Solve $y''=y^2$ Are there any 'basic' solutions to this differential equation (ie using polynomials, exponetials, trigonometric functions and logarithms)? I cannot figure it out at all using the techniques I know for solving differential equations.
 A: Multiplying this equation by $2y'$, we get
$$\left(y'^2\right)'=\frac23\left(y^3\right)',$$
which implies that 
$$y'^2=\frac23y^3+C_1.$$
Although the general solution of this can be written in terms of elliptic integrals (the equation is separable), there is a particularly simple solution corresponding to $C_1=0$: then
$$y'=\pm \sqrt{\frac23}y^{\frac32}\qquad \Longrightarrow \qquad -2y^{-\frac12}=\pm \sqrt{\frac23} x+C_2.$$
A: A hint: Multiply both sides of the equation by $y'$.
A: $$2y''y'=2y^2y'$$
$$(y')^2=\frac{2}{3}y^3+c_1$$
$$1=\frac{y'}{\sqrt{\frac{2}{3}y^3+c_1}}$$
$$x=\int\frac{dy}{\sqrt{\frac{2}{3}y^3+c_1}}+c_2$$ 
This elliptic integral cannot be expressed with a finite number of elementary functions. It can be reduced to an elliptic integral of the first kind, on the form of a complicated function $x(y)$. Furthermore, the inverse function $y(x)$ is not easy to derive.
It is better to refer to a particular case of Weierstrass's ODE :
$$(y')^2=\frac{2}{3}y^3+c_1$$
http://mathworld.wolfram.com/WeierstrassEllipticFunction.html
Eq.(34) where $g_2=0$ 
So, the simplest closed form of the general solution of $y''=y^2$ can be expressed on the form :
$$y(x)=6^{1/3}\wp \left(\frac{x+C_2}{6^{1/3} };0\: ,\: C_1\right)$$
$\wp(\:)$ is the Weierstrass-p function.
A: A physicist's substitution always helps when the independent variable is missing and you have second derivatives. Let $t$ be the independent variable ($y'=dy/dt$). Then use a new variable
$$y'=v$$
and substitute
$$y''=v\,dv/dy$$
and integrate with respect to $y$.
In this case:
$$v\,dv/dy=y^2$$
$$\int v\, dv=\int y^2 \, dy$$
$$v^2/2=y^3/3+C$$
Now put $v=dy/dt$  back in and integrate again to get $y(t)$.
In physics, you often have second derivatives over time (2nd Newton's law), so this is a routine bread-and-butter method for converting equations to first order and solving them routinely.
A: Alternative hint...note that $$y''=y'\frac{dy'}{dy},$$ which will make a separable variable DE
