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 $$2x^3 - x^2 y - 9x^2 + 3yx + C = 0$$