Exam (elaborations) TEST BANK FOR A First Course in Differential Equations with Modeling Applications 9th, Differential Equations with Boundary-Value Problems 7th By Dennis G. Zill (Complete Solution Manual)

Exam (elaborations) TEST BANK FOR A First Course in Differential Equations with Modeling Applications 9th, Differential Equations with Boundary-Value Problems 7th By Dennis G. Zill (Complete Solution Manual) Complete Solutions Manual A First Course in Differential Equations with Modeling Applications Ninth Edition Dennis G. Zill Loyola Marymount University Differential Equations with Boundary-Vary Problems Seventh Edition Dennis G. Zill Loyola Marymount University Michael R. Cullen Late of Loyola Marymount University By Warren S. Wright Loyola Marymount University Carol D. Wright * ; BROOKS/COLE C EN G A G E Learning- Australia • Brazil - Japan - Korea • Mexico • Singapore • Spain • United Kingdom • United States Table of Contents 1 Introduction to Differential Equations 1 2 First-Order Differential Equations 27 3 Modeling with First-Order Differential Equations 86 4 Higher-Order Differential Equations 137 5 Modeling with Higher-Order Differential Equations 231 6 Series Solutions of Linear Equations 274 7 The Laplace Transform 352 8 Systems of Linear First-Order Differential Equations 419 9 Numerical Solutions of Ordinary Differential Equations 478 10 Plane Autonomous Systems 506 11 Fourier Series 538 12 Boundary-Value Problems in Rectangular Coordinates 586 13 Boundary-Value Problems in Other Coordinate Systems 675 14 Integral Transforms 717 15 Numerical Solutions of Partial Differential Equations 761 Appendix I Gamma function 783 Appendix II Matrices 785 1 Introduction to Differential Equations 1. Second order; linear 2. Third order; nonlinear because of (dy/dx)4 3. Fourth order; linear 4. Second order; nonlinear bccausc of cos(r + u) 5. Second order; nonlinear because of (dy/dx)2 or 1 + (dy/dx)2 6. Second order: nonlinear bccausc of R~ 7. Third order: linear 8. Second order; nonlinear because of x2 9. Writing the differential equation in the form x(dy/dx) -f y2 = 1. we sec that it is nonlinear in y because of y2. However, writing it in the form (y2 — 1 )(dx/dy) + x = 0, we see that it is linear in x. 10. Writing the differential equation in the form u(dv/du) + (1 + u)v = ueu wc see that it is linear in v. However, writing it in the form (v + uv — ueu)(du/dv) + u — 0, we see that it, is nonlinear in ■Jill. From y = e-*/2 we obtain y' = — e~x'2. Then 2y' + y = —e~X//2 + e-x/2 = 0. 12. From y = | — |e-20* we obtain dy/dt = 24e-20t, so that % + 20y = 24e~m + 20 - |e_20t) = 24. clt 'o 5 / 13. R'om y = eix cos 2x we obtain y1 = 3e^x cos 2x — 2e3* sin 2a? and y” = 5e3,x cos 2x — 12e3,x sin 2x, so that y" — (k/ + l?y = 0. 14. From y = — cos:r ln(sec;r + tanrc) we obtain y’ — — 1 + sin.Tln(secx + tana:) and y" = tan x + cos x ln(sec x + tan a?). Then y" -f y = tan x. 15. The domain of the function, found by solving x + 2 0, is [—2, oo). From y’ = 1 + 2(x + 2)_1/2 we 1 Exercises 1.1 Definitions and Terminology have {y - x)y' = (y - ®)[i + (20 + 2)_1/2] = y — x + 2(y - x)(x + 2)-1/2 = y - x + 2[x + 4(z + 2)1/2 - a;] (a: + 2)_1/2 = y — x + 8(ac + 2)1;/i(rr + 2)~1/2 = y — x + 8. An interval of definition for the solution of the differential equation is (—2, oo) because y defined at x = —2. 