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!

Diff. EQ. Inhomogeneous Linear System (Double check please)

  1. Jun 23, 2005 #1
    Here is the problem:

    [tex]X'\,=\,\left(\begin{array}{cc} 0 & 2 \\ -1 & 3 \end{array}\right)\,X\,+\left(\begin{array}{c} 2 \\ e^{-3t}\end{array}\right)[/tex]

    Here is what I have:

    [tex]r_1\,=\,2,\,\,\,\,\,\,r_2\,=\,1[/tex]

    [tex]u_1'\,=\,-2\,e^{-2t}\,+2\,e^{-5t}[/tex]

    [tex]u_2'\,=\,2\,e^{-t}\,-\,e^{-4t}[/tex]

    [tex]\psi\,=\,\left(\begin{array}{cc} e^{2t} & 2\,e^t \\ e^{2\,t} & e^t \end{array}\right)[/tex]

    After integrating the [itex]u_1[/itex] and [itex]u_2[/itex] terms and multiplying the answer by [itex]\psi[/itex], I get:

    [tex]X\,=\,\left(\begin{array}{c} C_1\,e^{2t}\,+\,2\,C_2\,e^t\,+\,\frac{1}{10}\,e^{-3t}\,-\,3 \\ C_1\,e^{2t}\,+\,C_2\,e^t\,-\,\frac{3}{20}\,e^{-3t}\,-\,1 \end{array}\right)[/tex]

    Then seperating and combining terms:

    [tex]X\,=\,C_1\,e^{2t}\,\left(\begin{array}{c} 1 \\ 1 \end{array}\right)\,+\,C_2\,e^t\,\left(\begin{array}{c} 2 \\ 1 \end{array}\right)\,+\,\frac{1}{20}\,\left(\begin{array}{cc} 2\,e^{-3t}\,-\,60 \\ 3\,e^{-3t}\,-\,20 \end{array}\right)[/tex]

    Does this look correct? Thanks.
     
  2. jcsd
  3. Jun 24, 2005 #2
    Any ideas?

    Anyone have the same thing when solving this problem?
     
  4. Jun 24, 2005 #3

    saltydog

    User Avatar
    Science Advisor
    Homework Helper

    Hello VinnyCee. Takes a while. It's a challenging problem. Your answer is NOT correct: Missing a minus sign for the second equation. You know, you can back-substitute the answer into the original ODE to verify the answer or if you're lazy like me, just input the results into Mathematica and have it back-substitute and compare the results. :smile:

    I get:

    [tex]x_p(t)=-3+1/10 e^{-3t}[/tex]

    [tex]y_p(t)=-1-3/20e^{-3t}[/tex]

    Hope you didn't get my first (unedited) post where I said it was correct. Didn't notice the minus sign missing. :surprised
     
    Last edited: Jun 24, 2005
  5. Jun 24, 2005 #4
    Could you show me that?

    This backsubstitution that you mention. How do you type this? Can you show me the routine you did in Mathematica, so I can double check others also?
     
  6. Jun 24, 2005 #5

    saltydog

    User Avatar
    Science Advisor
    Homework Helper

    Code (Text):


    x[t_]:=-3+1/10 e^(-3 t);
    y[t_]:=-1-3/20 e^(-3 t);
    D[x[t],t]
    Simplify[2 y[t]+2]
    D[y[t],t]
    Simplify[-x[t]+3 y[t]+e^(-3 t)

     
    Just define the solutions (x[t] and y[t]), calculate the derivatives, then insert the x[t] and y[t] into the right hand side and compare the resuts.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Diff. EQ. Inhomogeneous Linear System (Double check please)
  1. Diff. eq. (Replies: 5)

  2. Diff Eqs (Replies: 7)

Loading...