How to solve the ODE $y(y'(x)+a)=bx$ I've been very frustrated attempting to separate this guy. Since there are three separate items when you multiply through by the y, it is very difficult to make all sides work out to forms $f(y)dy$ and $f(x)dx$. I've attempted the trick of getting it in the form $y'dx/y$ to get the integral to be $\log(y)$, but haven't successfully separated it yet. Any help would be appreciated, I want to understand your attempt, not just apply it if you know what I mean. 
Also, initial condition is: $ y(0)=0$
EDIT: we can try $y=Cx$ and find that it is a valid solution for specific C values. 

We have $$0 = y(y'+a) - bx = \frac{1}{2}\frac{d}{dx}(y^2 + ay - bx^2) - \frac{a}{2}x^2\frac{d}{dx}\left(\frac{y}{x}\right)$$
Let us therefore consider the function
$$h(x) = \frac{y(y'+a) - bx}{y^2 + axy - bx^2}$$
Then $h \equiv 0$ and by integrating the last expression we get
$$\int \frac{y(y'+a)-bx}{y^2 + axy - bx^2} dx = C$$
Lets rewrite this as
$$\frac{1}{2}\int \frac{\frac{d}{dx}(y^2 + axy - bx^2) - ax^2\frac{d}{dx}(y/x)}{y^2 + axy - bx^2} dx = C$$
This first part integrates to $$\frac{1}{2}\log(y^2 + axy - bx^2)$$ and for the second part we can write it as
$$-\frac{a}{2}\int \frac{\frac{d}{dx}(y/x + a/2)}{(y/x + a/2)^2 - (a^2/4 + b)} dx$$
which by making the substitution $z = \frac{y/x + a/2}{\sqrt{-a^2/4-b}}$ becomes
$$-\frac{a}{2\sqrt{a^2/4+b}}\int \frac{dz}{z^2 + 1} = -\frac{a}{2\sqrt{-a^2/4-b}}\arctan(z)$$
The solution can be written as the implicit equation
$$\frac{1}{2}\log(y^2 + axy - bx^2) -\frac{a}{2\sqrt{-a^2/4-b}}\arctan\left(\frac{y/x +a/2}{\sqrt{-a^2/4-b}}\right) = C$$
There might be some ways to simplify this further for particular choices of $a,b$, but in general it seems this is the closest you are going to get to a closed form solution.
${\bf Edit:}$
The solution above is for $4b + a^2 < 0$. In the opposite case the integral has a simpler representation. Making the substitution $z = \frac{y/x + a/2}{\sqrt{a^2/4+b}}$ instead we have
$$-\frac{a}{2\sqrt{a^2/4+b}}\int \frac{dz}{z^2 - 1} = -\frac{a}{4\sqrt{a^2/4+b}}\log \left(\frac{1-\frac{y/x + a/2}{\sqrt{a^2/4+b}}}{1+\frac{y/x + a/2}{\sqrt{a^2/4+b}}}\right)$$
so that we get
$$\frac{1}{2}\log(y^2 + axy - bx^2)-\frac{a}{4\sqrt{a^2/4+b}}\log \left(\frac{1-\frac{y/x + a/2}{\sqrt{a^2/4+b}}}{1+\frac{y/x + a/2}{\sqrt{a^2/4+b}}}\right) = C$$