16. Since tan:r is not defined for x = 7r/2 + mr, n an integer, the domain of y = 5t£v. An interval of definition for the solution of the differential equation is (—7r/10,7T/10 . A:, interval is (7r/10, 37t/10). and so on. 17. The domain of the function is {x 4 — x2 ^ 0} or {xx ^ —2 or x ^ 2}. Prom y' — 2.:: -= - we have An interval of definition for the solution of the differential equation is (—2, 2). Other (—oc,—2) and (2, oo). 18. The function is y — l/y/l — s in s. whose domain is obtained from 1 — sinx ^ 0 or . = 1 T An interval of definition for the solution of the differential equation is (tt/2. 5tt/2 A:.. . is (57r/2, 97r/2) and so on. 19. Writing ln(2X — 1) — ln(X — 1) = t and differentiating implicitly we obtain 2 dX 1 dX 2X - 1 dt X - l dt {a; | 5x ^ tt/2 + 7i-7r} or {;r | x ^ tt/IO + mr/5}. From y' — 25sec2 §x we have y' = 25(1 + tan2 5x) = 25 + 25 tan2 5a: = 25 + y2. the domain is {z | x ^ tt/2 + 2?i7r}. From y' = —1(1 — sin x) 2 (— cos.x) we have 2y' = (1 — sin;r)_ ‘?/’2 cos# = [(1 — sin:r)~1//2]3cos:r - f/3cosx. (2X - 1)(X - 1) dt IX — = -C2X - 1)(X - 1) = (X - 1)(1 - 2X . 2X - 2 - 2X + 1 dX _ 2 Exercises 1.1 Definitions and Terminology Exponentiating both sides of the implicit solution we obtain 2 X - 1 x ----- = el X - l 2X - 1 = Xel - ef (e* - 1) = (e‘ - 2)X ef' — 1 - 4 -2 X = e* - 2 ' -2 -4 Solving e* — 2 = 0 we get t = In2. Thus, the solution is defined on (—2) or on (In2, oo). The graph of the solution defined on (—oo,ln2) is dashed, and the graph of the solution defined on (In 2. oc) is solid. 20. Implicitly differentiating the solution, we obtain y —x2 dy — 2xy dx + y dy — 0 2 xy dx + (a;2 — y)dy = 0. Using the quadratic formula to solve y2 — 2x2y — 1 = 0 for y, we get y = (2x2 ± V4;c4 +4)/2 = a’2 ± vV 1 + 1. Thus, two explicit solutions are y = x2 + A'4 + 1 and y-2 = x2 — V.x4 + 1. Both solutions are defined on (—oo. oc). The graph of yj (x) is solid and the graph of y-2 is daalied. 21. Differentiating P = c?} } ( l + cie^ we obtain dP _ ( l + cie*) cie* - cie* • cie* _ Cie« [(l + cie‘) - cie4] eft (1 + cie*)" Ci 1 + CI&- CiC 1 + ci ef 1 + cief = P( 1 - P). 1 + cie( 2 PX ,2 ,2 22. Differentiating y = e~x / e: dt + c e~x we obtain Jo * f X *2 -r.2 / e dt — 2cxe = 1 Jo y / = e -*2e r2 ___ 2 r x J 2 2xe 2xe 2 rx +2 x e dt — 2cixe Jo —X Substituting into the differential equation, we have y' + 2xy = l — 2xe x I e* dt — 2cixe x +2xe x [ e* dt 4- 2cio;e x = 1. Jo Jo 3 Exercises 1.1 Definitions and Terminology 23. From y — ci e2x+c.2 xe2x we obtain ^ - (2c +C2 )e2x -r2c2xe2x and —| = (4cj + 4c;2)e2x + 4c2xe2j'. so that — 4 ^ + Ay = (4ci + 4co - 8ci - 4c2 + 4ci)e2x + (4c2 — Sc2 + 4e2)xe2x — 0. nd.Tx. 2 dd:xr 24. From y — Cix-1 + c^x + c%x ]n x + 4a;2 we obtain ^ = —cx 2 + C2 + c$ + C3 In x + 8rc, dx = 2cix,_3 + C3;r_1 -f 8. d2y and so that dx'3 _r “J/ dx2 J' dx dx2 = —6cix-4 - c3a r2, + 2a'2 ~ X + V ~ ^_6ci + 4ci + Cl + Cl^x 3 + ^_ °3 + 2cs ~~ 02 ~ C3 + C2^X t (—C3 + cz)x In a; + (16 - 8 + 4)x2 = 12x2. ( —x2, x 0 , f —2x, x 0 25. From y = ' we obtain y' = ^ ^ „ so that - 2/y = 0. t x . x 0 { 2x, x 0 26. The function y(x) is not continuous at x = 0 sincc lim y(x) = 5 and lim y(x) = —5. Thus. y’(x) x—0“ x—0+ does not exist at x = 0. 