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Methods of solving DE

  1. Jul 4, 2013 #1
    People I am fairly new with dealing with DEs (and here I was thinking I got it all :palm: )
    Any who: I am working on some problems here and the professor wants us to use 2 methods to solve this DE. My issue is I can't figure out what the 2nd method is!
    (dY)/dt=AY +F
    [itex]^{Y}[/itex] = [7,4,4; -6,-4,-7; -2,-1,2] and [itex]\textbf{}F[/itex] = [-3;0;3]e[itex]^{-iwt}[/itex]
    Initial contions: $$Y(0)=\begin{pmatrix} 1\\ -2\\ 3\end{pmatrix},\; $$

    I was able to find the eigenvalues & eigenvectors; hence the particular & general solutions. My issue i s I don't know of any other method to do this. Can anyone point me in the right direction?

    Thanks
     
    Last edited: Jul 4, 2013
  2. jcsd
  3. Jul 4, 2013 #2

    Simon Bridge

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    Is [7,4,4; -6,-4,-7; -2,-1,2] a matrix?

    So you wrote:
    $$Y=\begin{pmatrix} 7 & 4 & 4\\ -6 & -4 & -7\\ -2 & -1 & 2 \end{pmatrix},\; F=\begin{pmatrix} -3\\0\\-3 \end{pmatrix}e^{-i\omega t}$$ ... did I read that correcty?

    Anyway, did you try writing out the three DEs and solving them individually?
     
  4. Jul 4, 2013 #3
    Yes u read that correctly. Wait! Are u saying to create something like this:

    7y1 + 4y2 + 4y3 - 3eiωt
    -6y1 - 4y2 - 7y3
    -2y1 - y2 +2y3 + 3eiωt
     
  5. Jul 4, 2013 #4

    Simon Bridge

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  6. Jul 4, 2013 #5
    It can be solved by successive substitutions:

    [itex]\begin{cases}y_1'=7y_1+4y_2+4y_3-3e^{i\omega t}\\y_2'=-6y_1-4y_2-7y_3\\y_3'=-2y_1-y_2+2y_3+3e^{i\omega t}\end{cases}\\\\
    \begin{cases}y_1''=7y_1'+4y_2'+4y_3'-3i\omega e^{i\omega t}\\y_2'=-6y_1-4y_2-\frac{7}{4}\left(y_1'-7y_1-4y_2+3e^{i\omega t}\right)\\y_3'=-2y_1-y_2+\frac{1}{2}\left(y_1'-7y_1-4y_2+3e^{i\omega t}\right)+3e^{i\omega t}\end{cases}
    [/itex]

    Now we substitue for [itex]y_3'[/itex]. Next we can eliminate [itex]y_2[/itex] in a similar way. Finally we get equation of 3rd order with [itex]y_1[/itex] only.
     
  7. Jul 4, 2013 #6

    HallsofIvy

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    I would look for eigenvalues and eigenvectors of that coefficient matrix. I find that the characteristic equation is [itex]r^3- 5r^2- 45r+ 153= (r- 3)(r^2- 2r- 51)= 0[/itex]. The eigenvalues are 3 and [itex]1\pm\sqrt{51}[/itex] all of which are real numbers.

    An eigenvector corresponding to eigenvalue is (0, 1, 1).

    I haven't tried to find the Eigen vectors corresponding to the other two eigenvalues. I suspect they are rather messy.
     
    Last edited by a moderator: Jul 4, 2013
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