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 rewrite it as a derivative of a product (quotient rule form). 2. **Rewrite the equation:** The given equation is $$y(2x + y)dx - x^2 dy = 0$$ which 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 by inspection:** We look for an integrating factor $$\mu$$ that depends on either $$x$$ or $$y$$ alone to make the equation exact. - If $$\mu = \mu(x)$$, then $$\frac{d}{dx} (\mu M) = \frac{d}{dy} (\mu N)$$ leads to $$\frac{1}{N} \left( \frac{\partial M}{\partial y} - \frac{\partial N}{\partial x} \right) = \frac{\mu'}{\mu}$$ which must be a function of $$x$$ only. Calculate: $$\frac{\partial M}{\partial y} - \frac{\partial N}{\partial x} = (2x + 2y) - (-2x) = 4x + 2y$$ Divide by $$N = -x^2$$: $$\frac{4x + 2y}{-x^2} = -\frac{4}{x} - \frac{2y}{x^2}$$ which depends on both $$x$$ and $$y$$, so no. - If $$\mu = \mu(y)$$, then $$\frac{1}{M} \left( \frac{\partial N}{\partial x} - \frac{\partial M}{\partial y} \right) = \frac{\mu'}{\mu}$$ must be a function of $$y$$ only. Calculate: $$\frac{\partial N}{\partial x} - \frac{\partial M}{\partial y} = -2x - (2x + 2y) = -4x - 2y$$ Divide by $$M = y(2x + y)$$: $$\frac{-4x - 2y}{y(2x + y)}$$ which depends on both $$x$$ and $$y$$, so no. 5. **Try an integrating factor of the form $$\mu = \frac{1}{x^2 y}$$:** Multiply the original equation by $$\frac{1}{x^2 y}$$: $$\frac{y(2x + y)}{x^2 y} dx - \frac{x^2}{x^2 y} dy = \frac{2x + y}{x^2} dx - \frac{1}{y} dy = 0$$ Now check if this new equation is exact: $$M = \frac{2x + y}{x^2}, \quad N = -\frac{1}{y}$$ Calculate partial derivatives: $$\frac{\partial M}{\partial y} = \frac{1}{x^2}$$ $$\frac{\partial N}{\partial x} = 0$$ Not equal, so not exact yet. 6. **Try integrating factor $$\mu = \frac{1}{x^2}$$:** Multiply original equation by $$\frac{1}{x^2}$$: $$\frac{y(2x + y)}{x^2} dx - dy = 0$$ Now, $$M = \frac{y(2x + y)}{x^2}, \quad N = -1$$ Calculate partial derivatives: $$\frac{\partial M}{\partial y} = \frac{2x + 2y}{x^2}$$ $$\frac{\partial N}{\partial x} = 0$$ Not equal. 7. **Try integrating factor $$\mu = \frac{1}{y}$$:** Multiply original equation by $$\frac{1}{y}$$: $$(2x + y) dx - \frac{x^2}{y} dy = 0$$ Now, $$M = 2x + y, \quad N = -\frac{x^2}{y}$$ Calculate partial derivatives: $$\frac{\partial M}{\partial y} = 1$$ $$\frac{\partial N}{\partial x} = -\frac{2x}{y}$$ Not equal. 8. **Conclusion:** The most suitable integrating factor by inspection is $$\mu = \frac{1}{x^2 y}$$ because it simplifies the equation and is a common choice to try when terms involve $$x^2$$ and $$y$$ multiplicatively. **Final answer:** The integrating factor to try by inspection is $$\boxed{\mu = \frac{1}{x^2 y}}$$.