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Math Help - Differntial Equation Trouble With Integral

  1. #1
    Member zangestu888's Avatar
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    Differntial Equation Trouble With Integral

    the differintial equation given is:

     <br />
dx/dt=k(a-x)[sqrt(b-x)]<br /> <br />
(a) I did part a finding the x as a function of t when a=b

    part (B) is what am having trouble with it states;

    If a>b, find t as a function of x. { hint use substitution u=(sqrt(b-x))

    i tried the substituon i cant seem to simpify my work after sub, is for the left side


    -1/2 integral_ sqrt(b-x)du/(a-x)(u^(1/2))
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by zangestu888 View Post
    the differintial equation given is:

     <br />
dx/dt=k(a-x)[sqrt(b-x)]<br /> <br />
(a) I did part a finding the x as a function of t when a=b

    part (B) is what am having trouble with it states;

    If a>b, find t as a function of x. { hint use substitution u=(sqrt(b-x))

    i tried the substituon i cant seem to simpify my work after sub, is for the left side


    -1/2 integral_ sqrt(b-x)du/(a-x)(u^(1/2))
    ok, lets work on \int \frac 1{(a - x) \sqrt{b - x}}~dx

    Let u = \sqrt{b - x}

    \Rightarrow u^2 = b - x \implies \boxed{u^2 - b + a = a - x}

    \Rightarrow 2u~du = - dx

    \Rightarrow - 2 u~du = dx

    So our integral becomes:

    -2 \int \frac u{(u^2 - b + a)u}~du = -2 \int \frac 1{u^2 - b + a}~du

    which you should be able to handle. (since a > b, -b + a > 0, so think "arctangent")
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