I'm trying to solve the differential equation:(adsbygoogle = window.adsbygoogle || []).push({});

[tex]x^2y''+2xy'+(x^2-2)y=0[/tex]

using the Fröbenius method.

So I want a solution on the form

[tex]y=\sum_{n=0}^\infty a_{n}x^{n+s}[/tex]

After finding derivatives of y, inserting into my ODE, and after some rearranging:

[tex]\sum_{n=0}^\infty ((n+s)(n+s-1)+2(n+s)-2)a_{n}x^{n+s} + \sum_{n=2}^\infty a_{n-2}x^{n+s}[/tex]

Looking at n=0, assuming [tex]a_{0}[/tex] different from zero, I get two possible values for s:

[tex]s_{1}=1[/tex]

[tex]s_{2}=-2[/tex]

Both giving [tex]a_{1}=0[/tex]

Choosing s=1, I get

[tex]a_{n}=\frac{-a_{n-2}}{(n+s)(n+s-1)+2(n+s)-2}=\frac{-a_{n-2}}{n(n+3)}[/tex]

I know [tex]a_{n}=0[/tex] for odd n, so I'm interrested in finding [tex]a_{n}[/tex] for even n, expressed by [tex]a_{0}[/tex]. Best thing I can come up with is

[tex]a_{n}=\sum_{n=2,4,6,..}^\infty \frac{3(-1)^{n/2}a_{0}}{(n+1)!(n+3)}[/tex]

which seems overly complicated given the simple recursive formula. Any ideas?

Second problem; I want to find another linear independent solution, but not quite sure where to start given my ugly expression for [tex]a_{n}[/tex].

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# Fröbenius method

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