Hey guys. That's really an interesting problem in transformation theory and involves the solution of the general elliptic equation \(\displaystyle y''=A+By+Cy^2+Dy^3\). The following is adapted from "Introduction to Nonlinear Differential and Integral Equations" by H. Davis.

We wish to get the DE into the form:

\(\displaystyle \left(\frac{dz}{dx}\right)^2=(1-z^2)(1-k^2 z^2)=\Delta^2(z)\)

which has the solution \(\displaystyle z=\text{sn}(x,k)\) where \(\displaystyle \text{sn}\) is the Jacobi elliptic function. We can do that by writing \(\displaystyle y''=A-y+by^3\) in it's elliptic form \(\displaystyle \left(y'\right)^2=h^2(y-\alpha)(y-\beta)(y-\gamma)(y-\delta)=h^2\Delta_2^2(y)\) (multiply the general elliptic equation by y' and integrate) and then using the transformations:

\(\displaystyle z^2=\frac{\beta-\delta}{\alpha-\delta}\frac{y-\alpha}{y-\beta}\)

\(\displaystyle k^2=\frac{(\beta-\gamma)(\alpha-\delta)}{(\alpha-\gamma)(\beta-\delta)}\)

\(\displaystyle M^2=\frac{(\beta-\delta)(\alpha-\gamma)}{4}\)

and here is the challenging part. Show:

\(\displaystyle \frac{1}{\Delta(z)}\frac{dz}{dx}=\frac{M}{\Delta_2(y)}\frac{dy}{dx}\)

That expression is a little easier to show if the factored elliptic equation is only a cubic and involving simpler transformation rules and is the version I went through. I assume the quartic is just more messy. Anyway, it then follows:

\(\displaystyle z=\text{sn}(hMx,k)\)

and by inverting the expression for z above, the expression for y follows. Seems to me though the roots in general are complex which then means the solution is complex-analytic. I say give a good 45 minute presentation on this Polo and you don't have to take the final exam. Some nice plots too.