Solving Differential Equation with Laguerre Polynomials

[Logarythmic]

What differential equation does

$$\phi_n (x) := e^{-x/2} L_n (x)$$

solve? $$L_n$$ is a Laguerre polynomial.

Please give me a hint on this one. I haven't got a clue where to start.

 

[StatusX]

What differential equation do the Laguerre polynomials solve? If you don't know, look it up.

 

[Logarythmic]

The Laguerre differential equation is

$$xy'' + (1 - x)y' + ny = 0$$

and $$L_n (x)$$ is a solution to this but my $$\phi_n (x)$$ is not a solution to the Laguerre equation, is it?
I know that $$\phi_n (x)$$ should solve a self adjoint differential equation but I don't think the Laguerre eq. is?

 

[StatusX]

Find $\phi',\phi''$ in terms of the derivatives of L and use the differential equation relating the derivatives of L to get a DE relating the derivatives of $\phi$.

 

[Logarythmic]

Ok, so then I get

$$x \phi_n^{''} (x) + (1-x) \phi_n^{'} + (n + \frac{1}{2} - \frac{x}{4}) \phi_n (x) = 0$$

but this isn't self-adjoint?
In my case, self-adjoint means it can be written in the form

$$\frac{d}{dx} (p(x) \frac{d}{dx} \phi_n (x) ) + q(x) \phi_n (x)$$

 

[Logarythmic]

I made a mistake. The equation I get is

$$x \phi_n^{''} + \phi_n^{'} + (n + \frac{1}{2} - \frac{x}{4}) \phi_n = 0$$

and this is indeed self-adjoint. Thanks for your help!