Getting critical points of diff eq system

Question

I have to calculate the critical points of the following system. $$x'=cx+10x^2$$ $$y'=x-2y$$ So I solve the system $$cx+10x^2=0$$ $$x-2y=0$$ So if $x=2y$ I have $2yc+10*4y^2=2yc+40y^2=y(2c+40y)=0$ and I get $y=0$ and $y=-\frac{c}{20}$ f I substitute in $x=2y$ I get $x=0$ and $x=-\frac{c}{10}$ Then we have that the critical points are $(0,0)$ and $(-\frac{c}{10},-\frac{c}{20})$ Is this correct? I don't know if the procedure to get the critical points is like this. Thanks

Answer 1

Yes, indeed you have to solve the system

$$\begin{cases} cx + 10x^2 = 0 \\ x-2y = 0 \end{cases}$$

since a critical point $\mathbf{x_0}$ is a point such that $\mathbf{f'(x_0) = 0}$ (bold is used to denote vector notation since we have a 2-D system).

Your final calculations are correct as well. It would be useful to give a "name" to each critical point, such that it can be easily used during rest calculations etc.

For stability, consider the Jacobian matrix of your system :

$$J(x,y) = \begin{bmatrix} c+20x & 0 \\ 1 & -2\end{bmatrix}$$

Substitute now each of the critical points at a time and proceed with finding the corresponding eigenvalues to determine the kind of stability.

Answer 2

Yes, your answer is correct.

The critical points are where the derivatives are zero and you have found $(0,0)$ and $(\frac {-c}{10},\frac {-c}{20})$ which make both derivatives zero.