The solutions of $(y')^2=P(y)$ with $\deg P \in \{3,4\}$. Here is some background on my question. The Weierstraß function (as well as its translates) $\wp(x)$ solves the implicit first-order ODE
$$(y')^2=4y^3-g_2y-g_3.$$
Differentiating gives the second-order explicit ODE
$$y''=12y^2-g_2, $$
in which $g_3$ makes no appearance.
$\wp$ can be used to express the solution of any first-order equation of the form
$$(y')^2=a_3 y^3+a_2 y^2+a_1 y+a_0, $$
or equivalently, any second-order equation of the form
$$y''=b_2 y^2+b_1 y+b_0. $$
In both cases, all it takes is a simple change of variables to remove one of the powers, and get the leading coefficient right.
My question is about differential equations whose RHS is one degree larger, that is
$$(y')^2=a_4 y^4+a_3y^3+a_2y^2+a_1 y+a_0 , \tag{1}$$
or
$$y''=b_3 y^3+b_2y^2+b_1 y+b_0. \tag{2} $$
In some special cases, such as
$$y''=2k^2 y^3-(1+k^2) y, $$
and 
$$y''=-2y^3+(2-k^2)y,$$
the Jacobi elliptic functions $\operatorname{sn},\operatorname{dn}$ play a role in the solution. 
Here is my question: For which values of the coefficients in $(1)$ and $(2)$ can the ODEs be solved explicitly in terms of special functions? What are the solutions in those cases?
Thank you!
 A: The change of the function $y= \frac{\alpha z+\beta}{\gamma z+\delta}$ with $\alpha\delta-\beta\gamma=1$ transforms (1) into
$$\left(z'\right)^2=a_4\left(\alpha z+\beta\right)^4+
a_3\left(\alpha z+\beta\right)^3\left(\gamma z+\delta\right)
+a_2\left(\alpha z+\beta\right)^2\left(\gamma z+\delta\right)^2+
+a_1\left(\alpha z+\beta\right)\left(\gamma z+\delta\right)^3+
a_0\left(\gamma z+\delta\right)^4=\operatorname{poly}_4\left( z\right).$$
We can make the coefficient of $z^4$ disappear by the appropriate choice of $\alpha/\gamma$. The problem is thus completely equivalent to the previous differential equation with cubic right hand side and can be solved in terms of Weierstrass $\wp$-function. 
The change of variable above is exactly the one which transforms the integral $\int\frac{dy}{\sqrt{\operatorname{poly}_4\left( y\right)}}$ into $\int\frac{dz}{\sqrt{\operatorname{poly}_3\left( z\right)}}$. This has the following geometric interpretation: $w^2=\operatorname{poly}_4\left( y\right)$ is an elliptic curve realized as a $2$-sheeted covering of the Riemann sphere having $4$ branch points. Cubic Weierstrass form of the curve corresponds to moving one of the branch points to $\infty$ by a homography. 
