- #1
skyturnred
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Homework Statement
The problem is to solve:
y''+ty'+e[itex]^{t}[/itex]y=0, y(0)=0 and y'(0)=-1
Homework Equations
The Attempt at a Solution
My main issue is the following: I normally find the recursion relation, and then factor out the t[itex]^{whatever}[/itex] and I know that the coefficient to this must equal 0. However, the addition of the e[itex]^{t}[/itex] means I can't factor out ALL of the t[itex]^{whatever}[/itex] so I can't exactly equate the coefficient to 0.
So specifically with the question above, I get the following:
ty'=[itex]\sum[/itex][itex]^{inf}_{k=1}[/itex]ka[itex]_{k}[/itex]t[itex]^{k}[/itex]
e[itex]^{t}[/itex]=[itex]\sum[/itex][itex]^{inf}_{k=0}[/itex]t[itex]^{k}[/itex]/k!
e[itex]^{t}[/itex]y=[itex]\sum[/itex][itex]^{inf}_{k=0}[/itex]t[itex]^{2k}[/itex]a[itex]_{k}[/itex]/k!
y''=[itex]\sum[/itex][itex]^{inf}_{k=2}[/itex]k(k-1)a[itex]_{k}[/itex]t[itex]^{k-2}[/itex]
Then I plug those into the DE, write out the first few terms so that they all start at k=2, and put them all under the same sum to get as follows:
a[itex]_{0}[/itex]+a[itex]_{1}[/itex]t+a[itex]_{1}[/itex]t[itex]^{2}[/itex]+2a[itex]_{2}[/itex]+6a[itex]_{3}[/itex]t+[itex]\sum[/itex][itex]^{inf}_{k=2}[/itex][(k+2)(k+1)a[itex]_{k+2}[/itex]+ka[itex]_{k}[/itex]+a[itex]_{k}[/itex]t[itex]^{k}[/itex]]t[itex]^{k}[/itex]=0
From this you find that a[itex]_{2}[/itex]=-a[itex]_{0}[/itex]/2
and
a[itex]_{1}[/itex]=0
and
a[itex]_{3}[/itex]=0
Like I said, normally at this point I find the recursion relation and factor out a t[itex]^{k}[/itex] so that I know that the entire coefficient is 0. But in this case, after factoring out the t[itex]^{k}[/itex] one of the terms in the coefficient still have a t[itex]^{k}[/itex], so I don't know if I can't just equate this whole thing to 0.
Can anyone help?
Thank-you