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Math Help - Reduction to single higher-order linear ODE

  1. #1
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    Reduction to single higher-order linear ODE

    I need to reduce the following system to a single higher-order linear ODE for y

    x'(t) - y'(t) + z(t) = 0
    x(t) + y'(t) - z'(t) = 1
    y(t) + z'(t) = t3- 1

    I've tried setting ( z = y' - x' ) & (x = 1 - y' - z')
    However on substituting these it ends up as a big mess and I can't go anywhere. Could you someone tell me what substitution or technique I should use to work this problem?

    Thanks.
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  2. #2
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    Re: Reduction to single higher-order linear ODE

    Can get you to a cubic, don't know if it's possible to do better

    Differentiate eq(1), x'' - y''+z'=0 and substitute from eq(3), x''-y''+(t^{3}-1-y)=0..........(4)

    Substitute for z' in eq(2) also, x+y'-(t^{3}-1-y)=1...........(5).

    Now differentiate (5) twice and then subtract (4) to eliminate the x'' term.

    Note that you lose information during this process though, The 1 on the RHS of the second equation differentiates out and so could be any number at all.
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  3. #3
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    Re: Reduction to single higher-order linear ODE

    Yes that works, thanks a lot. I can't see a way of keeping the "1" unless I integrate the first equation and sub x(t) in the second. But I don't think that's possible. I believe this answer is sufficient. Thanks again
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