The population p(t) at time t of a certain mouse species satisfies the differential equation (d p(t))/(d t)=0.5 p(t)-450 . If p(0)=850 , then the time at which the population becomes zero is :

Correct answer: A. 2 18 Given differential equation is (d p(t))/(d t)=0.5 p(t)-450 (d p(t))/(d t)=(1)/(2) p(t)-450 (d p(t))/(d t)=(p(t)-900)/(2) 2 (d p(t))/(d t)=-[900-p(t)] 2 (d p(t))/(900-p(t))=-d t Integrate both the side, we get : -2 ∫ (d p(t))/(900-p(t))=∫ d t Let 900-p(t)=u -d p(t)=d u We have; 2 ∫ (d u)/(u)=∫ d t 2 u=t+c 2 [900-p(t)]=t+c when t=0, p(0)=8502 (50)=c 2[ ((900-p(t))/(50))]=t 900-p(t)=50 e^ (t)/(2) p(t)=900-50 e^ (t)/(2) let p(t_1)=00=900-50 e^ (t_1)/(2) t_1=2 18

BITSAT 2024Paper 1

Exam: BITSAT
Year: 2024
Paper: Paper 1
Subject: Mathematics
Chapter: Application of Derivatives

BITSAT 2024 · Paper 1 · Mathematics · Application of Derivatives

The population p(t) at time t of a certain mouse species satisfies...

The population

p(t)

at time

t

of a certain mouse species satisfies the differential equation

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

. If

p(0)=850

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

Options

A $2 \ln 18$
B $\ln 9$
C $\frac{1}{2} \ln 18$
D $\ln 18$

Correct answer

A. $2 \ln 18$

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

Integrate both the side, we get :

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

Let

900-p(t)=u \Rightarrow-d p(t)=d u\therefore \quad

We have;

2 \int \frac{d u}{u}=\int d t \Rightarrow 2 \ln u=t+c\Rightarrow \quad 2 \ln [900-p(t)]=t+c

when

t=0, p(0)=8502 \ln (50)=c \quad \Rightarrow 2\left[\ln \left(\frac{900-p(t)}{50}\right)\right]=t\Rightarrow \quad 900-p(t)=50 e^{\frac{t}{2}}\Rightarrow \quad p(t)=900-50 e^{\frac{t}{2}}

let

p\left(t_1\right)=00=900-50 e^{\frac{t_1}{2}} \quad \therefore t_1=2 \ln 18

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