I need a hint for the following problem.

"Solve $(x')^2=x^3 − 3x^2 − 4x + 12$ with the initial with initial condition $x(0)=3$".

I know I should somehow use Weierstrass's $P$ function because it satisfies the equation $(P')^2=4P^3+aP+b$. I tried first to obtain the lattice corresponding to the elliptic curve above but without and success.

  • $\begingroup$ Is this a trick question? $x=3$ is a critical point and thus the solution to the IVP should be $x(t) =3$. Right? $\endgroup$
    – MrSlunk
    Apr 23, 2014 at 7:36

2 Answers 2



  1. The translation $x \mapsto x+1$ transforms the cubic polynomial on the RHS to a depressed cubic.

  2. An appropriate scaling should produce the coefficient '4' for $x^3$.


Using the Weierstrass's function is the simplest way to a closed form. Nevertheless, the ODE can be solved in terms of Elliptic integrals, which are more commonly used.

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