How to solve this DE? Consider the ordinary differential equation
$$y''=xyy'$$
I'm pretty stumped, so any tips on how to proceed? It seems fairly simple but I'm drawing a blank.
 A: In dimensions, the two terms are $Y/X^2$ and $XY^2/X$.  To make them match, the dimensions are $Y=X^{-2}$.
Let $w=x^2y$, it becomes homogeneous in $x$.
$$\frac{dw}{dx}=2xy+x^2y'\\ \frac{d^2w}{dx^2}=2y+4xy'+x^2(xyy')\\
$$
You should convert the second equation to just $w$ and $x$, using the first equation to get rid of $y$.
Doubling $x$ now has no effect on the equation.  So adding a constant to $\ln x$ has no effect on the equation.  So convert to $t=\ln x$, find the equation for $w(t)$.  $t$ will not appear explicitly in the equation; just $w,\frac{dw}{dt}$ and $\frac{d^2w}{dt^2}$.
Do you know a method for reducing the order of a DE when $t$ doesn't appear explicitly?
A: I'm not sure if this really counts as an answer, but I'd like to see if this work according to @Michael's sketch is along the right track.
As he mentions, as step 1 we homogenize units by setting $w(x)=x^2y(x)$, and obtain a differential equation for $w$:
$$ x^2w'' = -6w-2w^2+4xw'+xww' .$$
Then (step 2) we change coordinates to eliminate the explicit dependent variable, setting $t= \ln x$. Notice that 
$$ \frac{\mathrm{d}w}{\mathrm{d}x}=\frac{\mathrm{d}w}{\mathrm{d}t}\frac{\mathrm{d}t}{\mathrm{d}x}=\frac{\mathrm{d}w}{\mathrm{d}t} \frac{1}{x} $$
and similarly
$$ \frac{\mathrm{d}^2w}{\mathrm{d}x^2}= -\frac{1}{x^2}\frac{\mathrm{d}w}{\mathrm{d}t}+\frac{1}{x^2}\frac{\mathrm{d}^2w}{\mathrm{d}t^2} .$$
So now with $'$ indicating differentiation with respect to $t$, we have 
$$w'' = -6w-2w^2+5w'+ww' .$$
For step 3, he suggests we reduce the order to a first order differential equation. Write $v:=w'$. Then
$$ \frac{\mathrm{d}^2w}{\mathrm{d}t^2} = \frac{\mathrm{d}v}{\mathrm{d}t} =\frac{\mathrm{d}v}{\mathrm{d}w}\frac{\mathrm{d}w}{\mathrm{d}t} = \frac{\mathrm{d}v}{\mathrm{d}w} v ,$$
So now letting $'$ indicate differentiation with respect to $w$, we obtain
$$ v'v-(w+5)v=-6w-2w^2 .$$
EDIT: I found an easy solution; I must have been making silly mistakes. Solutions of this differential equation are $v=2w$ and $v \equiv 0$ which yield solutions where $y$ is constant or $y$ is proportional to $x^{-2}$.
