1 Cce x2=21 ce x2= ce x2=2/. cxe x2=2/ .1 Cce x2=2/cxe x2= cxe x2=2/2D 2cxe x2= ce x2=2/2andy2 1D 1 Cce x2=21 ce x2=2!2 x2=2/2 .1 ce x2=2 ce x2=2/2D4ce x2= ce x2=2/2;12 Chapter 1 Basic 1/D ce x2=2/2D0:(e)IfyDtan x33Cc , theny0Dx2sec2 x33Cc Dx2 1 Ctan2 x33Cc (f) ; ; sinxC2;andy00 2C1 4C1/ sinx 2cosxCsinxC2 4xCx2D 2cosxCx2 4xC2:(g)IfyDc1exCc2xC2x, theny0Dc1exCc2 2x2andy00Dc1exC4x3, x/y00 Cxy0 xCx 1 x/x3 2x x x2/x3(h)IfyDc1sinxCc2cosxx1=2C4xC8theny0 Dc1cosx c2sinxx1=2 c1sinxCc2cosx2x3=2C4andy00D c1sinxCc2cosxx1=2 c1sinx c2cosxx3=2C34c1sinxCc2cosxx5=2, sox2y00 Cxy0C x2 14 yDc1 x 3=2sinx x1=2cosxC34x 1=2sinxCx1=2cosx 12x 1=2sinxCx3=2sinx 14x 1=2sinx Cc2 x 3=2cosxCx1=2sinxC34x 1=2cosx x1=2sinx 12x 1=2cosxCx3=2cosx 14x 1=2cosx C4xC x2 14.

4 4xC8/D4x3C8x2C3x (a)Ify0D xex, thenyD x/exCc, )1D1Cc, x/ex.(b)Ify0 Dxsinx2, thenyD 12cosx2Cc;y r 2 D1)1D0Cc, socD1andyD1 12cosx2.(c)Writey0 DtanxDsinxcosxD Integrating this yieldsyD lnjcosxjCc;y. =4/D3)3D =4//Cc, or3 Dlnp2Cc, socD3 lnp2, soyD lnp2D3 (d)Ify00Dx4, theny0Dx55Cc1; 1)325Cc1D 1)c1D 3715, soy0Dx55 3715. Therefore,yDx630 2/Cc2; 1)6430Cc2D 1)c2D 4715, soyD 4715 2/Cx630.(e)(A)Rxe2xdxDxe2x2 12Ze2xdxDxe2x2 e2x4. Therefore,y0 Dxe2x2 e2x4Cc1; ) 14Cc1D54)c1D54, soy0 Dxe2x2 e2x4C54; Using (A) again,yDxe2x4 e2x8 e2x8C54xCc2 Dxe2x4 e2x4C54xCc2; ) 14Cc2D7)c2D294, soyDxe2x4 e2x4C54xC294.(f)(A)Rxsinx dxD xcosxCRcosx dxD xcosxCsinxand (B)Rxcosx dxDxsinx RsinxdxDxsinxCcosx. Ify00D xsinx, then (A) implies thaty0 Dxcosx sinxCc1; 3)cD 3, soy0 Dxcosx sinx 3. Now (B) implies thatyDxsinxCcosxCcosx 3xCc2 DxsinxC2cosx 3xCc2; )2Cc2D1)c2D 1, soyDxsinxC2cosx 3x Concepts3(g)Ify000Dx2ex, theny00 DRx2exdxDx2ex 2 RxexdxDx2ex 2xexC2exCc1; )2Cc1D3)c1D1, so (A) 2xC2/exC1.

5 2xC2 2xC2/ex 2 2xC2/ex .2x 2 4xC6/ex,(A) implies 4xC6/exCxCc2; 2)6Cc2D 2)c2D 8, so (B) 4xC6/exCx 8; 4xC6 4xC6/ex 4 4xC6/ex .2x 4 6xC12/ex, (B) implies 6xC12/exCx22 8xCc3; )12Cc3D1)c3D 11, 6xC12/exCx22 8x 11.(h)Ify000D2 Csin2x, theny00D2x cos2x2Cc1; ) 12Cc1D3)c1D72,soy00D2x cos2x2C72. Theny0Dx2 sin2x4C72xCc2; 6)c2D 6, soy0Dx2 sin2x4C72x 6. ThenyDx33 Ccos2x8C74x2 6xCc3; )18Cc3D1)c3D78,soyDx33 Ccos2x8C74x2 6xC78.(i)Ify000D2xC1, theny00Dx2 CxCc1; )6Cc1D7)c1D1; 2/Cc2; 4)143Cc2D 4)c2D 263, 2/ 263. 2/2 2/Cc3; )83Cc3D1)c3D 53, 2/2 2/ (a) , ; , ; (A)y00D3C2C2lnxD5C2lnx. Now,3xy0 , from (A).(b)IfyDx23Cx 1, 1D13;y0D23xC1, ; (A)y00D23. Nowx2 xy0 CyC1Dx2 x 23xC1 Cx23Cx 1C1D23x2Dx2y00, from (A).(c) 1=2, 1=2D1;y0D 3=2, ; (A) 1/.1Cx2/ 5=2. Now,.x2 1/y 1/.1Cx2/ 1=2 x/.1Cx2/ 3= 1/.1Cx2/ 1= (A), 1/y (d)IfyDx21 x, 1=2D12;y0D 2/.1 x/2, 1=2/. 3=2/.1 1=2/2D3;(A) x/3. Now, (B)xCyDxCx21 xDx1 xand (C)xy0 yD 2/.

