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Coupled differential equations

  1. Mar 26, 2014 #1
    A bit related to my other topic but here goes:

    I have the set of differential equations:

    x1' = a-ax1-ax2
    x2' = b-bx1-bx2

    What is the solution to such a set? I could google systems of differential equations, but that turns up large texts on the theory. I just wonna know in a few lines what the solution to such a set is. I know how to diagonalize the coefficient matrix etc. if I need that.
     
  2. jcsd
  3. Mar 26, 2014 #2

    Mark44

    Staff: Mentor

    I did a search for "nonhomogeneous system of linear equations". Here is one link.
    http://tutorial.math.lamar.edu/classes/de/NonhomogeneousSystems.aspx
     
  4. Mar 26, 2014 #3

    Mark44

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    It takes more that "a few lines" to describe how to solve systems such as yours...
     
  5. Mar 26, 2014 #4

    Ray Vickson

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    Are the three 'a's in the first equation really supposed to be all the same (and ditto for the 'b's in the second equation)? If so, try looking at y = x1+x2.

    BTW: what does the word "wonna" mean?
     
  6. Mar 26, 2014 #5

    SteamKing

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    'wonna' is a variant of 'wanna' = 'want to'
     
  7. Mar 26, 2014 #6

    Mark44

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    Not like in "wonna these days"?:tongue:
     
  8. Mar 26, 2014 #7

    Ray Vickson

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    I know that of course; but it does no harm to try to encourage proper language.
     
  9. Mar 26, 2014 #8

    HallsofIvy

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    There several different ways to handle
    [itex]x_1'= a- ax_1- ax_2[/itex]
    [itex]x_2'= b- bx_1- bx_2[/itex]

    The more "elementary" method would be to differentiate the first equation again:
    [itex]x_1''= -ax_1'- ax_2'= ax_1'- a(b- bx_1- bx_2)= (a+ ab)x_1+ abx_2- ab[itex]
    but, from the first equation, multiplied by b, [itex]abx_2= ab- bx_1'- abx_1[itex] so
    [itex]x_1''= (a+ ab)x_1+ (ab- bx_1'- abx_1)- ab[/itex]
    [itex]x_1''+ bx_1'- ax_1= 0[/itex].
    That has characteristic equation [itex]r^2+ br- a= 0[/itex] with roots [itex]r= \frac{-b\pm\sqrt{b^2+ 4a}}{2}[/itex] so that [itex]x_1(t)= e^{-bt}(C_1e^{\sqrt{b^2+ 4a}t}+ C_2e^{-\sqrt{b^2+ 4a}t}[/itex] if [itex]b^2+ 4a>0[/itex], [itex]x_1(t)= e^{-bt}(C_1 cos(\sqrt{-b^2- 4a}t)+ C_2 sin(\sqrt{-b^2- 4a}t)[/itex] if [itex]b^2+ 4a< 0[/itex], or [itex]x_1(t)= e^{-bt}(C_1+ C_2t)[/itex] if [itex]b^2+ 4a= 0[/itex].

    In any case, [itex]x_2= (ab- bx_1'- abx_1)/ab[/itex].

    A more "sophisticated" method would be to write the differential equations as a matrix equation:
    [tex]\begin{pmatrix}x_1 \\ x_2\end{pmatrix}'= \begin{pmatrix}-a & -a \\ -b & -b\end{pmatrix}\begin{pmatrix}x_1 \\ x_2 \end{pmatrix}+ \begin{pmatrix}a \\ b\end{pmatrix}[/tex]

    The characteristic equation for that matrix is
    [tex]\left|\begin{array}{cc}-a- \lambda & -a \\ -b & -b- \lambda\end{array}\right|= (-a- \lambda)(-b- \lambda)- ab= \lambda^2+ (a+ b)\lambda= 0[/tex]
    which has roots [itex]\lambda= 0[/itex] and [itex]\lambda= -a- b[itex].
    We can find the eigenvectors corresponding to those eigenvectors and so diagonalize the matrix,
     
  10. Mar 27, 2014 #9

    Ray Vickson

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    The two DEs added together give ##y' = c - cy## where ##c = a+b## and ##y = x_1 + x_2##. Once we know ##y## we can get ##x_1, x_2## from ##x_1' = a - ay##, etc.
     
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