Question

A population grows at the rate of 8% per year. How long does it take for the population to double ? Use differential equation for it.

Answer 1

Let $P_o$ be the initial population and let the population after t years be P. Then,

$\frac{dP}{dt}=\frac{8P}{100}$

$\frac{dP}{dt}=\frac{2P}{25}$

$\frac{dP}{P}=\frac{2}{25}dt$

$\int \frac{1}{P}dP=\frac{2}{25}\int dt$

$\log P=\frac{2}{25}t+C$

At $t=0,P=P_o$

Therefore,

$\log P_o=\frac{2\times 0}{25}+C$

$C=\log P_o$

Substituting this value in equation 1, we get,

$\log P=\frac{2}{25}t+\log P_o$

$\log \frac{P}{P_o}=\frac{2}{25}t$

$t=\frac{25}{2}\log \left(\frac{P}{P_o}\right)$

When $P=2P_o$,

$t=\frac{25}{2}\log \left(\frac{2P_0}{P_o}\right)=\frac{25}{2}\log 2$

Thus, the population is doubled in $\frac{25}{2}\log 2$ years.