6 1 x/2 x21 x/2. From (B) and (C),.xCy/.xy0 x/3Dx32y00, (a) c/ais defined andx cDy1= ;1/; moreover, c/a 1Da y1=a a 1/=a.(b)ifa > 1ora < 0, theny 0is a solution of (B) on. 1;1/. (a)Sincey0 Dcwe must show that the right side of (B) reduces tocfor all values ofxin some4 Chapter 1 Basic Conceptsinterval. IfyDc2 CcxC2cC1, :Therefore,px2C4xC4yDxC2cC2and the right side of (B) reduces tocifx > 2c 2.(b)Ify1D , theny01D xC22andx2C4xC4yD0for allx. Therefore,y1satisfies(A) on. 1;1/.CHAPTER 2 First Order FIRST ORDER 3x2;jlnjyjD x3Ck;yDce .lnx/2= 3x; lnjyjD 3lnjxjCkD lnjxj3Ck; 1 CxxD 1x 1;jlnjyj D lnjxj xCk;yDce xx; )cDe;yDe .x 1 1x cotx;jlnjyj D lnjxj lnjsinxjCkD lnjxsinxjCk;yDcxsinx;y. =2/D2)cD ;yD kx;jlnjyj D klnjxjCk1 Dlnjx kjCk1;yDcjxj k; )cD3;yD3x 3; lnjy1j D 3x;y1De 3x;yDue 3x;u0e 3xD1;u0De3x;uDe3x3Cc;yD13 Cce 2x; lnjy1j D x2;y1De x2;yDue x2;u0e x2 Dxe x2;u0Dx;uDx22Cc;yDe x2 x22Cc . 1x; lnjy1j D lnjxj;y1D1x;yDux;u0xD7x2C3;u0D7xC3x;uD7l njxjC3x22Cc; Chapter 2 First Order 1x 2x; lnjy1j D lnjxj x2;y1De x2x;yDue x2x;u0e x2xDx2e x2;u0Dx3;uDx44Cc;yDe x2 x34 Ccx.

7 Tanx; lnjy1jDlnjcosxj;y1 Dcosx;yDucosx;u0cosxDcosx;u0D1;uDxCc; 2/.x 1/D5x 2 1x 1; lnjy1j D5lnjx 2j lnjx 1j Dln .x 2/5x 1 ; 2/5x 1; 2/5x 1; 2/5x 2/2x 1; 2/3;uD 2/2Cc;yD 2 1 2 1/. 3x; lnjy1j D 3lnjxj Dlnjxj 3;y1D1x3;yDux3;u0x3 Dexx2;u0 Dxex;uDxex exCc;yDexx2 4x1Cx2; lnjy1j D 2; ; ; ;u0D2;uD2xCc; ; )cD1; cotx; lnjy1j D lnjsinxj;y1D1sinx;yDusinx;u0sinxDcosx;u0 Dsinxcosx;uDsin2x2Cc;yDsinx2 Cccscx;y. =2/D1)cD12; 3x 1; lnjy1j D 3lnjx 1j Dlnjx 1j 3; 1/3; 1/3; 1 1 1/3;u0D1x 1 Csinx;uDlnjx 1j cosxCc;yDlnjx 1j 1/3; )cD0;yDlnjx 1j 1 2x; lnjy1j D2lnjxj ;y1Dx2;yDux2;u0x2D x;u0D 1x;uD lnjxjCc; lnjxj/; )cD1; lnx/. 3x 1; lnjy1j D 3lnjx 1j Dlnjx 1j 3; 1/3; 1/3; 1 1 1/4;u0D1x 1 Csec2x;uDlnjx 1jCtanxCc;yDlnjx 1/3; 1)cD1;yDlnjx 1 First Order 1; lnjy1j Dlnjx2 1j;y1Dx2 1; 1/; 1/Dx;u0 Dxx2 1;uD12lnjx2 1jCc; 1/ 12lnjx2 1jCc ; )cD 4; 1/ 12lnjx2 1j 4 . 2x; lnjy1jD x2;y1De x2;yDue x2;u0e x2Dx2;u0Dx2ex2;uDcCZx0t2et2dt;yDe x2 cCZx0t2et2dt ; )cD3;yDe x2 3 CZx0t2et2dt.