27. From y = emx we obtain y' = mernx. Then yf + 2y — 0 implies rnemx + 2emx = (m + 2)emx = 0. Since emx 0 for all x} m = —2. Thus y = e~2x is a solution. 28. From y = emx we obtain y1 = mernx. Then by' — 2y implies brriemx = 2e"lx or m = 5 Thus y = e2:c/5 0 is a solution. 29. From y = emx we obtain y' = memx and y" = rn2emx. Then y" — 5y' + Qy = 0 implies m2emx - 5rnemx + 6emx = (rn - 2)(m - 3)emx = 0. Since ema! 0 for all x, rn = 2 and m = 3. Thus y = e2x and y = e3:r are solutions. 30. From y = emx we obtain y1 = rnemx anl y" = rn2emx. Then 2y" + 7y/ — 4y = 0 implies 2m2emx + 7rnemx - 4ema: = (2m - l)(m + 4)ema' = 0. Exercises 1.1 Definitions and Terminology Since emx 0 for all x, rn ~ | and m = —4. Thus y — ex/2 and y = e ^ are solutions. 31. From y = xm we obtain y' = mxm~1 and y" = m(m — l)xm~2. Then xy" + 2y' = 0 implies xm(m, — l)xm~2 + 2mxm~l = [m(rn -1)4- = (m2 + m)xm_1 - rn(m + l).xm_1 = 0. Since a:"'-1 0 for ;r 0. m = 0 and m = — 1. Thus y = 1 and y — x~l are solutions. 32. From y = xm we obtain y' = mxm~1 and y" = m(m — l)xm~2. Then x2y" — 7xy' + 15y — 0 implies x2rn{rn — l)xrn~2 — lxmxm~A + 15:em = [m(m — 1) — 7m + 15]xm = (ro2 - 8m + 15)a,m = (m - 3) (to - 5)xm = 0. Since xm 0 for x 0. m = 3 and m = 5. Thus y — x^ and y = xa are solutions. In Problems 33-86 we substitute y = c into the differential equations and use y' — 0 and y" — 0 33. Solving 5c = 10 we see that y ~ 2 is a constant solution. 34. Solving c2 + 2c — 3 = (c + 3)(c — 1) = 0 we see that, y = —3 and y = 1 are constant solutions. 35. Since l/(c — 1) = 0 has no solutions, the differential equation has no constant solutions. 36. Solving 6c = 10 we see that y = 5/3 is a constant solution. 37. From x — e~2t + 3ec and y — —e~2t + 5ew we obtain ^ = —2e~2t + 18e6* and = 2e~2t + 30e6*. dt dt Then x- + 3y = (e~2t + 3e6t) + 3(-e'2* + oe6t) = -2e"2* + 18e6t = ^ Jub and 5:r + 3y = 5(e~2* + 3eet) + 3(-e~2* + 5e6') = 2e~2t + 30e6* = ^ . at 38. From x = cos 21 + sin 21 + and y — — cos 21 — sin 21 — we obtain — = —2 sin 2t -f 2 cos 22 + and ^ = 2 sin 22 — 2 cos 2t — -e* d.t 5 dt 5 and d2:r , „ . . ^ 1 ^ , ^2V , 1 / = —4 cos 2t — 4 sm 22 + re and -r-^- = 4 cos 2t + 4 sin 22-- e . dt2 Id d22 5 Then 1 1 cPx 4 y + et = 4(— cos 21 — sin 21 — pef) + el — —4 cos 21 — 4 sin 22 + -el = -7-^ 0 o dt and 5 Exercises 1.1 Definitions and Terminology 4x — ef = 4(cos 21 + sin 21 -I- ^e*) — e* = 4 cos 2£ + 4 sin 2t — ef — 39. (t/ )2 + 1 = 0 has no real solutions becausc {y')2 + 1 is positive for all functions y = 4(x). 40. The only solution of (?/)2 + y2 = 0 is y = 0, since if y ^ 0, y2 0 and (i/ ) 2 + y2 y2 0. 41. The first derivative of f(x) = ex is eT. The first derivative of f{x) = ekx is kekx. The differential equations are y' — y and y' = k.y, respectively. 42. Any function of the form y = cex or y = ce~x is its own sccond derivative. The corresponding differential equation is y" — y = 0. Functions of the form y = c sin x or y — c cos x have sccond derivatives that are the negatives of themselves. The differential equation is y" -+- y = 0. 43. We first note that yjl — y2 = /l — sin2 x = Vcos2 x = | cos.