1. **Problem:** Solve the differential equation $$2xy \, dx + (y^2 - x^2) \ dy = 0.$$ 2. **Rewrite in differential form:** $$2xy \, dx + (y^2 - x^2) \ dy = 0.$$ This can be thought of as $$M(x,y) dx + N(x,y) dy = 0$$ with $$M = 2xy, \quad N = y^2 - x^2.$$ 3. **Check if equation is exact:** Compute partial derivatives: $$\frac{\partial M}{\partial y} = 2x,$$ $$\frac{\partial N}{\partial x} = -2x.$$ Since $$\frac{\partial M}{\partial y} \neq \frac{\partial N}{\partial x},$$ the equation is not exact. 4. **Look for integrating factor:** Try an integrating factor dependent on $$x$$ or $$y$$. Calculate: $$\frac{\partial N}{\partial x} - \frac{\partial M}{\partial y} = -2x - 2x = -4x.$$ Divide by $$M = 2xy$$: $$\frac{-4x}{2xy} = \frac{-2}{y}.$$ Since the expression depends only on $$y$$ (single variable), the integrating factor $$\mu(y)$$ exists. 5. **Find integrating factor $$\mu(y)$$:** The formula for integrating factor depending on $$y$$ is: $$\mu(y) = e^{\int P(y) dy}$$ with $$P(y) = \frac{1}{M}\left( \frac{\partial N}{\partial x} - \frac{\partial M}{\partial y} \right) = -\frac{2}{y}.$$ Therefore, $$\mu(y) = e^{\int -\frac{2}{y} dy} = e^{-2 \ln |y|} = y^{-2}.$$ 6. **Multiply whole equation by integrating factor $$y^{-2}$$:** $$y^{-2} (2xy \, dx) + y^{-2} (y^2 - x^2) \, dy = 0$$ Simplify: $$2x y^{-1} \, dx + (1 - x^2 y^{-2}) \, dy = 0.$$ 7. **Check if new equation is exact:** New $$M = 2x y^{-1},$$ New $$N = 1 - \frac{x^2}{y^2}.$$ Partial derivatives: $$\frac{\partial M}{\partial y} = -2x y^{-2},$$ $$\frac{\partial N}{\partial x} = -2x y^{-2}.$$ Since these are equal, equation is now exact. 8. **Find potential function $$\psi(x,y)$$:** From $$\frac{\partial \psi}{\partial x} = M = 2x y^{-1},$$ Integrate w.r.t. $$x$$: $$\psi = \int 2x y^{-1} dx = y^{-1} x^2 + h(y).$$ 9. **Find $$h(y)$$ by differentiating $$\psi$$ w.r.t. $$y$$ and equating to $$N$$:** $$\frac{\partial \psi}{\partial y} = -y^{-2} x^2 + h'(y),$$ Set equal to $$N$$: $$-\frac{x^2}{y^2} + h'(y) = 1 - \frac{x^2}{y^2} \implies h'(y) = 1.$$ 10. **Integrate $$h'(y)$$:** $$h(y) = y + C,$$ where $$C$$ is constant. 11. **Compose general solution:** $$\psi(x,y) = y^{-1} x^2 + y = C.$$ Or equivalently, $$\frac{x^2}{y} + y = C.$$ **Final answer:** $$\boxed{\frac{x^2}{y} + y = C}.$$