How could we solve this differential equation $$y(y'+3)=ax^2+bx+c, \quad a,b,c \in \mathbb{R}$$
I really don't know how start. I am not familiar with this sort of differential equations (I know it is non-linear fist order, but don't see way to solve it).
I passed this exam, so this is just for fun (not homework or something like that).
Thanks.
Case $1$: $a=b=c=0$
Then $y(y'+3)=0$
$y=0$ or $y'=-3$
$y=0$ or $C-3x$
Case $2$: $a=b=0$ and $c\neq0$
Then $y(y'+3)=c$
$\dfrac{dy}{dx}=\dfrac{c}{y}-3$
$\dfrac{dx}{dy}=\dfrac{y}{c-3y}$
$x=\int\dfrac{y}{c-3y}dy$
$x=C-\dfrac{y}{3}-\dfrac{c\ln(3y-c)}{9}$
Case $3$: $a=0$ and $b\neq0$
Then $y(y'+3)=bx+c$
$\dfrac{dy}{dx}=\dfrac{bx+c}{y}-3$
Let $y=\left(x+\dfrac{c}{b}\right)u$ ,
Then $\dfrac{dy}{dx}=\left(x+\dfrac{c}{b}\right)\dfrac{du}{dx}+u$
$\therefore\left(x+\dfrac{c}{b}\right)\dfrac{du}{dx}+u=\dfrac{b}{u}-3$
$\left(x+\dfrac{c}{b}\right)\dfrac{du}{dx}=\dfrac{b}{u}-3-u$
$\dfrac{u}{b-3u-u^2}du=\dfrac{dx}{x+\dfrac{c}{b}}$
$\int\dfrac{u}{b-3u-u^2}du=\int\dfrac{dx}{x+\dfrac{c}{b}}$
$\begin{cases}-\dfrac{\ln(u^2+3u-b)}{2}-\dfrac{3}{\sqrt{4b+9}}\tanh^{-1}\dfrac{2u+3}{\sqrt{4b+9}}=\ln\left(x+\dfrac{c}{b}\right)+C&\text{when}~b\neq-\dfrac{9}{4}\\-\ln(2u+3)-\dfrac{3}{2u+3}=\ln\left(x+\dfrac{c}{b}\right)+C&\text{when}~b=-\dfrac{9}{4}\end{cases}$
Case $4$: $a\neq0$
Then $y(y'+3)=ax^2+bx+c$
$y\dfrac{dy}{dx}+3y=ax^2+bx+c$
This belongs to an Abel equation of the second kind.
Let $y=-3u$,
Then $\dfrac{dy}{dx}=-3\dfrac{du}{dx}$
$\therefore9u\dfrac{du}{dx}-9u=ax^2+bx+c$
$u\dfrac{du}{dx}-u=\dfrac{ax^2+bx+c}{9}$
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 or in http://www.iaeng.org/IJAM/issues_v43/issue_3/IJAM_43_3_01.pdf
For finding the special cases which has simpler form of the general solution,
Case $4$a:
$u\dfrac{du}{dx}-u=\dfrac{ax^2+bx+c}{9}$
$u\dfrac{du}{dx}-u=\dfrac{a}{9}\left(x^2+\dfrac{bx}{a}+\dfrac{c}{a}\right)$
$u\dfrac{du}{dx}-u=\dfrac{a}{9}\left(x^2+\dfrac{bx}{a}+\dfrac{b^2}{4a^2}+\dfrac{c}{a}-\dfrac{b^2}{4a^2}\right)$
$u\dfrac{du}{dx}-u=\dfrac{a}{9}\left(x+\dfrac{b}{2a}\right)^2+\dfrac{4ac-b^2}{36a}$
Let $s=x+\dfrac{b}{2a}$ ,
Then $u\dfrac{du}{ds}-u=\dfrac{as^2}{9}+\dfrac{4ac-b^2}{36a}$
According to http://science.fire.ustc.edu.cn/download/download1/book%5Cmathematics%5CHandbook%20of%20Exact%20Solutions%20for%20Ordinary%20Differential%20EquationsSecond%20Edition%5Cc2972_fm.pdf#page=138, we find a special case which has simpler form of the general solution when $4ac-b^2=\dfrac{2916}{625}$ .
Case $4$b:
$u\dfrac{du}{dx}-u=\dfrac{ax^2+bx+c}{9}$
$u\dfrac{du}{dx}-u=\dfrac{a}{9}\left(x+\dfrac{b+\sqrt{b^2-4ac}}{2a}\right)\left(x+\dfrac{b-\sqrt{b^2-4ac}}{2a}\right)$
Let $t=x+\dfrac{b\pm\sqrt{b^2-4ac}}{2a}$ ,
Then $u\dfrac{du}{dt}-u=\dfrac{a}{9}t\left(t\pm\dfrac{\sqrt{b^2-4ac}}{a}\right)$
$u\dfrac{du}{dt}-u=\dfrac{at^2}{9}\pm\dfrac{\sqrt{b^2-4ac}}{9}t$
According to http://science.fire.ustc.edu.cn/download/download1/book%5Cmathematics%5CHandbook%20of%20Exact%20Solutions%20for%20Ordinary%20Differential%20EquationsSecond%20Edition%5Cc2972_fm.pdf#page=133 and http://science.fire.ustc.edu.cn/download/download1/book%5Cmathematics%5CHandbook%20of%20Exact%20Solutions%20for%20Ordinary%20Differential%20EquationsSecond%20Edition%5Cc2972_fm.pdf#page=138, we find some special cases which has simpler form of the general solution when $b^2-4ac=\dfrac{2916}{625}$ .
