Solução_Calculo_Stewart_6e

F.
316 ¤ CHAPTER 18 SECOND-ORDER DIFFERENTIAL EQUATIONS ETX.10
CHAPTER 17
18.4 Series Solutions ET 17.4
1. Let y(x) = ∞ c nx n .Theny 0
(x) = ∞ nc nx n−1 and the given equation, y 0 − y =0, becomes
n=0
n=1
∞
nc n x n−1 − ∞
c n x n =0. Replacing n by n +1in the first sum gives ∞
(n +1)c n+1 x n − ∞ c n x n =0,so
n=1
n=0
∞
[(n +1)c n+1 − c n ]x n =0. Equating coefficients gives (n +1)c n+1 − c n =0, so the recursion relation is
n=0
c n+1 =
cn
n +1 , n =0, 1, 2,....Thenc 1 = c 0 , c 2 = 1 2 c 1 = c0 2 , c 3 = 1 3 c 2 = 1 3 · 1
2 c 0 = c0
3! , c 4 = 1 4 c 3 = c0
4! ,and
in general, c n = c 0
n! . Thus, the solution is y(x) = ∞
3. Assuming y(x) = ∞
−x 2 y = − ∞
n =0
n =0
c n x n ,wehavey 0 (x) =
n =2
n=0
∞
n =1
c n x n = ∞
n=0
nc n x n−1 =
n=0
c 0
∞
n! xn = c 0
∞
n =0
n=0
x n
n! = c 0e x .
(n +1)c n+1 x n and
c nx n+2
= − ∞ c n−2x n . Hence, the equation y 0 = x 2
y becomes ∞ (n +1)c n+1x n
− ∞ c n−2x n =0
or c 1 +2c 2 x + ∞ [(n +1)c n+1 − c n−2 ] x n =0. Equating coefficients gives c 1 = c 2 =0and c n+1 = cn−2
n +1
n =2
for n =2, 3, ....Butc 1 =0,soc 4 =0and c 7 =0and in general c 3n+1 =0. Similarly c 2 =0so c 3n+2 =0. Finally
c 3 = c0 3 , c 6 = c3 6 =
c0
6 · 3 = c0
3 2 · 2! , c 9 = c6 9 = c0
9 · 6 · 3 = c0
3 3 · 3! , ...,andc 3n = c0
. Thus, the solution
3 n · n!
is y (x) =
∞
c n x n =
∞
c 3n x 3n =
∞ c 0
∞ x 3n
3 n · n! x3n = c 0
3 n n! = c ∞ x 3 /3 n
0
= c 0 e x3 /3 .
n!
n =0
n =0
n =0
5. Let y (x) = ∞ c nx n ⇒ y 0
(x) = ∞ nc nx n−1 and y 00
(x) = ∞ (n +2)(n +1)c n+2x n . The differential equation
n=0
n=1
becomes ∞ (n +2)(n +1)c n+2 x n
+ x ∞ nc n x n−1
+ ∞ c n x n
=0or ∞ [(n +2)(n +1)c n+2 + nc n + c n ]x n =0
since
n=0
∞
n=1
n=1
n=0
nc nx n
= ∞ nc nx
. n Equating coefficients gives (n +2)(n +1)c n+2 +(n +1)c n =0, thus the
n=0
recursion relation is c n+2 =
n =0
n=0
n=0
n =0
n =0
−(n +1)cn
(n +2)(n +1) = − c n
, n =0, 1, 2,.... Then the even
n +2
coefficients are given by c 2 = − c 0
2 , c4 = − c 2
4 = c 0
2 · 4 , c6 = −c 4
6 = − c 0
, and in general,
2 · 4 · 6
c 2n =(−1) n c 0
2 · 4 ·····2n = (−1)n c 0
. The odd coefficients are c
2 n 3 = − c 1
n!
3 , c5 = −c 3
5 = c 1
3 · 5 , c7 = −c 5
7 = − c 1
3 · 5 · 7 ,
and in general, c 2n+1 =(−1) n c 1
3 · 5 · 7 ·····(2n +1) = (−2)n n! c 1
. The solution is
(2n +1)!
y (x) =c 0
∞
n=0
(−1) n
2 n n! x2n + c 1
∞
n=0
(−2) n n!
(2n +1)! x2n+1 .
7. Let y (x) = ∞ c nx n ⇒ y 0
(x) = ∞ nc nx n−1
= ∞ (n +1)c n+1x n and y 00
(x) = ∞ (n +2)(n +1)c n+2x n .Then
n=0
n=1
n=0
(x−1)y 00
(x) = ∞ (n+2)(n+1)c n+2 x n+1
− ∞ (n+2)(n+1)c n+2 x n
= ∞ n(n+1)c n+1 x n
− ∞ (n+2)(n+1)c n+2 x n .
n=0
n=0
n=1
n=0
n=0
n=0
n =2