Solution of differential equation - We find only one

I want to find all the solutions of the form $y(x)=x^m \sum_{n=0}^{\infty} a_n x^n, x>0 (m \in \mathbb{R})$ of the differential equation $x^2 y''+ xy'+x^2y=0$.

I have tried the following:

Since $x>0$ the differential equation can be written as:

$$y''+ \frac{1}{x}y'+y=0$$

$$p(x)=\frac{1}{x}, \ q(x)=1$$

The point $0$ is regular singular, i.e. the fuctions $xp(x)=1$,$x^2 q(x)=x^2$ can be written as power series with center $0$.

We suppose that there is a solution of the form $y(x)= x^m \sum_{n=0}^{\infty} a_n x^n$ for $0<x<R$, where $R$ is a suitable positive number and $m \in \mathbb{R}$.

Then we have:

$$x^2y=\sum_{n=0}^{\infty} a_n x^{n+m+2}= \sum_{n=2}^{\infty} a_{n-2}x^{n+m} \\ y'(x)= \sum_{n=0}^{\infty} (n+m) a_n x^{n+m-1} \Rightarrow xy'(x)= \sum_{n=0}^{\infty} (n+m) a_n x^{n+m} \\ y''(x)= \sum_{n=0}^{\infty} (n+m)(n+m-1) a_n x^{n+m-2} \Rightarrow x^2 y''(x)= \sum_{n=0}^{\infty} (n+m)(n+m-1) a_n x^{n+m}$$

$x^2 y''+ xy'+x^2y=0 \Rightarrow \sum_{n=0}^{\infty} (n+m) (n+m-1) a_n x^{n+m}+ \sum_{n=0}^{\infty} (n+m) a_n x^{n+m}+ \sum_{n=2}^{\infty} a_{n-2} x^{n+m}=0$

$\Rightarrow m(m-1)a_0 x^m+(1+m) m a_1 x^{m+1}+ma_0 x^m +(1+m) a_1 x^{m+1}+ \sum_{n=2}^{\infty} \left[ (n+m) (n+m-1) a_n+(n+m) a_n+a_{n+2}\right] x^{n+m}=0$

$\Rightarrow (m(m-1)a_0+ma_0) x^m+((1+m)ma_1+(1+m)a_1) x^{m+1}+ \sum_{n=2}^{\infty} \left[ (n+m)(n+m-1) a_n +(n+m) a_n+ a_{n-2}\right] x^{n+m}=0$

$\Rightarrow m^2 a_0 x^m +(1+m)^2 a_1 x^{m+1}+ \sum_{n=2}^{\infty} \left[ (n+m)^2 a_n+a_{n-2}\right] x^{n+m}=0$

It has to hold: $\left\{\begin{matrix} m^2 a_0=0 & \\ (1+m)^2 a_1=0 & \\ (n+m)^2 a_n=-a_{n-2} \Rightarrow a_n=-\frac{a_{n-2}}{(n+m)^2} &, n=2,3, \dots \end{matrix}\right.$

For $a_0 \neq 0$, it has to hold: $m=0$.

For $m=0$:

$$a_1=0 \\ a_2=-\frac{a_0}{4} \\ a_3=0 \\ a_4=\frac{a_0}{4^3} \\ a_5=0 \\ a_6=-\frac{a_0}{4^3 6^2}$$

Is it right so far?

But in this way we find only one solution for the differential equation, since we find only one possible $m$.

So do we maybe have to do somethig else?

EDIT: If we would want to find a formula for $a_n$, for $n=2k+1$ it would hold $a_{2k+1}=0$, right?  Is the following right or could we right $a_{2k}$ in a better way? $$a_{2k}=(-1)^k \frac{a_0}{ 4^3 \cdot \prod_{j=3}^{k} (2j)^2 }, k=2, \dots \text{ and } a_2=-\frac{a_0}{4}$$

Also couldn't we find two linealy independent solutions of the given differential equation?

• This seems (to me) to be normal since the solution of the equation is $$y=c_1 J_0(x)+c_2 Y_0(x)$$ where appear Bessel functions. The first one can be expended as a Taylor series at $x=0$ (and this is the solution you nicely obtained) while this is not feasible for the second one. – Claude Leibovici May 30 '15 at 15:06
• @ClaudeLeibovici So if we would want to find a formula for $a_n$, for $n=2k+1$ it would hold $a_{2k+1}=0$, right?  Is the following right or could we right $a_{2k}$ in a better way? $$a_{2k}=(-1)^k \frac{a_0}{ 4^3 \cdot \prod_{j=3}^{k} (2j)^2 }, k=2, \dots \text{ and } a_2=-\frac{a_0}{4}$$ Also couldn't we find two linealy independent solutions of the given differential equation? – evinda May 30 '15 at 15:38
• I must confess that I prefer to stay with the recurrence equation. – Claude Leibovici May 30 '15 at 15:54
• @ClaudeLeibovici I thought to find a formula for $a_n$ in order to be able to write the general form of the solution of the differential equation.. Is the formula I found wrong? – evinda May 30 '15 at 15:56
• It seems to be $\frac{(-1)^k 2^{-2 k}}{(k!)^2}$ – Claude Leibovici May 30 '15 at 17:40