Exercicios resolvidos James Stewart vol. 2 7ª ed - ingles

348 0 CHAPTER 17 SECOND-ORDER DIFFERENTIAL EQUATIONS ,
7. The a~xi liary equation is r 2 + 1 = 0 with roots 1' = ±i, so the complementary solution is Yc(x) = C1 cosx + c 2 sinx.
For y"-+: y = e"' try yp 1 (x) = Ae"'. Then y~ 1 = y; 1
= Ae"' and substitution gives Ae"' +'Ae"' = e"' => A=~.
so yp 1 (x) = ~ e"'. For y" + y = x 3 try y 112 (x) = Ax 3 + Bx 2 + Cx + D. Then y~2 = 3Ax 2 + 2Bx + f:! and
y; 2
= 6Ax + 2B. Substituting, we have 6Ax + 2B + Ax 3 + Bx 2 + Cx + D = x 3 , so A = 1, B = 0,
6A + C.= 0 => C = - 6, and 2B + D = 0 => D = 0. Thus Yv 2 ( x) = x 3 - 6x and the general solution is
y(x) = Yc(x) + Yp 1 (x) + YP 2 (x) = c1 cosx + C2 sinx + ~e"' + x 3 - 6x. But 2 = y(O) = c1 + 4 =>
c1 = ~ and 0 = y' (0) = c2 + ~ - 6 => c2 .= ¥. Th11s the solution to the initial-value problem is
y(x) = ~ cosx + ¥ sinx + %e"' + x 3 - 6x.
9. The auxiliary equation is r 2 - r = 0 with roots r = 0, r = 1 so the complementary solution is Yc(x) = c 1 + c2 e"'.
Try yp(x) = x(Ax + B)e"' so that no term in YP is a solution of the complementary equation. Then
y~ = (Ax 2 + (2A + B)x + B)e"' andy;= (Ax 2 + (4A + B )x + (2A + 2B))e"'. Substitution into the differential equation
gives (Ax 2 + (4A + B)x + (2A + 2B))e"'- (Ax 2 + (2A + B )x + B)e"' = xe"' => (2Ax + (2A + B))c"' = xe"' =>
A =~. B = -1. Thus yp(x) = (!x 2 -
x)e"' and ~he general solution is y(x) = c1 + c2e"' + Gx 2 - x)e"'. But
2 = y(O) = c 1 + c 2 and 1 = y' (0) = c2 - 1, so c2 = 2 and c1 = 0. The solution to the initial-value problem is
y(x) = 2e"' + (~x 2 - x)e"' = e"'(!x 2 - x + 2)'.
11. The auxiliary equation is r 2 + 3r + 2 = (r + 1)(r + 2) = 0, so r = - 1, r = - 2 and Yc(x) = c1e- "' + c2e- 2 "'.
Try YP = A cos x + B sin x => y~ = - A sin x + B cos x, y; = - A cos x - B sin x . Substituting into the differential
equation gives (- A cosx - B sin x) + 3( - A sinx + B cos x) + 2(Acosx + B sinx) = cosx or
(A + 3B) cos x + ( -3A + B) sin x = cos x. Then solving the equations
A + 3B ·= 1, - 3A + B = 0 gives A = 1
10
, B = fo and the general
solution is y(x ) = c1e-"' -j- c2e- 2 "' + 1
10
cos x + fo sinx. The graph
shows yp and several other solutions. Notice that all solutions are
asymptotic to YP as x -+ oo. Except for y 11 , all solutions approach either oo
or -oo as x -+ - oo.
-3 ~~~7-?~~~~~ s
Yp
- 3
13. Here Yc(x) = cre 2 "' + c2e-"', and a trial solution is 1h•(x) = (Ax+ B)e"' cosx + (Cx + D )e"' sin x.
15. Here Yc(x) = c1e 2 "' + c2e"'. For y"- 3y' + 2y = e"' try y 1 , 1
(x) = Axe"' (since y = Ae"' is a solution of the complementary
equation) and fo ~ i/' - 3y' + 2y = sin x try yp 2 (x) = B cos x + C sin x. Thus a trial solution is
Yp (x) = Yv 1 (x) + Yp 2 (x) =Axe"' + B cosx + Csinx.
17. Since Yc(x) = e- x (c1 cos 3x + c2 sin 3x) we try Yv(x) = x(Ax 2 + Bx +C) e- x cos 3x + x(Dx 2 +Ex+ F) e-x sin 3x
(so that no term ofyp is a solution of the complementary equation).
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