Exact solution of $y''=(ay+b)y'+cy^2+dy$

Question

Consider the second order nonlinear ode of the following form $y''=(ay+b)y'+cy^2+dy$, where $a, b, c, d$ are real constants. Can we find the exact solution of this equation?

Answer 1

$$\frac{d^2y}{dx^2}=(ay+b)\frac{dy}{dx}+cy^2+d\,y$$ This is an autonomous ODE. In order to reduce the order, the usual change of function is : $$\frac{dy}{dx}=u(y)\quad;\quad \frac{d^2y}{dx^2}=\frac{d}{dx}\frac{dy}{dx}=\frac{du}{dx}=\frac{du}{dy}\frac{dy}{dx}=\frac{du}{dy}u$$ $$u\frac{du}{dy}=(ay+b)u+cy^2+d\,y$$ This is a first order non-linear ODE.

Let $\quad u(y)=\frac{1}{v(y)}\quad ;\quad \frac{du}{dy}= -\frac{dv}{dy}\frac{1}{v^2}$ $$\frac{1}{v}\left(-\frac{dv}{dy}\frac{1}{v^2}\right)=(ay+b)\frac{1}{v}+cy^2+d\,y$$ $$\frac{dv}{dy}=-(ay+b)v^2-(cy^2+d\,y)v^3$$ This is an Abell ODE of the first kind. In general, there is no closed form solution, except in cases of particular values of the parameters. To go further, see : https://arxiv.org/ftp/arxiv/papers/1503/1503.05929.pdf

Answer 2

$y''=(ay+b)y'+cy^2+dy$

$\dfrac{d^2y}{dx^2}=(ay+b)\dfrac{dy}{dx}+cy^2+dy$

Let $u=\dfrac{dy}{dx}$ ,

Then $\dfrac{d^2y}{dx^2}=\dfrac{du}{dx}=\dfrac{du}{dy}\dfrac{dy}{dx}=u\dfrac{du}{dy}$

$\therefore u\dfrac{du}{dy}=(ay+b)u+cy^2+dy$

This belongs to an Abel equation of the second kind.

Case $1$: $a\neq0$

Let $s=y+\dfrac{b}{a}$ ,

Then $u\dfrac{du}{ds}=asu+c\left(s-\dfrac{b}{a}\right)^2+d\left(s-\dfrac{b}{a}\right)$

$u\dfrac{du}{ds}=asu+cs^2+\dfrac{(ad-2bc)s}{a}+\dfrac{b^2c-abd}{a^2}$

Let $t=\dfrac{as^2}{2}$ ,

Then $\dfrac{du}{ds}=\dfrac{du}{dt}\dfrac{dt}{ds}=as\dfrac{du}{dt}$

$\therefore asu\dfrac{du}{dt}=asu+cs^2+\dfrac{(ad-2bc)s}{a}+\dfrac{b^2c-abd}{a^2}$

$u\dfrac{du}{dt}=u+\dfrac{cs}{a}+\dfrac{ad-2bc}{a^2}+\dfrac{b^2c-abd}{a^3s}$

$u\dfrac{du}{dt}=u\pm\dfrac{c\sqrt{2t}}{a\sqrt{a}}+\dfrac{ad-2bc}{a^2}\pm\dfrac{b^2c-abd}{a^2\sqrt{a}\sqrt{2t}}$

This belongs to an Abel equation of the second kind in the canonical form.

Please follow the method in https://arxiv.org/ftp/arxiv/papers/1503/1503.05929.pdf

For the special case $4c^2=a^2d-2abc$ , this exactly belongs to the ODE of the form http://science.fire.ustc.edu.cn/download/download1/book%5Cmathematics%5CHandbook%20of%20Exact%20Solutions%20for%20Ordinary%20Differential%20EquationsSecond%20Edition%5Cc2972_fm.pdf#page=137.