How to solve nonlinear differential equation system?

Question

I'm trying to solve this problem about nonlinear differential system:

Consider the first order nonlinear differential equation system given by $$ \left\{ \begin{array}{} x'& = \ 1-x-y \\ y'& = \ x(y^2-1)(1-x-y). \end{array} \right. \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (*) $$ a) Calculate the equilibrium points of the system.
b) Find the solution that passes through the point $(0,2)$ and determine its orbit.

The item a) its clear for me. I find that all points of the line $y=-x+1$ are equilibrium points of $(*)$. My problem comes with item b). I don't really know how to solve the problem, because since it is a non-linear problem I don't know how to find the solution :( In what way does it affect that the solution must pass through the point $(0,2)$?

Any help is welcome, thank you!!

Answer 1

you start by substituting $x'$ in the second equation

$y'=x(y^2-1)x'$

$\dfrac{dy}{dt}=x(y^2-1)\dfrac{dx}{dt}$

$\dfrac{dy}{y^2-1}=xdx$

$\displaystyle\int\dfrac{dy}{y^2-1}=\displaystyle\int xdx$

$-\dfrac{1}{2}(ln(y+1)-ln(y-1))=\dfrac{x^2}{2}+c$

$\dfrac{1}{2}\left(ln \left(\dfrac{y+1}{y-1}\right)\right)=-\dfrac{x^2}{2}+c$

$ln \left(\dfrac{y+1}{y-1}\right)=-x^2+c$

$\dfrac{y+1}{y-1}=Ce^{-x^2}$

$y+1=(y-1)Ce^{-x^2}$

$y(1-Ce^{-x^2})=-1-Ce^{x^2}$

$y=-\dfrac{1+Ce^{-x^2}}{1-Ce^{-x^2}}$