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Math Help - General Solution

  1. #1
    Junior Member
    Joined
    Oct 2009
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    48

    General Solution

    Find the general solution of:

    du/dt = u^2-U

    I found the integral of 1/u(u-1) du = int 1 dt to be -ln(u)+ln(u-1) = t using partial fractions so:

    u = -1/(e^t-1) but u(0) should = 2 and it would leave -1/(1-1) in this case so undefined

    Could anyone tell me where I've gone wrong, thanks
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  2. #2
    Member
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    Sep 2009
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    I found the integral of 1/u(u-1) du = int 1 dt to be -ln(u)+ln(u-1) = t using partial fractions so:
    Well you forgot to add a constant.

    -ln(u)+ln(u-1)=t+C

    Then you get:

    u=\frac{1}{1-e^te^C}

    So just solve for C when u(0)=2.
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