Critical points of a diff eq system

[Phys pilot]

Homework Statement

I have to calculate the critical points of the following system.
$$x'=cx+10x^2$$
$$y'=x-2y$$

The Attempt at a Solution

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

 

[RPinPA]

I'm not sure what you mean by the procedure. Critical points are where the derivatives are 0. This led you to a system of nonlinear equations, which you solved. There isn't one set way to solve a system of equations (though with nonlinear systems you'll usually end up with some form of substitution as you did), so there isn't one "procedure".

Yes, your solution appears to be correct.

 

[Mark44]

Homework Statement

I have to calculate the critical points of the following system.
$$x'=cx+10x^2$$
$$y'=x-2y$$

The Attempt at a Solution

So I solve the system
$$cx+10x^2=0$$
$$x-2y=0$$

If you solve your first equation for x, you should get ##x = \frac{-c \pm \sqrt{c^2}}{10} = \frac{-c \pm |c|}{10}##.
If c ≥ 0, you get one value, but if c < 0, you get another value. What all this is saying is that there are two critical points.

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

 

[Phys pilot]

If you solve your first equation for x, you should get ##x = \frac{-c \pm \sqrt{c^2}}{10} = \frac{-c \pm |c|}{10}##.
If c ≥ 0, you get one value, but if c < 0, you get another value. What all this is saying is that there are two critical points.

Yes, I get the critical point (0,0) and then another critical point depending on the variable c so i should study the case when c is positive, negative and zero. So i will get 3 critical points in total.
thank you
P.S. The problem is that when c=0 I obtain a null eigenvalue and I can't relate this to any type of critical point

 

[pasmith]

Homework Statement

I have to calculate the critical points of the following system.
$$x'=cx+10x^2$$
$$y'=x-2y$$

The Attempt at a Solution

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

Substituting [itex]x = 2y[/itex] into [itex]cx + 10x^2 = 0[/itex] is not necessary. You can immediately solve [itex]cx + 10x^2 = x(c + 10x) = 0[/itex] to obtain [itex]x = 0[/itex] or [itex]x = -c/10[/itex]. Then [itex]x - 2y= 0[/itex] requires that [itex]y = \frac12 x[/itex].

Thus the fixed points are at (0,0) and (-c/10, -c/20). Now when [itex]c = 0[/itex] these fixed points coincide, and there is a bifurcation.

 

[Phys pilot]

Substituting [itex]x = 2y[/itex] into [itex]cx + 10x^2 = 0[/itex] is not necessary. You can immediately solve [itex]cx + 10x^2 = x(c + 10x) = 0[/itex] to obtain [itex]x = 0[/itex] or [itex]x = -c/10[/itex]. Then [itex]x - 2y= 0[/itex] requires that [itex]y = \frac12 x[/itex].

Thus the fixed points are at (0,0) and (-c/10, -c/20). Now when [itex]c = 0[/itex] these fixed points coincide, and there is a bifurcation.

Do you know how to get the axis from the ker of the jacobian matrix?