For $y''-2\alpha y'+\frac{2}{x} y=0$ , provided that $y(x)=\sum_{n=0}^{\infty} a_nx^n$ show you can determine $a_{n+1}$ from $a_n$

Question

Given the differential equation $y''-2\alpha y'+\frac{2}{x} y=0$, 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 $n\geq1$

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

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

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

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

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

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

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

Answer 1

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

A power series $y(x):=\sum_{n=0}^\infty a_n x^n$ with positive radius of convergence and $a_0\ne0$ cannot be the solution of the given ODE, because $y'(x)$ and $y''(x)$ would be bounded in the neighborhood of $x=0$, and ${1\over x}y(x)$ 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 $$y'(x)=\sum_{n=1}^\infty n a_nx^{n-1},\quad y''(x)=\sum_{n=2}^\infty n(n-1) a_nx^{n-2},\quad{2\over x}y(x)=\sum_{n=1}^\infty 2a_n x^{n-1}\ .$$ It follows that $$y''-2\alpha y'+{2\over x}y=\sum_{n=0}^\infty\bigl((n+2)(n+1)a_{n+2}-2\alpha(n+1)a_{n+1}+2a_{n+1}\bigr)x^n\ .$$ The resulting equation now is $$(n+2)(n+1)a_{n+2}-2\alpha(n+1)a_{n+1}+2a_{n+1}=0\qquad(n\geq0)\ ,$$ resp., $$(n+1)n a_{n+1}+2(1-n\alpha) a_n=0\qquad(n\geq1)\ .$$

Answer 2

$$\sum_{n=0}^{\infty}(n+2)(n+1)a_{n+2}x^n -\sum_{n=0}^{\infty}2\alpha (n+1)a_{n+1}x^n+\sum_{n=0}^{\infty}a_n\frac{x}{2}x^n=0 \quad\text{is not correct.}$$

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

$$\sum_{n=0}^{\infty}(n+2)(n+1)a_{n+2}x^n -\sum_{n=0}^{\infty}2\alpha (n+1)a_{n+1}x^n+\sum_{n=1}^{\infty}a_n 2x^{n-1}=0$$ $a_0\frac{2}{x}=0$

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