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 Po,
\(( \dfrac{dP}{dt})\) = 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 = Po unit when t = 0,
20 log (Po) = 0 + c
20log\( ( \dfrac{dP}{P_o})\) = c
Then, equation (i) becomes,
20 log (P) = t + 20 log (Po)
By cross multiplication we get,
20 log \( ( \dfrac{P}{P_o})\)= t
Let time is t, when P = 2Po,
Then, 20 log\( \dfrac{2P}{2P_o}\)= t1
t1 = 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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