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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

    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


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


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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


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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!) :
    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


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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


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    You can write it in matrix form:
    a & b & 0 & 0\\
    0 & c & d & 0\\
    0 & 0 & e & f\\
    h & 0 & 0 & g


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

    The solution of the scalar equation:

    [tex]\frac{dv}{dt} = av[/tex]
    [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]
    [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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