# How to solve nonlinear differential equation system?

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!!

• Try substituting the first equation into the second via the chain rule: $y'_t=y'_x x'_t$ Commented Jan 6, 2023 at 23:27
• The orbits of the system $\dot x=f(x,y),\dot y=g(x,y)$ are the integral curves of $\frac{dy}{dx}=\frac{g(x,y)}{f(x,y)}$. In your case you end up with a separable equation. Commented Jan 7, 2023 at 3:00
• Than you so much @Artem. Commented Jan 7, 2023 at 13:48

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

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

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

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

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

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

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

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

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

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

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

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

• Wow!! You are a the best. Thank you so much. Very elegant solution :-) I tryed to do something similar but I didn’t know how to "eliminate" the differentials. I appreciate a lot your response! Commented Jan 7, 2023 at 13:43