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For the ODE $u''(x) + x^{-n}u(x) = 0$ can I use Frobenius method for $n > 2$? I think not since we need $x^2 x^{-n}$ to be analytic but is there something else I can use to find a solution? Some generalisation?

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Assume you only want to know the cases that $n$ is integer:

Try let $u(x)=\sum\limits_{k=0}^\infty a_kx^{k+r}$ ,

Then $u'(x)=\sum\limits_{k=0}^\infty(k+r)a_kx^{k+r-1}$

$u''(x)=\sum\limits_{k=0}^\infty(k+r)(k+r-1)a_kx^{k+r-2}$

$\therefore\sum\limits_{k=0}^\infty(k+r)(k+r-1)a_kx^{k+r-2}+x^{-n}\sum\limits_{k=0}^\infty a_kx^{k+r}=0$

$\sum\limits_{k=0}^\infty(k+r)(k+r-1)a_kx^{k+r-2}+\sum\limits_{k=0}^\infty a_kx^{k+r-n}=0$

$\sum\limits_{k=0}^\infty(k+r)(k+r-1)a_kx^{k+r-2}+\sum\limits_{k=2-n}^\infty a_{k+n-2}x^{k+r-2}=0$

$\sum\limits_{k=2-n}^{-1}a_{k+n-2}x^{k+r-2}+\sum\limits_{k=0}^\infty((k+r)(k+r-1)a_k+a_{k+n-2})x^{k+r-2}=0$

which fails to solve by the conventional version of Frobenius method as we can't get any indicial equations.

But when we try let $u(x)=\sum\limits_{k=0}^\infty a_kx^{r-k}$ ,

Then $u'(x)=\sum\limits_{k=0}^\infty(r-k)a_kx^{r-k-1}$

$u''(x)=\sum\limits_{k=0}^\infty(r-k)(r-k-1)a_kx^{r-k-2}$

$\therefore\sum\limits_{k=0}^\infty(r-k)(r-k-1)a_kx^{r-k-2}+x^{-n}\sum\limits_{k=0}^\infty a_kx^{r-k}=0$

$\sum\limits_{k=0}^\infty(r-k)(r-k-1)a_kx^{r-k-2}+\sum\limits_{k=0}^\infty a_kx^{r-k-n}=0$

$\sum\limits_{k=0}^\infty(r-k)(r-k-1)a_kx^{r-k-2}+\sum\limits_{k=n-2}^\infty a_{k-n+2}x^{r-k-2}=0$

$\sum\limits_{k=0}^{n-3}(r-k)(r-k-1)a_kx^{r-k-2}+\sum\limits_{k=n-2}^\infty((r-k)(r-k-1)a_k+a_{k-n+2})x^{r-k-2}=0$

which can solve by this ''modified version'' of ''Frobenius method'' .

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    Sorry, last comment should have gone to @doraemonpaul2012-11-19
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If $n>2$, then $x=0$ is not a regular singular point. However, you can still get a solution in terms of Bessel functions,

$ y \left( x \right) ={\it c_1}\,\sqrt {x}\, {{\rm J_{ \left( 2-n \right) ^{-1}}}\left(2\,{\frac {{x}^{-\frac{n}{2}+1}}{n-2}}\right)} +{\it c_2}\,\sqrt {x}\, {{\rm Y_{(2-n)^{-1}}}\left(2\,{\frac {{x}^{-\frac{n}{2}+1}}{n-2}}\right)}\,,$

where $J$ is Bessel function of the first kind and $Y$ is Bessel function of second kind.

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    @markclops How is this answer helping you to understand the problem and the approach(es) available to solve it? I am curious.2012-12-07