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A simple differential equation

  1. Aug 1, 2007 #1
    The depth of water x at any time t in a tank is given by
    [tex] A\frac{dx}{dt}= B-c\sqrt{x}\\ [/tex],
    where A, b and C are constants. Initially x=0; show that [tex] C^2t=2AB\ln{\frac{B}{B_C\sqrt{x}}}-2AC\sqrt{x}\\ [/tex]
    To solve this let [tex]x= y^2\\ [/tex].
    Then [tex] \int \frac{A}{B-C\sqrt{x}} = \int dt = \int \frac{2Ay}{B-Cy}dy \\ [/tex].
    Let u=B-Cy, therefore [tex] y=\frac{B-u}{c}\\ [/tex], therefore the integral is
    [tex] -2A \int \frac{B-u}{C^2u} du = \frac{-2A}{C^2} \int \frac{B-u}{u} du \\ [/tex].
    which [tex] = \frac{-2AB}{C^2} \int \frac{1}{u} du + \frac{2A}{C^2} \int du \\ [/tex].
    Therefore we have [tex]\frac{-2AB}{C^2}\ln(u) + \frac{2A}{C^2}u \\[/tex]. We have [tex] \frac{-2AB}{C^2} \ln{B_Cy} + \frac{2A}{C^2}(B-Cy) \\[/tex].
    Which implies [tex] C^2t = -AB\ln{\frac{B}{C\sqrt{x}}} + 2A(B-C\sqrt{x})\\ [/tex].
    As you can see I get a different answer. I integrated by parts also to get a different result. Please help. Thanks.
     
    Last edited: Aug 1, 2007
  2. jcsd
  3. Aug 1, 2007 #2

    dextercioby

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    Just make [itex] B-C\sqrt{x} =t [/itex]
     
  4. Aug 3, 2007 #3
    I tried solving it dextercioby's way I used y instead of his t, and I got the wrong answer, I got the integral [tex] = \frac{ -AB}{2C} \int dy + \frac{AB}{2C} \int y dy \\ [/tex]. Where [tex] \sqrt{x} = \frac{B-y}{C} \\[/tex] and [tex] dx=-dy \frac{B-y}{2C} \\ [/tex]. Thanks for helping.
     
  5. Aug 7, 2007 #4
    Sorry I wrote down the wrong answer to dextercioby's method of solving this. What I actually got was the following: [tex] t = \frac{-AB}{2C} \int \frac{1}{y}dy + \frac{A}{2C} \int dy [/tex]. Which is not the right answer.
     
  6. Aug 10, 2007 #5
    I was thinking someone might have a second opinion on what I did originally, i.e., #1. Thanks for the help.
     
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