Solução_Calculo_Stewart_6e

F.
TX.10
SECTION 18.4 SERIES SOLUTIONS ET SECTION 17.4 ¤ 317
Since ∞ n(n +1)c n+1 x n
= ∞ n(n +1)c n+1 x n , the differential equation becomes
n=1
n=0
∞
n(n +1)c n+1 x n
− ∞ (n +2)(n +1)c n+2 x n
+ ∞ (n +1)c n+1 x n =0 ⇒
n=0
n=0
n=0
∞
[n(n +1)c n+1 − (n +2)(n +1)c n+2 +(n +1)c n+1]x n =0 or
n=0
∞
[(n +1) 2 c n+1 − (n +2)(n +1)c n+2]x n =0.
Equating coefficients gives (n +1) 2 c n+1 − (n +2)(n +1)c n+2 =0for n =0, 1, 2, .... Then the recursion relation is
c n+2 =
(n +1) 2
n +1
cn+1 =
(n +2)(n +1) n +2 cn+1,sogivenc0 and c1,wehavec2 = 1 c1, 2 c3 = 2 3 c2 = 1 c1, 3 c4 = 3 4 c3 = 1 c1,and
4
in general c n = c 1
n , n =1, 2, 3, .... Thus the solution is y(x) =c ∞
0 + c 1
c 0 − c 1 ln(1 − x) for |x| < 1.
9. Let y(x) = ∞
y 00 (x) =
becomes
c n+2 =
y(0) = ∞
∞
n =0
∞
n =0
n =0
c n x n .Then−xy 0 (x) =−x ∞
n =1
nc n x n−1 = − ∞
n =1
n=0
(n +2)(n +1)c n+2x n , and the equation y 00 − xy 0 − y =0
n=1
x n
. Note that the solution can be expressed as
n
nc n x n = − ∞
n =0
[(n +2)(n +1)c n+2 − nc n − c n ]x n =0. Thus, the recursion relation is
nc n x n ,
nc n + c n
(n +2)(n +1) = c n(n +1)
(n +2)(n +1) = c n
for n =0, 1, 2, .... One of the given conditions is y(0) = 1. But
n +2
n=0
c n(0) n = c 0 +0+0+···= c 0,soc 0 =1. Hence, c 2 = c 0
2 = 1 2 , c4 = c 2
4 = 1
2 · 4 , c6 = c 4
6 = 1
2 · 4 · 6 , ...,
c 2n = 1
2 n n! . The other given condition is y0 (0) = 0. Buty 0 (0) = ∞
By the recursion relation, c 3 = c 1
3
problem is y(x) =
∞
n =0
11. Assuming that y(x) = ∞
y 00 (x)=
c n x n =
n =0
∞
n =2
n=1
nc n (0) n−1 = c 1 +0+0+··· = c 1 ,soc 1 =0.
=0, c5 =0, ..., c2n+1 =0for n =0, 1, 2, .... Thus, the solution to the initial-value
∞
n =0
c 2n x 2n =
∞
n =0
c nx n ,wehavexy = x ∞
n(n − 1)c n x n−2 =
x 2n
2 n n! = ∞
n =0
∞
n=−1
c nx n =
=2c 2 + ∞ (n +3)(n +2)c n+3 x n+1 ,
n =0
and the equation y 00 + x 2 y 0 + xy =0becomes 2c 2 + ∞
recursion relation is c n+3 =
n =0
(x 2 /2) n
= e x2 /2 .
n!
∞
n =0
c nx n+1 , x 2 y 0 = x 2
∞
n =1
nc nx n−1 =
(n +3)(n +2)c n+3 x n+1 [replace n with n +3]
n =0
∞
n =0
nc nx n+1 ,
[(n +3)(n +2)c n+3 + nc n + c n] x n+1 =0.Soc 2 =0and the
−nc n − c n (n +1)cn
= − , n =0, 1, 2, ....Butc0 = y(0) = 0 = c2 and by the
(n +3)(n +2) (n +3)(n +2)
recursion relation, c 3n = c 3n+2 =0for n =0, 1, 2, ....Also,c 1 = y 0 (0) = 1, soc 4 = − 2c 1
4 · 3 = − 2
4 · 3 ,
c 7 = − 5c4
2 · 5
2 2 5 2
7 · 6 =(−1)2 7 · 6 · 4 · 3 =(−1)2 , ..., c 3n+1 =(−1) n 2 2 5 2 ·····(3n − 1) 2
. Thus, the solution is
7!
(3n +1)!
y(x) =
∞ c n x n
= x +
(−1) ∞ n 2 2 5 2 ·····(3n − 1) 2 x 3n+1
.
(3n +1)!
n =0
n =1