(B) Let $P$ be the population at time $t$. Given $\frac{dP}{dt} = kP$. Integrating,we get $\ln P = kt + C$,or $P(t) = P_0 e^{kt}$. At $t = 0$,$P = 20$

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(B) Let $P$ be the population at time $t$. Given $\frac{dP}{dt} = kP$. Integrating,we get $\ln P = kt + C$,or $P(t) = P_0 e^{kt}$. At $t = 0$,$P = 20$ lakhs,so $P_0 = 20$. At $t = 30$,$P = 40$ lakhs,so $40 = 20 e^{30k} \Rightarrow e^{30k} = 2 \Rightarrow e^{15k} = \sqrt{2}$. We need to find the population after another $15$ years,i.e.,at $t = 30 + 15 = 45$ years. $P(45) = 20 e^{45k} = 20 (e^{15k})^3 = 20 (\sqrt{2})^3 = 20 (2 \sqrt{2}) = 40 \sqrt{2}$ lakhs.

Class 12 Mathematics Differential Equations Application of differential equations

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The population of a city increases at a rate proportional to the population at that time. If the population of the city increases from $20$ lakhs to $40$ lakhs in $30$ years,then after another $15$ years,the population is:

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$10 \sqrt{2}$ lakhs
B
$40 \sqrt{2}$ lakhs
C
$30 \sqrt{2}$ lakhs
D
None of these

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The population of a city increases at a rate proportional to the population at that time. If the population of the city increases from $20$ lakhs to $40$ lakhs in $30$ years,then after another $15$ years,the population is:

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