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Differential equations

  1. Aug 22, 2012 #1
    Would someone be able to help me with the quesiton


    Life cycle of butterfly is egg -> caterpillar -> butterfly.
    Each week 5% of the eggs will hatch and 2% of the caterpillars will turn to butterflies.
    You start with 100 eggs and 60 caterpillars.

    Using calculus to obtain an equation which will refine your model for the amount of caterpillars at any time.
    How many caterpillars are there after 40 weeks?
     
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  3. Aug 22, 2012 #2

    HallsofIvy

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    How is this not homework? And what attempt have you made to do it yourself?

    Let C(t) be the number of caterpillars at time t (in weeks) and E(t) the number of eggs. Then dE/dt is the rate of change of number of eggs. What is it equal to each week? ("Each week 5% of the eggs will hatch".) dC/dt is the rate at which the number of caterpillars changes. What is that equal to? ("2% of the caterpillars will turn to butterflies". And, of course, the hatched eggs become caterpillars.)
     
  4. Aug 22, 2012 #3

    Simon Bridge

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    Hi 106267;
    You will get the most out of these forums if you attempt the problem yourself and show us the attempt. That way we can help you with the particular area where you get stuck.
     
  5. Aug 23, 2012 #4
    Dw, i figured it. I just found a function of e with regards to time and subbed that into the equation dc/ct=0.05e-0.02c. After that i integrated it, solved for a homogenous equation and than a partial solution
     
  6. Aug 23, 2012 #5

    Simon Bridge

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    Any old function e(t)?
    Don't butterflies lay eggs?
     
  7. Aug 27, 2012 #6
    Ok, so i got the equation C(t)= 226.667e^(-0.02(t)) - 166.667e^(-0.05(t))
    and at t=40 the equation gives the answer 79.2921

    however, if the relationship stated in the initial post about the question is model in excel it gives a value of 78.55. What assumptions are made by the differential equations at proves such a small difference?

    60 t=0
    63.7
    67.081
    70.16163
    72.9595349
    75.49142483
    77.77312292
    79.81961073
    81.64507127
    83.26292995
    84.68589346
    85.9259866
    86.99458732
    87.90246
    88.65978701
    89.27619866
    89.76080172
    90.12220636
    90.36855188
    90.507531
    90.54641303
    90.49206579
    90.35097644
    90.12927129
    89.83273451
    89.46682604
    89.03669843
    88.54721292
    88.0029547
    87.40824735
    86.76716655
    86.08355316
    85.36102555
    84.60299131
    83.81265844
    82.99304589
    82.14699355
    81.27717183
    80.38609064
    79.47610796
    78.54943798 t=40
     
  8. Aug 28, 2012 #7
    It looks like you used a first order forward integration formula with time steps of 1 week in excel to approximate the solution to the differential equations. The problem formulation implies "continuous compounding" of the production and destruction rates, rather than first order forward difference approximation. This accounts for most of the difference in the results. If you had used a higher order integration formula, the difference in results would have been less.
     
  9. Aug 30, 2012 #8
    Is this relationship like compound interest?
     
  10. Aug 31, 2012 #9
    It's not exactly the same as compound interest, but the comparison between the exact solution of the differential equations and the (approximate) solution you got in your spreadsheet algorithm is analogous to the comparison between "continuous compounding" of interest, and compounding only at discrete time intervals (like say, monthly).
     
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