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ODE dilution problem

  1. Jan 22, 2010 #1
    A tank initially contains 200 gal of fresh water with 0 lbs of salt. Brine whose salt concentration is 2lb/gal flow in at 2 gal/min and the mix flows out at the same rate.

    x = salt content, t = time.

    So I figure that dx = 4dt - 2xdt/200 or dx = (400 - x).01dt
    dx/(400-x) = .01dt

    [tex]\int0x[/tex] [tex]\frac{dx}{400-x}[/tex]] = [tex]\int0100[/tex]1000.01dt

    ln([tex]\frac{400-x}{400}[/tex] = 1
    400 - x = 400e
    x = 400-400e

    which obviously isn't correct.

    Can somebody show me what I'm doing wrong here?
  2. jcsd
  3. Jan 22, 2010 #2
    I forgot to preview my original post, hence the horrible LaTeX job. But no matter, I've figured out the problem in my solution. I'll restate the problem since my first attempt is next to illegible:

    200 gal fresh water, brine with 2lb/gal flows in @ 2 gal/min, and the mix flows out at the same rate. Find the salt content of the water after 100 minutes:
    x= salt content
    t= time

    dx = 4dt - [tex]\frac{2xdt}{200}[/tex] = (400-x).01dt

    [tex]\frac{dx}{400-x}[/tex] = .01dt

    [tex]\int[/tex][tex]^{x}_{0}[/tex][tex]\frac{dx}{400-x}[/tex] = [tex]\int[/tex][tex]^{100}_{0}[/tex].01dt

    And this is where I went wrong. Forgetting to apply the chain rule on the left side I would get:

    ln([tex]\frac{400-x}{400}[/tex]) = 1

    when I should've gotten:

    -ln([tex]\frac{400-x}{400}[/tex]) = 1

    which gives the correct answer of 252.8 lbs when solving for x.
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