# Finding eigenvalues and eigenfunctions for a boundary problem.

Let the boundary problem be: $$x^2y''+\lambda y=0$$ with $$1 and boundary conditions $$y(1) = 0 = y(2)$$. Assume $$\lambda = \mu^2>0$$I am trying to solve for $$r$$ by first dividing $$x^2y''+\lambda y=0$$ by $$x^2$$ to get $$y''+ \frac{\lambda}{x^2} y =0$$. This will give me $$r = \pm i \frac{\mu}{x}$$.

However, this will give me $$y(x) = C_1\cos(\mu)+C_2\sin(\mu)$$. I'm not quite sure how to proceed from here because it seems like the $$x$$ terms just cancel out unless I am doing something wrong.

This is a Cauchy-Euler equation, so we assume $$y=x^r.$$ Then we have $$y'=rx^{r-1}$$ and $$y''=r(r-1)x^{r-2},$$ which we insert into the original DE as \begin{align*} r(r-1)x^r+\lambda x^r&=0\\ r^2-r+\lambda&=0\\ r&=\frac{1\pm\sqrt{1-4\lambda}}{2}. \end{align*} Let \begin{align*} r_+&=\frac{1+\sqrt{1-4\lambda}}{2}\\ r_-&=\frac{1-\sqrt{1-4\lambda}}{2}. \end{align*} Then $$y=c_+ x^{r_+}+c_- x^{r_-}.$$ The boundary conditions yield \begin{align*} 0&=c_+ +c_- \\ 0&=c_+ 2^{r_+}+c_- 2^{r_-}. \end{align*} Unfortunately, the only solution is the trivial solution $$c_-=c_+=0,$$ implying that there are no solutions to this eigenvalue problem (since eigenvectors are nonzero by definition).

Since $$y$$ is, in general, a function of $$x$$, we will first need to convert the equation into an equation with constant coefficients so that your method can be applied. Notice that this can be done by replacing $$x$$ by $$e^{z}$$, where since $$1 < x < 2$$, we have $$0 < z < \ln 2$$.

Also, if we denote $$D \equiv \dfrac{d}{dx}$$ and $$\theta \equiv \dfrac{d}{dz}$$, then

$$x^2 y'' = \theta \left( \theta - 1 \right) y$$

Hence, we get the boundary value problem

$$\theta^2 y - \theta y + \lambda y = 0, \ 0 < z < \ln 2$$

and the boundary conditions $$y \left( 0 \right) = y \left( \ln 2 \right) = 1$$.

Now, you can proceed by your method to find the eigenvalues and eigenfunctions for the BVP. However, in the end, do not forget to replace $$z$$ by suitable function of $$x$$, so that the solution is in terms of what was given in the problem statement.