Solve differential equation $y'''(t)=y(t) y'(t)$. Solve following diferential equations
$$y'''(t)=y(t) y'(t)$$
I would appreciate some help with this problem. Thanks in advance.
 A: Integrating once, we have
$$y'' = \frac{1}{2}y^2 + C$$
Multiplying both sides by $y'$ and integrating again, we have
$$\frac{1}{2}(y')^2 = \frac{1}{6}y^3 + Cy + D \ \ \ \  \text{ or } \ \ \ \ \   y' = \pm \sqrt{\frac{y^3}{3} +2 Cy + D}$$
This equation is separable. Now, how amenable the integral $\displaystyle \int \frac{dy}{\sqrt{\frac{y^3}{3} + 2Cy + D}}$ is will depend on the constants. E.g., if $C = D = 0$, then this straightforward; not so in most other cases unfortunately!
A: The general solution of $y'''(t)=y(t)y'(t)$ can be analytically expressed thanks to the Jacobi amplitude function (see the page below). Alternatively, an equivalent form using the Weierstrass elliptic function could be used.

A: A surprising amount of the time, you can get solutions to ODE problems just by trying common solutions, and $y = A t^b$ is often a good first guess. Plugging this in gives
$$A b(b-1)(b-2) t^{b-3}= A^2b t^{2b - 1}$$
The exponents have to agree, so $b - 3 = 2b - 1$ or $b = -2$. So we now have
$$-24At^{-5} = -2A^2t^{-5}$$
This forces $A$ to be $0$ or $12$. Thus $y = 12 t^{-2}$ is one solution. Since the differential equation is expressed in terms of $y$ only, $y = 12(t - a)^{-2}$ is a solution for any $a$. 
This doesn't give all solutions since it's a third order equation which is expected to have three constants showing up.. but as can be seen from the other answers, all the other solutions are expressed as elliptic integrals. So for example if you're just looking for a solution with a certain value of $y(b)$ for some $b$ the above would give the most reasonable option.
