1. **State the problem:** Find the general solution of the differential equation $$x^2 \frac{dy}{dx} = y - xy$$. 2. **Rewrite the equation:** Divide both sides by $x^2$ (assuming $x \neq 0$): $$\frac{dy}{dx} = \frac{y}{x^2} - \frac{y}{x} = y\left(\frac{1}{x^2} - \frac{1}{x}\right) = y\left(\frac{1 - x}{x^2}\right).$$ 3. **Separate variables:** $$\frac{dy}{y} = \left(\frac{1 - x}{x^2}\right) dx.$$ 4. **Integrate both sides:** $$\int \frac{1}{y} dy = \int \left(\frac{1}{x^2} - \frac{1}{x}\right) dx = \int x^{-2} dx - \int x^{-1} dx.$$ 5. **Calculate integrals:** $$\int x^{-2} dx = -x^{-1} + C_1 = -\frac{1}{x} + C_1,$$ $$\int x^{-1} dx = \ln|x| + C_2.$$ 6. **Combine results:** $$\ln|y| = -\frac{1}{x} - \ln|x| + C,$$ where $C = C_1 - C_2$ is a constant. 7. **Simplify:** $$\ln|y| + \ln|x| = -\frac{1}{x} + C,$$ which implies $$\ln|xy| = -\frac{1}{x} + C.$$ 8. **Exponentiate both sides:** $$|xy| = e^{C} e^{-\frac{1}{x}} = Ce^{-\frac{1}{x}},$$ where $C = e^{C}$ is a positive constant. 9. **Final general solution:** $$xy = Ce^{-\frac{1}{x}}.$$ **Answer choice matching:** This corresponds to option (A) if we rewrite $e^{1/x}$ as $e^{-1/x}$ with a constant adjustment. Since the problem's option (A) is $xy = Ce^{1/x}$, the correct solution is actually $$xy = Ce^{-\frac{1}{x}}$$ which matches none exactly but closest to (A) with a sign difference in the exponent. **Note:** The correct general solution is $$xy = Ce^{-\frac{1}{x}}$$.