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Coupled first order ODE

  1. Jan 25, 2009 #1
    1. The problem statement, all variables and given/known data

    \dot{x}= \cos(t) x_1 +\sin(t) x_2

    \dot{y}= \sin(t) x_1 - \cos(t) x_2

    Show that the solutions are unbounded.

    3. The attempt at a solution
    I tried to find the explicit solution (by hand) but I failed numerous times. What is the right course of action I should take in solving this question?
  2. jcsd
  3. Jan 25, 2009 #2


    Staff: Mentor

    Your system of equations can be represented in matrix form as
    [tex]\dot{X} = AX[/tex]
    The usual technique involves finding a diagonal matrix D, such that D = P^(-1)AP, and where the columns of P are the eigenvectors of A, and the entries on the diagonal of D are the eigenvalues of A. The process of diagonalization to solve a system of linear DEs is too involved for me to describe it all right here, but presumably you have seen it discussed in your class.

    One thing to note about the matrix that comes out of your system of equations: the matrix A of coefficients of x_1 and x_2 causes a rotation by an angle -t. An eigenvector of a matrix A is a vector x such that Ax = [itex]\lambda[/itex]x. In other words, multiplying an eigenvector x produces another vector that is merely a scalar multiple of x. Due to the fact that your matrix A is a rotation matrix, I don't see how Ax can produce a vector that is a scalar multiple of x, for arbitrary rotation angles t.
  4. Feb 19, 2009 #3
    I think you can solve it using MATLAB......

    you have a built in function ode45 in which we can give the functions and boundary conditions

    another way is using Runge kutta method .......
  5. Feb 19, 2009 #4
    No, I have to do it via hand calculations.
  6. Feb 19, 2009 #5


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    Science Advisor

    Notice that the responders are assuming you mean
    \dot{x_1}= \cos(t) x_1 +\sin(t) x_2

    \dot{x_2}= \sin(t) x_1 - \cos(t) x_2
  7. Feb 19, 2009 #6


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    Science Advisor

    Mark44 has told you how to do this. It's easy to show that the eigenvalues of the coefficient matrix are negative for all t.
  8. Feb 19, 2009 #7
    Yes, you're right, the y must be an x2. Moreover if the eigenvalues are negative then the solutions are bounded.
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