Question

The solution of the differential equation \( (x^2 - 4xy - 2y^2)dx + (y^2 - 4xy - 2x^2)dy = 0 \), is

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

Hint:

When an equation \(Mdx + Ndy=0\) is exact, the solution can be found using the formula: \( \int M dx \) (treating y as a constant) + \( \int (\text{terms in N not containing x}) dy = K \). In this case, \( \int (x^2 - 4xy - 2y^2) dx + \int (y^2) dy = K \), which gives \( \frac{x^3}{3} - 2x^2y - 2xy^2 + \frac{y^3}{3} = K \). This is a very fast method.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
The given differential equation is of the form \( M(x,y)dx + N(x,y)dy = 0 \). We first check if it is an exact differential equation by verifying if \( \frac{\partial M}{\partial y} = \frac{\partial N}{\partial x} \). If it is exact, the solution is of the form \( F(x,y) = C \), 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 \( N(x,y) = y^2 - 4xy - 2x^2 \). We compute the 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) can be found by integrating \( M \) with respect to \( x \) and adding a function of \( y \), \( g(y) \). \[ F(x,y) = \int M(x,y) dx = \int (x^2 - 4xy - 2y^2) dx = \frac{x^3}{3} - 4\frac{x^2}{2}y - 2xy^2 + g(y) \] \[ F(x,y) = \frac{x^3}{3} - 2x^2y - 2xy^2 + g(y) \] To find \( g(y) \), we differentiate \( F(x,y) \) with respect to \( y \) and set it equal to \( N(x,y) \). \[ \frac{\partial F}{\partial y} = 0 - 2x^2 - 4xy + g'(y) \] We are 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: \[ g(y) = \int y^2 dy = \frac{y^3}{3} \] So, the general solution is \( \frac{x^3}{3} - 2x^2y - 2xy^2 + \frac{y^3}{3} = K \). To match the given options, we can multiply the entire 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 of the differential equation is \( x^3 - 6x^2y - 6xy^2 + y^3 + C = 0 \).