1. **Problem statement:** Verify that the straight lines $y = \pm \frac{5}{3}x$ are solution curves to the differential equation $$\frac{dy}{dx} = \frac{25x}{9y}$$ for $x \neq 0$. 2. **Recall the formula:** The differential equation is given by $$\frac{dy}{dx} = \frac{25x}{9y}$$. 3. **Check the proposed solution $y = \frac{5}{3}x$: ** Calculate $\frac{dy}{dx}$ from $y = \frac{5}{3}x$: $$\frac{dy}{dx} = \frac{5}{3}$$ Substitute $y = \frac{5}{3}x$ into the right side of the differential equation: $$\frac{25x}{9y} = \frac{25x}{9 \cdot \frac{5}{3}x} = \frac{25x}{15x} = \frac{25}{15} = \frac{5}{3}$$ Since both sides equal $\frac{5}{3}$, the line $y = \frac{5}{3}x$ satisfies the differential equation for $x \neq 0$. 4. **Check the proposed solution $y = -\frac{5}{3}x$: ** Calculate $\frac{dy}{dx}$ from $y = -\frac{5}{3}x$: $$\frac{dy}{dx} = -\frac{5}{3}$$ Substitute $y = -\frac{5}{3}x$ into the right side of the differential equation: $$\frac{25x}{9y} = \frac{25x}{9 \cdot \left(-\frac{5}{3}x\right)} = \frac{25x}{-15x} = -\frac{25}{15} = -\frac{5}{3}$$ Both sides equal $-\frac{5}{3}$, so the line $y = -\frac{5}{3}x$ also satisfies the differential equation for $x \neq 0$. **Final answer:** The straight lines $y = \pm \frac{5}{3}x$ are indeed solution curves to the differential equation $\frac{dy}{dx} = \frac{25x}{9y}$ for $x \neq 0$.