- #1
Sol-chan
- 6
- 0
Solve by the method of series
y''-xy'+y=0
From what I understand, I need to substitute in
y(x)=SUM(Cn*xn) from n=0 to infinity
I get that:
y'(x)=SUM(n*Cn*xn-1) from n=1 to infinity
and
y''(x)=SUM(n*(n-1)*Cn*xn-2) from n=2 to infinity
When I substitute in and collect the terms, I get:
2C2 + C0 + SUM[(Cn+2*(n+2)*(n+1)-Cn*n+Cn)*xn] from n=1 to infinity = 0
So then I get that
C2 = -C0/2
and
Cn+2 = Cn/(n+2)
And I don't know what to do after that. Any hints on where to go from there would be much appreciated...
y''-xy'+y=0
From what I understand, I need to substitute in
y(x)=SUM(Cn*xn) from n=0 to infinity
I get that:
y'(x)=SUM(n*Cn*xn-1) from n=1 to infinity
and
y''(x)=SUM(n*(n-1)*Cn*xn-2) from n=2 to infinity
When I substitute in and collect the terms, I get:
2C2 + C0 + SUM[(Cn+2*(n+2)*(n+1)-Cn*n+Cn)*xn] from n=1 to infinity = 0
So then I get that
C2 = -C0/2
and
Cn+2 = Cn/(n+2)
And I don't know what to do after that. Any hints on where to go from there would be much appreciated...