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What the Hell. Differential Eq

  1. Apr 19, 2008 #1
    [SOLVED] What the Hell. Differential Eq

    Okay, so I used Method of undetermined coefficients for this one. I got a solution that when differentiated and plugged back in almost works, bit is off by a little.


    [​IMG]


    When I plug back into y'=4x+3y+(-5t+6)
    I get y'=........+9 not 6. So I have a feeling it has to do with my D=11/7 term..... but I have no idea what.

    Anyone see what I am doing wrong? Is my matrix incorrect? Or my assumption of Xp? Or neither?

    Thanks
     
  2. jcsd
  3. Apr 20, 2008 #2
    Sorry, I forgot to change the matrix....this is what I used to get A,B,C,D

    [tex]\left[\begin{array}{cccc}6 & 1& 0 & 0\\4 & 3 & 0 & 0\\ 1 & 0 & 6 & 1\\
    0 & -1 & 4 & 3\end{array}\right]*\left[\begin{array}{c}A\\B\\C\\D\end{array}\right]=\left[\begin{array}{c}-3\\5\\0\\0\\
    \end{array}\right][/tex]
     
  4. Apr 20, 2008 #3
    well after solving:
    t: 6A+B+3=0
    4A+3B-5=0
    you get 14A+14=0
    B=3 after solving for the next equation of t^0 you get that:
    C=0 and D=-1.
    I think in this simple case it's better to just solve it without matrix.
     
  5. Apr 20, 2008 #4

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    Your matrix is not correct. For example, your third row says "A+ 6C+ D= 0" but the corresponding equation (A= 6At+ 6C+ Bt+ D+ 3t with t= 000) is "A= 6B+ D" which is "-A+ 6C+ D= 0".
     
  6. Apr 20, 2008 #5

    exk

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    have you tried taking the laplace x-form of your system? (sometimes that simplifies the problem)
     
  7. Apr 20, 2008 #6
    Right. I actually just use a CAS to solve the matrix for me...so that is easier. I just could not figure out where my matrix was wrong, but I have got the same numbers as you now.

    Now I am having trouble getting his to plug back in again!
     
  8. Apr 20, 2008 #7
    GOT IT! Thanks!
     
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