-r|. This prompts us to consider values of x for which cos x 0, such as x = tt. In this case % dx i {sklx) = c o s x l^ , . = COS7T = —1. X=7T but /l - y2x=7r = V1 - sin2 7r = vT = 1. Thus, y = sin re will only be a solution of y' - y l — y2 when cos x 0. An interval of definition is then (—tt/2, tt/2). Other intervals are (3tt/2, 5tt/2), (77t/2, 9tt/2). and so on. 44. Since the first and second derivatives of sint and cos t involve sint and cos t, it is plausible that a linear combination of these functions, Asint+B cos t. could be a solution of the differential equation. Using y' — A cos t — B sin t and y" = —A sin t — B cos t and substituting into the differential equation we get y" + 2y' + 4y = —A sin t — B cos t + 2A cos t — 2B sin t + 4A sin t + 4B cos t = (3A — 2B) sin t + (2A + 3B) cos t = 5 sin t. + TT7« ^ITirl A -- --- 13 Thus 3A — 2B = 5 and 2A + 3B = 0. Solving these simultaneous equations we find A = j# and B = — . A particular solution is y = sint — ^ cost. 45. One solution is given by the upper portion of the graph with domain approximately (0,2.6). The other solution is given by the lower portion of the graph, also with domain approximately (0.2.6). 46. One solution, with domain approximately (—oo, 1.6) is the portion of the graph in the second quadrant together with the lower part of the graph in the first quadrant. A second solution, with domain approximately (0,1.6) is the upper part of the graph in the first quadrant. The third solution, with domain (0, oo), is the part of the graph in the fourth quadrant. 6 Exercises 1.1 Definitions and Terminology 47. Differentiating (V1 + y^)/xy = 3c we obtain xy(3x2 + 3y2y') - (a?3 + y*)(xi/ + y) x?y2 = 0 3 x3y + 3 xy^y' — x'^y' — x% — xy^y’ — yA — 0 (3:ry3 - xA - xyz}i/ = -3x3y + xi y + y4 , = y4 - 2x3y _ y(y[i - 2x3) ^ 3 — x4 rt:(2y3 — a:3) 48. A tangent line will be vertical where y' is undefined, or in this case, where :r(2y3 — x3) = 0. This gives x = 0 and 2y3 = a:3. Substituting y?J — a;3/2 into ;r3 + y3 = 3xy we get x3+ h 3 = 3x { w x) -x3 = — r2 2 2V3a a:3 = 22/ V z2(.x - 22/3) = 0. Thus, there are vertical tangent lines at x = 0 and x = 22/3, or at (0,0) and (22/3,21'/3). Since 22/3 ~ 1.59. the estimates of the domains in Problem 46 were close. 49. The derivatives of the functions are ^(.x) — —xf a/25 — x2 and ^{x) = x//25 — x2, neither of which is defined at x = ±5. 50. To determine if a solution curve passes through (0,3) we let 2 = 0 and P = 3 in the equation P = c-ie1/ (1 + eye*). This gives 3 = c j/ ( l + ci) or c = — | . Thus, the solution curve (—3/2)e* = —3e* 1 - (3/2)eL 2 - 3e{ passes through the point, (0,3). Similarly, letting 2 = 0 and P = 1 in the equation for the oneparameter family of solutions gives 1 = ct/(l + ci) or ci = 1 + c-|. Since this equation has no solution, no solution curve passes through (0. 1). 51. For the first-order differential equation integrate f(x). For the second-order differential equation integrate twice. In the latter case we get y = f ( f f(x)dx)dx + cja: + C2 - 52. Solving for y’ using the quadratic formula we obtain the two differential equations y = — ^2 + 2J 1 + 3ar®^ and y1 = — ^2 — 2y 1 4-3a?^^ , so the differential equation cannot be put in the form dy/dx = f(x,y). 7 Exercises 1.1 Definitions and Terminology 53. The differential equation yy'—xy = 0 has normal form dy/dx = x. These are not equivalent because y = 0 is a solution of the first differential equation but not a solution of the second. 