Life cycle of butterfly Differential equations

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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?
 

Answers and Replies

  • #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.)
 
  • #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.
 
  • #4
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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
 
  • #5
Simon Bridge
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Any old function e(t)?
Don't butterflies lay eggs?
 
  • #6
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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
 
  • #7
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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.
 
  • #8
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Is this relationship like compound interest?
 
  • #9
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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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