8 1; lnjy1j D x;y1De x;yDue x;u0e xDe xtanxx;u0 Dtanxx;uDcCZx1tanttdt;yDe x cCZx1tanttdt ; )cD0;yDe 1 1x; lnjy1j D x lnjxj;y1De xx;yDue xx;u0e xxDex2x;u0 Dexex2;uDcCZx1etet2dt;yDe xx cCZx1etet2dt ; )cD2e;yD1x 2e .x 1/Ce xZx1etet2dt . (b)Eqn. (A) is equivalent toy0 2xD 1;0 ;1/. Herey01y1D2x; lnjy1j D2lnjxj;y1Dx2;yDux2;u0x2D 1x;u0D 1x3;uD12x2Cc, soyD12 Ccx2is the general solution of (A) on. 1;0 ;1/.(c)From the proof of(b), any solution of (A) must be of the formyD8 < :12Cc1x2; x 0;12Cc2x2; x < 0;.C/forx 0, and any function of the form (C) satisfies (A) forx 0. To complete the proof we must showthat any function of the form (C) is differentiable and satisfies (A) atxD0. By definition, ! 0 Dlimx! 1=2xif the limit exists. 1=2xD c1x; x > 0c2x; x < 0; 0 1, any function of the form (C) satisfies (A) atxD0.(d)From(b)any solutionyof (A) on. 1;1/is of the form (C), Chapter 2 First Order Equations(e)Ifx0> 0, then every function of the form (C) withc1Dy0 1=2x20andc2arbitrary is a solutionof the initial value problem on.

9 1;1/. Since these functions are all identical ;1/, this does notcontradict Theorem , which implies that (B) (so (A)) has exactly one solution ;1/such A similar argument applies ifx0< (a)LetyDc1y1Cc2y2. :(b)Letf1Df2 Dfandc1D c2D1.(c)Letf1Df,f2D0, (a)If Dtany, then , so 0 3 D 1; 1De3x; Due3x;u0e3xD 1;u0D e 3x;uDe 3x3Cc; D13 Cce3xDtany;yDtan 1 13 Cce3x .(b)If Dey2, then 0D2yy0ey2, so 0C2x D1x2; 1D1x2; Dux2;u0x2D1x2;u0D1;uDxCc; D1xCcx2 Dey2;yD ln 1xCcx2 1=2.(c)Rewrite the equation asy0yC2xlnyD4x. If Dlny, then 0Dy0y, so 0C2x D4x; 1D1x2; Dux2;u0x2D4x;u0D4x3;uDx4Cc; Dx2 Ccx2 Dlny;yDexp x2 Ccx2 .(d)If D 11Cy, then , so 0C1x D 3x2; 1D1x; Dux;u0xD 3x2;u0D 3x;uD 3lnjxj c; D 3lnjxjCcxD 11Cy;yD inspection,y k (kDinteger) is a constant solution. Separate variables to find others: cosysiny y0D sinx; 0is a constant solution. Separate variables to find others: lnyy y0D x2;.lny/22D 1andy 1are constant SOLUTIONS .

10 For others, separate variables:.y2 1/ 3=2yy0D1x2; .y2 1/ 1=2D 1x cD 1 Ccxx ;.y2 1/1=2D x1 Ccx ;.y2 1/D x1 Ccx 2;y2D1C x1 Ccx 2;yD 1C x1 Ccx 2!1= inspection,y 0is a constant solution. Separate variables to find others:y0yD x1Cx2;lnjyjD ;yDcp1Cx2, which includes the constant solutiony 1/2y0D2xC3;.y 1/33Dx2C3xCc;.y 1/3D3x2C9xCc;yD1C 3x2C9xCc/1= x; 1y 1yC1 y0D x; ln yyC1 D x22Ck;yyC1 Dce x2=2; )cDe22; x2=2; ce x2=2/Dce x2=2;yDce x2=21 ce x2=2; settingcDe22yieldsyDe .x2 4/=22 e .x2 4/= 1/.y 2/D 1xC1; 161yC1 121y 1C131y 2 y0D 1xC1; 1yC1 3y 1C2y 2 y0D 6xC1; lnjyC1j 3lnjy 1jC2lnjy 2j D 6lnjxC1jCk;.yC1/.y 2 1 ; )cD 256;.yC1/.y 2 1/3D ; 1y yy2C1 y0D2x; ln jyjpy2C1!Dx2Ck;ypy2C1 Dcex2; )cD1p2;ypy2C1 Dex2p2; ; ex2/De2x2;yD1p2e 2x2 1/.y 2/D 2x; 1y 2 1y 1 y0D 2x; ln y 2y 1 D x2Ck;y 2y 1 Dce x2; )cD12;y 2y 1De x22;y 2De 1/;y 1 e x22!D2 e x22;yD4 e x22 e interval of validity is. 1;1/. 2/D 1;12 1y 2 1y y0D 1; 1y 2 1y y0D 2; ln y 2y D 2xCk;y 2yDce 2x; )cD 1;y 2yD e 2x;y 2D ye 2x; 2x/D2;yD21Ce 2x.