54. Differentiating we get y' = c + 3c2%2 and y" = 602x. Then C2 - y"/(x and ~ 1/ — xy"f 2, so v=iy'-^-)x+{t)x3=xy'-rv and the differential equation is x2y" — 3xy' + Sy = 0. 2 55. (a) Since e~x is positive for all values of x. dy/dx 0 for all x, and a sohition. y(x), of the differential equation must be increasing on any interval. (b) lim ^ = lim e~x‘ = 0 and lim ^ = lim e~x = 0. Since dy/dx approaches 0 as x v ' x^-cc dx x-+-x dx approaches —oc and oc, the solution curve has horizontal asymptotes to the left and to the right. (c) To test concavity we consider the second derivative d2y d (dy d { * _ 2 dr.)-dxe' Since the sccond derivative is positive for x 0 and negative for x 0, the solution curve is concave up on (—00.0) and concave down 011 (0. 00). x 56. (a) The derivative of a constant solution y — c is 0, so solving 5 — c = 0 we see that, c — 5 and so y = 5 is a constant sohition. (b) A solution is increasing where dyjdx = 5 — y 0 or y 5. A solution is decreasing where dy/dx = 5 — y 0 or y 5. 57. (a) The derivative of a constant solution is 0, so solving y(a — by) = 0 we see that y = 0 and y = a/b are constant solutions. (b) A solution is increasing where dy/dx = y(a — by) = by(a/b — y) 0 or 0 y a/b. A solution is decreasing where dy/dx = by(a/b — y) 0 or y 0 or y a/b. (c) Using implicit differentiation we compute = y(-by') + y'{a - by) = y'(a - 2by). Solving d2y/dx2 = 0 we obtain y = a/2b. Since dly/dx2 0 for 0 y a/2b and d2y/dx2 0 for a/26 y a/b, the graph of y = p(x) has a point of inflection at y = a/26. 8 Exercises 1.1 Definitions and Terminology (d) 58. (a) If y = c is a constant solution I lien y' = 0. but c2 + 4 is never 0 for any real value of c. (b) Since y* = y2 + 4 0 for all x where a solution y = o(x) is defined, any solution must be increasing on any interval on which it is defined. Thus it cannot have any relative extrema. 59. In Mathematica use Clear [y] y[x_]:= x Exp[5x] Cos[2x] y[xl y""[x] — 20y’ "[x] + 158y"[x] — 580y'[x] +84 ly[x]//Simplify The output will show y{x) = e0Xx cos 2x. which verifies that the corrcct function was entered, and 0, which verifies that this function is a solution of the differential equation. 60. In Mathematica use Clear [y] y[x_]:= 20Cos[5Log[x]]/x — 3Sin[5Log[x]]/x y(x3 x~3 y'"[x] + 2x~2 y"[x] + 20x y'[x] — 78y[x]//Simplify The output will show y(x) = 20 cos(o In x)/x—Z sin(5 In x)/x. which verifies that the correct function was entered, and 0, which verifies that this function is a solution of the differential equation. x 9 Exercises 1.2 Initial-Value Problems . problems 1. Solving —1/3 = 1/(1 + ci) we get c — —4. The solution is y = 1/(1 — 4e~x). 2. Solving 2 = 1/(1 + ce) we get c = - ( l /2)e_1. The solution is y — 2/(2 - e " ^ -1)) . 3. Letting x = 2 and solving 1/3 = 1/(4 + c) we get c = —1. The solution is y — 1 / (x2 — 1). This solution is defined on the interval (l,oc). 4. Letting x = -2 and solving 1/2 = 1/(4 + c) we get c = —2. The solution is y = 1/(.:;;2 — 2). This solution is defined on the interval (—oo, —y/2). 5. Letting x = 0 and solving 1 = 1/e we get c — 1. The solution is y = l/(a;2 + 1). This solution is defined on the interval (—oo, oo). 