In a bank, principal increases continuously at the rate of $5 %$ per year.

Solution:

Let p and t be the principal and time respectively.

It is given that the principal increases continuously at the rate of 5% per year.

$\Rightarrow \frac{d p}{d t}=\left(\frac{5}{100}\right) p$

$\Rightarrow \frac{d p}{d t}=\frac{p}{20}$

$\Rightarrow \frac{d p}{p}=\frac{d t}{20}$

Integrating both sides, we get:

$\int \frac{d p}{p}=\frac{1}{20} \int d t$

$\Rightarrow \log p=\frac{t}{20}+\mathrm{C}$

$\Rightarrow p=e^{\frac{1}{20}+\mathrm{C}}$      ...(1)

Now, when t = 0, p = 1000.

$\Rightarrow 1000=e^{C} \ldots(2)$

At t = 10, equation (1) becomes:

$p=e^{\frac{1}{2}+\mathrm{C}}$

$\Rightarrow p=e^{0.5} \times e^{\mathrm{C}}$

$\Rightarrow p=1.648 \times 1000$

$\Rightarrow p=1648$

Hence, after 10 years the amount will worth Rs 1648.