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AIEEE 2012 Solved Paper

Section: Mathematics

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Question : 79 of 89

Marks: +1, -0

The population

p(t)

t

of a certain mouse species satisfies the differential equation

\frac{d p(t)}{d t}=0.5 p(t)-450

p(0)=850

, then the time at which the population becomes zero is :

$2 \ln 18$
$\ln 9$
$\frac{1}{2} \ln 18$
$\ln 18$

Solution:

Given differential equation is

\frac{d p(t)}{d t}=0.5 p(t)-450

\Rightarrow \frac{d p(t)}{d t}=\frac{1}{2} p(t)-450

\Rightarrow \frac{d p(t)}{d t}=\frac{p(t)-900}{2}

\Rightarrow 2 \frac{d p(t)}{d t}=[900-p(t)]

\Rightarrow 2 \frac{d p(t)}{900-p(t)}=-dt

Integrate both sides, we get

-2 \int \frac{d p(t)}{900-p(t)}=\int dt

\text{ Let } 900-p(t)=u

\Rightarrow -d p(t)=du

\therefore \text{ We have, }

\therefore

We have,

2 \int \frac{du}{u}=\int dt \Rightarrow 2 \ln u=t+c

\Rightarrow 2 \ln [900-p(t)]=t+c

when

t=0, p(0)=850

2 \ln (50)=c

\Rightarrow 2 \left[ \ln \left( \frac{900-p(t)}{50} \right) \right] =t

\Rightarrow 900-p(t)=50 e^{\frac{t}{2}}

\Rightarrow p(t)=900-50 e^{\frac{t}{2}}

Let

p(t_1)=0

0=900-50 e^{\frac{t_1}{2}}

\therefore t_1=2 \ln 18

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