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Homework Help: 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

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    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

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    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.
     
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