Differential Equations - Power Series problem with e^t

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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
 
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skyturnred said:

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:

[itex]ty'=\sum^{\infty}_{k=1}ka_{k}t^{k}[/itex]

You don't need to put the itex tags around every item. Just put them around the whole equation. Right click on the above expression and see how I changed it for you.

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!

You don't multiply two series like that. You need the Cauchy Product. See:
http://planetmath.org/cauchyproduct