I have the diffeq $y'^2=\frac23 y^3+\alpha$ for some $\alpha\in\mathbb R$. This has two solutions in the form of a Weierstrass P elliptic function which are $$6^{1/3}\wp\left(\frac{x\pm C}{6^{1/3}};0, \alpha\right)$$
The latter two arguments make sense, since they are just the $g$ constants in the weierstrass diffeq $y'^2=4y^3-g_2y-g_3$, but I am having trouble getting the $6^{1/3}$ factors and the main argument.
What I attempted to do was use the substitution $u(x)=6^{1/3}y(x)$. This generates the differential $dy=\frac{du}{6^{1/3}}$ which I can substitute into my diffeq to get $$\left(\frac{dy}{dx}\right)^2=\frac23 y^3+\alpha\Longleftrightarrow \left(\frac{du}{6^{1/3}dx}\right)^2=\frac23\left(\frac{u}{6^{1/3}}\right)^3+\alpha\Longleftrightarrow\frac{1}{6^{2/3}}u'^2=4u^3+\alpha$$ This is close, but not quite correct as there is an extra factor on the LHS.
Any hints as to how to proceed? Thanks
Alternatively, (since the diffeq is seperable) if anyone is able to reduce the integral (which is the square root of a cubic) down to an elliptic integral of the first kind and invert it, that works too, but I think this would take more work so I'm not too sure.
Edit: Additionally, I have also tried solving the diffeq given here using the inverse Weierstrass P function $$\wp^{-1}(z; g_2, g_3)=\int_{\infty}^z\frac{dt}{\sqrt{4t^3-g_2t-g_3}}$$ which after seperation gives me something like $$\int\frac{du}{6^{1/3}\sqrt{4u^3+\alpha}}=\int\pm1dx$$ $$\wp^{-1}(u; 0, \alpha) = 6^{1/3}(C\pm x)\tag{???}$$ This is quite close but still not quite right since the factor of $6^{1/3}$ is not in the correct part of the equation. Furthermore, I'm not sure how to relate the bounds of the definite integral definition to the indefinite one (for example, there could be a negative I am missing or smth). Help would be appreciated!
You can instead just do this $$y'^2 = \frac{2}{3}y^3+\alpha$$ $$\implies \left(\frac{\text{d}y}{\text{d}x}\right)^2 = 4\left(\frac y{\sqrt[3]{6}}\right)^3+\alpha$$ This way you can use the substitution $u(x) = \frac{y(x)}{\sqrt[3]6}$ (which has the differential $\text{d}y=\text{d}u\sqrt[3]6$). This returns the equation $$\left(\frac{\text{d}u\cdot \sqrt[3]6}{\text{d}x}\right)^2 = 4u^3+\alpha $$ $$\implies \sqrt[3]6^2\, u'^2 = 4u^3+\alpha$$ which is in the form of a Weierstrass differential equation $y'^2=4y^3-g_2y-g_3$.
Indeed, if we square root both sides, we can see that $$\sqrt[3]6^\, u' = \pm\sqrt{4u^3+\alpha}$$ $$\implies \sqrt[3]6\int\frac{\text{d}u}{\sqrt{4u^3+\alpha}}=\pm x+C$$ I checked using mathematica (which spewed out a ton of barely legible hypergeometric spam but whatever) that the indefinite integral and the definite integral definition of the inverse Weierstrass elliptic function have a positive sign in front of the main function along with some constants being added, which means that $$\implies \sqrt[3]6\,\wp^{-1}(u; 0, -\alpha) =\pm x+C$$ Solving for $u$ gives $$u=\wp\left(\frac{C\pm x}{\sqrt[3]6}; 0, -\alpha\right)$$ To get the solution you give, we just solve for $y$, redefine $C$, and rid ourselves of the $\pm$ using the fact that the Weierstrass elliptic function is even.
Also note that both of your solutions are technically not correct? Your alpha has the wrong sign. Furthermore, the two solutions you give are actually just one solution, since you can collect the $\pm$ into the constant.