[Math] Show that the solution of $2y”=3y^2$ with $y(0)=0$ and $y'(0)=1$ is given implicitly by:

Question

$$\int_0^y \left(\frac{dt}{\sqrt {1+t^3}}\right) \, = x $$

This is an example of an elliptic integral.

This problem was extracted from http://ocw.mit.edu/courses/mathematics/18-034-honors-differential-equations-spring-2009/recitations/

I can not figure out how should I solve it. I tried to solve it by integrating (indefinite integrals), and I obtained:

$$y^2=(3/2)(y^2)(x^2)$$

However, $$y^2$$ would cancel out.

I appreciate your help

Answer 1

A start: That it is a solution can be verified by differentiating the given function twice (use the Fundamental Theorem of Calculus). We also need to check that the initial conditions are satisfied. The calculations are straightforward.

For finding a solution without being told what it is, the standard trick is to multiply through by $y'$. Then integrate. We get $(y')^2=y^3+C$. Continue.