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Differential equations |
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| Aug22-12, 06:11 AM | #1 |
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Differential equations
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? |
| Aug22-12, 07:37 AM | #2 |
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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.) |
| Aug22-12, 08:08 AM | #3 |
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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. |
| Aug23-12, 03:35 AM | #4 |
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Differential equations
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
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| Aug23-12, 08:14 AM | #5 |
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Any old function e(t)?
Don't butterflies lay eggs? |
| Aug27-12, 07:08 AM | #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 |
| Aug28-12, 08:35 PM | #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.
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| Aug30-12, 08:58 PM | #8 |
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Is this relationship like compound interest?
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| Aug31-12, 11:51 AM | #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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