TEST BANK FOR A First Course in Differential Equations with Modeling Applications 9th, Differential

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 {x\x ^ —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 .