Question

Find the function \( f \) which satisfies the equation \( \frac{df}{dx} = 2f \), given that \( f(0) = e^3 \)

• A. \( 2x + 3 \)
• B. \( \log(2x + 3) \)
• C. \( e^{2x+3} \)
• D. \( \frac{x^2}{2} \)

Hint:

For equations of the form \( \frac{df}{dx} = kf \), the solution is always exponential \( f = Ce^{kx} \)Use initial condition to find \( C \).

Answer 1

The Correct Option is C

Step 1: Write the differential equation.
\[ \frac{df}{dx} = 2f \]
This is a separable differential equation.

Step 2: Separate variables.
\[ \frac{df}{f} = 2dx \]

Step 3: Integrate both sides.
\[ \int \frac{df}{f} = \int 2dx \]
\[ \ln |f| = 2x + c \]

Step 4: Remove logarithm.
\[ f = e^{2x+c} = Ce^{2x} \]
where \( C = e^c \).

Step 5: Use initial condition.
\[ f(0) = Ce^0 = C \]
Given \( f(0) = e^3 \), so:
\[ C = e^3 \]

Step 6: Substitute value of \( C \).
\[ f = e^3 \cdot e^{2x} = e^{2x+3} \]

Step 7: Final conclusion.
\[ \boxed{f(x) = e^{2x+3}} \]