$\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.

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- Jun 3rd 2009, 09:49 PMszpengchaoURGENT! power series solution of differential equation
$\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. - Jun 3rd 2009, 11:00 PMchisigma
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

$\displaystyle \chi$ $\displaystyle \sigma$