Step 1: Understanding the Concept:
The differential equation is in the form \( M(x,y)dx + N(x,y)dy = 0 \). Verify if it's exact by checking \( \frac{\partial M}{\partial y} = \frac{\partial N}{\partial x} \). If exact, the solution is \( F(x,y) = C \), where \( \frac{\partial F}{\partial x} = M \) and \( \frac{\partial F}{\partial y} = N \).
Step 2: Check for Exactness:
Let \( M(x,y) = x^2 - 4xy - 2y^2 \) and \( N(x,y) = y^2 - 4xy - 2x^2 \).
Compute partial derivatives:
\[ \frac{\partial M}{\partial y} = \frac{\partial}{\partial y}(x^2 - 4xy - 2y^2) = -4x - 4y \]
\[ \frac{\partial N}{\partial x} = \frac{\partial}{\partial x}(y^2 - 4xy - 2x^2) = -4y - 4x \]
Since \( \frac{\partial M}{\partial y} = \frac{\partial N}{\partial x} \), the equation is exact.
Step 3: Find the Solution:
The solution \( F(x,y) = K \) (where K is a constant) is found by integrating \( M \) with respect to \( x \) and adding a function of \( y \), \( g(y) \).
\[ F(x,y) = \int M(x,y) dx = \int (x^2 - 4xy - 2y^2) dx = \frac{x^3}{3} - 4\frac{x^2}{2}y - 2xy^2 + g(y) \]
\[ F(x,y) = \frac{x^3}{3} - 2x^2y - 2xy^2 + g(y) \]
Find \( g(y) \) by differentiating \( F(x,y) \) with respect to \( y \) and setting it equal to \( N(x,y) \).
\[ \frac{\partial F}{\partial y} = 0 - 2x^2 - 4xy + g'(y) \]
Given \( N(x,y) = y^2 - 4xy - 2x^2 \).
\[ -2x^2 - 4xy + g'(y) = y^2 - 4xy - 2x^2 \]
\[ g'(y) = y^2 \]
Integrating with respect to \( y \) gives:
\[ g(y) = \int y^2 dy = \frac{y^3}{3} \]
The general solution is \( \frac{x^3}{3} - 2x^2y - 2xy^2 + \frac{y^3}{3} = K \).
Multiply the equation by 3:
\[ x^3 - 6x^2y - 6xy^2 + y^3 = 3K \]
Let \( C = -3K \). The solution is \( x^3 - 6x^2y - 6xy^2 + y^3 + C = 0 \). This matches option (B).
Step 4: Final Answer:
The solution is \( x^3 - 6x^2y - 6xy^2 + y^3 + C = 0 \).