# Numerical Integration

Gold Member

## Homework Statement

Given the system
$$x'(t)=-ax(t)+ky(t)+g$$
$$y'(t)=lx(t)-by(t)+h$$

If g=h=0,

a) Find the equilibrium
b) Show that if ab-lk does not equal 0, the steady state found in (a) is the only solution
c) choose a,b,l,k such that ab-lk > 0. Find numerically the solution of the system starting in a neighborhood of the equilibrium.

2. The attempt at a solution
a) If x'(t)=y'(t)=0, then ax(t)=ky(t) and lx(t)=by(t). This is true for ab = lk, i.e. ab-lk=0.

and then I run into trouble. I don't know how to explicity show (b), and have even less of an idea on how to start (c)

Help!
Josh

Related Calculus and Beyond Homework Help News on Phys.org
Should

"b) Show that if ab-lk does not equal 0, the steady state found in (a) is the only solution"

"b) Show that if ab-lk does not equal 0, the steady state found in (a) is the only equilibrium solution"

Now, consider that an equilibrium or "steady state" solution correspond to x'=y'=0.
Finding such a solution here leads you to an algebraic linear system of equations.
Go back to and reuse the theory for linear systems of equations.

Gold Member
I don't understand how I'm supposed to get values for x(t) and y(t).

Can you show me an example?

HallsofIvy
Homework Helper
You have, as you said in your first post, the two equations ax= ky and lx= by. Are you saying you do not know how to solve two linear, simultaneous equations?

Gold Member
The part that confuses is me is that in (b) we prove that if ab-lk does not equal zero then x and y have only one equilibrium point..(0,0). Then part (c) says to solve the system for ab-lk>0, which would imply that (x,y)=(0,0).

Is this correct or am i missing something?

HallsofIvy
Homework Helper
One more time: can you solve the two equations ax= ky and lx= by for x and y? If you can, what happens to that solution if ab- lk DOES equal 0?

Yes, if ab-lk> 0 then (x,y)= (0,0) is the only EQUILIBRIUM solution. But you are asked to find a solution for an intial value in a neighorborhood of (0,0), not (0,0) itself.

Gold Member
I don't think you understand my extreme inability to solve any part of this problem. Here is an attempt at (c):

a=0.3
b=0.2
l=0.1
k=0.4

Then ab-lk=0.02 > 0.

0.3 x(t) = 0.4 y(t)
0.1 x(t) = 0.2 y(t)

So, x(t) = 1.333 y(t)
y(t) = 0.5 x(t)

And I don't know what to do after this since it doesn't even seem to be true. Should I just plug in values of x(t),y(t) close to 0? I am not given any initial values. I really need help.

Last edited:
HallsofIvy
Homework Helper
I don't think you understand my extreme inability to solve any part of this problem.
I have asked you repeatedly to solve the linear equations ax= ky and lx= by. If you cannot do that then you should not be attempting differential equations!
From ax= ky, y= (a/k)x. Putting that into kx= by we have lx= (ab/k)x so that lkx= abx or (lk-ab)x= 0. If lk- ab is not 0 what is x equal to? What is y equal to? If lk- ab= 0 what can you say about x?

Here is an attempt at (c):

a=0.3
b=0.2
l=0.1
k=0.4

Then ab-lk=0.02 > 0.

0.3 x(t) = 0.4 y(t)
0.1 x(t) = 0.2 y(t)

So, x(t) = 1.333 y(t)
y(t) = 0.5 x(t)

And I don't know what to do after this since it doesn't even seem to be true. Should I just plug in values of x(t),y(t) close to 0? I am not given any initial values. I really need help.
Yes, it is true: x= (4/3)y and y= (1/2)x so x= (4/3)(1/2)X= (2/3)x. That is, (1/3)x= 0 so x= 0 and then y= 0. The only equilibrium solution is (0,0). Now choose a starting value for (x,y) that is "close to" (0,0) and numerically solve the system of equations.

Gold Member
Thanks a lot for your help ivy, I have no other source to turn to on this problem. I have never done differential equations before and my teacher just gave us these projects without covering any of the material in class.

Did I solve the system correctly for these starting values? (I assume that I am to plug in one value in each equation and solve for the other variable.)

(x,y) = (0.01, 0.02)

0.3 * 0.01 = 0.4 * y(t) => y(t) = 0.0075

0.1 * x(t) = 0.5 * 0.02 => x(t) = 0.1

edit: Since I am no longer solving for the equilibrium, am I supposed to keep x' and y' in the system?

Last edited:
Gold Member
Now choose a starting value for (x,y) that is "close to" (0,0) and numerically solve the system of equations.
This part is my biggest problem. I'm not exactly sure what constitutes a solution here. Am I to use Euler's method and approximate some values of x and y? Or am I trying to come up with a function that satisfies the diff eq? Any help at all here is appreciated!

Gold Member
I really need help guys, any takers?

J77
Yes -- use Eulers method.

Gold Member
When I choose values of x,y close to the equilibrium does t=0?