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Math Help - System of two 4th order diff equations

  1. #16
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    Quote Originally Posted by derdack View Post
    It is completely clear. But first step from initial equations. It should be to express the x'''' and y'''' from initial equation. I am not sure that it is correct in first your post where you express x'''' and y'''' after some algebra.
    No the very first post is incorrect the matrix should be correct but still double chick it
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  2. #17
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    Ok. Thank you a lot for your attention Mr Set.
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  3. #18
    Behold, the power of SARDINES!
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    When I got 8 (first order) differential equations. I need particular solutions, and in past problem with wronskian, constants had differentiation. I should integrate all of them. When I integrate 4th and 8th equation, I have f1[t] and f2[t]. I know that solution of diff equation of 4th order had form x=C1*e^(i*p1*t)+C2*e^(-i*p1*t)+C3*e^(i*p2*t)+C4*e^(-i*p2*t)+C5*e^(p3*t)+C6*e^(-p3*t)+C7*e^(p4*t)+C8*e^(-p4*t) or x=C1*sin(p1*t)+C2*cos(p2*t)+C3*sinh(p3*t)+C4*cosh( p4*t). How to find particular solutions of system? Revise me about premise above...
    This is a direct generalization of solving inhomogeneous ODE's

    To solve the system

    \mathbf{X}'=A\mathbf{X}+\mathbf{F}(t)

    First we need to solve the Homogenous system

    \mathbf{X}'=A\mathbf{X}

    To get the complimentary solution \mathbf{X}_c

    Now we will use the variation of parameters suppose the particular solution is of the form

    \displaystyle \mathbf{X}_p=\mathbf{X}_c\mathbf{U}(t), \text{ where } \mathbf{U}(t)=\begin{pmatrix} u_1(t) \\ u_2(t) \\ u_3(t) \\ u_4(t) \\ u_5(t) \\ u_6(t) \\ u_7(t) \\ u_8(t)\end{pmatrix}

    Now if we plug this into the ODE we get

    \mathbf{X}'_c\mathbf{U}(t)+\mathbf{X}_c\mathbf{U}'  (t)=A\mathbf{X}_c\mathbf{U}(t)+\mathbf{F}(t)

    This gives

    [\mathbf{X}'_c-A\mathbf{X}_c]\mathbf{U}(t)+\mathbf{X}_c\mathbf{U}'(t)=\mathbf{F  }(t) \iff \mathbf{X}_c\mathbf{U}'(t)=\mathbf{F}(t)

    The first group is zero because it is a solution to the homogenous system

    Since the complimentary solution is a fundamental matrix it's determinant is not always zero so it is invertable this gives

    \displaystyle \mathbf{X}_c\mathbf{U}'(t)=\mathbf{F}(t) \iff \mathbf{U}'(t)=\mathbf{X}^{-1}_c\mathbf{F}(t) \iff \mathbf{U}(t)=\int \mathbf{X}^{-1}_c\mathbf{F}(t)dt

    Where the integration is done entrywise.

    So finally the particular solution is

    \displaystyle \mathbf{X}_p=\mathbf{X}_c\int \mathbf{X}^{-1}_c\mathbf{F}(t)dt
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  4. #19
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    Thank you again and I will try to find it in complete form as you show me above.

    Best wishes Mr Set!
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