TITLE: Solving $\frac{dy}{dx} = ay^2 + (b + qe^{rx})y$
QUESTION [1 upvotes]: I have been attempting to solve this differential equation:
$\frac{dy}{dx} = ay^2 + (b + qe^{rx})y$
I attempt solving the Bernoulli equation using $v = y^{-1}$.  However when I do this, I get an unsolvable integrating factor $e^{bx+qe^{rx}}$ that prevents me from solving this differential equation.
Is there another way to go about this problem that I may be overlooking?

REPLY [0 votes]: This is Bernoulli type ODE:
$$\frac{dy}{dx}-(b+qe^{rx})y=ay^2 \implies y^{-2}\frac{dy}{dx}-(b+q e^{rx})y^{-1}=a.$$
Let $y^{-1}=v \implies -y^{-2}\frac{dy}{dx}=\frac{dv}{dx}$, then we get, the linear ODE
$$\frac{dv}{dx}+(b+qe^{rx})v=-a$$
whose integrating factor is $I=\exp[bx+(q/r)e^{rx})]$
so the solution is
$$v=y^{-1}=\exp[-(bx+(q/r)e^{rx})] \int  -a \exp[bx+(q/r)e^{rx})] dx+ C \exp[-(bx+(q/r)e^{rx})]$$
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