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Homework Help: Diff.equation homework

  1. Nov 8, 2005 #1
    The following diff.equation determines the number of people with a spesific illness:
    [tex]
    \frac{{dx}}{{dt}} = k\left( {150000 - x\left( t \right)} \right)
    [/tex]
    At t=0, 30000 people are infected, and at t=15, 60000. How long will it take for 120000 are infected? Here is my work:
    [tex]
    \begin{array}{l}
    x\left( t \right) = 30000 \cdot e^{k \cdot t} \\
    x\left( {15} \right) = 30000 \cdot e^{k \cdot 15} = 60000 \\
    \Rightarrow k = 0,046 \\
    x\left( t \right) = 30000 \cdot e^{0,046 \cdot t} = 120000 \\
    \end{array}
    [/tex]
    Problem is that this leads to no good, the answer is supposed to be about t=72 (days), but I'm not sure how to implement the 150000 in the beginning (at least I think that's the prob)...anyone?
     
  2. jcsd
  3. Nov 8, 2005 #2
    dx/dt = k*(p - x) has solutions of the form x = Aexp(-kt) + p,
    not just x = A*exp(-k*t).
     
  4. Nov 8, 2005 #3

    TD

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    Homework Helper

    Solving your DE will give an extra constant and you already had the constant k. You can determine the values of these two constants with x(0) = 30000 and x(15) = 60000. Are you sure of your solution of the DE?
     
  5. Nov 8, 2005 #4
    You mean am I sure it's supposed to be 72? That's just what it says in the back of the book that's all...
    I just want to know where in my attempt I turned wrong...
     
  6. Nov 8, 2005 #5

    TD

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    Homework Helper

    No, I'm talking about the solution of the differential equation.
     
  7. Nov 11, 2005 #6
    No, I'm not at all sure about my solution becasue I'm not taking the 150000 into effect since I don't know where or how to do so. How would you go about solving it?
     
  8. Nov 11, 2005 #7

    saltydog

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    Homework Helper

    TSN, as Qbert stated, the solution to the DE is:

    [tex]x(t)=Ae^{-kt}+150000[/tex]

    right?

    You arrange it into the form:

    [tex]x^{'}+kx=150000k[/tex]

    obtain an integrating factor, integrate, done deal.

    So you have two unknowns in the solution above, k and A, but you have two conditions:

    x(0)=30000

    x(15)=60000

    Substituting the first condition into the equation yields:

    [tex]30000=Ae^{-0k}+150000[/tex]

    you can solve for A right?

    now that you have A, substitute the second condition:

    [tex]60000=Ae^{-15k}+150000[/tex]

    That allows you to find k then. You know, rearrange, take the ln of both sides, solve for k.

    Then once you have these values, you can easily solve for what t yields 120000:

    [tex]120000=Ae^{-kt}+150000[/tex]

    I bet you can do this.:smile:
     
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