1. **State the problem:** We want to find the most suitable integrating factor by inspection for the differential equation $$y(2x + y)dx - x^2 dy = 0$$ to make it separable or exact. 2. **Rewrite the equation:** The given equation can be written as $$M(x,y)dx + N(x,y)dy = 0$$ where $$M = y(2x + y)$$ and $$N = -x^2$$. 3. **Check if the equation is exact:** Calculate $$\frac{\partial M}{\partial y}$$ and $$\frac{\partial N}{\partial x}$$. $$\frac{\partial M}{\partial y} = 2x + 2y$$ $$\frac{\partial N}{\partial x} = -2x$$ Since $$\frac{\partial M}{\partial y} \neq \frac{\partial N}{\partial x}$$, the equation is not exact. 4. **Find integrating factor candidates:** - If $$\frac{\partial}{\partial y} \left(\frac{M}{N}\right)$$ depends only on $$x$$, then an integrating factor $$\mu(x)$$ might work. - If $$\frac{\partial}{\partial x} \left(\frac{N}{M}\right)$$ depends only on $$y$$, then an integrating factor $$\mu(y)$$ might work. 5. **Try integrating factor depending on $$x$$:** Calculate $$\frac{\partial}{\partial y} \left(\frac{M}{N}\right) = \frac{\partial}{\partial y} \left(\frac{y(2x + y)}{-x^2}\right) = \frac{\partial}{\partial y} \left(-\frac{y(2x + y)}{x^2}\right)$$ Simplify inside: $$-\frac{y(2x + y)}{x^2} = -\frac{2xy + y^2}{x^2} = -\frac{2y}{x} - \frac{y^2}{x^2}$$ Derivative with respect to $$y$$: $$-\frac{2}{x} - \frac{2y}{x^2}$$ which depends on both $$x$$ and $$y$$, so no integrating factor $$\mu(x)$$. 6. **Try integrating factor depending on $$y$$:** Calculate $$\frac{\partial}{\partial x} \left(\frac{N}{M}\right) = \frac{\partial}{\partial x} \left(\frac{-x^2}{y(2x + y)}\right)$$ Rewrite denominator: $$y(2x + y) = 2xy + y^2$$ Derivative with respect to $$x$$: Use quotient rule or rewrite as $$-x^2 / (2xy + y^2)$$. Derivative: $$\frac{d}{dx} \left(-\frac{x^2}{2xy + y^2}\right) = - \frac{2x(2xy + y^2) - x^2(2y)}{(2xy + y^2)^2}$$ Simplify numerator: $$2x(2xy + y^2) - x^2(2y) = 4x^2 y + 2x y^2 - 2x^2 y = 2x^2 y + 2x y^2 = 2x y (x + y)$$ So derivative is: $$- \frac{2x y (x + y)}{(2xy + y^2)^2}$$ which depends on both $$x$$ and $$y$$. 7. **Try integrating factor of the form $$\mu = \frac{1}{x^2 y^2}$$ or inspect for simpler forms:** Rewrite the equation as: $$y(2x + y) dx = x^2 dy$$ Divide both sides by $$x^2 y$$: $$\frac{2x + y}{x^2} dx = \frac{1}{y} dy$$ This suggests substitution or integrating factor involving $$\frac{1}{x^2 y}$$. 8. **Conclusion:** The most suitable integrating factor by inspection is $$\mu = \frac{1}{x^2 y}$$ which simplifies the equation to a separable form. **Final answer:** The integrating factor to try by inspection is $$\boxed{\mu = \frac{1}{x^2 y}}$$.