1. **State the problem:** Solve the differential equation $$2x\,dx + \frac{x^2}{y}\,dy = 0$$ by finding an integrating factor and then solving the equation. 2. **Rewrite the equation:** The given equation can be written as $$M(x,y)\,dx + N(x,y)\,dy = 0$$ where $$M = 2x$$ and $$N = \frac{x^2}{y}$$. 3. **Check if the equation is exact:** Calculate $$\frac{\partial M}{\partial y}$$ and $$\frac{\partial N}{\partial x}$$. $$\frac{\partial M}{\partial y} = 0$$ $$\frac{\partial N}{\partial x} = \frac{2x}{y}$$ Since $$\frac{\partial M}{\partial y} \neq \frac{\partial N}{\partial x}$$, the equation is not exact. 4. **Find the integrating factor:** We check if an integrating factor depends only on $$x$$ or only on $$y$$. Calculate $$\frac{\frac{\partial N}{\partial x} - \frac{\partial M}{\partial y}}{M} = \frac{\frac{2x}{y} - 0}{2x} = \frac{2x/y}{2x} = \frac{1}{y}$$ which depends only on $$y$$. Therefore, the integrating factor $$\mu(y) = e^{\int \frac{1}{y} dy} = e^{\ln|y|} = y$$. 5. **Multiply the entire differential equation by the integrating factor $$y$$:** $$y(2x) dx + y \left( \frac{x^2}{y} \right) dy = 2xy\,dx + x^2\,dy = 0$$ 6. **Check if the new equation is exact:** New $$M = 2xy$$, new $$N = x^2$$. Calculate $$\frac{\partial M}{\partial y} = 2x$$ and $$\frac{\partial N}{\partial x} = 2x$$. Since $$\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}$$, the equation is exact. 7. **Find the potential function $$\psi(x,y)$$ such that $$\frac{\partial \psi}{\partial x} = M = 2xy$$:** Integrate with respect to $$x$$: $$\psi = \int 2xy\,dx = x^2 y + h(y)$$ 8. **Find $$h(y)$$ by differentiating $$\psi$$ with respect to $$y$$ and equating to $$N$$:** $$\frac{\partial \psi}{\partial y} = x^2 + h'(y) = N = x^2$$ So, $$h'(y) = 0 \Rightarrow h(y) = C$$ (constant). 9. **Write the implicit solution:** $$\psi(x,y) = x^2 y = C$$ This is the implicit solution to the differential equation. **Final answer:** $$x^2 y = C$$