Step 1: Understanding the Concept:
When an amount increases continuously at a certain rate, it follows the continuous compound interest formula. This can be modeled by a simple first-order differential equation.
Step 2: Key Formula or Approach:
The rate of change of principal $P$ with respect to time $t$ is given by $\frac{dP}{dt} = rP$, where $r$ is the rate of interest per year.
The solution to this differential equation is $P(t) = P_0 e^{rt}$, where $P_0$ is the initial principal.
Step 3: Detailed Explanation:
Given rate $r = 10% = 0.10$.
The differential equation is:
\[ \frac{dP}{dt} = 0.10 P \]
Separating variables and integrating:
\[ \int \frac{dP}{P} = \int 0.10 dt \]
\[ \ln P = 0.10t + C \]
\[ P = e^{0.10t + C} = A e^{0.10t} \]
Initially, at $t = 0$, the deposited amount is Rs. 2000. So, $P_0 = 2000$.
\[ 2000 = A e^0 \implies A = 2000 \]
The equation for the principal at any time $t$ is:
\[ P(t) = 2000 e^{0.10t} \]
We need to find the amount after $t = 5$ years.
\[ P(5) = 2000 \cdot e^{0.10 \times 5} \]
\[ P(5) = 2000 \cdot e^{0.5} \]
Given that $e^{0.5} = 1.648$:
\[ P(5) = 2000 \times 1.648 \]
\[ P(5) = 3296 \]
Step 4: Final Answer:
The amount will become Rs. 3296.