Question

The particular solution of \[ \log \frac{dy}{dx} = 3x + 4y, \quad y(0) = 0 \] is:

• A. \( e^{3x} + 3e^{-4y} = 4 \)
• B. \( 4e^{3x} - 3e^{-4y} = 3 \)
• C. \( 3e^{3x} + 4e^{4y} = 7 \)
• D. \( 4e^{3x} + 3e^{-4y} = 7 \)

Hint:

For differential equations, first separate variables, then integrate both sides.

Answer 1

The Correct Option is D

Step 1: Converting the equation.
Rewriting the given equation: \[ \frac{dy}{dx} = e^{3x + 4y} \] Separating variables: \[ e^{-4y} dy = e^{3x} dx \] Integrating both sides: \[ \int e^{-4y} dy = \int e^{3x} dx \] \[ \frac{e^{-4y}}{-4} = \frac{e^{3x}}{3} + C \] Multiplying by -4: \[ e^{-4y} = -\frac{4}{3} e^{3x} + C \] Using \( y(0) = 0 \), solving for \( C \), we get: \[ 4e^{3x} + 3e^{-4y} = 7 \] Thus, the correct answer is (D).