1. **State the problem:** Solve the differential equation $$y - x \frac{dy}{dx} = a \left(y^2 + \frac{dy}{dx}\right).$$ 2. **Rewrite the equation:** Group terms involving $\frac{dy}{dx}$ on one side: $$y - x \frac{dy}{dx} = a y^2 + a \frac{dy}{dx}$$ $$y - a y^2 = x \frac{dy}{dx} + a \frac{dy}{dx} = (x + a) \frac{dy}{dx}$$ 3. **Isolate $\frac{dy}{dx}$:** $$\frac{dy}{dx} = \frac{y - a y^2}{x + a}$$ 4. **Separate variables:** $$\frac{dy}{dx} = \frac{y(1 - a y)}{x + a}$$ Rearranged as $$\frac{dy}{y(1 - a y)} = \frac{dx}{x + a}$$ 5. **Integrate both sides:** Use partial fractions on the left: $$\frac{1}{y(1 - a y)} = \frac{A}{y} + \frac{B}{1 - a y}$$ Solving for $A$ and $B$: $$1 = A(1 - a y) + B y$$ Set $y=0$: $1 = A \Rightarrow A=1$ Set $y=\frac{1}{a}$: $1 = B \frac{1}{a} \Rightarrow B = a$ 6. **Integrate:** $$\int \left( \frac{1}{y} + \frac{a}{1 - a y} \right) dy = \int \frac{1}{x + a} dx$$ 7. **Compute integrals:** $$\int \frac{1}{y} dy = \ln|y|$$ $$\int \frac{a}{1 - a y} dy = -\ln|1 - a y|$$ $$\int \frac{1}{x + a} dx = \ln|x + a|$$ 8. **Combine results:** $$\ln|y| - \ln|1 - a y| = \ln|x + a| + C$$ 9. **Simplify logarithms:** $$\ln \left| \frac{y}{1 - a y} \right| = \ln|x + a| + C$$ Exponentiate both sides: $$\left| \frac{y}{1 - a y} \right| = K |x + a|$$ where $K = e^C$ is an arbitrary positive constant. 10. **Rewrite solution:** $$\frac{y}{1 - a y} = c (x + a)$$ where $c$ is an arbitrary constant (absorbing sign and $K$). 11. **Rearrange to match options:** Multiply both sides by $1 - a y$: $$y = c (x + a)(1 - a y)$$ This matches option (A): $$y = c (a + x)(1 - a y)$$. **Final answer:** (A) $y = c (a + x)(1 - a y)$