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Math Help - Can't integrate this differential equation

  1. #1
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    [SOLVED] Integration of D.E.

    Hi, having difficulty with this D.E. that relates to a tank of initially 400L of pure water, with 1L of salt water (concentration 4g/L) added per min, and 3L of the well-mixed tank solution removed per min. This is what I have so far;

    Q'(t)+\frac{3Q(t)}{400-2t}=4

    I'm trying to find an integrating factor though I keep ending up with a final differentiated x(t) function that doesn't make sense (based on t-200, resulting in the function only existing from t > 200 when it should be from t > 0)

    P(t)=\frac{3}{2(200-t)} leading to

    I(t)=-(200-t)^{3/2}

    but then this doesn't seem to make the LHS equal to the derivative of Q(t)I(t)? Without the negative sign it would work fine but I'm sure that's the integral of P(t)... a little help? Maybe I've done something stupid...


    Last edited by McChickenb; May 4th 2010 at 10:14 AM.
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  2. #2
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    Hello, McChickenb!

    \frac{dQ}{dt} +\frac{3}{400-2t}\,Q\;=\;4

    We have: . \frac{dQ}{dt} + \frac{3}{2}\cdot\frac{1}{200-t}\,Q \;=\;4


    Integrating factor: . I \;=\;e^{\frac{3}{2}\!\int \!\frac{dt}{200-t}} \;=\; e^{-\frac{3}{2}\ln(200-t)} \;=\;e^{\ln(200-t)^{-\frac{3}{2}}} \;=\;(200-t)^{-\frac{3}{2}}


    Multiply by I\!:\;\;(200-t)^{-\frac{3}{2}}\,\frac{dQ}{dt} + \frac{3}{2}\!\cdot\!\frac{1}{200-t}\!\cdot\!(200-t)^{-\frac{3}{2}} \,Q     \;=\;4(200-t)^{-\frac{3}{2}}


    . . . . . . . . . . . . . . . . . . (200-t)^{-\frac{3}{2}}+  \frac{3}{2}\,(200-t)^{-\frac{5}{2}}\,Q \;=\;4(200-t)^{-\frac{3}{2}}


    We have: . \frac{d}{dt}\bigg[(200-t)^{-\frac{3}{2}}\,Q\bigg] \;=\;4(200-t)^{-\frac{3}{2}}


    Integrate: . (200-t)^{-\frac{3}{2}}\,Q \;=\;8(200-t)^{-\frac{1}{2}} + C


    Multiply by (200-t)^{\frac{3}{2}}\!:\;\;\;Q \;=\;8(200-t) + C(200-t)^{\frac{3}{2}}


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  3. #3
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    D'oh! That's perfect. I was confusing myself on the simple stuff. Thanks a lot!
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