So I'm a little unsure about how to solve this system of differential equations. I missed the first five minutes of my instructor's lecture the day we went overt this, so I feel like I didn't quite get the whole picture.
Inasmuch, I feel like the chapter on Systems of Diff Eq's in Thomas' Calculus is not very helpful.
The question is:
"Find and classify (as stable or unstable) the equilibrium solutions of the following differential equations. Use the Differential Equa option in Winplot to investigate stability."
$dx/dt= -0.01x + 0.02y \\ dy/dt= 0.02xy - 0.03y$
So this is what I started doing.
The equilibrium points are the points that satisfy the equation when each are set to $0$ correct?
So...
$1/100(-x+2y)=0 \\ y/100(2x-3)=0$
So then I believe we solve for one equation and substitute those values into the other.
Then for equation two to we would have $y=0, x=3/2$
Do we then substitute these values into equation 1?
What I have in my notes doesn't match what I thought I had to do, so I am a little confused.
I am running late for work, otherwise I would provide more information.
Any help would be great!
You have the first step correct:
We have that:
$$\frac{dx}{dt}=\frac{1}{100}(-x+2y)=0 $$ $$\frac{dy}{dt}=\frac{y}{100}(2x-3)=0$$
Now we check the zeros:
$$\frac{dx}{dt}=\frac{1}{100}(-x+2y)=0,\ x = 2y$$ $$\frac{dy}{dt}=\frac{y}{100}(2x-3)=0, \ y = 0,\ x =3/2$$
Now for each ODE, we check the values above and below each equilibrium point.
For $\frac{dx}{dt}$, when $x > 2y$, $\frac{dx}{dt} <0$. When $x < 2y$, $\frac{dx}{dt} >0$. Therefore the equilibrium point is stable.
For $\frac{dy}{dt}$, when $y > 0$, $\frac{dy}{dt} >0$. When $y < 0$, $\frac{dy}{dt} >0$. Therefore the equilibrium point is unstable.
Note that in each case, we are only analyzing the variable corresponding to the derivative that is taken. For the second equation, we are not worried about the value of $x$, only $y$.