1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Solving x'=Ax+g using undetermined coefficients

  1. May 8, 2010 #1
    1. The problem statement, all variables and given/known data

    Problem #4 from here: http://www.math.washington.edu/~jtittelf/ExtraCredit.pdf


    2. Relevant equations

    ?

    3. The attempt at a solution

    There's only one example of using undetermined coefficients in my textbook. I'm not sure where to start. x = t2a + tb + c ?????? That's just a ball park guess, since then x' will have just have constant vectors and vectors multiplied by t, like in the problem.
     
  2. jcsd
  3. May 8, 2010 #2

    gabbagabbahey

    User Avatar
    Homework Helper
    Gold Member

    I'd start by finding the eigenvalues and eigenvectors of [tex]\begin{pmatrix}2 & 2 \\ 1 & 3\end{pmatrix}[/tex]....
     
  4. May 8, 2010 #3
    Fine. det(AI)=0 ----> (2-ß)*(3-ß)-2*1 = 0 ----> ß2 - 5ß + 4 = 0 ----> (ß-4)(ß-1)=0 ----> ß=4, 1

    -----> ß(1) = (1 1)T, B(2)=(-2 1)T

    Ah-ha! So I know part of the solution is e4t(1 1)T + et(-2 1)T.

    Maybe I assume the full solution is e4t(1 1)T + et(-2 1)T + at2 + bt + c????
     
  5. May 8, 2010 #4

    gabbagabbahey

    User Avatar
    Homework Helper
    Gold Member

    You don't get the zero vector for your first eigenvector...show your calculations for that part
     
  6. May 8, 2010 #5
    (see edit)
     
  7. May 8, 2010 #6

    gabbagabbahey

    User Avatar
    Homework Helper
    Gold Member

    Close...I think you want to assume the solution is of the form

    [tex]\textbf{x}(t)=c_1e^{4t}\begin{pmatrix} 1 \\ 1\end{pmatrix}+c_2e^{t}\begin{pmatrix} -2 \\ 1\end{pmatrix}+\textbf{a}t+\textbf{b}[/tex]

    The [itex]c_1[/itex] and [itex]c_2[/itex] are needed since any linear combination of your two independent homogeneous solutions will also be a solution to the homogeneous equation. And I don't think you need a [itex]t^2[/itex] term for your particular solution.
     
  8. May 8, 2010 #7
    If you just have at + b then you'll just have a when you take the derivative. Ultimately I'm trying up an x whose derivative looks something like it does in the problem. Right? (t t)T = t(1 1)T.
     
  9. May 8, 2010 #8

    gabbagabbahey

    User Avatar
    Homework Helper
    Gold Member

    Right, but [tex]\begin{pmatrix}2 & 2 \\ 1 & 3\end{pmatrix}\textbf{x}(t)[/tex] will also have a [itex]t[/itex] in it which should cancel the at for an appropriate choice of a.
     
  10. May 8, 2010 #9
    a and b have constant entries, right?
     
  11. May 8, 2010 #10

    gabbagabbahey

    User Avatar
    Homework Helper
    Gold Member

    They'd better!
     
  12. May 8, 2010 #11

    gabbagabbahey

    User Avatar
    Homework Helper
    Gold Member

    Keep in mind, that if you get an equation like Ma=-a for some matrix M, either -1 is one of its eigenvalues or a=0:wink:
     
  13. May 8, 2010 #12
    Thanks, buddy. One more quick question!

    On problem 3, I have a 3x3 matrix eigenvalues -2,-1 and corresponding eigenvectors (1 0 0)T and (0 1 0)T. Now, I know what with a 2x2 matrix, if I have only one eigenvector, giving me a solution x=eat$, then I guess a second solution x=teat$ + eat#. With a 3x3 matrix, do I try that with each of my two eigenvectors? Doing so seems to give me a solution for one but not the other.
     
  14. May 8, 2010 #13

    gabbagabbahey

    User Avatar
    Homework Helper
    Gold Member

    The easiest way to solve problem 3 is to not do it in matrix form...just let [tex]\textbf{x}(t)=\begin{pmatrix}x_1(t) \\ x_2(t) \\ x_3(t) \end{pmatrix}[/tex] and solve the 3 DEs you get.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook