Question

The solution of the differential equation \( \log \left( \frac{dy}{dx} \right) = 9x - 6y + 6 \) is (given that \( y = 1 \) when \( x = 0 \))

• A. \( 3 e^{6y} = 2 e^{9x-6} + e^6 \)
• B. \( 3 e^{6y} = 2 e^{9x+6} + e^6 \)
• C. \( 3 e^{6y} = 2 e^{9x+6} - e^6 \)
• D. \( 3 e^{6y} = 2 e^{9x-6} - e^6 \)

Hint:

For differential equations, remember to separate the variables and integrate, then apply initial conditions to find the constant of integration.

Answer 1

The Correct Option is B

Step 1: Solve the differential equation.
The given equation is \( \log \left( \frac{dy}{dx} \right) = 9x - 6y + 6 \). Taking the exponential of both sides, we get: \[ \frac{dy}{dx} = e^{9x - 6y + 6} \] This simplifies to: \[ \frac{dy}{dx} = e^{9x} e^{-6y} e^6 \]
Step 2: Separate variables and integrate.
Rewriting the equation: \[ e^{6y} dy = 2 e^{9x} dx \] Integrating both sides gives: \[ 3 e^{6y} = 2 e^{9x} + C \] Using the initial condition \( y = 1 \) when \( x = 0 \), we find \( C \).
Step 3: Conclusion.
Substituting the value of \( C \) gives the solution: \[ 3 e^{6y} = 2 e^{9x+6} + e^6 \]