Question:
The particular solution of
\[
\log \frac{dy}{dx} = 3x + 4y, \quad y(0) = 0
\]
is:
The particular solution of
\[
\log \frac{dy}{dx} = 3x + 4y, \quad y(0) = 0
\]
is:
Show Hint
For differential equations, first separate variables, then integrate both sides.
Updated On: Nov 25, 2025
- \( e^{3x} + 3e^{-4y} = 4 \)
- \( 4e^{3x} - 3e^{-4y} = 3 \)
- \( 3e^{3x} + 4e^{4y} = 7 \)
- \( 4e^{3x} + 3e^{-4y} = 7 \)
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The Correct Option is D
Solution and Explanation
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).
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).
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