Having trouble with an Eigenvalue Differential Equation

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:

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

• 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. – John Wayland Bales Mar 29 '16 at 1:50
• Oops, my bad. It's not Cauchy-Euler. There is no $y^\prime$ term. – John Wayland Bales Mar 29 '16 at 2:01

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 $$y_n(x)=x\,\sin\left(\frac{n\pi}{\log L}\,\log x\right).$$
• Sorry I have one quick question about this - why are you considering $\lambda >1$? When looking for eigenvalues, I thought you just look at $\lambda <0,\quad \lambda = 0,\quad \lambda >0$. – Spuds Mar 29 '16 at 20:00
• 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 $\lambda>1$. – Martin Argerami Mar 29 '16 at 22:54