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Differential Equations Tank problem

  1. Feb 5, 2010 #1
    1. The problem statement, all variables and given/known data
    The problem states that at time 0 the tank has 10 lb of salt dissolved in 100 gallons of water. The capacity is 200. Assume that water containing 1/8 lb of salt per gallon is entering the tank at a rate of 2 gal/min and the mizture is draining from the tank at a rate of 1 gal/min.
    a) set up the initial value problem
    b)solve using method of integrating factors.


    2. Relevant equations
    t:time
    y: amount of salt in tank (lbs)
    v:volume of water (lbs)
    v(t) = 100+t



    3. The attempt at a solution

    a) for the equation dy/dt = 1/4 - y/(100-t) y(0) = 10
    b) *this is where i think i'm messing up...

    dy/dt +(1/(100-t))y = 1/4

    [tex]\mu[/tex](t) = [tex]e^{\int\frac{1}{100-t}dt}[/tex]

    [tex]e^{-ln|100-t|}[/tex] = [tex]e^{ln|100-t|^{-1}}[/tex] = [tex](100-t)^{-1}[/tex]

    = [tex]\frac{1}{100-t}[/tex]

    so i multiply through and get

    ([tex]\frac{1}{100-t}[/tex] * y)' = [tex]\frac{1}{4(100-t)}[/tex]
    [tex]\frac{1}{100-t}[/tex] * y = [tex]\int[/tex][tex]\frac{1}{4(100-t)}[/tex]dt
    [tex]\frac{1}{100-t}[/tex] * y = ln|4(100-t)| +C
    then solving for y
    y = (100-t)*(ln|4(100-t)| +C)

    This doesn't seem right to me for some reason, I just want to make sure i'm doing this problem right. Can someone please let me know? Thank you.
     
  2. jcsd
  3. Feb 5, 2010 #2

    tiny-tim

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    Hi SpiffyEh! :smile:
    Technique looks fine, but

    i] shouldn't it be 100+t ?

    ii] your 4 should be outside the ln :wink:
     
  4. Feb 5, 2010 #3
    oops, sorry yeah it should be plus t. Should it be outside the ln? or should i just multiply the values by 4?
     
  5. Feb 6, 2010 #4

    tiny-tim

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    (just got up :zzz: …)

    Not following you :redface: … aren't they the same? :confused:
     
  6. Feb 6, 2010 #5
    yeah, i just thought it would be easier to multiply it through
     
  7. Feb 6, 2010 #6
    with the initital condition i got c = 1/10 - ln400
    so.. the whole equation is....
    y(t) = (100+t)*(ln|400+4t|+1/10-ln400)

    does that look right?
     
    Last edited: Feb 6, 2010
  8. Feb 6, 2010 #7

    tiny-tim

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    Nooo :redface:

    your integrating factor was for 100 - t.

    Now it's 100 + t, the integrating factor is completely different.
     
  9. Feb 6, 2010 #8
    it was supposed to be 100+t, i accidently changed it in the middle of all the equations
     
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