6. Lotting x = 1/2 and solving —4 = l/ ( l/4 + c) we get c = —1/2. The solution is y = lj(x 2 — 1/2) = 2/(2x2 — 1). This solution is defined on the interval (—l/y/2 , l//2 )- In Problems 7-10 we use x — c cos t + 0 2 sin t and x' — —c sin t + C2 cos t to obtain a system, of two equations in the two unknowns ei and C2 ■ 7. From the initial conditions we obtain the system C2 = 8. The solution of the initial-value problem is x = — cost + 8sint. 8. From the initial conditions we obtain the system 0 2 = 0 -ci - 1. The solution of the initial-value problem is x = — cos t. 9. From the initial conditions we obtain /3 1 I -7 T C] + - C2 = - VS Solving, we find c = V3/4 and C2 = 1/4. The solution of the initial-value problem is x = (a/3/4) cost + (1/4) sint 10 Exercises 1.2 Initial-Value Problems 10. From the initial conditions we obtain /2 /2 r - T Q T 2 = / 2 2~ ~2~ = Solving, we find ci = — 1 and c2 = 3. The solution of the initial-value problem is x = — cost+3 sin t. Problems 11-14 we use y = cax + C2 e~x and if — ce£ — C2 e~x to obtain a system of two equations the two unknowns c and 0 9 - 11. From the initial conditions we obtain Ci + C2 = 1 ci - c2 = 2. Solving, wo find c = ^ and C2 = — 5 . The solution of the initial-value problem is y = |ex — ^e~x. 12. From the initial conditions we obtain ec + e-1C2 = 0 ec — e~lC2 = e. Solving, we find ci = and C2 = — ^e2. The solution of the initial-value problem is :j = ex - e2e~x = ex - e2~x. 13. From the initial conditions we obtain e-1ci 4- ec2 = 5 — ec2 = —5. Solving, we find ci = 0 and C2 = 5e 1. The solution of the initial-value problem is y = -5e 1e x = - — 1 — t* Of - 14. From the initial conditions we obtain ci + C2 = 0 Cl - c2 = 0. Solving, we find ci = C2 = 0. The solution of the initial-value problem is y = 0. 15. Two solutions are y = 0 and y = x%i. I ' . Two solutions are y — 0 and y = x2. (Also, any constant multiple of x2 is a solution.) d f 2 / 1 For fix, y) = y2/3 we have ~ Thus, the differential equation will have a unique solution ay 3' any rectangular region of the plane where y ^ 0. 11 Exercises 1.2 Initial-Value Problems 18. For f'(x,y) = yjxy we have d f /dy - jx/y ■ Thus, the differential equation will have a unique solution in any region where x 0 and y 0 or where x 0 and y 0. ? d f l 19. For fix. y) = — we have = —. Thus, the differential equation will have a unique solution in x ay x any region where x ^ 0. 20. For f(x,y) = x + y we have = 1. Thus, the differential equation will have a unique solution in the entire plane. 21. For f(x, y) - x2/{& — y2) we have d f /dy — 2x‘1y/(A. — y2)2. Thus the differential equation will have a unique solution in any region where y —2, —2 y 2, or y 2. d f _3 x2y2 22. For f(x. y) ~ —-—* we have -1- *- ----- h- . Thus, the differential equation will have a unique V J 1 + y3 dy (l + y3)2 solution in any region where y ^ — 1. y2 d f 2x^y 23. For f(x, y) = —tt--rr we have — = ---- :—k . Thus, the differential equation will have a unique M x2 + y2 dy (x2 + y2)2 solution in any region not containing (0,0). 24. For / (x, y) = (y + x)/(y — x) we have df/dy = —2x/(y — x)2. Thus the differential equation will have a unique solution in any region where y x or where y x. In Problems 25-28 we identify f{x,y) = jy2 — 9 and df/dy = y/jy2 — 9. We see that f and df/dy are both continuous in the regions of the plane determined by y — 3 and y 3 with no restrictions on x. 25. Since 4 3, (1,4) is in the region defined by y 3 and the differential equation has a unique solution through (1,4). 