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  • Need solution for RD Sharma maths class 12 chapter Differential Equations exercise 21.11 question 7

Need solution for RD Sharma maths class 12 chapter Differential Equations exercise 21.11 question 7

Answers (1)

Answer:  312500

Given:  Population of city in 1990=2,00,000

            Population on city in 2000=2,50,000

To find: Population of city in 2010

Hint: Use \frac{d P}{d t}=\lambda P and find the value of constant lambda.

Solution: Let P be the population of the city at time t

\text { Then, } \begin{aligned} \frac{d P}{d t} \propto P \end{aligned}

        =\frac{d P}{d t}=\lambda P

Where lambda is the proportionality constant

        =\frac{d P}{p}=\lambda d t

Integrating on both sides we get,

        \begin{aligned} &=>\int \frac{d P}{P}=\int \lambda d t \\\\ &=>\log P=\lambda t+\log k \ldots(i) \end{aligned}

Where k is integral constant

\begin{aligned} \text { Whent } &=1990 \text {, we have } P=2,00,000 \\\\ &=>\log 2,00,000=\lambda \times 1990+\log k \ldots(i i) \end{aligned}

[utting t=0 and A=A0 in equation (i)]

At

        \begin{aligned} t=& 2000, \text { we have } P=2,50,000 \\\\ &=>\log 2,50,000=\lambda \times 2000+\log k \ldots(i i i) \end{aligned}

Subtracting equation (ii) and (iii)

        \begin{aligned} &=>\log 2,50,000-\log 2,00,000=2000 \lambda-1990 \lambda \\\\ &=>\log \left(\frac{2,50,000}{2,00,000}\right)=10 \lambda \\\\ &=>\log \left(\frac{5}{4}\right)=10 \lambda \end{aligned}

        =>\lambda=\frac{1}{10} \log \left(\frac{5}{4}\right)

Putting the value of lambda in equation (ii) we get,

        \begin{aligned} &=>\log 2,00,000=\frac{1990}{10} \log \left(\frac{5}{4}\right)+\log k \\ &=>\log k=\log 2,00,000-199 \log \left(\frac{5}{4}\right) \end{aligned}

Substituting the value of lambda in equation (ii) we get

        \begin{aligned} &=>\log P=\left(\frac{1}{10} \log \frac{5}{4}\right) 2010+\log 2,00,000-199 \log \frac{5}{4} \\\\ &=>\log P=201 \log \frac{5}{4}+\log 2,00,000-199 \log \frac{5}{4} \\\\ &=>\log P=\log \left(\frac{5}{3}\right)^{201}+\log 2,00,000-\log \left(\frac{5}{4}\right)^{199} \end{aligned}

        \begin{aligned} &=>\log P=\log \left[\frac{\left(\frac{5}{3}\right)^{201}}{\left(\frac{5}{4}\right)^{199}}\right]+\log 2,00,000 \\\\ &=>\log P=\log \left[\left(\frac{5}{4}\right)^{201-199} \times 2,00,000\right] \\\\ &=>\log P=\log \left[\left(\frac{5}{4}\right)^{2} \times 2,00,000\right] \end{aligned}

        \begin{aligned} &=>\log P=\log \left[\frac{25}{16} \times 2,00,000\right] \\\\ &=>\log P=\log 3,12,500 \\\\ &=>P=3,12,500 \end{aligned}

Thus, the population of city in 2010=3,12,500

 

Posted by

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