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I am suppose to find the power series solutions to some diffeqs.

[tex]y'=xy[/tex]

The method is to assume that

[tex]y=\sum_{n=0}^{\infty}{c_nx^n}[/tex]

is a solution to the diffeq.Then since we can differenitate term by term we have

[tex]

\frac{d}{dx} \left[ \sum_{n=0}^{\infty}{c_nx^n} \right]=\sum_{n=1}^{\infty}{nc_nx^{n-1}}

[/tex]

Then we sub these back into our diffeq to get

[tex]

\sum_{n=1}^{\infty}{c_{n}nx^{n-1}}-x\sum_{n=0}^{\infty}{c_nx^n}=

[/tex]

[tex]

\sum_{n=1}^{\infty}{c_{n}nx^{n-1}}-\sum_{n=0}^{\infty}{c_nx^{n+1}}=0

[/tex]

So now I need to try to get the powers of [tex]x^n[/tex] and the starting index the same for both sums. But I cannot seems to figure how to push things into that form.

If I try adding one to the index of the left most series so I have the same starting index I get to

[tex]

\sum_{n=0}^{\infty}c_{n+1}(n+1)x^{n-1} -\sum_{n=0}^{\infty}{c_nx^{n+1}}=0

[/tex]

But without having the same powers for x I cannot factor all the x's. How will I get my recurrance relation then?

Thanks for the help

Jeremy

[tex]y'=xy[/tex]

The method is to assume that

[tex]y=\sum_{n=0}^{\infty}{c_nx^n}[/tex]

is a solution to the diffeq.Then since we can differenitate term by term we have

[tex]

\frac{d}{dx} \left[ \sum_{n=0}^{\infty}{c_nx^n} \right]=\sum_{n=1}^{\infty}{nc_nx^{n-1}}

[/tex]

Then we sub these back into our diffeq to get

[tex]

\sum_{n=1}^{\infty}{c_{n}nx^{n-1}}-x\sum_{n=0}^{\infty}{c_nx^n}=

[/tex]

[tex]

\sum_{n=1}^{\infty}{c_{n}nx^{n-1}}-\sum_{n=0}^{\infty}{c_nx^{n+1}}=0

[/tex]

So now I need to try to get the powers of [tex]x^n[/tex] and the starting index the same for both sums. But I cannot seems to figure how to push things into that form.

If I try adding one to the index of the left most series so I have the same starting index I get to

[tex]

\sum_{n=0}^{\infty}c_{n+1}(n+1)x^{n-1} -\sum_{n=0}^{\infty}{c_nx^{n+1}}=0

[/tex]

But without having the same powers for x I cannot factor all the x's. How will I get my recurrance relation then?

Thanks for the help

Jeremy

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