- #1
zenterix
- 708
- 84
- Homework Statement
- Consider the system
$$x'=x-2y+\frac{x^2}{4}$$
$$y'=5x-y-y^2$$
- Relevant Equations
- To find the critical points we need to find ##(x,y)## combinations that make the above equations equal to zero.
From the first equation we can write
$$y=\frac{x}{2}+\frac{x^2}{8}$$
Subbing into the rhs of the second equation and equating to zero we find (after some algebra) that
$$x(x-4)(x^2+12x+72)=0$$
This equation has roots ##0##, ##4##, and ##-6\pm 6i##.
Then, ##x=0\implies y=0## and ##x=4\implies y=4##. So ##(0,0)## and ##(4,4)## are critical points.
My question is about the cases where ##x## is a complex root.
$$x=-6+6i\implies y=-3-6i$$
$$x=-6-6i\implies y=-3+6i$$
Why aren't ##(-6+6i,-3-6i)## and ##(-6-6i,-3+6i)## considered critical points?
Note
Here is a problem set with solutions from MIT OCW's 18.03 "Differential Equations" course. The first problem asks us to find the critical points of the system above. In the solution presented, they simply say that "only the real roots give critical points".
My question is simply why this is.
$$y=\frac{x}{2}+\frac{x^2}{8}$$
Subbing into the rhs of the second equation and equating to zero we find (after some algebra) that
$$x(x-4)(x^2+12x+72)=0$$
This equation has roots ##0##, ##4##, and ##-6\pm 6i##.
Then, ##x=0\implies y=0## and ##x=4\implies y=4##. So ##(0,0)## and ##(4,4)## are critical points.
My question is about the cases where ##x## is a complex root.
$$x=-6+6i\implies y=-3-6i$$
$$x=-6-6i\implies y=-3+6i$$
Why aren't ##(-6+6i,-3-6i)## and ##(-6-6i,-3+6i)## considered critical points?
Note
Here is a problem set with solutions from MIT OCW's 18.03 "Differential Equations" course. The first problem asks us to find the critical points of the system above. In the solution presented, they simply say that "only the real roots give critical points".
My question is simply why this is.