Consider a lake whose fish population is governed by the equation: dP/dt= 0.1P[(

Question

Consider a lake whose fish population is governed by the equation: dP/dt= 0.1P[(P/10)-1] if the initial fish population is 3,000 and assuming no fishing is allowed in the lake, what is the fish population after 10 days?

Explanation / Answer

dP/dt= 0.1P[(P/10)-1] =>dP/dt = 0.01P(P-10) Let 1/P(P-10) = A/P + B/P-10 =>1/P(P-10) = (AP-10A+BP)/P(P-10) =>1 = AP+BP - 10A =>A+B = 0, -10A = 1 =>A = -1/10, B = 1/10 =>1/P(P-10) = -1/10P + 1/10(P-10) dP/dt = 0.01P(P-10) =>dP/P(P-10) = 0.01dt =>-dP/10P + dp/10(P-10) = 0.01dt Integrating we get, -0.1lnP + 0.1ln(P-10) = 0.01t + C, where C is integration constant. =>-lnP+ln(P-10) = 0.1t + C =>ln((P-10)/P) = 0.1t + C Given t = 0 => P = 3000 =>ln((3000-10)/3000) = 0 + C =>C = -0.003338901 Therefore ln((P-10)/P) = 0.1t-0.003338901 If t = 10 days = 10/365 = 0.02739726 =>ln((P-10)/P) = 0.024058359 =>P-10/P = 1.024350096 =>P-10 = 1.024350096P =>0.024350096P = 10 =>P = 410.6759956

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