Series solution to first order differential equation I need to find a series solution to the following simple differential equation
$$x^2y'=y$$
Assuming the solution to be of the form $y=\sum a_nx^n$  and equating the coefficients on both the sides, all the coefficients turn out to be zero which is definitely wrong.
Please guide me to the correct solution.
Any help is appreciated. Thanks!
 A: if you let $y = \sum_{ n = 0}^{\infty} a_nx^n$ 
$x^2y' = \sum_{ n = 0}^{\infty} a_nnx^{n+1} = \sum_{ n = 0}^{\infty} a_nx^n$ 
so equating coefficients on each side gives you $a_{0} = 0$ and a relation
$$
a_{n+1} = na_n
$$
which does indeed lead to all coefficients being zero
What if you supposed that $y = \sum_{ -\infty}^{\infty} a_nx^n$
then $a_{-0}$ could equal A say and then you would get $-1a_{-1} = A$
You should then be able to recover the answer that Autolatry found using separation of variables 
A: It could be useful to remark that $y(x)=0$ is the only solution of $x^2 y^\prime (x)=y(x)$ which is defined and analytic on the whole real line. In fact, the other solutions of your ODE are of the type:
$$y(x)=C \exp \left( -\frac{1}{x}\right)$$
and none of them is both defined and analytic in $\mathbb{R}$.
Therefore, it is really obvious that the power series method yields only the zero solution.

Moreover, if you let:
$$y(x) := u\left( \frac{1}{x}\right)$$
(for a suitable differentiable $u(t)$) you can reduce your ODE to a simpler one: in fact, you have:
$$y^\prime (x) = -\frac{1}{x^2}\ \dot{u}\left( \frac{1}{x}\right)$$
(where the dot denotes derivation with respect to the $t$ variable) that is:
$$\dot{u} \left( \frac{1}{x}\right) = -x^2\ y^\prime (x)\; .$$
Therefore $y(x)$ solves your ODE away from zero iff $u(t)$ solves:
$$\dot{u}(t) = - u(t)\; .$$
Hence $u(t)=Ce^{-t}$ and $y(x)=Ce^{-1/x}$.
A: Well,
\begin{eqnarray}
\int \frac{dy}{y} &=& \int \frac{dx}{x^{2}}\\
\implies \ln y    &=& -\frac{1}{x}+c
\end{eqnarray}
Hence,
\begin{equation}
y=Ae^{-\frac{1}{x}}
\end{equation}
Is a series solution necessary?
