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2nd order differential equation (from M4) watch

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    Hi, I've been having a bit of trouble solving this equation for a while now, I can't get one of the constants right, although my general solution has the right terms in it. The equation is:
    a + 2wv + w^2*x = 2wu (a is acceleration, v is velocity, w and u are constants). My general solution is x = A*exp(-wt) + Bt*exp(-wt) + 2u/w. Initial conditions are t,x,v all zero. I keep getting A=-2u/w which is right, but then B=-2u, which should just be B=-u, according to the answer. (It's heinemann M4, ex4A, q10 if that helps)
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    D^2 + 2w D + w^2 = 0
    (D + w)^2 = 0
    Complementary solution: x = (A + Bt) e^(-wt).

    For a particular solution try x = k. That works if w^2 k = 2wu. So k = 2u/w.

    General solution: x = (A + Bt) e^(-wt) + 2u/w.

    Since x = 0 when t = 0 we have 0 = A + 2u/w. So A = -2u/w.

    So x = (Bt - 2u/w) e^(-wt) + 2u/w. The derivative of that wrt t is
    dx/dt = (B - Bwt + 2u) e^(-wt). Since dx/dt = 0 when t = 0 we have B + 2u = 0. So B = -2u.

    Required solution: x = (2u/w) (1 - (1 + wt) e^(-wt)).

    The book's wrong!
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    CF:

    d²x/dt² + 2w dx/dt + w²x = 0
    (d/dt + w)²x = 0
    "repeated root" of -w => x = (A+Bt)e^(-wt)

    PI:

    x = (A+Bt)e^(-wt) + C
    dx/dt = -w(A+Bt)e^(-wt) + Be^(-wt)
    d²x/dt = w²(A+Bt)e^(-wt) -2wBe^(-wt)

    d²x/dt² + 2w dx/dt + w²x = [w²(A+Bt) -2wB - 2w²(A+Bt) +2wB + w²(A+Bt)]e^(-wt) + w²C === 2wu

    => C = 2u/w

    => x = (A+Bt)e^(-wt) + 2u/w


    t=0:

    x = A + 2u/w = 0 => A = -2u/w
    v = -w(-2u/w) + B => B = -2u

    i get the same answers as you
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    I get the same results as you bezza.
    • Thread Starter
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    thanks a lot everyone, I've spend the last week going back to it and trying different approaches, all getting the same answer. I kinda suspected the book could be wrong (it is heinemann after all!), but couldn't remember anyone else mentioning a problem. Thank you
 
 
 

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