Question

The population p (t) at time t of a certain mouse species satisfies the differential equation$\dfrac{{dp\left( t \right)}}{{dt}} = 0.5p\left( t \right) - 450$, if p (0) = 850, then the time at which the population become zero is:
$\left( a \right)2\log 18$
$\left( b \right)\log 9$
$\left( c \right)\dfrac{1}{2}\log 18$
$\left( d \right)\log 18$

Answer 1

Hint: In this particular question use the concept that the solution of any differential equation is the sum of complementary function and particular integral, so find the complementary function by equating the L.H.S of the equation to zero and find it roots, then the general solution of the complementary function is written as, ${C_1}{e^{at}} + {C_2}{e^{bt}}$, where a and b are the roots so use these concepts to reach the solution of the question.

Complete step-by-step answer:
Given differential equation
$\dfrac{{dp\left( t \right)}}{{dt}} = 0.5p\left( t \right) - 450$
Now written in standard form we have,
$ \Rightarrow \dfrac{{dp\left( t \right)}}{{dt}} - 0.5p\left( t \right) = - 450$
$ \Rightarrow \left( {D - 0.5} \right)p\left( t \right) = - 450$, where $D = \dfrac{d}{{dt}}$
Now as we know that the solution of the above equation is the sum of complementary function and particular integral.
Therefore, y = C.F + P.I
Now first find out complementary functions.
So for this equation L.H.S of the above equation to zero so we have,
$ \Rightarrow \left( {D - 0.5} \right) = 0$
$ \Rightarrow D = 0.5$
So the solution of the complementary function is given as,
C.F = $C{e^{\left( {0.5} \right)t}}$, where $C$ is some arbitrary constant.
Now for Particular integral we have,
$ \Rightarrow \left( {D - 0.5} \right)p\left( t \right) = - 450{e^{0t}}$, $\left[ {\because {e^{0t}} = 1} \right]$
$ \Rightarrow p\left( t \right) = \dfrac{{ - 450{e^{0t}}}}{{D - 0.5}}$
Now in the above equation the coefficient of t in the power of exponential is 0, so substitute in place of D 0, so the particular integral is.
Therefore, P.I = $\dfrac{{ - 450}}{{0 - 0.5}} = 900$
$ \Rightarrow P.I = 900$
So the solution of the given differential equation is,
$ \Rightarrow p\left( t \right) = C{e^{0.5t}} + 900$
So this is the required solution of the given differential equation.
Now it is given that p (0) = 850
So substitute in place of t in the above equation, 0 we have,
$ \Rightarrow p\left( 0 \right) = C{e^{0.5\left( 0 \right)}} + 900$
$ \Rightarrow p\left( 0 \right) = C + 900$, $\left[ {\because {e^0} = 1} \right]$
$ \Rightarrow 850 = C + 900$
$ \Rightarrow C = 850 - 900 = - 50$
So the solution of the differential equation become,
$ \Rightarrow p\left( t \right) = - 50{e^{0.5t}} + 900$
Now we have to find out the value of time at which the population of mouse species become zero, so substitute p (t) = 0 in the above equation we have,
$ \Rightarrow 0 = - 50{e^{0.5t}} + 900$
$ \Rightarrow 50{e^{0.5t}} = 900$
\[ \Rightarrow {e^{0.5t}} = \dfrac{{900}}{{50}} = 18\]
Now take log on both sides we have,
\[ \Rightarrow \log {e^{0.5t}} = \log 18\]
Now as we know that $\log {a^b} = b\log a,\log e = 1$ so we have,
\[ \Rightarrow 0.5t\left( {\log e} \right) = \log 18\]
\[ \Rightarrow 0.5t = \log 18\]
\[ \Rightarrow t = \dfrac{{\log 18}}{{0.5}} = 2\log 18\]
So this is the required time at which the population of mouse species becomes zero.
Hence option (a) is the correct answer.

Note: Whenever we face such types of questions the key concept we have to remember is that always recall that there are multiple methods to solve the particular integral depending on the R.H.S of the differential equation, if the R.H.S is in the form of ${e^{pt}}$, then substitute in place of D i.e. (d/dt), p and solve as above then add these two solutions as above we will get the solution for the differential equation, then according to given initial condition find out the value of constant term as above, then substitute p (t) = 0, and calculate the time at which the population of mouse species become zero, which is the required answer.