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Math Help - Setting up and solving a DE

  1. #1
    gur
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    Setting up and solving a DE

    If you have a quantity h increasing by 160% every 4 hours, but 50000 units are subtracted every hour, how would you write the differential equation for the rate of change of h?
    Is it

    dh/dt = c*h -20*t where c is a constant and t is the number of hours

    or is it just

    dh/dt = c*h -20
    ?
    And would you solve them using sep. of variables, integrating factor or another method?
    Thank you for any help.
    Last edited by mr fantastic; March 8th 2010 at 11:56 AM. Reason: Changed post title
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  2. #2
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    Quote Originally Posted by gur View Post
    If you have a quantity h increasing by 160% every 4 hours, but 50000 units are subtracted every hour, how would you write the differential equation for the rate of change of h?
    Is it

    dh/dt = c*h -20*t where c is a constant and t is the number of hours

    or is it just

    dh/dt = c*h -20
    ?
    And would you solve them using sep. of variables, integrating factor or another method?
    Thank you for any help.
    Look at the units of all the different parameters, p.e. the rate of change:

    \left[\frac{dh}{dt}\right]=\frac{units}{hour}

    with "units" the "unit" of the quantity h. Now you have a rise of 160% every 4 hours, so you have:

    \left[\frac{h\cdot 1.6}{4hours}=\frac{2\cdot h}{5}\right]=\frac{units}{hour}

    And finally the amount extracted is:

    -50000\frac{units}{hour}

    This means you have for the DE:

    \frac{dh}{dt}=\frac{2 \cdot h}{5}-50000

    This is one way of finding the DE.

    Coomast
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  3. #3
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    The 50000 units per hour is a constant flow, not dependent on h or t, so the equation would be your second choice (but I don't see how you got k=20):

    \frac{dh}{dt} = ch - k

    You can use either method - separation of variables is a more direct method:

    \frac{dh}{ch-k} = dt

    I'll let you solve it from here - integrate both sides (remember constant of integration), then solve for h. You'll need an initial condition for h - if the problem doesn't give a number, I would just use h_0 at t=0.

    To get c, I think you have to set k=0 and then set c so that h increases to 2.6 times its initial value in 4 hours.

    Post again if you're still having trouble.
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  4. #4
    gur
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    With the condition h=1000000, t=0
    I found
    kh-50000=(1000000k-50000)e^(kt)
    How do I find k now?
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  5. #5
    Flow Master
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    Quote Originally Posted by gur View Post
    With the condition h=1000000, t=0
    I found
    kh-50000=(1000000k-50000)e^(kt)
    How do I find k now?
    The DE to solve was given to you in post #2 ....
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