solving differential equations : $ay' + by^2 + cy = u$

Question

solving differential equations: $ay' + by^2 + cy = u$ with $a,b,c$ are positive constants and $u$ is arbitrary constant.

I can solve this equation:

$ay' + by^2 + cy = 0$

$\Rightarrow -a\frac{dy}{y(by+c)}=dx $

$\Rightarrow \frac{a}{c}(\frac{bdy}{by+c}-\frac{dy}{y})=dx$

$\Rightarrow ln(\frac{by+c}{y})=\frac{c}{a}x$

$\Rightarrow y = \frac{c}{e^{\frac{c}{a}x}-b}$

But with an arbitrary constant at the righthand side, I have no idea.

Answer 1

With a constant in the RHS, the equation is still separable and you solve it the same way !

$$\frac{a\,dy}{by^2+cy-u}=-dx.$$

After integration,

$$\frac{2a}{\sqrt{c^2+4 b u}}\arctan\frac{c + 2 b y}{\sqrt{c^2 + 4 b u}}=c-x$$ from which you can draw $y$.

Answer 2

This is an equation of the Riccati type. One can set $y=\frac{az'}{bz}$, then $$ u=ay'+by^2+cy=\frac{a^2z''}{bz}+\frac{caz'}{bz}\\~\\ \iff a^2z''+acz'-ubz=0 $$ This now is a second order linear DE with constant coefficients, so standard solution methods apply. The characteristic roots are solutions of the quadratic polynomial $$ 0=4a^2\lambda^2+4acλ-4ub=(2aλ-c)^2-(c^2+4ub) $$