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Here is the problem:

$$ x^2y''-xy+\lambda y = 0,\quad y(1)=0,\quad y(L)=0,\quad L>0 $$

I am asked to find the Eigenvalues and Eigenfunction.

I can't figure out how to get a general equation for y. I tried integrating factors but that was a mess, I'm not sure if there's a better way to do this. It might be a Sturm Liouville equation, but I'm not sure to to solve those. Any help would be appreciated.

Here is the solution given:

$ \lambda_n=1+(n\pi/ln(L))^2,\quad y_n(x)=xsin(n\pi ln(x)/ln(L));\quad n=1,2,3... $

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    $\begingroup$ This is a second order Cauchy-Euler equation so it has two solutions of the form $y=x^m$ where $m$ satisfies the auxiliary equation $m(m-1)-m+\lambda=0$. In the event that the discriminant is negative $m$ will be complex with $m=a\pm bi$ for some $a,b$. In that case the general solution will be of the form $y=e^{ax}(c_1\sin(b\ln(x))+c_2\cos(b\ln(x)))$. You will have to use your boundary conditions to get the exact answer. $\endgroup$ – John Wayland Bales Mar 29 '16 at 1:50
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    $\begingroup$ Oops, my bad. It's not Cauchy-Euler. There is no $y^\prime$ term. $\endgroup$ – John Wayland Bales Mar 29 '16 at 2:01
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There is likely a typo in the equation. The solution you give solves $$ x^2y''-xy'+\lambda_ny=0. $$ As John mentioned, this is an Euler equation with characteristic equation $m(m-1)-m+\lambda =0$. For $\lambda\leq1$, $m$ is real and there is no solution with $y(L)=0$. For $\lambda>1$ we get solutions $$ y(x)=c_1 x \cos(\sqrt{\lambda-1}\,\log x) + c_2 x \sin(\sqrt{\lambda-1}\,\log x). $$ The condition $y(0)=0$ forces $c_1=0$. The condition $y(L)=0$ forces $$ \sin(\sqrt{\lambda-1}\,\log L)=0, $$so $$\sqrt{\lambda-1}\log L=n\pi,\ \ \ n\in\mathbb N$$ This gives us eigenvalues $$ \lambda_n=1+\left(\frac{n\pi}{\log L}\right)^2 $$ and eigenvectors $$ y_n(x)=x\,\sin\left(\frac{n\pi}{\log L}\,\log x\right). $$

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  • $\begingroup$ You da bomb dot com. May your limits never be undefined. $\endgroup$ – Spuds Mar 29 '16 at 18:37
  • $\begingroup$ Sorry I have one quick question about this - why are you considering $\lambda >1$? When looking for eigenvalues, I thought you just look at $\lambda <0,\quad \lambda = 0,\quad \lambda >0$. $\endgroup$ – Spuds Mar 29 '16 at 20:00
  • $\begingroup$ You look at what you have. Here the characteristic equation is $m^2-2m+\lambda$. This has real solution for $\lambda\geq1$ (which you can check cannot satisfy $y(0)=y(L)=0$), and non-real solutions for $\lambda>1$. $\endgroup$ – Martin Argerami Mar 29 '16 at 22:54
  • $\begingroup$ @Spuds: since you are new to the site, please consider upvoting the answer if you found it useful (not just mine, but any answer that you find useful; that's how this site works). $\endgroup$ – Martin Argerami Mar 29 '16 at 23:01

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