MATH 202 Useful File Solution Manual, Spring 2011

COMPLETE SOLUTIONS MANUAL FOR: — ZILL’S A FIRST COURSE IN DIFFERENTIAL EQUATIONS WITH MODELING APPLICATIONS 7TH EDITION AND ZILL & CULLEN’S DIFFERENTIAL EQUATIONS WITH BOUNDARY-VALUE PROBLEMS STH EDITION BROOKS/COLE S| ete THOMSON LEARNING Australia * Canada * Mexico * Singapore + Spain + United Kingdom + United States Table of Contents an VW eF WO N wo on 10 11 12 13 14 15 Appendix I Appendix II Introduction to Differential Equations First-Order Differential Equations Modeling with First-Order Differential Equations Higher-Order Differential Equations Modeling with Higher-Order Differential Equations Series Solutions of Linear Equations The Laplace Transform Systems of Linear First-Order Differential Equations Numerical Solutions of Ordinary Differential Equations Plane Autonomous Systems and Stability Orthogonal Functions and Fourier Series Partial Differential Equations and Boundary-Value Problems in Rectangular Coordinates Boundary-Value Problems in Other Coordinate Systems Integral Transform Method Numerical Solutions of Partial Differential Equations Gamma function Introduction to Matrices 22 71 104 194 240 308 370 430 458 491 538 616 654 695 717 718 1 Introduction to Differential Equations 10. 11. 12. 13. 14. Exercises 1.1 . Second-order; linear. . Third-order; nonlinear because of (dy/dz)*. . The differential equation is first-order. Writing it in the form 2(dy/dx) + y? = 1, we see that it is nonlinear in y because of y*. However, writing it in the form (y? — 1)(dz/dy) + 2 = 0, we see that it is linear in z. The differential equation is first-order. Writing it in the form u(dv/du) + (1+ u)v = ue” we see that it is linear in v. However, writing it in the form (v + uv — ue“)(du/dv) + u = 0, we see that it is nonlinear in u. . Fourth-order; linear . Second-order; nonlinear because of cos(r + u) Second-order; nonlinear because of ,/1 + (dy/dz)? . Second-order; nonlinear because of 1/R? . Third-order; linear Second-order; nonlinear because of <* From y = e~*/? we obtain y! = —heot/?, Then 2y’+y= —eWt/2 4 eH 2/2 = 0), From y = £ — $670 we obtain dy/dt = 24e~?°", so that dy 208 (2 6 “2 =+ = + 20{--= = 24. pt 20y = 24e 7 + 20 (= — ze From y = e** cos 2x we obtain y! = 3e%* cos 2x — 2e2* sin 2x and y” = 5e*" cos 2z — 12e3” sin 2z, so that y” — 6y’ + 13y =0. From y = — cos zIn(secz + tanz) we obtain y’ = —1 + sinx|n(secz + tanz) and y! = tanz+coszIn(secr+tanz). Then y” +y = tang. 1 Exercises 1.1 15. Writing In(2X ~ 1) — In(X — 1) = ¢ and differentiating implicitly ~ x we obtain 2 aX 1 dX _, ‘ 2X-1 dt X-1 dt 2 2 1 ) ax _, ~-- +e Comer “dt —4 2 N24 iF \ 2X -2~-2X+1dX _ -2 (2X —1)(X-1) dt — | -4 Xx | af = -(2X ~1(X ~1) =(X 1) - 2X), Exponentiating both sides of the implicit solution we obtain xX — fay 2X) Lt <y ox 1 = Xel—e! => (e—1) =(e'-2)X x= S, X-1 et -2 Solving e’ — 2 = 0 we get t = In2. Thus, the solution is defined on (—oo, 1n2) or on (In2,00). The graph of the solution defined on (—oo, In 2) is dashed, and the graph of the solution defined on (In 2, co) is solid. 