The population $$p\left( t \right)$$ at time $$t$$ of a certain mouse species satisfies the differential equation $$\frac{{dp\left( t \right)}}{{dt}} = 0.5\,p\left( t \right) - 450.$$     If $$p\left( 0 \right) = 850,$$   then the time at which the population becomes zero is :

Solution :
Given differential equation is
$$\eqalign{ & \frac{{dp\left( t \right)}}{{dt}} = 0.5\,p\left( t \right) - 450 \cr & \Rightarrow \frac{{dp\left( t \right)}}{{dt}} = \frac{1}{2}\,p\left( t \right) - 450 \cr & \Rightarrow \frac{{dp\left( t \right)}}{{dt}} = \frac{{p\left( t \right) - 900}}{2} \cr & \Rightarrow 2\frac{{dp\left( t \right)}}{{dt}} = - \left[ {900 - \,p\left( t \right)} \right] \cr & \Rightarrow 2\frac{{dp\left( t \right)}}{{900 - \,p\left( t \right)}} = - dt \cr} $$
Integrate both sides, we get
$$\eqalign{ & - 2\int {\frac{{dp\left( t \right)}}{{900 - \,p\left( t \right)}}} = \int {dt} \cr & {\text{Let }}900 - \,p\left( t \right) = u \cr & \Rightarrow - dp\left( t \right) = du \cr & \therefore {\text{We have, }}2\int {\frac{{du}}{u} = \int {dt} } \,\,\,\, \Rightarrow 2\,\ln \,u = t + c \cr & \Rightarrow 2\,\ln \left[ {900 - \,p\left( t \right)} \right] = t + c \cr & {\text{when }}t = 0,\,\,\,\,p\left( 0 \right) = 850 \cr & 2\,\ln \,\left( {50} \right) = c \cr & \Rightarrow 2\left[ {\ln \left( {\frac{{900 - \,p\left( t \right)}}{{50}}} \right)} \right] = t \cr & \Rightarrow 900 - \,p\left( t \right) = 50{e^{\frac{t}{2}}} \cr & \Rightarrow p\left( t \right) = 900 - \,50{e^{\frac{t}{2}}} \cr & {\text{Let }}p\left( {{t_1}} \right) = 0 \cr & 0 = 900 - - \,50{e^{\frac{{{t_1}}}{2}}} \cr & \therefore {t_1} = \,2\,\ln \,18 \cr} $$