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System of Two Dimensional First Order ODE

  1. Nov 6, 2011 #1
    1. The problem statement, all variables and given/known data

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

    2. Relevant equations

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

    3. 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..
     
  2. jcsd
  3. Nov 6, 2011 #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.
     
  4. Nov 7, 2011 #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.
     
  5. Nov 7, 2011 #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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