A population grows at the rate of 5% per year. If x=x(t) denotes the number of individuals in the population after t years, then the rate of change of x is equal to 5% of x. Form the desired differential equation . Find the time period

Asked by Sakshi | 1 year ago |  239

##### Solution :-

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

t= 20 log 2

Therefore, time period required is 20 log 2 years.

Answered by Aaryan | 1 year ago

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