Solving the Differential Equation: (x² + 2xy − y²) dx + (y² + 2xy − x²) dy = 0
This article will guide you through the process of solving the given differential equation:
(x² + 2xy − y²) dx + (y² + 2xy − x²) dy = 0
This equation is classified as an exact differential equation. To verify this, we need to check if the following condition holds:
∂M/∂y = ∂N/∂x
Where:
 M(x,y) = x² + 2xy − y²
 N(x,y) = y² + 2xy − x²
Let's calculate the partial derivatives:
 ∂M/∂y = 2x  2y
 ∂N/∂x = 2y  2x
Since ∂M/∂y = ∂N/∂x, the equation is indeed exact.
Solving the Exact Differential Equation
To solve an exact differential equation, we follow these steps:

Find a potential function, ϕ(x, y), such that:
 ∂ϕ/∂x = M(x, y)
 ∂ϕ/∂y = N(x, y)

Integrate ∂ϕ/∂x = M(x, y) with respect to x:
ϕ(x, y) = ∫(x² + 2xy − y²) dx = (x³/3) + x²y − xy² + g(y)
g(y) is an arbitrary function of y.

Differentiate ϕ(x, y) with respect to y and equate it to N(x, y):
∂ϕ/∂y = x²  2xy + g'(y) = y² + 2xy − x²

Solve for g'(y):
g'(y) = y² + 4xy

Integrate g'(y) with respect to y:
g(y) = (y³/3) + 2xy² + C
where C is an arbitrary constant.

Substitute the value of g(y) back into the expression for ϕ(x, y):
ϕ(x, y) = (x³/3) + x²y − xy² + (y³/3) + 2xy² + C

The general solution of the differential equation is given by:
ϕ(x, y) = C
(x³/3) + x²y − xy² + (y³/3) + 2xy² = C
Conclusion
Therefore, the general solution of the given exact differential equation (x² + 2xy − y²) dx + (y² + 2xy − x²) dy = 0 is (x³/3) + x²y − xy² + (y³/3) + 2xy² = C, where C is an arbitrary constant.