<s>[INST] Differential equation Laguerre $xy''+(1-x)y'+ay=0, a \in \mathbb{R}$ The differential equation Laguerre $xy''+(1-x)y'+ay=0, a \in \mathbb{R}$ is given.


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*Show that the equation has $0$ as its singular regular point .

*Find a solution of the differential equation of the form $x^m \sum_{n=0}^{\infty} a_n x^n (x>0) (m \in \mathbb{R})$

*Show that if $a=n$, where $n \in \mathbb{N}$ then there is a polynomial solution of degree $n$.

*Let $L_n$ the polynomial $L_n(x)=e^x \frac{d^n}{dx^n} (x^n \cdot e^{-x})$ (show that it is a polynomial), $n=1,2,3, \dots$. Show that $L_n$ satisfies the equation Laguerre if $a=n(n=1,2, \dots)$.


That's what I have tried:


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*For $x \neq 0$ the differential equation can be written as: $y''+ \frac{1-x}{x}y'+ \frac{a}{x}y=0$.
$p(x)= \frac{1-x}{x}, q(x)= \frac{a}{x}$
The functions $x \cdot p(x)= 1-x, \ x^2q(x)=ax$ can be written as power series in a region of $0$.
Thus, $0$ is a singular regular point.

*We suppose that there is a solution of the form $y(x)=x^m \sum_{n=0}^{\infty} a_n x^n= \sum_{n=0}^{\infty} a_n x^{n+m}$.
Then $y'(x)= \sum_{n=0}^{\infty} a_n(n+m) x^{n+m-1} \Rightarrow -xy'(x)= \sum_{n=0}^{\infty} -a_n(n+m) x^{n+m}$ and $y''(x)= \sum_{n=0}^{\infty} a_n(n+m)(n+m-1) x^{n+m-2} \Rightarrow xy''(x)= \sum_{n=0}^{\infty} a_n(n+m)(n+m-1) x^{n+m-1}$


So we have 
$$\sum_{n=0}^{\infty} a_n (n+m)(n+m-1) x^{n+m-1}+ \sum_{n=0}^{\infty} a_n (n+m) x^{n+m-1} + \sum_{n=1}^{\infty} -a_{n-1} (n+m-1) x^{n+m-1}+ \sum_{n=1}^{\infty} a a_{n-1} x^{n+m-1}=0 \\ \Rightarrow a_0 m (m-1) x^{m-1}+ a_0 m x^{m-1}+ \sum_{n=1}^{\infty} \left[ a_n (n+m) (n+m-1)+ a_n(n+m)-a_{n-1}(n+m-1)+ a a_{n-1} \right] x^{n+m-1}=0$$
It has to hold:
$$a_0 m^2=0 \overset{a_0 \neq 0}{ \Rightarrow } m=0$$
$$a_n (n+m) (n+m-1)+ a_n (n+m)- a_{n-1} (n+m-1)+ a a_{n-1}=0$$
For $m=0$: $a_{n} n (n-1)+ a_n n-a_{n-1} (n-1)+ a a_{n-1}=0 \Rightarrow a_n n^2+ a_{n-1} (a-n+1)=0 \Rightarrow a_n=- \frac{a_{n-1}(a-n+1)}{n^2}$
For $n=1: \ a_1=-aa_0$
For $n=2: \ a_2= \frac{aa_0(a-1)}{2^2} $
For $n=3: \ a_3=- \frac{aa_0(a-1) (a-2)}{2^2 3^2} $
For $n=4: \ a_4= \frac{aa_0(a-1)(a-2)(a-3)}{2^2 3^2 4^2} $
For $n=5: \ a_5= -\frac{aa_0(a-1)(a-2)(a-3)(a-4)}{2^2 3^2 4^2 5^2} $
We see that $a_n=(-1)^n a_0 \frac{\prod_{i=0}^{n-1} (a-i)}{\prod_{i=2}^n i^2}$
$$\frac{a_{n+1}}{a_n}=(-1) \frac{a-n}{(n+1)^2} \to 0$$
So the series $\sum_{n=0}^{\infty} a_n x^n$ converges and its radius of convergence is equal to $+\infty$.
Is it right so far? Could you give me a hint what we could do in order to answer the other two questions? 
EDIT:
Do we differentiate the Leguerre polynomial as follows?
$$$$
$$\frac{d}{dx} L_n(x)=e^x \frac{d^n}{dx^n} (x^n e^{-x})+\frac{d^{n+1}}{dx^{n+1}}(x^n e^{-x})$$
$$\frac{d^2}{dx^2} L_n(x)=e^x \frac{d^n}{dx^n} (x^n e^{-x})+ e^{x} \frac{d^{n+1}}{dx^{n+1}}(x^n e^{-x})+\frac{d^{n+2}}{dx^{n+2}}(x^n e^{-x})$$
EDIT: Substituting the above, we cannot show that $L_n$ satisfies the equation Laguerre if $a=n$.
Do you maybe know how else we could show this?
 [/INST] You have already done that third part, notice that if $a = n \in \mathbb{N}$ then $a_{i} = 0$ for all $i \ge n+1$. So you have a polynomial of degree $n$ as a solution. 
For the last part, just some ideas - try differentiating the Leguerre polynomial and substituting it into the equation. Or, conversely, try to expand the derivative to show the coefficients of the power series are your $a_i$s?
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