26. Since (5,3) is not in cither of the regions defined by y —3 or y there is no guarantee of a unique solution through (5,3). 27. Since (2, —3) is not in either of the regions defined by y — 3 or y 3, there is no guarantee of a unique solution through (2, —3). 28. Since (—1,1) is not in either of the regions defined by y —3 or y 3, there is no guarantee of a unique solution through (—1, 1). 29. (a) A one-parameter family of solutions is y = ex. Since y' = c, xy' = xc = y and y(0) = c ■ 0 = 0. (b) Writing the equation in the form y' — y/x, we see that R cannot contain any point on the y-axis. Thus, any rectangular region disjoint from the y-axis and containing (xq, ijq) will determine an 12 Exercises 1.2 Initial-Value Problems interval around xg and a unique solution through (so- yo). Since ;i’o = 0 in part (a), we are not guaranteed a unique solution through (0.0). (c) The piecewise-defined function which satisfies y(0) = 0 is not a solution sincc it is not differentiable at x - 0. (I 9 9 30. (a) Since — tan (a; + c) = sec-(a: + c) = 1 + tan"(x -i- c), we see that y = tan(x + c) satisfies the LlX differential equation. (b) Solving y(0) = tan : — 0 we obtain c = 0 and y = tan x. Since tan:r is discontinuous at x — ±7t/2; the solution is not defined on (—2,2) because it contains ±tt/2. (c) The largest interval on which the solution can exist is (—tt/2, 7t/2). d 1 1 1 31. (a) Since — (-----) = 7---- = i f . we see that y = ------- is a solution of the differential v ' dx ^ rx. A+- ryJ ((xr . A+- c)"‘ X - h C equation. (b) Solving y(0) = —1 jc = 1 we obtain c = — 1 and y — 1/(1 — x). Solving y(0) = —1/c = — 1 wc obtain c — 1 and y = —1/(1 + x-). Being sure to includc x = 0, we see that the interval of existence of y — 1/(1 — x) is (—oc, 1), while the interval of existence of y = —1/(1 + x) is ( 1, oc). (c) By inspection we see that y = 0 is a solution 011 (—00, 00). 32. (a) Applying y(l) = 1 to y = —l/(x -f c) gives 1 1 = 1 + c or 1 + c = —1. Thus c = —2 and 1 y x — 2 2 — x (b) Applying y(3) = — 1 to y = — 1/ (;r + c) gives 1 -1 = Thus c = —2 and y or 3 + c = 1. x — 2 2 — x (c) No, they are not the same solution. The interval I of definition for the solution in part (a) is (—00, 2); whereas the interval I of definition for the solution in part (b) is (2. 00). See the figure. , 1 } ; / (1,1) (3, -1) --- 4 ‘ 13 Exercises 1.2 Initial-Value Problems 33. (a) Differentiating 3x2 — y2 = c we get 6x — 2yyf = 0 or yy' = 3a;. (b) Solving 3a:2 — y2 = 3 for y we get y = ©1 (a:) = /3{x2 - 1). y = h(x) = - ‘3(x- - 1) y = M x) = 1 a; oo, 1 x oo. —oc x —1. y = ?4 (x) = ~j3{x2 - 1), —oo x —1. (c) Only y = pz{x) satisfies y{—2) = 3. 34. (a) Setting x = 2 and y = —4 in 3;r2 — y2 = c we get 12 — 16 = —4 = c, so the explicit solution is y = —y 3a:2 + 4, —oo x oo. (b) Setting c. = 0 we have y = /3a: and y = — /3^, both defined on (—oo;oc). y i 4: Jn Problems 35-38 we consider the points on the graphs with x-coordinates xq = —1. xq = 0, a ;Z*0 = 1. 77?,e slopes of the tangent lines at these points are compared with the slopes