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Solve a differential equation

  1. Sep 15, 2009 #1
    1. The problem statement, all variables and given/known data

    Water is pumped into a cylindrical tank with cross section area A at a constant rate k, and
    leaks out through a hole of area a in the bottom of the tank at the rate
    αa (2gh(t))^1/2
    where g is the acceleration due to gravity, h(t) is the depth of water in the tank at time t,
    and α is a constant with 0.5 ≤ α ≤ 1.0. It follows that
    lim h(t)
    t→∞



    2. Relevant equations

    A h'(t) = k-αa (2gh(t))^1/2



    3. The attempt at a solution

    all i got to is [ A/ k-αa (2gh(t))^1/2 ] dh = dt

    I can't seem to solve the differential equation to get h(t)
     
  2. jcsd
  3. Sep 15, 2009 #2

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

    Hi vorse! :smile:

    (shouldn't your "a" should be the same as "A"? oh, and have a square-root: √ :wink:)

    The LHS is of the form (P + Q√h)dh … so just integrate it. :wink:
     
  4. Sep 15, 2009 #3
    A and a are different; they are both constant however, so it shouldn't matter much in the integration;


    "The LHS is of the form (P + Q√h)dh … so just integrate it." what do you mean by this form?


    If i let u = P+Q√h
    then du = 1/2Q(h)^-1/2 right? and this substitution doesn't work. I don't have anything to substitute for the du; can you clarify a little bit?
     
  5. Sep 16, 2009 #4

    Mark44

    Staff: Mentor


    (P + Q√h)dh = Pdh + Qh1/2dh
    Can't you integrate these two expressions without resorting to a substitution?
     
  6. Sep 16, 2009 #5
    well, I tried separating the denominator into the form f(x) = A/cx+d +B/dx+e, etc... i think it's called partial separating integration or something, not sure, but that didn't work out. I think I'm here is because I don't know how to integrate the following equation.
     
  7. Sep 16, 2009 #6

    tiny-tim

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    Science Advisor
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    What
    denominator?

    I don't see a fraction. :confused:
     
  8. Sep 16, 2009 #7
    all i got to is [ A/ k-αa (2gh(t))^1/2 ] dh = dt


    see the A is divided by ( k-αa (2gh(t))^1/2)

    so, in a clearer way to write it [A / ( k-αa (2gh(t))^1/2)]dh = dt

    btw, what programs are out there where I can type math equations on the comp?
     
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