Question

In a bank, principal increases continuously at the rate of $5\%$ per year. An amount of $Rs.1000$ is deposited with this bank, how much will it worth after $10years(e^{0.5}=1.648)$.

Answer 1

$\frac{dp}{dt}=\left(\frac{5}{100}\right)p20.\frac{dp}{p}=dt20lnp=t+cp=e^{(\frac{t}{20}+k)}$
at t$=$0 p$=$1000
$1000=e^k$
at t$=$10
$p=e^{(\frac{1}{2}+k)}=e^{0.5}.e^k$
Given that $e^{0.5}=1.648$
$p=1.648\times 1000=1648$