I need to solve this ode' system $$ \begin{cases} \dot x=y\\\dot y=-x+x^2=x(x-1) \end{cases} $$ To solve it as a function $x(y)$ or $y(x)$ is trivial, but I need the solution as a function of time: $\left(x(t),y(t)\right)$.
The system has a first integral: $H(x,y)=\frac{1}{2}\left(x^2+y^2\right)-\frac{1}{3}x^3$. In particular, if no general solution is possible, I'm searching for a solution for $H=\frac{1}{6}$.
Any idea? Thank you!
Multiplying the 2nd order equation $\ddot x=x(x-1)$ by $\dot x$ on both sides and integrating with respect to time, we get a nonlinear 1st order equation for $x$:
$$\ddot x=x(x-1)\\ \ddot x \dot x = x(x-1)\dot x\\ \int_0^{t} \dot x(\tau)\ddot x(\tau)d\tau = \int_0^{t}x(\tau)(x(\tau)-1)\dot x(\tau)d\tau\\ \int_{\dot x_0}^{\dot x(t)} \dot x\,d\dot x = \int_{x_0}^{x(t)}x(x-1)\, dx$$
$$\frac12(\dot x^2 - \dot x_0^2)=\frac13(x-1)^3+\frac12(x-1)^2-\frac13(x_0-1)^3-\frac12(x_0-1)^2\\ \dot x^2=\dot x_0^2+\frac23(x-1)^3+(x-1)^2-\frac23(x_0-1)^3-(x_0-1)^2=f(x;x_0,\dot x_0)$$
Once we have $\dot x^2=f(x;x_0,\dot x_0)$, we can at least find an implicit equation for $x$:
$$\frac{dx}{dt}=\sqrt{f(x;x_0,\dot x_0)}\\ \int_{x_0}^x \frac{d\chi}{\sqrt{f(\chi;x_0,\dot x_0)}}=\int_0^t d\tau=t.$$
The final integral will in general ony be computable in terms of elliptic integrals.