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Differential Equation Problem?

Differential equation problem (population equation)?

  • so here are the details: The population doubles every week(excluding other factors) 20,000 are eaten a day by predators the initial population is 200,000. find an equation for the population of mosquitoes at any time t. i came up with the differential equation: dp/dt = 2P - 140,000 (changing the # of mosquitoes eaten per day to per week to match the other factor) either this equation or my method of solving it went wrong. i used integrating factors to make the problem look like a chain rule and then integrated both sides to find p(t). any corrections to my initial equation would be helpful plus what you found to be the integrating factor.

  • Answer:

    P(t) = 200,000 * 2^t-140000 The derivative of this would tell you the rate of change of the population: dP/dt = 2^t(ln(200,000))

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P(t) = 200,000 * 2^t-140000 The derivative of this would tell you the rate of change of the population: dP/dt = 2^t(ln(200,000))

ireadlot...

Use definite integrals. On the left integrate from t=1940 to t=t to get k(t-1940). Also, remember that ∫ (1/t)dt = ln |t| + C, but you don't need the absolute value bars as long as 0 < P < 220. Also, shouldn't that second term on the right be ln(220 - P) instead of ln(P - 220)? On the left, integrate for P = 100 (millions is implied by the def'n of P) to P = P: ln(P) - ln(220-P) One simplification of the result is: ln(P) - ln((220 - P) - ln(100) + ln(120) = 220*k*(t - 1940) ln( P / (220 - P) ) = 220*k*(t - 1940) + ln(100/120) From here, take the exponential of both sides and solve for P.

Mary

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