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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!

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Yes, I was there too.... I was thinking for some reason there may be some complications due to the sign of the square root. Anyway, thanks! – nate Feb 1 '12 at 0:30
"I have seen some solutions (but did not understand them) where the polynomial was "factored" into 3 roots" - and that is precisely the Weierstrass approach. – J. M. Feb 8 '12 at 18:17
up vote 2 down vote accepted

The equation is not too hard to solve.

$$\begin{align*} \left(\frac{dy}{dz}\right)^2 &= cy^3+y^2\\ \frac{dy}{y \sqrt{cy+1}} &= dz\\ cy+1 &= u\\ 2 \frac{du}{u^2-1} &= dz\\ - 2 \tanh^{-1} u + C&= z\\ u &= \tanh \frac{-z+C}{2}\\ \sqrt{cy+1} &= -\tanh \frac{z-C}{2}\\ cy+1 &= \tanh^2 \frac{z-C}{2}\\ y &= \frac{\tanh^2 \frac{z-C}{2}-1}{c}\\ y &= -\left({c\times \cosh^2 \frac{z-C}{2}}\right)^{-1} \end{align*}$$

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Aaahhh, I think that the $g^2 = 3\,c-f$$ must have been a hint towards a substitution in the integration. Turns out it was just confusing the way it was suggested in the problem. Now I see it as a straight-forward problem. Thanks all! – nate Feb 1 '12 at 1:16

There is this famous differential equation ... $$ [\wp'(z)]^2 = 4[\wp(z)]^3 - g_2\wp(z) - g_3 $$ Weierstrass Elliptic Functions

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Ahh. Thank you - I was also wondering how I was going to solve this pde if I wasn't able to get rid of 2 BC's - would lead to a third-order polynomial in p(z) just like you mentioned. Thanks! – nate Feb 1 '12 at 23:31

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


(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


We then compute the discriminant


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


and thus


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.

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