How to solve $a \frac{d^2 y}{d x}+b \frac{d y}{d x} = f(y)$? Let $a,b$ be real numbers and $y$ is a function of $x$.
$f$ is a given function.
How to solve the ODE : 
$a \dfrac{d^2 y}{d x}+b \dfrac{d y}{d x} = f(y)$ ?
Can it be done in closed form ?
 A: I can only suggest a partial solution for $b=0$. In this case, set $a=1$ and multiply the equation by $2y'$. It then transforms into
$$\left((y')^2-2F(y)\right)'=0,$$
where $F(y)$ denotes the antiderivative of $f(y)$. Then
$$y'=\pm\sqrt{C_1+2F(y)}\quad\Longrightarrow\quad \int\frac{dy}{\sqrt{C_1+2F(y)}}=\pm x+C_2.$$
Though we have solved the equation in quadratures, the integral on the left is very rarely computable in closed form. Essentially, $F(y)$ should be a rational function of sufficiently small degree or something reducible to that.
A: Assume $a\neq0$ and $f(y)$ is a non-linear function for convenient to study the key meaning of this question.
in fact this ODE is the special case of http://eqworld.ipmnet.ru/en/solutions/ode/ode0317.pdf.
Let $u=\dfrac{dy}{dx}$ and consider $y$ as the independent variable, you will get $au\dfrac{du}{dy}+bu=f(y)$ .
When $b\neq0$ , Let $u=\dfrac{1}{v}$ , you will get $\dfrac{dv}{dy}=-\dfrac{f(y)v^3}{a}+\dfrac{bv^2}{a}$ , which is an Abel equation of the first kind.
Then you can study http://www.hindawi.com/journals/ijmms/2011/387429/#sec2 for solving this Abel equation of the first kind analytically.
