1. **Problem:** Write the given differential equations in differential form. 2. **Recall:** The differential form of an equation involving derivatives $y'$ or $\frac{dy}{dx}$ is typically expressed as $M(x,y)dx + N(x,y)dy = 0$. 3. **Solutions:** a) Given $dx = y$, rewrite as $dx - y = 0$. Since $dx$ is already differential, the differential form is: $$dx - y\,dy = 0$$ But here $dx = y$ means $\frac{dx}{dy} = y$, so the differential form is simply: $$dx - y\,dy = 0$$ b) Given $y' = y^2 + x$, recall $y' = \frac{dy}{dx}$, so: $$\frac{dy}{dx} = y^2 + x \implies dy = (y^2 + x) dx$$ Rearranged to differential form: $$(y^2 + x) dx - dy = 0$$ c) Given $xy' + y^2 = 0$, rewrite $y' = \frac{dy}{dx}$: $$x \frac{dy}{dx} + y^2 = 0 \implies x dy + y^2 dx = 0$$ d) Given $-x y' - x = y'$, rearranged: $$-x y' - x = y' \implies -x y' - y' = x \implies y'(-x - 1) = x \implies y' = \frac{x}{-x - 1}$$ Writing in differential form: $$\frac{dy}{dx} = \frac{x}{-x - 1} \implies (-x - 1) dy - x dx = 0$$ e) Given $x^2 y - y x y = 2 x y$, simplify the left side: $$x^2 y - y x y = x y (x - y) = 2 x y$$ Rearranged: $$x y (x - y) - 2 x y = 0 \implies x y (x - y - 2) = 0$$ This is an algebraic expression, not a differential equation, so no differential form applies here. **Final answers:** - a) $$dx - y dy = 0$$ - b) $$(y^2 + x) dx - dy = 0$$ - c) $$x dy + y^2 dx = 0$$ - d) $$(-x - 1) dy - x dx = 0$$ - e) No differential form (algebraic expression)