16. Implicitly differentiating the solution we obtain y d d: — 22? — day + 2y 4 =0 = -2° dy -2cydz+ydy =0 4 => 2xydz + (x? — y)dy = 0. 2 Using the quadratic formula to solve y?—2x2y—1 = 0 for y, we get y = (22? + Var? +4)/2 = 2? + Vet +1. Thus, two explicit _ solutions are yy = 22 + V2t +71 and y = x? — Vz? +1. Both ° solutions are defined on (—oo, 00). The graph of y;(z) is solid and ~4 the graph of yg is dashed. 17. Differentiating P = cye’/ (a + ce’) we obtain dP (1 + ce’) ce’ — cyet - cyet dt (1 +cyet)” cyet [(1 + ce’) - ce] ~ 1+cyet 1l+cyet = P(1- P). 2 fe 42 2 18. Differentiating y =e I eb dt+cje-* we obtain 2 / —2" 2 a2? [* e ~2? ~2? = 2 y =e *e* —2zre f eat —2c126 =1-2ze » ° dt — 2c\xe~” . 2 19. 20. 21. 22. 23. Exercises 1.1 Substituting into the differential equation, we have © 2st 42 2 y +22y=1- aze~*" [ edt — 2eyne-?" + 2re7* | edt +2cre7*_ = 1. 0 = ert Qe . Qa Qe dy — (Ap Qe Qe From y = c)e** + coze** we obtain a (2c) +c2)e** +2core~* and ie (4c1 +4c2)e** +4coxe™*, so that dy dy Qe on aaa 7 45m + 4y = (41 + 42 — 8¢1 — 4c + 4c )e* + (4c2 — 8c2 + 4c9)re* = 0. dz dx From y = cjz7! + cox + c3z nz + 42? we obtain d YX -qa? +co+c3 +c3lnzr + 8g, dz a a = 2e,273 + c3z7} +8, and fia pad =-6cjr * — cx”, so that , , d°y d*y dy 3 2 q <f 429? Fg Sy Pa a a TY = (—6c + 4c) + cy +c )a7! + (—c3 + 203 — co — 03 + co) 0 +(—c3 +c3)elnz+(16-8+ 4)a? = 1227. (a) From ¢) = 2? we obtain ¢} = 2z, so 1 — 2g) = x(2x) — 22? = 0. From ¢) = —2? we obtain $4 = —2z, so xb — 27 = 2(—2x) — 2(—2?) = 0. Thus, ¢; and @» are solutions of the differential equation on (—00, 00). 2 -2", 2£<0 -2r, £<0 b) Fr - ’ " in y= , that ry’ — 2y = 0. (b) From y {2 2>0 we obtain y/ { 2>0 so that ry’ — 2y The function y(x) is not continuous at x = 0 since lim y(x) = 5 and lim, y(xz) = —5. Thus, y/(x) 2-07 rr does not exist at x = 0. (a) The domain of the function, found by solving z + 3 > 0, is [—3, 00). 3 24. 25. 26. 27. Exercises 1.1 (b) From y!=1+(z+3)7'/? we have . (y—2)y -yt+2-2= [2+2Vr43—a)[14 (14 (x - 3) -[c+2Vr+3]+2-2 =2/Vr+34+2-2-2Vr4+34+2-2=0. Since y(zx) is not differentiable at x = —3, y is a solution of the differential equation on (—3, 00). (a) An interval on which tan 5t is continuous is —1/2 < 5t < m/2, so 5 tan 5¢ will be a solution on (—1/10, 7/10). (b) For (1—sint)~!/? to be continuous we must have 1—sint > 0 or sint < 1. Thus, (1—sint)7!/? will be a solution on (7/2, 57/2). (a) From y = e™ we obtain y’ = me™. Then y’ + 2y = 0 implies me™ +4 2 = (m+2)e™ = 0. ~2 is a solution. Since e™ > 0 for all t, m = —2. Thus y =e (b) From y = e™ we obtain y! = me™ and y” = m2e™. Then y” — 5y! + 6y = 0 implies me™ — 5me™ + 6e™ = (m — 2)(m—3)e™ =0. 3t are solutions. Since e > 0 for all t, m= 2 and m=3. Thus y = e* andy =e (a) From y = t™ we obtain y’ = mt™~! and y” = m(m — 1)t™~?. Then ty” + 2y’ = 0 implies tm(m —1)t™-? + 2met™! = [m(m — 1) + 2m]t™") = (Mm? + m)t™! = m(m+1)t™-! =0. Since t”-! > 0 fort > 0, m=0 and m= —1. Thus y = 1 and y = ¢7! are solutions. (b) From y = t” we obtain y! = mt™~! and y” = m(m — 1)t™-?. Then é?y” — 7ty! + l5y = 0 implies t?m(m — 1)t™-? — Teme! + 15t™ = [m(m — 1) — 7m + 15]t™ = (m? — 8m + 15)t™ = (m — 3)(m —5)t™ =0. Since t” > 0 for t > 0, m=3 and m=5. Thus y = # and y = @° are solutions. From x = e7 + 3e% and y = —e~* + 5e% we obtain dy dz ~2t 6t ! ~2t 6t — =~ _2 = a 2e~“* + 18e" and at 2e~~ + 30e"". Then x +3y = (e~* + 3e%) + 3(-e7* + 528) dz = —2e7! 