How to solve an "almost quadratic" ODE? Suppose for some constants $\alpha,\beta,\gamma$ that we're given the following ODE: $$\alpha y''+\beta xy'+\gamma y=0.$$ Now, I know how to find the general solution for $y(x)$ if any of $\alpha,\beta,\gamma$ should turn out to be $0$, but I've just ended up with the ODE $$2y''+xy'+y=0.$$ Can anybody give me the first (few) step(s) of a general procedure one can use for such ODEs?
 A: Write $xy'+y$ as $(xy)'$ and integrate to get:
$$
y' + \frac{x}{2}y = c_1
$$
Which can be solved using the integrating factor $\exp\left(\int \frac{x}{2} \, dx\right) = \exp\left(\frac{x^2}{4}\right)$. The solution cannot be written in terms of elementary functions though:
\begin{align*}
\exp\left(\frac{x^2}{4}\right)y' + \dfrac{x}{2}\exp\left(\frac{x^2}{4}\right)y &= c_1 \exp\left(\frac{x^2}{4}\right) \\
\exp\left(\frac{x^2}{4}\right)y &= c_1 \int \exp\left(\frac{x^2}{4}\right) \, dx + c_2
\end{align*}
Thus:
$$
y = c_1 \exp\left(-\frac{x^2}{4}\right) \int \exp\left(\frac{x^2}{4}\right) \, dx + c_2\exp\left(-\frac{x^2}{4}\right)
$$
This only works if $\beta = \gamma$, but it does work for the ODE you have.
A: Assume your solution 
$$ y(x)=\sum_{k=0}^{\infty} a_k x^{k+\alpha} \,,$$
and plug into the differential equation and try to find a recurrence relation in $a_k$. Off course, you need to determine $\alpha$ as a first step. The well known power series method for second order ode is Frobenius method.
A: Maple gives the general solution using the Kummer M and U functions.
$$ y \left( x \right) =c_{{1}}{{\rm e}^{-{\frac {\beta\,{x}^{2}}{
2 \alpha}}}}
{{\rm M}\left(-{\frac {-2\,\beta+\gamma}{2\beta}},\frac{3}{2},\,{\frac {\beta\,{x}^{2}}{2\alpha}}\right)}
x+c_{{2}}{{\rm e}^{-{\frac {\beta\,{x}^{2}}{2\alpha}}}}
{{\rm U}\left(-{\frac {-2\,\beta+\gamma}{2\beta}},\,\frac{3}{2},\,{\frac {\beta\,{x}^{2}}{2\alpha}}\right)}
x
$$
It could also be written in terms of hypergeometric functions.
A: The general form as you have it, is a hypergeometric differential equation. You can manipulate it into a standard form and then apply the Frobenius method. It's already worked out here for several cases:
http://en.wikipedia.org/wiki/Frobenius_solution_to_the_hypergeometric_equation
