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Equation solving

  1. Feb 2, 2006 #1
    I have 4 difftiate equations now I want to solve them but i have no idea
    dx/dt=ax+by
    dy/dt=cy+dz
    dz/dt=ez+fu
    du/dt=gu+hx

    Given that a,b,c,d,e,f,g,h are constants.
    x,y,z,u are functions(t)
    This problme will appraer in thenext exam, I am sure, my taecher emphasized it many times .
    Hitn me plz.
     
  2. jcsd
  3. Feb 2, 2006 #2

    Hurkyl

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    Can you solve a pair of such equations?
     
  4. Feb 2, 2006 #3

    VietDao29

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    Homework Helper

    Please do not double post!!! :grumpy:
    One post in Precalculus Mathematics is enough!
    But do you study this in precalculus by the way?
     
  5. Feb 2, 2006 #4

    HallsofIvy

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    4 gets a bit complicated!

    A way to get a handle on them is to use "operator" notation. Replace the derivative by the symbol "D" (for derivative of course!) :
    Dx=ax+by
    Dy=cy+dz
    Dz=ez+fu
    Du=gu+hx
    and treat the "D" as if it were a constant (as long as you are dealing with "linear equations with constant coefficients" that works!) and solve the equations for x, y, z, u: the result will involve powers of D. Replace the D by the derivative again (i.e. Dx= dx/dt, D2x= d2/dt, etc.) and solve the resulting differential equations in a single function.
     
  6. Feb 2, 2006 #5
    Is problem easy to you ?
     
  7. Feb 3, 2006 #6

    HallsofIvy

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    IF I were given specific numbers in the four equations, yes, it would be easy for me. If I were required to write a general solution including the coefficients, a- h, it would be tedious be nothing especially difficult.

    Hurkyl asked before, "Can you solve a pair of such equations?". In other words is just that there are so many equations or do you not understand the concepts involved?
     
  8. Feb 3, 2006 #7

    SGT

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    You can write it in matrix form:
    [tex]
    \left[
    \begin{array}{cc}
    \frac{dx}{dt}\\
    \frac{dy}{dt}\\
    \frac{dz}{dt}\\
    \frac{du}{dt}
    \end{array}
    \right]
    =
    \left[
    \begin{array}{cccc}
    a & b & 0 & 0\\
    0 & c & d & 0\\
    0 & 0 & e & f\\
    h & 0 & 0 & g
    \end{array}
    \right]
    \cdot
    \left[
    \begin{array}{cc}
    x\\
    y\\
    z\\
    u
    \end{array}
    \right]
    [/tex]

    or

    [tex]\vec {\frac{dv}{dt}} = A\cdot\vec{v}[/tex]

    The solution of the scalar equation:

    [tex]\frac{dv}{dt} = av[/tex]
    is
    [tex]v=e^{at}\cdot v_0[/tex]

    Similarly, the solution of the matrix differential equation is:
    [tex]\vec{v} = e^{At}\cdot \vec{v_0}[/tex]
    where
    [tex]e^{At} = I + At + \frac{A^2t^2}{2!} + \frac{A^3t^3}{3!}+ ...[/tex]
     
    Last edited by a moderator: Feb 3, 2006
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