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Quick Question: Matrix added to Differential

  1. Nov 3, 2009 #1
    Hey guys,
    I have a quick question about adding a vector to a system of differential equations.

    Like U'(t)= Ax+V

    Say A is a 2x2 matrix and V is a 2x1 vector.

    Can you all explain how I could handle that V vector? Should somehow include it into A?
     
  2. jcsd
  3. Nov 4, 2009 #2

    Mark44

    Staff: Mentor

    You're not adding a vector to a system of differential equations. Ax is a 2x1 vector, and V is a 2x1 vector, so Ax + V is a 2x1 vector, as is U', which is probably a function of x, not t.
     
  4. Nov 4, 2009 #3

    lanedance

    User Avatar
    Homework Helper

    As mentioned you haven't explained the problem very well... but guessing, do you mean
    [tex] \textbf{x}(t) = (x_1(t), x_2(t)) [/tex]
    [tex] \textbf{x}'(t) = (x_1'(t), x_2'(t)) [/tex]

    and the differential equation is
    [tex] \textbf{x}'(t) = A \textbf{x}(t) +\textbf{v} [/tex]...?

    if so, and if A is in invertible, try writing it as
    [tex] \textbf{x}'(t) = A \textbf{x}(t) +A(A^{-1}\textbf{v}) [/tex]
    and consider a variable change
     
  5. Nov 4, 2009 #4
    I wrote the question a little wrong....
    I meant
    U'(t)=A u(t)+V
    Where A is just a regular 2X2 matrix and V is a 2x1 vector.
    like this

    U'(t)=(1,2;3,4)U(t)+(1;e^t)

    I already know how to solve systems such as
    U'(t)=Au(t) by using the eigenvalues & eigenvectors of A but I haven't learned what to do in the case which you add a V into the equation
     
  6. Nov 4, 2009 #5

    Mark44

    Staff: Mentor

    U'(t) = AU(t) is a homogeneous system.
    U'(t) = AU(t) + V is a nonhomogeneous system. What you choose for a particular solution depends a lot on what the solution is for the homogeneous problem. I don't have my DE resources handy right now, so I can't provide any more details these.
     
  7. Nov 4, 2009 #6
    so what if it included a initial solution, would that change anything?
     
  8. Nov 4, 2009 #7

    Mark44

    Staff: Mentor

    Do you mean initial conditions?
     
  9. Nov 4, 2009 #8
    Such as a vector U(0)=(1;1)
     
  10. Nov 4, 2009 #9

    Mark44

    Staff: Mentor

    That would enable you to find the constants associated with your homogeneous solution, that's all. IOW, having initial conditions wouldn't really change what I said in post #5.
     
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