The population P = P(t) at time 't' of a certain species follows the differential equation = 0.5P – 450. If P(0) = 850, then the time at which population becomes zero is
Question
The population P = P(t) at time 't' of a certain species follows the differential equation = 0.5P – 450. If P(0) = 850, then the time at which population becomes zero is
Solution
The given differential equation is dP/dt = 0.5P - 450.
This is a first order linear differential equation and can be solved using an integrating factor.
First, we rewrite the equation in standard form: dP/dt - 0.5P = -450.
The integrating factor e^∫-0.5 dt = e^-0.5t.
Multiplying every term of the equation by the integrating factor, we get:
e^-0.5t dP/dt - 0.5e^-0.5t P = -450e^-0.5t.
The left hand side of this equation is the derivative of (e^-0.5t P) with respect to t. So, we can write:
d/dt (e^-0.5t P) = -450e^-0.5t.
Integrating both sides with respect to t, we get:
e^-0.5t P = -900e^-0.5t + C.
Solving for P, we get:
P = -900 + Ce^0.5t.
Given that P(0) = 850, we can solve for C:
850 = -900 + C*1 => C = 1750.
So, the solution to the differential equation is:
P = -900 + 1750e^0.5t.
Setting P = 0 to find the time at which the population becomes zero, we get:
0 = -900 + 1750e^0.5t.
Solving for t, we get:
t = 2 ln(1750/900) ≈ 1.34.
So, the population becomes zero at approximately t = 1.34.
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