Question

The number of arbitrary constants in the particular solution of the differential equation \[ \log \left( \frac{dy}{dx} \right) = 3x + 4y; \quad y(0) = 0 \] is/are:

• A. \( 2 \)
• B. \( 1 \)
• C. \( 0 \)
• D. \( 3 \)

Hint:

A particular solution to a differential equation satisfies the equation and all given initial or boundary conditions. Unlike the general solution, it does not contain any arbitrary constants.

Answer 1

The Correct Option is C

{Step 1:} Rewrite the Differential Equation\\ We start with the given differential equation: \[ \log \left( \frac{dy}{dx} \right) = 3x + 4y \] To simplify, we exponentiate both sides to eliminate the logarithm: \[ \frac{dy}{dx} = e^{3x + 4y} \] \[ \frac{dy}{dx} = e^{3x} \cdot e^{4y} \] 

{Step 2: Separate the Variables} The equation is separable, meaning we can rearrange terms to group all \( y \)-terms on one side and \( x \)-terms on the other: \[ \frac{dy}{e^{4y}} = e^{3x} dx \] \[ e^{-4y} dy = e^{3x} dx \] 

{Step 3: Integrate Both Sides} We integrate both sides to find the general solution: \[ \int e^{-4y} dy = \int e^{3x} dx \] \[ -\frac{1}{4} e^{-4y} = \frac{1}{3} e^{3x} + C \] where \( C \) is the constant of integration. 

{Step 4: Solve for \( y \)} Multiply both sides by \(-4\) to simplify: \[ e^{-4y} = -\frac{4}{3} e^{3x} - 4C \] Let \( C' = -4C \), then: \[ e^{-4y} = -\frac{4}{3} e^{3x} + C' \] Taking the natural logarithm of both sides: \[ -4y = \ln \left( -\frac{4}{3} e^{3x} + C' \right ) \] \[ y = -\frac{1}{4} \ln \left( -\frac{4}{3} e^{3x} + C' \right ) \] This expression represents the general solution of the differential equation, containing one arbitrary constant \( C' \). 

{Step 5: Apply the Initial Condition} We are given the initial condition \( y(0) = 0 \). Substitute \( x = 0 \) and \( y = 0 \) into the general solution to find \( C' \): \[ 0 = -\frac{1}{4} \ln \left( -\frac{4}{3} e^{0} + C' \right ) \] \[ 0 = -\frac{1}{4} \ln \left( -\frac{4}{3} + C' \right ) \] Multiply both sides by \(-4\): \[ 0 = \ln \left( -\frac{4}{3} + C' \right ) \] Exponentiate both sides: \[ 1 = -\frac{4}{3} + C' \] \[ C' = 1 + \frac{4}{3} = \frac{7}{3} \] Substitute \( C' \) back into the general solution: \[ y = -\frac{1}{4} \ln \left( -\frac{4}{3} e^{3x} + \frac{7}{3} \right ) \] \[ y = -\frac{1}{4} \ln \left( \frac{7}{3} - \frac{4}{3} e^{3x} \right ) \] \[ y = -\frac{1}{4} \ln \left( \frac{7 - 4 e^{3x}}{3} \right ) \] \[ y = -\frac{1}{4} \left( \ln (7 - 4 e^{3x}) - \ln 3 \right ) \] \[ y = \frac{1}{4} \ln 3 - \frac{1}{4} \ln (7 - 4 e^{3x}) \] 

This is the particular solution satisfying the initial condition \( y(0) = 0 \). 

{Step 6: Determine the Number of Arbitrary Constants} In the general solution, there was one arbitrary constant \( C' \). However, after applying the initial condition, this constant was determined uniquely. 

Therefore, the particular solution has: \[ \boxed{0} \] arbitrary constants.