ordinary differential equation of third order using substitution How one can solve ODE in the following form?
$$y''' y +(y'')^{2} =0$$
It looks like some kind of substitution. I did try some substitution but they were useless.
thanks in advance.
 A: With reference to http://eqworld.ipmnet.ru/en/solutions/ode/ode0503.pdf,
Let $u=\left(\dfrac{dy}{dx}\right)^2$ ,
Then $\dfrac{du}{dx}=2\dfrac{dy}{dx}\dfrac{d^2y}{dx^2}$
$\dfrac{du}{dy}\dfrac{dy}{dx}=2\dfrac{dy}{dx}\dfrac{d^2y}{dx^2}$
$2\dfrac{d^2y}{dx^2}=\dfrac{du}{dy}$
$2\dfrac{d^3y}{dx^3}=\dfrac{d}{dx}\left(\dfrac{du}{dy}\right)=\dfrac{d}{dy}\left(\dfrac{du}{dy}\right)\dfrac{dy}{dx}=\mp\sqrt u\dfrac{d^2u}{dy^2}$
$\therefore\mp y\sqrt u\dfrac{d^2u}{dy^2}+\dfrac{du}{dy}=0$
$\dfrac{d^2u}{dy^2}=\pm\dfrac{1}{y\sqrt u}\dfrac{du}{dy}$
Which reduces to a generalized Emden-Fowler equation 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=362
WLOG, just consider $\dfrac{d^2u}{dy^2}=\dfrac{1}{y\sqrt u}\dfrac{du}{dy}$
The general solution is $\begin{cases}u=\dfrac{1}{t^2}\\y=\pm C_2e^{\int\frac{dt}{t^3\left(-\frac{1}{t}-\frac{1}{2t^2}+C_1\right)}}\end{cases}$
$\begin{cases}\left(\dfrac{dy}{dx}\right)^2=\dfrac{1}{t^2}\\y=\pm C_2e^{\int\frac{dt}{C_1t^3-t^2-\frac{t}{2}}}\end{cases}$
$\begin{cases}\dfrac{dy}{dx}=\pm\dfrac{1}{t}\\y=\pm C_2e^{\int\frac{dt}{C_1t^3-t^2-\frac{t}{2}}}\end{cases}$
$\begin{cases}\dfrac{\dfrac{dy}{dt}}{\dfrac{dx}{dt}}=\pm\dfrac{1}{t}\\y=\pm C_2e^{\int\frac{dt}{C_1t^3-t^2-\frac{t}{2}}}\end{cases}$
$\begin{cases}\dfrac{dx}{dt}=\pm\frac{C_2}{C_1t^2-t-\frac{1}{2}}e^{\int\frac{dt}{C_1t^3-t^2-\frac{t}{2}}}\\y=\pm C_2e^{\int\frac{dt}{C_1t^3-t^2-\frac{t}{2}}}\end{cases}$
$\begin{cases}x=\pm\int\frac{C_2}{C_1t^2-t-\frac{1}{2}}e^{\int\frac{dt}{C_1t^3-t^2-\frac{t}{2}}}~dt+C_3\\y=\pm C_2e^{\int\frac{dt}{C_1t^3-t^2-\frac{t}{2}}}\end{cases}$
