Solve non-linear ode system as a function of t. { x ˙ = y y...

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 $x$:
$$\begin{gathered} \ddot{x}=x(x-1) \\ \ddot{x}\dot{x}=x(x-1)\dot{x} \\ {\int }_0^t\dot{x}(\tau )\ddot{x}(\tau )d\tau ={\int }_0^{t}x(\tau )(x(\tau )-1)\dot{x}(\tau )d\tau \\ {\int }_{{\dot{x}}_0}^{\dot{x}(t)}\dot{x}d\dot{x}={\int }_{x_0}^{x(t)}x(x-1)dx \end{gathered}$$
$$\begin{gathered} \frac{1}{2}({\dot{x}}^2-{\dot{x}}_0^2)=\frac{1}{3}(x-1{)}^3+\frac{1}{2}(x-1{)}^2-\frac{1}{3}(x_0-1{)}^3-\frac{1}{2}(x_0-1{)}^2 \\ {\dot{x}}^2={\dot{x}}_0^2+\frac{2}{3}(x-1{)}^3+(x-1{)}^2-\frac{2}{3}(x_0-1{)}^3-(x_0-1{)}^2=f(x;x_0,{\dot{x}}_0) \end{gathered}$$
$$\begin{gathered} \frac{1}{2}({\dot{x}}^2-{\dot{x}}_0^2)=\frac{1}{3}(x-1{)}^3+\frac{1}{2}(x-1{)}^2-\frac{1}{3}(x_0-1{)}^3-\frac{1}{2}(x_0-1{)}^2 \\ {\dot{x}}^2={\dot{x}}_0^2+\frac{2}{3}(x-1{)}^3+(x-1{)}^2-\frac{2}{3}(x_0-1{)}^3-(x_0-1{)}^2=f(x;x_0,{\dot{x}}_0) \end{gathered}$$
Once we have ${\dot{x}}^2=f(x;x_0,{\dot{x}}_0)$, we can at least find an implicit equation for $x$:
$$\begin{gathered} \frac{dx}{dt}=\sqrt{f(x;x_0,{\dot{x}}_0)} \\ {\int }_{x_0}^x\frac{d\chi }{\sqrt{f(\chi ;x_0,{\dot{x}}_0)}}={\int }_0^{t}d\tau =t. \end{gathered}$$
The final integral will in general ony be computable in terms of elliptic integrals.