Q.Solve the following differential equation: 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 10 years $(e^{0.5} = 1.648)$.

The key idea is that the principal grows according to an exponential growth rate: the rate of increase is proportional to the current principal.

Step 1 – Set up the differential equation

Let $P$ be the principal at time $t$ (in years). The rate of increase is $5\%$ per year, so

Step 2 – Solve

Separate variables:

Integrate:

Given $P_0 = 1000$ at $t=0$. …

The key idea here is exponential growth at a continuous rate. When a quantity increases "continuously at a rate of $5\%$ per year," it means that at every instant, the rate of increase is proportional to the current amount. This is not the same as adding $5\%$ of the original principal each year (simple interest) or even compounding annually. It's the mathematical ideal of compounding every infinitesimal moment — which is exactly what the differential equation captures.

Why does this matter? Because in real banks, interest might compound monthly or daily, but here the problem explicitly says "continuously." That forces us to use the exponential model. The given value $e^{0.5}=1.648$ is a strong hint: after 10 years, the exponent will be $0.05 \times 10 = 0.5$, so the answer will be $1000 \times 1.648$.

Let's walk through it step by step.

1. Set up the differential equation. Let $P(t)$ be the principal (in rupees) at time $t$ years. The phrase "increases continuously at the rate of $5\%$ per year" means the instantaneous rate of change $\frac{dP}{dt}$ is $5\%$ of the current principal. In mathematical terms:

This is a first-order linear differential equation — and it's also separable.

2. Solve the differential equation. Separate variables:

Integrate both sides:

Since principal is positive, we drop the absolute value. Exponentiate:

Let $e^C = A$, so $P(t) = A e^{0.05 t}$.

3. Apply the initial condition. We are told Rs 1000 is deposited initially, meaning at $t=0$, $P(0)=1000$. Substitute:

So $A = 1000$, and the particular solution is: …

Method: Continuous growth — finding the future amount

Use this when the continuous growth rate is known and you must find the amount after a given time.

Steps

Step 1: Set up the proportional-rate model

$\dfrac{dP}{dt}=\dfrac{r}{100}P$, which integrates to $P=P_0\,e^{rt/100}$.

Step 2: Substitute the known rate, principal and time

Plug in the given values (here $r=5$, $P_0=1000$, $t=10$) to get $P=1000\,e^{0.5}$.

Step 3: Evaluate using the supplied constant …

Common Mistakes

Mistake 1: Forgetting the $\dfrac{1}{100}$ in the rate

Why it's wrong: $5\%$ means $\frac{dP}{dt}=\frac{5}{100}P=0.05P$; using $5$ blows up the exponent. Correct approach: exponent is $\frac{rt}{100}=0.5$.

Mistake 2: Miscomputing the exponent $\dfrac{rt}{100}$ …