JEE MAIN - Mathematics (2012 - No. 8)
The population $$p$$ $$(t)$$ at time $$t$$ of a certain mouse species satisfies the differential equation $${{dp\left( t \right)} \over {dt}} = 0.5\,p\left( t \right) - 450.\,\,$$ If $$p(0)=850,$$ then the time at which the population becomes zero is :
$$2ln$$ $$18$$
$$ln$$ $$9$$
$${1 \over 2}$$$$ln$$ $$18$$
$$ln$$ $$18$$
Explanation
Given differential equation is
$${{dp\left( t \right)} \over {dt}} = 0.5p\left( t \right) - 450$$
$$ \Rightarrow {{dp\left( t \right)} \over {dt}} = {1 \over 2}p\left( t \right) - 450$$
$$ \Rightarrow {{dp\left( t \right)} \over {dt}} = {{p\left( t \right) - 900} \over 2}$$
$$ \Rightarrow 2{{dp\left( t \right)} \over {dt}} = \left[ {900 - p\left( t \right)} \right]$$
$$ \Rightarrow 2{{dp\left( t \right)} \over {900 - p\left( t \right)}} = - dt$$
Integrate both sides, we get
$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, - 2\int {{{dp\left( t \right)} \over {900 - p\left( t \right)}}} = \int {dt} $$
$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,Let\,\,900 - p\left( t \right) = u$$
$$ \Rightarrow - dp\left( t \right) = du$$
$$\therefore$$ $$\,\,\,\,\,\,$$ We have,
$$2\int {{{du} \over u}} = \int {dt \Rightarrow 2\,\ln \,u = t + c} $$
$$ \Rightarrow 2\ln \left[ {900 - p\left( t \right)} \right] = t + c$$
$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,$$ when $$t = 0,p\left( 0 \right) = 850$$
$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,$$ $$2\ln \left( {50} \right) = c$$
$$ \Rightarrow 2\left[ {\ln \left( {{{900 - p\left( t \right)} \over {50}}} \right)} \right] = t$$
$$ \Rightarrow 900 - p\left( t \right) = 50{e^{{t \over 2}}}$$
$$ \Rightarrow p\left( t \right) = 900 - 50{e^{{t \over 2}}}$$
Let $$p\left( {{t_1}} \right) = 0$$
$$0 = 900 - 50{e^{{{{t_1}} \over 2}}}$$
$$\therefore$$ $$\,\,\,\,\,\,{t_1} = 2\ln 18$$
$${{dp\left( t \right)} \over {dt}} = 0.5p\left( t \right) - 450$$
$$ \Rightarrow {{dp\left( t \right)} \over {dt}} = {1 \over 2}p\left( t \right) - 450$$
$$ \Rightarrow {{dp\left( t \right)} \over {dt}} = {{p\left( t \right) - 900} \over 2}$$
$$ \Rightarrow 2{{dp\left( t \right)} \over {dt}} = \left[ {900 - p\left( t \right)} \right]$$
$$ \Rightarrow 2{{dp\left( t \right)} \over {900 - p\left( t \right)}} = - dt$$
Integrate both sides, we get
$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, - 2\int {{{dp\left( t \right)} \over {900 - p\left( t \right)}}} = \int {dt} $$
$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,Let\,\,900 - p\left( t \right) = u$$
$$ \Rightarrow - dp\left( t \right) = du$$
$$\therefore$$ $$\,\,\,\,\,\,$$ We have,
$$2\int {{{du} \over u}} = \int {dt \Rightarrow 2\,\ln \,u = t + c} $$
$$ \Rightarrow 2\ln \left[ {900 - p\left( t \right)} \right] = t + c$$
$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,$$ when $$t = 0,p\left( 0 \right) = 850$$
$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,$$ $$2\ln \left( {50} \right) = c$$
$$ \Rightarrow 2\left[ {\ln \left( {{{900 - p\left( t \right)} \over {50}}} \right)} \right] = t$$
$$ \Rightarrow 900 - p\left( t \right) = 50{e^{{t \over 2}}}$$
$$ \Rightarrow p\left( t \right) = 900 - 50{e^{{t \over 2}}}$$
Let $$p\left( {{t_1}} \right) = 0$$
$$0 = 900 - 50{e^{{{{t_1}} \over 2}}}$$
$$\therefore$$ $$\,\,\,\,\,\,{t_1} = 2\ln 18$$
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