ordinary differential equation: $(f'(z))^2 = c\,f(z)^3 + f(z)^2$ I swear I have seen this type of ODE before, but I can't remember how to attack it. In general, I would like to know how to solve $$\left(f'(z)\right)^m = c\,G(z)^n$$ where $m,\;n \in \mathbb{N}$ and $G(z)$ is just a polynomial in $f(z)$.
This sounds hard, so I would be happy with, $$\left(f'(z)\right)^2 = c\,G(z)^n,$$ though this latter equation may be too difficult too. For my homework, though, I need to know, $$\left(f'(z)\right)^2 = \left( c\,f^3 + f^2 \right).$$ It was also suggested in the homework question to utilize $$g^2 = 3\,c-f.$$ If I do this without thinking I get an equation $$\left(f'(z)\right)^2 = c\,f(z)^2,$$ which seems much easier, though I am still a little rattled by the plus-minus.
In case it matters, this is a related to a method for solving the Korteweg-deVries equation, $$u_t + u\,u_x + u_{xxx} = 0.$$ I have seen some solutions (but did not understand them) where the polynomial was "factored" into 3 roots... something like that. I just don't want to know the answer, but how to get it. Please keep in mind that this is my first class in PDEs.  
Thanks for any help!
 A: The equation is not too hard to solve.
$$\begin{align*}
\left(\frac{dy}{dz}\right)^2 &= cy^3+y^2\\
\frac{dy}{y \sqrt{cy+1}} &= dz\\
cy+1 &= u\\
2 \frac{du}{u^2-1} &= dz\\
- 2 \tanh^{-1} u + C&= z\\
u &= \tanh \frac{-z+C}{2}\\
\sqrt{cy+1} &= -\tanh \frac{z-C}{2}\\
cy+1 &= \tanh^2 \frac{z-C}{2}\\
y &= \frac{\tanh^2 \frac{z-C}{2}-1}{c}\\
y &= -\left({c\times \cosh^2 \frac{z-C}{2}}\right)^{-1}
\end{align*}$$ 
A: There is this famous differential equation ...
$$
[\wp'(z)]^2 = 4[\wp(z)]^3 - g_2\wp(z) - g_3
$$
Weierstrass Elliptic Functions
A: Here's how to complete GEdgar's Weierstrass solution. Start with the differential equation
$$(f^\prime (z))^2 = c\,f(z)^3 + f(z)^2$$
and introduce a new function $g(z)$ satisfying the relation
$$f(z)=\frac4{c}g(z)-\frac1{3c}$$
(This is essentially equivalent to "depressing" a cubic equation plus a rescaling.)
We can then derive a differential equation for $g(z)$ through this substitution:
$$(g^\prime (z))^2=4g(z)^3-\frac1{12}g(z)+\frac1{216}$$
and comparing this with the Weierstrass differential equation, we find that
$$g(z)=\wp\left(z;\frac1{12},-\frac1{216}\right)$$
We then compute the discriminant
$$\Delta=\left(\frac1{12}\right)^3-27\left(-\frac1{216}\right)^2=0$$
and find that the underlying cubic is degenerate. Abramowitz and Stegun give appropriate formulae for the case of $\Delta=0$, $g_2 > 0$, and $g_3 < 0$ (see formula 18.12.3); applying the formula listed there yields
$$\wp\left(z;\frac1{12},-\frac1{216}\right)=\frac1{12}+\frac14\mathrm{csch}^2\left(\frac{z}{2}\right)$$
and thus
$$f(z)=\frac4{c}\left(\frac1{12}+\frac14\mathrm{csch}^2\left(\frac{z}{2}\right)\right)-\frac1{3c}=\frac1{c}\mathrm{csch}^2\left(\frac{z}{2}\right)$$
Note that I only derived a particular solution; the modifications necessary for a general solution (i.e., taking the constant(s) of integration into account) is left as an exercise to the reader.
