Consider $$ \frac{d}{dt}x(t) =\frac{x^3(t)-x^2(t)}{n} \\ x(0) =x_0 ~~~~~~~~~~~~~~~ $$ $n$ is a positive constant. How to get the solution ?
I get this question from modification a proof of mean curvature flow. I have little knowledge about ODE, so I get stuck.
$$\frac{d}{dt}x(t)=\frac{x^3(t)-x^2(t)}{n}$$ This ODE is separable : $$dt=n\frac{dx}{x^3-x^2}=n\left(\frac{1}{x-1}-\frac{1}{x}-\frac{1}{x^2} \right)dx$$ $$t=n \left(\ln|x-1|-\ln|x|+\frac{1}{x}\right)+c$$ $x(0)=x_0\quad\to\quad c=-n\left(\ln|x_0-1|-\ln|x_0|+\frac{1}{x_0}\right)$
$$t(x)=n \left(\ln\left|\frac{x-1}{x_0-1}\right|-\ln\left|\frac{x}{x_0}\right|+\frac{1}{x} -\frac{1}{x_0}\right)$$
The solution of the ODE is obtained on the form of $t$ as a function of $x$. There is no closed form for the inverse function $x(t)$. It has to be solved by numerical calculus.