Question

The population $P = P(t)$ at time 't' of a certain species follows the differential equation $\frac{dP}{dt} = 0.5P - 450$. If $P(0) = 850$, then the time at which population becomes zero is :

• A. $\log_e 9$
• B. $\frac{1}{2}\log_e 18$
• C. $\log_e 18$
• D. $2\log_e 18$

Hint:

In equations of the form $\frac{dy}{dt} = ay - b$, the equilibrium point is $y = b/a$. If the initial value is below this point, the value will eventually reach zero.

Answer 1

The Correct Option is D

Step 1: Rearrange the equation: $\frac{dP}{dt} = \frac{P - 900}{2} \Rightarrow \frac{dP}{P-900} = \frac{1}{2}dt$.
Step 2: Integrate both sides: $\ln|P-900| = \frac{t}{2} + C$.
Step 3: Use $P(0) = 850$: $\ln|850-900| = 0 + C \Rightarrow C = \ln 50$.
Step 4: For $P(t) = 0$: $\ln|0-900| = \frac{t}{2} + \ln 50$.
Step 5: $\frac{t}{2} = \ln 900 - \ln 50 = \ln \left(\frac{900}{50}\right) = \ln 18$.
Step 6: $t = 2\ln 18$.