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)

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 .