# Critical points of system of DE

Gold Member

## Homework Statement

Find all the critical points of the system of DE:
$x'=x-x^2-2xy$ and $y'=2y-2y^2-3xy$.

## Homework Equations

$y'=f(x, \vec y)$ has critical point if $f(\vec x )=0$.

## The Attempt at a Solution

I'm self studying DE's and I'm not understanding well what critical points are.
It looks like I must equate x' to 0 and isolate x (which I've done and it gave me $x=0$ or $x=1-2y$) and equal y' to 0 and isolate y, which I've done and got $y=0$ or $y=1-\frac{3x}{2}$.
Now I don't know what to do.

Dick
Homework Helper

## Homework Statement

Find all the critical points of the system of DE:
$x'=x-x^2-2xy$ and $y'=2y-2y^2-3xy$.

## Homework Equations

$y'=f(x, \vec y)$ has critical point if $f(\vec x )=0$.

## The Attempt at a Solution

I'm self studying DE's and I'm not understanding well what critical points are.
It looks like I must equate x' to 0 and isolate x (which I've done and it gave me $x=0$ or $x=1-2y$) and equal y' to 0 and isolate y, which I've done and got $y=0$ or $y=1-\frac{3x}{2}$.
Now I don't know what to do.

I think you've got a fine start. The critical points are just where (x',y')=(0,0). (x,y)=(0,0) is certainly one. Combine your two conditions for x and y in pairs and solve each pair. You'll get four critical points. The one I gave you combines x=0 and y=0. Now combine x=0 and y=1-3x/2. There are two more combinations to try.

Gold Member
Ah ok thank you.
So for example for $(x,y)=(0,1-\frac{3x}{2})$.
Gives $y'=3x-\frac{9x^2}{2}$ evaluated in x=0 gives 0. But on the other hand since $y=1-\frac{3x}{2}$, $y'=3/2$ which differ from the previous answer, 0. So that this isn't a critical point? Is the reasoning good? Or it's because 3/2 isn't equal to $3x-\frac{-9x^2}{2}$ that it isn't a critical point?

Edit: I'm confused about the notation. Is x'=dx/dy and y'=dy/dx?

Edit2: I tried the other combinations and none gave me critical points. Only the (0,0) is critical point.

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vela
Staff Emeritus
Homework Helper
Both x and y are functions of a parameter, t, and the derivative is with respect to this parameter. The differential equations mean
\begin{align*}
\frac{dx}{dt} &= x-x^2-2xy \\
\frac{dy}{dt} &= 2y-2y^2-3xy
\end{align*}
For the case x=0 and y=1-(3/2)x, you have y = 1-(3/2)(0) = 1. If you plug (x,y)=(0,1) back into the original differential equations, you will indeed find that dx/dt=0 and dy/dt=0.

Gold Member
Ah I see vela, thank you. In this case I obtain that these 4 points are indeed critical.
Would it have been possible to obtain a point that wasn't critical?
Edit: nevermind the answer is yes, in another problem I get that not all of the points are critical.

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