Solving the Weierstrass differential equation for $g_2^3=27g_3^2$ I am asked to solve the Weierstrass Differential Equation:
$$(y')^2=4y^3-g_2y-g_3.$$
in the case that $g_2^3=27g_3^2$. This seems easy, but I can't quite figure it out. Any ideas?
 A: The solution $y(z)=\wp(z+c;\omega_1,\omega_2)$ of this differential equation in the general case is called Weierstrass $\wp$-function:
$$\wp(z;\omega_1,\omega_2)=\frac{1}{z^2}+\sum_{(m,n)\in\mathbb{Z}^2\backslash\{(0,0)\}}\left\{\frac{1}{(z+m\omega_1+n\omega_2)^2}-\frac{1}{(m\omega_1+n\omega_2)^2}\right\},$$
where
\begin{align}
&g_2=60\sum_{(m,n)\in\mathbb{Z}^2\backslash\{(0,0)\}}\frac{1}{(m\omega_1+n\omega_2)^4},\\
&g_3=140\sum_{(m,n)\in\mathbb{Z}^2\backslash\{(0,0)\}}\frac{1}{(m\omega_1+n\omega_2)^6}.
\end{align}
So I wouldn't call this differential equation easy.
However, for precisely your constraint on parameters ($g_2^3=27g_3^2$) it can be solved in terms of elementary functions. Indeed, parameterizing $g_{2,3}$ as $g_2=3\alpha^2$, $g_3=\alpha^3$ we see that the cubic polynomial on the right has a double root and we get
$$(y')^2=(y-\alpha)(2y+\alpha)^2.$$
This can be integrated (in terms of elementary functions), since because of the double root we can write
$$\int\frac{dy}{(2y+\alpha)\sqrt{y-\alpha}}=\pm z+\mathrm{const}.$$
I suppose you can take it from here.