6t 2e~™ + 18e Zi and 28. 29. 30. 31. 32. 33. 34. Exercises 1.1 5a + 3y = 5(e7*! + 3e%) + 3(-e7-% + 5e%) d = 27% + 30e% = 2, dt From x = cos 2t + sin 2t + de’ and y = —cos2t — sin 2t — fe! we obtain dz 1 d . 1 a —2sin 2t + 2cos 2t + ze and ra 2sin 2t — 2cos 2t — ze and 5 5 d-x 1 d* . 1 mo —4cos 2t — 4sin 2t+ ze and a = 4cos 2t + 4sin 2t — ze Then t_ _¢; li t 4y +e = 4(—cos 2t — sin 2t — 5e )+e 1 a = —4cos 2t — 4sin 2t + ze = and ; 4x —e' = 4(cos2t +sin 2¢ + ze) -eé 1 d? = 4cos 2t + 4sin 2t — ze = ae (y')? + 1 =0 has no real solution. The only solution of (y’)? + y? = 0 is y = 0, since if y #0, y2 > O and (y')? +y2 >y? >0. The first derivative of f(t) = et is et. The first derivative of f(t) = e* is ke. The differential equations are y' = y and y’ = ky, respectively. * is its own second derivative. The corresponding Any function of the form y = ce’ or y = ce7 differential equation is y’ — y = 0. Functions of the form y = csint or y = ccost have second derivatives that are the negatives of themselves. The differential equation is y” + y = 0. Since the nth derivative of 6(x) must exist if 6(x) is a solution of the nth order differential equation, all lower-order derivatives of ¢(x) must exist and be continuous. {Recall that a differentiable function is continuous.} Solving the system e1yi(0) + coyo(0) = 2 ery} (0) + coy9(0) = for cy and cp we get _ 2uh(0) wi(0)u8(0) — vj(0)u2(0) 2y;,(0) and = ~FOwhl0) — v4 (Ou0(0) c] 35. 36. 37. 38. 39. 40. 41. 42. Exercises 1.1 Thus, a particular solution is _ 2yh(0) 7 2y4 (0) b nil0yvh(0) — vi(Oyya0) 2 ~ yr (Oy (0) — viO)uaC0) where we assume that yi (0)y$(0) — y}(0)ye(0) 4 0. 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(t)dt)dt + cit +c. Solving for y’ using the quadratic formula we obtain the two differential equations v => (2+2yi+38) and y => (2-2V1+38), so the differential equation cannot be put in the form dy/dt = f(t, y). The differential equation yy’ — ty = 0 has normal form dy/dt = t. These are not equivalent because y = 0 is a solution of the first differential equation but not a solution of the second. Differentiating we get y’ = c) + 3cot” and y” = 6cgt. Then cp = y"/6t and c, = y’ — ty"/2, so , ty’ y" 3 J 1 2, = -—> —jt=ty--> y (v tle y aby and the differential equation is t?y” — 3ty’ + 3y = 0. When g(t) = 0, y = 0 is a solution of a linear equation. (a) Solving (10 — 5y)/3x = 0 we see that y = 2 is a constant solution. (b) Solving y* + 2y —3 = (y+3)(y—1) =0 we see that y = —3 and y = 1 are constant solutions. (c) Since 1/(y — 1) = 0 has no solutions, the differential equation has no constant solutions. (d) Setting y’ = 0 we have y” = 0 and 6y = 10. Thus y = 5/3 is a constant solution. 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). 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, 00), is the part of the graph in the fourth quadrant. 6