Dtjytyk

Given the differential equation $y''-2alpha y'+frac{2}{x} y=0$,
show that if $y(x)=sum_{n=0}^{infty} a_nx^n$ is a solution of the
differential equation, then $a_0=0$ and that we can determine $a_{n+1}$ from $a_n$ for $ngeq1$

So this is the first part of a longer question and I'm already stuck because of the $frac{2}{x}$. I do know that since $a_0=y(0)=sum_{n=0}^{infty}a_n0=0$.

The first thing I did was find $y''$ and $y'$ and basically plugged everything in. This gave me the equation

$$sum_{n=0}^{infty}(n+2)(n+1)a_{n+2}x^n -sum_{n=0}^{infty}2alpha (n+1)a_{n+1}x^n+sum_{n=0}^{infty}a_nfrac{x}{2}x^n=0$$
Put everything in one summation

$$sum_{n=0}^{infty}(n+2)(n+1)a_{n+2}x^n - 2alpha (n+1)a_{n+1}x^n+a_nfrac{x}{2}x^n=0$$

Usually we factor out the $x^n$ and get

$$sum_{n=0}^{infty}Big((n+2)(n+1)a_{n+2} - 2alpha (n+1)a_{n+1}+a_nfrac{x}{2}Big)x^n=0$$
Since $x^n$ can't be $0$
$$(n+2)(n+1)a_{n+2} - 2alpha (n+1)a_{n+1}+a_nfrac{x}{2}=0$$

We can write this as
$$a_{n+2}=frac{-2a_n}{x(n+2)(n+1)}+frac{2alpha a_{n+1}}{n+2}$$

But now you can only find $a_{n+2}$ if you have $a_{n+1}$ and $a_{n}$
I don't really know what to do now

Your main mistake was that you did not treat the sum ${2over x}sum_{n=1}^infty a_n x^n$ correctly.

A power series $y(x):=sum_{n=0}^infty a_n x^n$ with positive radius of convergence and $a_0ne0$ cannot be the solution of the given ODE, because $y'(x)$ and $y''(x)$ would be bounded in the neighborhood of $x=0$, and ${1over x}y(x)$ would be unbounded. We may therefore assume $y(x)=sum_{n=1}^infty a_n x^n$ and then see whether something acceptable results. We compute
$$y'(x)=sum_{n=1}^infty n a_nx^{n-1},quad y''(x)=sum_{n=2}^infty n(n-1) a_nx^{n-2},quad{2over x}y(x)=sum_{n=1}^infty 2a_n x^{n-1} .$$
It follows that
$$y''-2alpha y'+{2over x}y=sum_{n=0}^inftybigl((n+2)(n+1)a_{n+2}-2alpha(n+1)a_{n+1}+2a_{n+1}bigr)x^n .$$
The resulting equation now is
$$(n+2)(n+1)a_{n+2}-2alpha(n+1)a_{n+1}+2a_{n+1}=0qquad(ngeq0) ,$$
resp.,
$$(n+1)n a_{n+1}+2(1-nalpha) a_n=0qquad(ngeq1) .$$

$$sum_{n=0}^{infty}(n+2)(n+1)a_{n+2}x^n -sum_{n=0}^{infty}2alpha (n+1)a_{n+1}x^n+sum_{n=0}^{infty}a_nfrac{x}{2}x^n=0 quadtext{is not correct.}$$

$$sum_{n=0}^{infty}(n+2)(n+1)a_{n+2}x^n -sum_{n=0}^{infty}2alpha (n+1)a_{n+1}x^n+sum_{n=0}^{infty}a_nfrac{2}{x}x^n=0$$

$$sum_{n=0}^{infty}(n+2)(n+1)a_{n+2}x^n -sum_{n=0}^{infty}2alpha (n+1)a_{n+1}x^n+sum_{n=1}^{infty}a_n 2x^{n-1}=0$$
$a_0frac{2}{x}=0$

$$sum_{n=0}^{infty}(n+2)(n+1)a_{n+2}x^n -sum_{n=0}^{infty}2alpha (n+1)a_{n+1}x^n+sum_{n=0}^{infty}a_{n+1} 2x^{n}=0$$
$$(n+2)(n+1)a_{n+2}+left( -2alpha(n+1)+2 right)a_{n+1}=0$$
$$a_{n+2}=frac{2alpha(n+1)-2}{(n+2)(n+1)}a_{n+1}$$
$$a_{n+1}=frac{2alpha n-2}{(n+1)n}a_{n}$$