From the question it is given that,

Population grows at the rate of 5% per year.

So, let us assume population after time t be p and initial population be P_{o},

\(( \dfrac{dP}{dt})\) = 5% × P

Then,

\( \dfrac{dp}{dt}= \dfrac{P}{20}\)

By cross multiplication we get,

20 \( ( \dfrac{dP}{P})\)= dt

Integrating on both side we get,

20 ∫\( ( \dfrac{dP}{P})\) = ∫ dt

20 log P = t + c …. … [equation (i)]

From question, P = P_{o} unit when t = 0,

20 log (P_{o}) = 0 + c

20log\( ( \dfrac{dP}{P_o})\) = c

Then, equation (i) becomes,

20 log (P) = t + 20 log (P_{o})

By cross multiplication we get,

20 log \( ( \dfrac{P}{P_o})\)= t

Let time is t, when P = 2P_{o},

Then, 20 log\( \dfrac{2P}{2P_o}\)= t_{1}

t_{1 }= 20 log 2

Therefore, time period required is 20 log 2 years.

Answered by Aaryan | 1 year agoFind the particular solution satisfying the given condition \( (\dfrac{dy}{dx}) = (x – 1)\dfrac{dy}{dx} = 2x^3y\)

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