Question

The population \[p\left( t \right)\] a time \[t\] of a certain mouse species satisfies the differential equation \[\dfrac{{dp\left( t \right)}}{{dt}} = \dfrac{1}{2}p\left( t \right) - 450\]. If \[p\left( 0 \right) = 850\], then the time at which the population becomes zero is:
A. \[2\ln 18\]
B. \[\ln 9\]
C. \[\dfrac{1}{2}\ln 18\]
D. \[\ln 18\]

Answer 1

Hint: First we will first integrate the given equation by separating variables and then use \[p\left( 0 \right) = 850\] in the obtained equation to find the value of \[C\]. Then we will substitute the value of \[C\] in equation and take \[p\left( t \right) = 0\] in the obtained equation. Then Use the logarithm property, \[\dfrac{{\ln a}}{{\ln b}} = \ln a - \ln b\] to find the required value.

Complete step by step answer:

We are given that the population \[p\left( t \right)\] a time \[t\] of a certain mouse species satisfies the differential equation \[\dfrac{{dp\left( t \right)}}{{dt}} = \dfrac{1}{2}p\left( t \right) - 450\].
Rewriting the given differential equation, we get
\[ \Rightarrow \dfrac{{dp\left( t \right)}}{{dt}} = \dfrac{{p\left( t \right) - 900}}{2}\]

Integrating the above equation by separating variables, we get
\[ \Rightarrow 2\int {\dfrac{{dp\left( t \right)}}{{p\left( t \right) - 900}} = } \int {dt} \]

Using the formula of integral \[\int {\dfrac{1}{x}dx} = \ln \left| x \right| + c\] in the above equation, we get
\[ \Rightarrow 2\ln \left| {p\left( t \right) - 900} \right| = t + C{\text{ ......eq.(1)}}\]
Using\[p\left( 0 \right) = 850\] in the above equation, we get
\[
   \Rightarrow 2\ln \left| {850 - 900} \right| = 0 + C \\
   \Rightarrow 2\ln \left| { - 50} \right| = C \\
   \Rightarrow C = 2\ln 50 \\
 \]

Substituting the value of \[C\] in equation (1), we get
\[ \Rightarrow 2\ln \left| {p\left( t \right) - 900} \right| = t + 2\ln 50\]

Taking \[p\left( t \right) = 0\] as at the time when the population becomes zero in the above equation, we get
\[
   \Rightarrow 2\ln \left| {0 - 900} \right| = t + 2\ln 50 \\
   \Rightarrow 2\ln 900 = t + 2\ln 50 \\
 \]

Subtracting the above equation by \[2\ln 50\] on both sides, we get
\[
   \Rightarrow 2\ln 900 - 2\ln 50 = t + 2\ln 50 - 2\ln 50 \\
   \Rightarrow 2\ln 900 - 2\ln 50 = t \\
   \Rightarrow 2\left( {\ln 900 - \ln 50} \right) = t \\
   \Rightarrow t = 2\left( {\ln 900 - \ln 50} \right) \\
 \]

Using the logarithm property, \[\dfrac{{\ln a}}{{\ln b}} = \ln a - \ln b\] in the above equation, we get
\[
   \Rightarrow t = 2\ln \dfrac{{900}}{{50}} \\
   \Rightarrow t = 2\ln 18 \\
 \]

Hence, option A is correct.

Note: In solving this type of question, one should know that the separation of the variable is any of the several methods for solving ordinary and partial differential equations, in which algebra allows one to rewrite an equation so that each of two variables occurs on a different side of the equation. The basic properties of logarithmic functions should be well known to the students. Students should substitute values properly and avoid calculation mistakes.