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Am having difficulty obtaining a solution for this ODE. $ (\omega^2-x^{\alpha})\psi^{''} - \alpha x^{\alpha-1}\psi^{'} - \gamma \psi = 0 $ where $ \omega $ and $ \gamma $ are positive constants, $ x $ is the independent variable and $ \psi $ is a function of $ x $. I need the solution for this ODE for general $ \alpha $. I tried using the series method of solution to solve (frobenius) but I wasn't successful.

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  • $\begingroup$ While you may already be aware, the $\alpha=2$ case is simply the Legendre Equation. $\endgroup$ – Kajelad Mar 22 '17 at 9:49
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    $\begingroup$ any ideas as to solving for general $ \alpha $ $\endgroup$ – U. Daisy Mar 23 '17 at 7:53
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    $\begingroup$ I'm not sure. For what it's worth, it can be written in Sturm-Liouville Form, but unfortunately the integrating factor is rather cumbersome. $\endgroup$ – Kajelad Mar 23 '17 at 7:57
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    $\begingroup$ The integrating factor would be$$\frac 1{\omega^2-x^\alpha}e^{\int\frac{-\alpha x^{\alpha-1}}{\omega^2-x^\alpha}}$$ which doesn't have a very nice solution. According to Wolfram, it can be expressed in terms of Hypergeometric functions. Additionally, while S-L form tells you several useful things about the solution set, it won't necessarily help you obtain the general solution in a case such as this. $\endgroup$ – Kajelad Mar 27 '17 at 20:54
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    $\begingroup$ I tried Frobenius method as well, but I got a really bad recursion for which an explicit formula will not easily be found. I can share what I got if you want, however. $\endgroup$ – Isaac Browne Mar 31 '17 at 3:18
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It's not pretty, but here's what I found when $\alpha$ is a positive integer.

The first step I took was using product rule, a fairly standard procedure in ODEs. $$(\omega^2-x^{\alpha})\psi^{''} - \alpha x^{\alpha-1}\psi^{'} - \gamma \psi = 0$$ $$((\omega^2-x^{\alpha})\psi^{'})' - \gamma \psi = 0$$ $$((\omega^2-x^{\alpha})\psi^{'})'=\gamma \psi$$ Now applying the series method we set $\psi=\sum_{n=0}^{\infty}a_nx^n$, and $a_n=0$ if $n$ is negative $$((\omega^2-x^{\alpha})\sum_{n=1}^{\infty}na_nx^{n-1})'=\sum_{n=0}^{\infty}a_nx^n$$ $$\bigg(\sum_{n=1}^{\infty}\omega^2na_nx^{n-1}-\sum_{n=1}^{\infty}na_nx^{\alpha+n-1}\bigg)'=\sum_{n=0}^{\infty}a_nx^n$$

$$\sum_{n=2}^{\infty}\omega^2(n)(n-1)a_{n}x^{n-2}-\sum_{n=2}^{\infty}n(\alpha+n-1)a_nx^{n+\alpha-2}=\sum_{n=0}^{\infty}a_nx^n$$

$$\sum_{n=0}^{\infty}\omega^2(n+1)(n+2)a_{n+2}x^{n}-\sum_{n=\alpha}^{\infty}(n+1)(n+2-\alpha)a_{n+2-\alpha}x^{n}=\sum_{n=0}^{\infty} a_nx^n$$

Finally equating these coefficients gives us the following recursion for $n\geq \alpha$ $$\omega^2(n+1)(n+2)a_{n+2}-(n+1)(n+2-\alpha)a_{n+2-\alpha}=a_n$$ And for $n<\alpha$, we have $$\omega^2(n+1)(n+2)a_{n+2}=a_n$$ So, all power series with sequences $a_n$ that satisfy these recursions will work.

If anyone solves these recursions generally, I will be surprised to say the least.

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  • $\begingroup$ I forgot the $\gamma$, but that is just a constant multiplier over on the right side. I'll update it later. $\endgroup$ – Isaac Browne May 9 '17 at 20:06
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Hint:

$(\omega^2-x^\alpha)\psi''-\alpha x^{\alpha-1}\psi'-\gamma\psi=0$

$(x^\alpha-\omega^2)\psi''+\alpha x^{\alpha-1}\psi'+\gamma\psi=0$

Which relates to an ODE of the form http://science.fire.ustc.edu.cn/download/download1/book%5Cmathematics%5CHandbook%20of%20Exact%20Solutions%20for%20Ordinary%20Differential%20EquationsSecond%20Edition%5Cc2972_fm.pdf#page=267.

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