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Differential equations- Pop Growth

  1. Sep 9, 2008 #1
    dy/dt=(.5+sint)y/5 what is time t when the population has doubled?
  2. jcsd
  3. Sep 9, 2008 #2
    [tex] \frac{dy}{dt}=\frac{(.5+sint)y}{5}[/tex] Well, you first need to solve this one, and come up with a function that predicts population at any point in time t. Do you know how to solve this diff, eq?

    [tex] \frac{dy}{dt}=\frac{(.5+sint)y}{5}=>\frac{dy}{y}=\frac{1}{5}(.5+sint)dt[/tex]

    Now integrate on both sides, and solve for y. I guess Y=f(t) or sth like that.

    SO there is another info. that you have been provided with, Y(t)=2Yo, where Yo is your initial population. Have you been provided with initial population?or with another info. on this problem?
  4. Sep 9, 2008 #3


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    You don't need the initial population. The ode is linear. But you do need to write down the solution, as stupidmath points out. The general solution will have the form C*f(t). So you just need to solve C*f(t)=2*C*f(0) for t. You don't need to know C. It cancels.
  5. Sep 9, 2008 #4
    Got it!

    Thank both of you so much for your help.
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