Why do we need sturm liouville form to solve ODE? What is the reason that we have to recast a 2nd order ODE into SL form to find its eigenfunctions?
for example, let $Ly=y''+y'+\frac{y}{4}=-ky$, boundary conditions $y=0$ at $x=0$ and $y-2y'=0$ at $x=\pi$, with eigenvalues $k$ and eigenfunctions $y$. This is not in SL form. So using an integrating factor $e^x$, we have $[e^x y']'+\frac{e^x}{4}=-ke^xy$, where ' representing differentiation wrt $x$. Now it's in SL form.
However to find the eigenfunctions we can perfectly use the original equation.
So why the extra work?
 A: You're right, you don't have to write it in the Sturm-Liouville form to find its eigenfunctions. But in that case you've got nothing more than that, and we wish to find useful properties. So would like to use this thereome that says the following.
If you have an equation of the type:
$$(py')'-qy=-\lambda  w y$$
One usually assumes some regularity on $p, q$ and $w$. With boundary conditions:
$$\alpha_1 y(a) + \alpha_2 y'(a)=0 \quad \text{and} \quad \beta_1y(b)+\beta_2 y'(b)=0$$
Then, we have a regular Sturm-Lioville problem with the following properties:


*

*Its eigenvalues are infinite countable, non-degenerate and bounded from below: $\lambda_0<\lambda_1<\dots$

*It's eigenfuctions are orthogonal wrt the inner product: $\int_a^by_n(x) y_m(x) w(x) dx=\delta_{nm}$ and they form a complete set. Here you can see why using the equation as given wouldn't be enough. You need to know the weight function $w$ , to define the appropriate inner product.


In your case, the eigenfunctions would be orthogonal wrt $e^x$.
