# 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?

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$$y(x)=\sum_{k=0}^{\infty} a_k x^{k+\alpha} \,,$$

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.

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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.

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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.

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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

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