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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?

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    $\begingroup$ 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. $\endgroup$ – Claude Leibovici May 30 '15 at 15:06
  • $\begingroup$ @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? $\endgroup$ – evinda May 30 '15 at 15:38
  • $\begingroup$ I must confess that I prefer to stay with the recurrence equation. $\endgroup$ – Claude Leibovici May 30 '15 at 15:54
  • $\begingroup$ @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? $\endgroup$ – evinda May 30 '15 at 15:56
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    $\begingroup$ It seems to be $\frac{(-1)^k 2^{-2 k}}{(k!)^2}$ $\endgroup$ – Claude Leibovici May 30 '15 at 17:40
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Essentially what you are doing is using the Frobenius method for solving ODEs locally, NOT GLOBALLY!

You were able to find the first solution using the usual approach but the second solution is a little trickier but not that tricky once you see it. Essentially the second solution is another power series plus ln(x) * the first solution.

Here's a wikipedia link for the method and the part I am talking about is under the section title Z-seperate roots.

http://en.wikipedia.org/wiki/Frobenius_method

Hope this helps.

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  • $\begingroup$ You mean the formula that is at: "Z-separate roots" ? $\endgroup$ – evinda Jun 4 '15 at 14:52

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