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I am having a difficulty solving this second order linear ODE with a variable coefficient. Below is the equation:

$$x^\beta\,y'' - \beta\,x^{\beta-1}y' - \gamma\,y = 0 $$

With the following boundary conditions $ y(0) = 0 $ and $y(L) = 0$. Where $ \gamma $ is a constant. I have employed the series method of solution, the Frobenius method to be precise. I also tried to transform the equation to a Bessel kind or something more familiar, but I have been unsuccessful. I tried the Sturm-Liouville approach for solving Second ODE.

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closed as off-topic by GNUSupporter 8964民主女神 地下教會, Juniven, user91500, Claude Leibovici, JonMark Perry Mar 1 '17 at 9:58

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  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – GNUSupporter 8964民主女神 地下教會, Juniven, user91500, Claude Leibovici, JonMark Perry
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  • $\begingroup$ One can wrangle this into the form of a modified Bessel equation with a couple of substitutions. I'm not sure if that's quite what you want, but for a general $\beta$, one can't do any better. $\endgroup$ – Chappers Feb 28 '17 at 1:42
  • $\begingroup$ Actually am looking at a general case of $\beta$ hmmmmm $\endgroup$ – daisy Feb 28 '17 at 6:04
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Let $$\xi=\int\frac{1}{x^{\beta/2}}dx=\frac{1}{(1-\beta/2)x^{\beta/2-1}}$$ the equation becomes $$\xi\frac{d^{2}}{d\xi^{2}}y(\xi)-\frac{3\beta}{2-\beta}\frac{d}{d\xi}y(\xi)-\gamma\xi{y(\xi)}=0$$ And the solution is $$y(\xi)=c_{1}\xi^{\frac{1+\beta}{2-\beta}}J_{\frac{\beta+1}{\beta-2}}(-i\sqrt{\gamma}\xi)+c_{2}\xi^{\frac{1+\beta}{2-\beta}}Y_{\frac{\beta+1}{\beta-2}}(-i\sqrt{\gamma}\xi)$$

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