# How to solve $y(y'+3)=ax^2+bx+c, \quad a,b,c \in \mathbb{R}$

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.

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 $b^2-4ac=\dfrac{2916}{625}$ .

• Thanks, that was helpful. But, I can't see that special case in that link, maybe I am blind Mar 1, 2013 at 21:15
• Dear doraemonpaul, there is one big problem. Method hindawi.com/journals/ijmms/2011/387429/#sec2 is wrong, look at en.wikipedia.org/wiki/… : "The solutions proposed by these two people is wrong! So, Abel equation is still unsolved!" And I was reading mention paper and note same mistakes, so I think is true that it's wrong. Mar 2, 2013 at 10:29