How to make Laguerre's equation into Sturm-Liouville form how do I put Laguerre's diff equation into Sturm-Liouville form?
$$xy'' + (1 - x)y' + λy = 0$$
I have to use the integrating factor method, for which I obtained $e^{-x}$
Is the integrating factor correct? How do I go about completing it? Many thanks.
 A: That's correct. We just multiply by $e^{-x}$:
$$
\underbrace{e^{-x}xy'' + e^{-x}(1 - x)y'}_{=\left(xe^{-x}y'\right)'} + λe^{-x}y = 0,
$$
and then we get
$$
\left(xe^{-x}y'\right)'+λe^{-x}y = 0.
$$
A: The general second order ODE 
$$f(x)y''(x)+g(x)y'(x)+h(x)y(x)=0\tag 1$$
is transformed to a Sturm-Liouville ODE 
$$\left(p(x)y'(x)\right)'+q(x)y(x)=0\tag 2$$
thanks to the change of functions :
$$\begin{cases}
p(x)=\exp\left(\int \frac{g(x)}{f(x)}dx \right) \\
q(x)=\frac{h(x)}{f(x)}\exp\left(\int \frac{g(x)}{f(x)}dx \right)
\end{cases} \tag 3$$
This is easy to prove from Eq.$(2):\quad py''+p'y'+qy=0$ compared to Eq.$(1)$
$$y''+\frac{p'}{p}y'+\frac{q}{p}y=y''+\frac{g}{f}y'+\frac{h}{f}y\quad\implies\quad\begin{cases}
\frac{p'}{p}=\frac{g}{f} \\
\frac{q}{p}=\frac{h}{f}
\end{cases}$$
In the case of $\quad xy''+(1-x)y'+\lambda y=0$
$$f(x)=x\quad\;\quad g(x)=1-x\quad\;\quad h(x)=\lambda$$
$$\begin{cases}
p(x)=\exp\left(\int \frac{1-x}{x}dx \right)=xe^{-x} \\
q(x)=\frac{h(x)}{f(x)}\exp\left(\int \frac{g(x)}{f(x)}dx \right)=\frac{\lambda}{x}xe^{-x}=\lambda e^{-x}
\end{cases}$$
$$\boxed{\left(xe^{-x}y' \right)'+\lambda e^{-x}y=0}$$
Of course, in the present very simple ODE, there is no need for the general calculus. Only inspection is sufficient to find the above result.
In case of Tchebycheff's and Hermite's differential equations use the equations $(3)$ to compute $p(x)$ and $q(x)$. 
A: You can also transform to normal form by a substitution
$$
          y = e^{x/2}f.
$$
The equation for $f$ is$$
   x\left(e^{x/2}f''+e^{x/2}f'+\frac{1}{4}e^{x/2} f\right) \\+(1-x)\left(e^{x/2}f'+\frac{1}{2}e^{x/2}f\right)\\+\lambda e^{x/2}f = 0,
$$
or
$$
       xf''+f'+(\frac{1}{2}-\frac{1}{4}x)f+\lambda f = 0 \\
        (xf')'+(\frac{1}{2}-\frac{1}{4}x+\lambda)f = 0.
$$
This is not what you asked for, but it's an alternative way to approach the problem that leads to a simpler equation with polynomial coefficients in self-adjoint form.
