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

352 D CHAPTER 17 SECOND-ORDER DIFFERENTIAL EQUATIONS
Q' (t) = I (t) = e-lot[( -10c1 + 20c2) cos 20t + ( - lOc2- 20ct) sin 20t) but 0 = Q'(O) = - 10c1 + 20c~. Thus the charge
is Q(t) = - 2 ~ 0 e- tot(6 cos 20t + 3 sin 20t) + 1 ~ 5
and the current is J(t) = e-lOt (~)sin 20t.
15. As in Exercise 13, Qc(t) = e-tot(ci. cos 20t + c2 sin20t) but E(t) = 12sin lOt so try
Qp(t) = A cos lOt+ B sin lOt. Substituting into the differential equation gives
(-lOOA + 200B + 500A)cos lOt+ (-lOOB- 200A + 500B) sin lOt = ·12sin lOt =>
400A + 200B = 0 and 400B - 200A = 12. Thus A= - 2 ~ 0 , B = 1 ~ 5 and the general solution is
Q(t) = e- lOt(c1 cos 20t + c2 sin 20t) - 2 ~ 0 cos lOt+ 1 ~ 5 sin lOt. But 0 = Q(O) = c1 - 2 ~ 0 so c1 = 2~ 0 .
Also Q' (t) = fs sin lOt+ 2
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cos lOt+ e-lOt[( - lOc1 + 20c2) cos 20t + ( -lOcz - 20ct) sin 20t] and
0 = Q' (0) = fg - lOc1 + 20c2 so c2 = - 5 ~ 0 . Hence the charge is given by
Q(t) = e-wt ( 2 ~ 0 cos 20t- 5 ~ 0 sin 20t] -
2 ~ 0 cos lOt + 1 ~ 5 sin lOt.
11. x(t)=Acos(wt +8) ¢:? x(t)=A[coswtcos8-sinwtsin5] ¢:? x(t)=A(~coswt+~sinwt)where
cos 8 = c1 / A and sin 8 = -c2/ A ¢:? x(t) = c1 cos wt + c2 sin wt. [Note that cos 2 8 + sin 2 8 = 1 => c~ + c~ = A 2 .)
17.4 Series Solutions
00 00
1. Let y(x) = 2: C,:.x" . Then y'(x) = 2: nenx"- 1 and the given equation, y' - y = 0, becomes
n=O ·
n=l
00 00 00 00
2: nenxn-l- 2: CnX" = 0. Replacing n by n + 1 in the first sum gives 2: (n + l)cn+tX"- L enx" = 0, so
n=l n=O n=O n=O
00
2: [(n + l)Cn+I- en]x" = 0. Equating coefficients gives (n + l )cn+l - en= 0, so the recursion relation is
n=O
oo oo · oon
in general, en= ~.Thus, the solution is y(x) = 2: cnx'' = 2: c~ x" = eo 2: ; = eoe"' .
n. n=O n=O n. n=O n.
00 00 00
3. Assuming y(x) = 2: enx", we have y'(x) = 2: ncnx"- 1 = 2: (n + l)Cn+tx" and
n = O n =l n=O
~ 00 00 00
-x 2 y =- I:: Cnx"+ 2 =- I:: Cn- 2x". Hence, the equation y' = x 2 y becomes I:: (n + l )cn+tx"- 2: Cn- 2x" = 0
n=O n=2 n=O n=2
or Ct + 2c2x + f: [(n + l)cn+I - Cn-2) x" = 0. Equating coefficients gives Ct = c2 = 0 and Cn+t = Cn- 2 1
n=2 n+
for n = 2, 3, ... . But c1 = 0, so C