(x2 + 2xy − Y2) Dx + (y2 + 2xy − X2) Dy = 0

Jasonbradley

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Jun 17, 2024

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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:

1. Find a potential function, ϕ(x, y), such that:

   • ∂ϕ/∂x = M(x, y)
   • ∂ϕ/∂y = N(x, y)

2. 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.

3. Differentiate ϕ(x, y) with respect to y and equate it to N(x, y):

   ∂ϕ/∂y = x² - 2xy + g'(y) = y² + 2xy − x²

4. Solve for g'(y):

   g'(y) = y² + 4xy

5. Integrate g'(y) with respect to y:

   g(y) = (y³/3) + 2xy² + C

   where C is an arbitrary constant.

6. 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

7. 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.