What is the name of this special class of Ordinary Differential Equations? Using a suitable substitution, transform the following equation into a first order Linear differential equation:
$y' = f_0(x) +f_1(x)*y + f_2(x) *y^2$  for $f_2(x)\not\equiv 0$
Attempt at a solution: I've found that if $y_1(x)$ is a solution, we can use $y(x) = y_1(x) + \frac{1}{u(x)}$ but I'm unfamiliar with proving stuff for this special class of ODEs. Please help.
 A: For 
$$y' = f_0(x) +f_1(x)*y + f_2(x) *y^2 \hspace{5mm}  \text{for} \, f_2(x) \not\equiv 0$$
then, suppose $q_{0}(x)$ is a solution, let $y(x) = q_{0}(x) + p(x)$ to obtain
\begin{align}
q_{0}' + p' &= f_{0} + f_{1} q_{0} + f_{1} p + f_{2} (q_{0}^{2} + 2 q_{0} p + p^{2}) \\
p' &= f_{2} p^{2} + (f_{1} + 2 q_{0} f_{2}) p + (- q_{0}^{'} + f_{2} q_{0}^{2} + f_{1} q_{0} + f_{0}) \\
p' &= f_{2} p^{2} + Q_{0} p,   
\end{align}
where $Q_{0} = f_{1} + 2 q_{0} f_{2}$. 
Now 
\begin{align}
\frac{p'}{p^{2}} - \frac{Q_{0}}{p} &= f_{2} \\
- \frac{d}{dx} \left(\frac{1}{p}\right) - Q_{0} \, \frac{1}{p} &= f_{2}
\end{align}
Let $v(x) = 1/p(x)$ to obtain, with use of integrating factor,
\begin{align}
v' + Q_{0} v &= - f_{2} \\
\frac{d}{dx} \left[ v \, e^{\int^{x} Q_{0} dt} \right] &= - f_{2} \, e^{ \int^{x} Q_{0} dt} \\
v &= e^{- \int^{x} Q_{0} dt} \, \left[c - \int^{x} f_{2}(u) \, e^{\int^{u} Q_{0} dt} \, du \right] 
\end{align}
Placing everything together provides
$$y(x) = q_{0}(x) + e^{\int^{x} (f_{1}(t) + 2 q_{0}(t) f_{2}(t)) dt} \, \left(c + \int^{x}  f_{2}(u) \, e^{- \int^{u} (f_{1}(t) + 2 q_{0}(t) f_{2}(t)) dt} \, du \right)^{-1} $$
