. . . ohhhhhh . . . you guys give up too easy. Assume we are given that the solution to:
\left(\frac{dz}{dx}\right)^2=(1-z^2)(1-k^2z^2)=\Delta^2(z)
is:
y=sn(z,k)
where sn is the Jacobi elliptic function. Then we seek a transformation z=z(y) that transforms:
<br />
\begin{align*}<br />
\left(\frac{dy}{dx}\right)^2&=a+by+cy^2+dy^3+ey^4\\<br />
&=h^2(y-\alpha)(y-\beta)(y-\gamma)(y-\delta) \\<br />
&=h^2 \Delta_2^2(y)<br />
\end{align*}<br />
into this standard form. To this end we let:
z^2=\frac{(\beta-\gamma)}{(\alpha-\delta)}\frac{(y-\alpha)}{(y-\beta)}=s\frac{(y-\alpha)}{(y-\beta)},\quad k^2=\frac{(\beta-\delta)}{(\alpha-\gamma)}\frac{(\alpha-\delta)}{(\beta-\delta)},\quad M^2=\frac{(\beta-\delta)(\alpha-\delta)}{4}
for which we obtain:
\frac{1}{\Delta(z)}\frac{dz}{dx}=\frac{M}{\Delta_2(y)}\frac{dy}{dx}=Mh
so that:
\frac{dz}{dx}=Mh\Delta(z)=Mh\sqrt{(1-z^2)(1-k^2z^2)}
and therefore:
z=sn(hMv,k),\quad v=x-x_0
or:
y=\frac{z^2\beta-s\alpha}{z^2-s}
I believe though the actual implementation of this would be difficult as I have never worked a real problem using this method but I think would be a nice project for someone taking non-linear DEs next semester. :)
