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} The initial differential equation is \( \log \left( \frac{dy}{dx} \right) = 3x + 4y \). To remove the logarithm, we exponentiate both sides: \( \frac{dy}{dx} = e^{3x + 4y} \). This can be further simplified as \( \frac{dy}{dx} = e^{3x} \cdot e^{4y} \).

{Step 2: Separate the Variables} The differential equation is separable. We rearrange the terms to group all \( y \)-dependent terms on one side and all \( x \)-dependent terms on the other: \( \frac{dy}{e^{4y}} = e^{3x} dx \), which is equivalent to \( e^{-4y} dy = e^{3x} dx \). 

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

{Step 4: Solve for \( y \)} To isolate \( y \), we first multiply the equation by \(-4\): \( e^{-4y} = -\frac{4}{3} e^{3x} - 4C \). We can consolidate the constant term by setting \( C' = -4C \), resulting in \( e^{-4y} = -\frac{4}{3} e^{3x} + C' \). Next, we take the natural logarithm of both sides: \( -4y = \ln \left( -\frac{4}{3} e^{3x} + C' \right ) \). Finally, we solve for \( y \): \( y = -\frac{1}{4} \ln \left( -\frac{4}{3} e^{3x} + C' \right ) \). This equation represents the general solution of the differential equation, featuring one arbitrary constant \( C' \). 

{Step 5: Apply the Initial Condition} We use the given initial condition \( y(0) = 0 \). Substituting \( x = 0 \) and \( y = 0 \) into the general solution allows us to determine the value of \( C' \): \( 0 = -\frac{1}{4} \ln \left( -\frac{4}{3} e^{0} + C' \right ) \), which simplifies to \( 0 = -\frac{1}{4} \ln \left( -\frac{4}{3} + C' \right ) \). Multiplying by \(-4\) gives \( 0 = \ln \left( -\frac{4}{3} + C' \right ) \). Exponentiating both sides yields \( 1 = -\frac{4}{3} + C' \). Solving for \( C' \), we get \( C' = 1 + \frac{4}{3} = \frac{7}{3} \). Substituting this value back into the general solution, we obtain the particular solution: \( y = -\frac{1}{4} \ln \left( -\frac{4}{3} e^{3x} + \frac{7}{3} \right ) \). This can be rewritten as \( y = -\frac{1}{4} \ln \left( \frac{7}{3} - \frac{4}{3} e^{3x} \right ) \), \( y = -\frac{1}{4} \ln \left( \frac{7 - 4 e^{3x}}{3} \right ) \), \( y = -\frac{1}{4} \left( \ln (7 - 4 e^{3x}) - \ln 3 \right ) \), and finally \( y = \frac{1}{4} \ln 3 - \frac{1}{4} \ln (7 - 4 e^{3x}) \). 

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

{Step 6: Determine the Number of Arbitrary Constants} The general solution contained one arbitrary constant, \( C' \). However, applying the initial condition resulted in a unique value for this constant. 

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