Finding eigenvalues and eigenfunctions of differential equation I need to write the following d.e. in Sturm-Liouville form and find the eigenvalues and eigenfunctions.
$$\frac {\mathrm d ^2 y}{\mathrm d x^2} + 7 \frac {\mathrm d y} {\mathrm d x} + (e^{3x} + \lambda)y = 0$$
where $y(0)=y(1)=0$.
I have written it as $-(p(x)y')' + q(x)y = \lambda r(x) y$ with $p(x) = e^{7x}$ , $q(x) = -e^{10x}$ and $r(x) = e^{7x}$.
I am not sure how to find the eigenvalues though. I know that the boundary conditions are relevant but I'm not sure where to start in terms of finding the eigenfunctions. How do I know by looking at the d.e. when it is possible to find them?
 A: First let's get rid of the first derivative term. Let $y(x)=e^{-7x/2}g(x)$. Then
$$\frac{dy}{dx}=e^{-7x/2}\frac{dg}{dx}-\frac72e^{-7x/2}g$$
$$\frac{d^2y}{dx^2}=e^{-7x/2}\frac{d^2g}{dx^2}-7e^{-7x/2}\frac{dg}{dx}+\frac{49}4e^{-7x/2}g$$
So
$$\frac{d^2g}{dx^2}+\left(e^{3x}+\lambda-\frac{49}4\right)g=0$$
Now we can wipe out that butt-ugly $e^{3x}$ business. Let $u=e^{3x/2}$. Then
$$\frac{dg}{dx}=\frac{dg}{du}\frac{du}{dx}=\frac32u\frac{dg}{du}$$
$$\frac{d^2g}{du^2}=\frac32u\left(\frac32\frac{dg}{du}+\frac32u\frac{d^2g}{du^2}\right)$$
So
$$u^2\frac{d^2g}{du^2}+u\frac{dg}{du}+\left(\frac49u^2+\frac49\lambda-\frac{49}9\right)g=0$$
This is Bessel's differential equation with general solution
$$g(u)=C_1J_{\sqrt{49-4\lambda}/3}\left(\frac23u\right)+C_2Y_{\sqrt{49-\lambda}/3}\left(\frac23u\right)$$
That brings us back to
$$y(x)=e^{-7x/2}\left(C_1J_{\sqrt{49-4\lambda}/3}\left(\frac23e^{3x/2}\right)+C_2Y_{\sqrt{49-\lambda}/3}\left(\frac23e^{3x/2}\right)\right)$$
Now you have to find $\lambda$ such that $y(0)=y(1)=0$ for $C_1$ and $C_2$ not both zero.
