$\displaystyle xy''+(1-x)y'+\lambda y=0 $
find the power series solution about x=0 with y(0)=1 and show:
y is a polynomial iff lambda is a non-negative integer n.
We start supposing that the second order linear ODE...
$\displaystyle x\cdot y^{''} + (1-x)\cdot y^{'} + \lambda\cdot y =0$ , $\displaystyle y(0)=1$ (1)
... has an analytic solution of the form...
$\displaystyle y(x)= \sum_{n=0}^{\infty} a_{n}\cdot x^{n}$ (2)
If we substitute the expression (2) directly in (1) we obtain...
$\displaystyle \sum_{n=0}^{\infty} \{(\lambda - n)\cdot a_{n} + (n+1)^{2} \cdot a_{n+1}\}\cdot x^{n} =0$ (3)
... which immediately gives us...
$\displaystyle a_{n+1} = \frac{n-\lambda}{(n+1)^2}\cdot a_{n}$ (4)
If we set $\displaystyle a_{0}=1$ the (4) allow us to derive all the $\displaystyle a_{n}$. It is evident from (4) that if $\displaystyle \lambda= k$ with $\displaystyle k$ an integer, all the $\displaystyle a_{n}$ with $\displaystyle n>k$ vanish, so that $\displaystyle y(*)$ is a polynomial of degree $\displaystyle k$...
Kind regards
$\displaystyle \chi$ $\displaystyle \sigma$