First slide

Applications of differential equations

Question

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

Moderate

A

$\frac{1}{2} \log_{e} 18$
B

$\log_{e} 9$
C

$\log_{e} 18$
D

$2 \log_{e} 18$

Solution

The given differential equation
$\frac{d P (t)}{d t} = \frac{P (t) - 900}{2}$
integrate both sides
$\begin{matrix} \int_{0}^{t} \frac{d P (t)}{P (t) - 900} = \int_{0}^{t} \frac{d t}{2} \\ {\{\ln |P (t) - 900|\}}_{0}^{t} = {\{\frac{t}{2}\}}_{0}^{t} \\ \ln |P (t) - 900| - \ln |P (0) - 900| = \frac{t}{2} \\ \ln |P (t) - 900| - \ln 50 = \frac{t}{2} \end{matrix}$
When population becomes zero
it gives
$\begin{matrix} \ln (900) - \ln (50) = \frac{t}{2} \\ t = 2 (\ln (\frac{900}{50})) \\ = 2 \ln 18 \end{matrix}$
therefore, $t = 2 \ln 18$

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