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Math Help - differential equation problem

  1. #1
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    Exclamation differential equation problem

    I almost have the answer but I can't quite get there.

    Differential equation is

    dH/dt= (40-H)^2/100

    I solved it through here:

    H= 40 - 200/(2t + 5)
    However, when I plug in a value t=5 I get the wrong answer. My professor has this but the following step says:

    H=80t/(2t+5)

    How do I get to that?
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  2. #2
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    Quote Originally Posted by swimmergirl View Post
    I almost have the answer but I can't quite get there.

    Differential equation is

    dH/dt= (40-H)^2/100

    I solved it through here:

    H= 40 - 200/(2t + 5)
    However, when I plug in a value t=5 I get the wrong answer. My professor has this but the following step says:

    H=80t/(2t+5)

    How do I get to that?
    \frac{dt}{dH} = \frac{100}{(40 - H)^2}

    \Rightarrow t = \int \frac{100}{(40 - H)^2} \, dH = \frac{100}{40 - H} + C

    where C is a constant whose value cannot be calculated without additional data (such as H = 0 at t = 0 ....?)
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  3. #3
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    correct...initial conditions are H(0)=0. Sorry these weren't given in the problem but its about the height of a tree over time. Therefore, I just put that initially it isn't growing which suffices H(0)=0 and I forgot to mention that.
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  4. #4
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    Quote Originally Posted by swimmergirl View Post
    correct...initial conditions are H(0)=0. Sorry these weren't given in the problem but its about the height of a tree over time. Therefore, I just put that initially it isn't growing which suffices H(0)=0 and I forgot to mention that.
    t = \frac{100}{40 - H} + C

    Substitute H = 0, ~ t = 0: 0 = \frac{100}{40} + C \Rightarrow C = - \frac{5}{2}.

    Therefore:

    t = \frac{100}{40 - H} - \frac{5}{2} \Rightarrow 2t + 5 = \frac{200}{40 - H}

     \Rightarrow \frac{1}{2t + 5} = \frac{40 - H}{200} \Rightarrow \frac{200}{2t + 5} = 40 - H

    \Rightarrow H = 40 - \frac{200}{2t + 5} = \, ....
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