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

SECTION 17.4 SERIES SOLUTIONS 0 353
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5. Let y (x) = I: c.,x" '* y' (x) = I: nenx"- 1 and y" (x) = I: (n + 2){n + l)cn+2X". The differential equation
n = O n=l n =O
00 . 00 00 00
becomes L: (n + 2){n + 1)cn+2x" + x I: nc.,.x"- 1 + L.: CnX 11 = 0 or L.: [(n + 2){n + 1)cn+2 + nen + en]x" = 0
n=O n=l n = O n=O
[since n~1 nenx" = n~O ncnx"] . Equating coefficients gives (n + 2)(n + 1)Cn+2 + (n + 1)Cn = 0, thus the
. I . . - (n + l )c,. Cn 0 1 2 Th th
recurs1onreat1ontscn+2= (n+ 2
)(n+l) =--n-+- 2
,n=,, , .... en eeven
coe ffi ctents . are gtven . b y c2 = - Co C2 co C4 Co d . 1
2 , C4 = - 4 = 2 . , co= -6 =-~,an m genera,
4
n Co (- 1)"Co . C1 C3 C1 C5 C1
c2n = (-1)
=
• Theoddcoeffi.ctents are C3 = - - , cs = - - = - ,
2 2 1
c7 = - - = - --- ,
2 · 4 · · · · · n n n. 3 5 3 · 5 7 3 · 5 · 7
d
. I ( l )n Cl . (-2)nn!ct Th I . .
an m genera, C2n+l = -
(Z ) = (
3 5 7
2
l )l . e so utton IS
· · ····· n+ 1 n+ .
00 (-1)" ~
00
{-2)n n l
1 {x) = c 0 "' - - x-" + c 1 "' x 2 "+
y
1
n=O L- 2 11 n.
'
n L- =O {2 n + 1)1 ·
00 00 00 00
7. Let y (x) = L.: CnX" :::} y' (x) = I.: ncnx"- 1 = I.: (n + l)Cn+lx".and y" (x) = L.: (n + 2){n + l )c,+2xn. Then
n=O n=l n=O n=O
00 00 00 00
(x-1)y"(x) = L (n+2){n+l )Cn+2Xn+l _ L.: (n+2)(n+l)cn+2X" = L n(n+1)Cn+1x"- I.: (n+2)(n+l )cr•+2x".
n = O n=O n =l n =O
DO 00
Since L.: n(n + 1)cn+1x" = L.: n(n + 1)Cn+J.'t", the differential equation becomes
n = l
n=O
DO 00 00
L n(n + l)Cn+lX"- L.: (n + 2)(n + 1)Cn+2X" + 2: (n + 1)Cn+!X" = 0 :::}
n=O n=O n =D
00 00
L: [n(n + 1)Cn+l - (n + 2)(n + l )Cn+2 + (n + 1)cn+1]x" = 0 or 2: [(n + 1) 2 cn+I - (n + 2)(n + 1)Cn+2]x" = 0.
u=O
Equating coefficients gives (n + 1) 2 c,,+I - (n + 2)(n + 1)cn+2 = 0 for n = 0, 1, 2, .... Then the recursion relation is
n=O
in general en = Ct , n = 1, 2, 3, .... Thus the solution is y(x) =Co + c1 f xn. Note that the solution can be expressed as
n
n =1 n
co- c1 ln{l - x) for JxJ < 1.
00 00
9. Let y(x) = 2::.: c .. x".-Then -xy'(x) = - x L.: nc,.xn-I
~ ~
2: nc.,x" =- L.: ncnx",
n=O n=l n= l n =O
~
y"(x) = L.: (n + 2){n + l )cn+2X" , and the equation y"- xy' - y = 0 becomes
n = O
00
L: [(n + 2)(n + 1)Cn+2- nen - en]x" = 0. Thus, the recursio11 relation is
n = O
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