Question

The solution of the differential equation $\frac{dy}{dt} = ky$, $k > 0$ is given by}

• A. $y(t) = C e^{kt}$
• B. $y(t) = C e^{-kt}$
• C. $y(t) = C e^{2kt}$
• D. $y(t) = C e^{-2kt}$

Hint:

The equation $\frac{dy}{dt} = ky$ is the fundamental model for unlimited growth (like initial bacterial growth or continuous compounding interest) where the rate of change is proportional to the current amount.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
This represents the classical first-order separable differential equation governing exponential growth.

Step 2: Detailed Explanation:
Let us solve the equation by separating the variables:
\[ \frac{dy}{dt} = ky \]
Divide both sides by $y$ and multiply by $dt$:
\[ \frac{1}{y} \, dy = k \, dt \]
Integrate both sides:
\[ \int \frac{1}{y} \, dy = \int k \, dt \]
\[ \ln|y| = kt + C_1 \]
Exponentiate both sides to solve for $y(t)$:
\[ y(t) = e^{kt + C_1} = e^{C_1} e^{kt} \]
Let $C = e^{C_1}$ be an arbitrary constant:
\[ y(t) = C e^{kt} \]
Therefore, the solution to the differential equation is $y(t) = C e^{kt}$.

Step 3: Final Answer
The correct option is (A).