Question

The general solution of the differential equation: \[ (6x^2 - 2xy - 18x + 3y) dx - (x^2 - 3x) dy = 0 \] 

• A. \( 2x^3 - x^2 y - 9x^2 + 3y + C = 0 \)
• B. \( 4x^3 - 2x^2 y - 6x^2 + 6xy + C = 0 \)
• C. \( 2x^2 - 4xy - y^2 - x + 3y + C = 0 \)
• D. \( 3x^2 + 5xy - 2y^2 - 4x - 2y + C = 0 \) ]

Hint:

For exact differential equations, check if \( \frac{\partial M}{\partial y} = \frac{\partial N}{\partial x} \), then integrate \( M(x, y) \) with respect to \( x \), match terms using \( N(x, y) \), and solve for the general solution.

Answer 1

The Correct Option is A

Step 1: Checking for Exactness Given: \[ M(x, y) = 6x^2 - 2xy - 18x + 3y, \quad N(x, y) = -(x^2 - 3x) \] Compute: \[ \frac{\partial M}{\partial y} = -2x + 3, \quad \frac{\partial N}{\partial x} = -2x + 3 \] Since \( \frac{\partial M}{\partial y} = \frac{\partial N}{\partial x} \), the equation is exact. 

Step 2: Solving for \( F(x, y) \) Integrate \( M(x, y) \) with respect to \( x \): \[ F(x, y) = \int (6x^2 - 2xy - 18x + 3y) dx \] \[ = 2x^3 - x^2 y - 9x^2 + 3yx + g(y) \] Differentiate with respect to \( y \): \[ \frac{dF}{dy} = -x^2 + 3x + g'(y) \] Since \( \frac{dF}{dy} = N(x, y) \), equating: \[ -x^2 + 3x + g'(y) = -x^2 + 3x \] \[ g'(y) = 0 \Rightarrow g(y) = C \] 

Step 3: Final Equation \[ 2x^3 - x^2 y - 9x^2 + 3yx + C = 0 \]