1. **State the problem:** We are given the differential equation $$y(2x + y)dx - x^2 dy = 0$$ and asked to find the most suitable integrating factor by inspection. 2. **Rewrite the equation:** The 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:** Common integrating factors depend on either $$x$$ or $$y$$ alone. - If $$\mu = \mu(x)$$, then $$\frac{1}{N}(\frac{\partial M}{\partial y} - \frac{\partial N}{\partial x})$$ must be a function of $$x$$ only. Calculate: $$\frac{1}{N}(\frac{\partial M}{\partial y} - \frac{\partial N}{\partial x}) = \frac{1}{-x^2}((2x + 2y) - (-2x)) = \frac{1}{-x^2}(2x + 2y + 2x) = \frac{4x + 2y}{-x^2}$$ This depends on both $$x$$ and $$y$$, so $$\mu(x)$$ is not suitable. - If $$\mu = \mu(y)$$, then $$\frac{1}{M}(\frac{\partial N}{\partial x} - \frac{\partial M}{\partial y})$$ must be a function of $$y$$ only. Calculate: $$\frac{1}{M}(\frac{\partial N}{\partial x} - \frac{\partial M}{\partial y}) = \frac{1}{y(2x + y)}(-2x - (2x + 2y)) = \frac{-2x - 2x - 2y}{y(2x + y)} = \frac{-4x - 2y}{y(2x + y)}$$ This also depends on both $$x$$ and $$y$$, so $$\mu(y)$$ is not suitable. 5. **Try an integrating factor of the form $$\mu = x^m y^n$$:** By inspection, try $$\mu = \frac{1}{x^2 y^2}$$ or similar to simplify terms. 6. **Conclusion:** The most suitable integrating factor by inspection is $$\mu = \frac{1}{x^2 y^2}$$ or a function involving powers of $$x$$ and $$y$$ to make the equation exact. This is a common approach when neither $$\mu(x)$$ nor $$\mu(y)$$ works alone. **Final answer:** The most suitable integrating factor to try by inspection is $$\mu = \frac{1}{x^2 y^2}$$ or a similar power function of $$x$$ and $$y$$.