URGENT! power series solution of differential equation

**szpengchao** · June 3rd 2009, 09:49 PM

$xy''+(1-x)y'+\lambda y=0$
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.

**chisigma** · June 3rd 2009, 11:00 PM

We start supposing that the second order linear ODE...

$x\cdot y^{''} + (1-x)\cdot y^{'} + \lambda\cdot y =0$ , $y(0)=1$ (1)

... has an analytic solution of the form...

$y(x)= \sum_{n=0}^{\infty} a_{n}\cdot x^{n}$ (2)

If we substitute the expression (2) directly in (1) we obtain...

$\sum_{n=0}^{\infty} \{(\lambda - n)\cdot a_{n} + (n+1)^{2} \cdot a_{n+1}\}\cdot x^{n} =0$ (3)

... which immediately gives us...

$a_{n+1} = \frac{n-\lambda}{(n+1)^2}\cdot a_{n}$ (4)

If we set $a_{0}=1$ the (4) allow us to derive all the $a_{n}$ . It is evident from (4) that if $\lambda= k$ with $k$ an integer, all the $a_{n}$ with $n>k$ vanish, so that $y(*)$ is a polynomial of degree $k$ ...

Kind regards

$\chi$ $\sigma$

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