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System of Two Dimensional First Order ODE

  • #1

Homework Statement



Find all solutions to [dx/dt; dy/dt] = [1, 2; 0, 1]*[x; y]

Homework Equations



the eigenvalue characteristic equation: det(A-λ*I)=0

The Attempt at a Solution



This results in real, repeated eigen values: λ1,2 = 1

for λ1 = 1, (1-1)k1 + 2k2 = 0

choose k1 = 1 then k2 = 0 gives us the eigenvector K1 = [1, 0]

this gives part of the general solution X1 = [1, 0]*e^t

to get a second solution we create a new vector P to avoid duplication

with K1 = [1,0] and P = [p1; p2] we solve (A-λ2*I)*P=K

(1-1)*p1 + 2*p2 = 1

choose p1 = 0 then p2 = 1/2

this gives us the solution X2 = [1;0]*t*e^t + [0;1/2]*e^t

putting it together X = X1 + X2

[x; y] = c1*[1, 0]*e^t + c2*([1;0]*t*e^t + [0;1/2]*e^t)
or
x = c1*e^t + c2*t*e^t
y = (1/2)*c2*e^t

Have I found all the solutions to this system? I think this is the general solution, is this all they are asking for?

If we multiply an eigenvalue by some multiple to get a new eigenvalue, is it really possible to find all the solutions? It seems like there would be infinitely many..
 

Answers and Replies

  • #2
LCKurtz
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You have the general solution. You have two arbitrary constants which is what you would expect. There are always infinitely many solutions because in a homogeneous system a constant times a solution is still a solution.
 
  • #3
HallsofIvy
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The "fundamental theorem" here is that the set of all solutions to an n order linear differential equation or system of differential equations forms a vector space of dimension n. That is, there exist n independent solutions such that the general solution is a linear combination of those n solutions.

Here, you have two first order differential equations which is equivalent to a second order equation. Any solution can be written as a linear combination of two independent solutions. Since you have two distinct eigenvalues, solutions corresponding to those eigenvalues are independent and the general solution is a linear combination of them.

By the way, you are NOT "multiplying an eigenvalue by some multiple to get a new eigenvalue", you are multiplying an eigenvector. The set of all eigenvectors, corresponding to a given eigenvalue, forms a vector space- yes, you can get all eigenvectors by multiplying.

"It seems like there would be infinitely many". Yes, and you have an infinite number of constants, c1 and c2, to multiply by.
 
  • #4
OK that makes sense.

So does this seem like a complete solution? My prof stresses showing all of our work, do you think I can go in more detail? I am afraid that the more detail I provide the more ammunition I am giving the grader to take points away since the theory behind the procedure isn't 100% clear to me..

For instance we create a new vector P to avoid duplication, I know we do this because it works but I don't understand why it works. I would think that multiplying by t would change the solution.. It must have something to do with setting: (A-λ2*I)*P=K
 

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