1. **State the problem:** Solve the differential equation $$y - x \left(\frac{dy}{dx}\right) = a \left(y^2 + \frac{dy}{dx}\right).$$ 2. **Rewrite the equation:** Group terms involving $\frac{dy}{dx}$ on one side: $$y - x \frac{dy}{dx} = a y^2 + a \frac{dy}{dx}$$ Move all $\frac{dy}{dx}$ terms to the left and others to the right: $$- x \frac{dy}{dx} - a \frac{dy}{dx} = a y^2 - y$$ 3. **Factor out $\frac{dy}{dx}$ on the left:** $$\left(-x - a\right) \frac{dy}{dx} = a y^2 - y$$ 4. **Isolate $\frac{dy}{dx}$:** $$\frac{dy}{dx} = \frac{a y^2 - y}{-x - a} = \frac{a y^2 - y}{-(x + a)} = -\frac{a y^2 - y}{x + a}$$ 5. **Rewrite as:** $$\frac{dy}{dx} = -\frac{a y^2 - y}{x + a}$$ 6. **Separate variables:** $$\frac{dy}{a y^2 - y} = -\frac{dx}{x + a}$$ 7. **Factor denominator in $y$ integral:** $$a y^2 - y = y (a y - 1)$$ So, $$\frac{dy}{y (a y - 1)} = -\frac{dx}{x + a}$$ 8. **Use partial fractions for the left side:** Assume $$\frac{1}{y (a y - 1)} = \frac{A}{y} + \frac{B}{a y - 1}$$ Multiply both sides by $y (a y - 1)$: $$1 = A (a y - 1) + B y = A a y - A + B y = y (A a + B) - A$$ Equate coefficients: - Constant term: $-A = 1 \Rightarrow A = -1$ - Coefficient of $y$: $A a + B = 0 \Rightarrow -1 \cdot a + B = 0 \Rightarrow B = a$ 9. **Rewrite integral:** $$\int \frac{dy}{y (a y - 1)} = \int \left(-\frac{1}{y} + \frac{a}{a y - 1}\right) dy = \int -\frac{1}{y} dy + \int \frac{a}{a y - 1} dy$$ 10. **Integrate each term:** - $$\int -\frac{1}{y} dy = -\ln|y|$$ - For $$\int \frac{a}{a y - 1} dy$$, substitute $u = a y - 1$, so $du = a dy$, $dy = \frac{du}{a}$: $$\int \frac{a}{u} \cdot \frac{du}{a} = \int \frac{1}{u} du = \ln|u| = \ln|a y - 1|$$ 11. **Combine integrals:** $$\int \frac{dy}{y (a y - 1)} = -\ln|y| + \ln|a y - 1| + C_1 = \ln \left|\frac{a y - 1}{y}\right| + C_1$$ 12. **Integrate right side:** $$\int -\frac{dx}{x + a} = -\ln|x + a| + C_2$$ 13. **Set integrals equal:** $$\ln \left|\frac{a y - 1}{y}\right| + C_1 = -\ln|x + a| + C_2$$ Combine constants $C = C_2 - C_1$: $$\ln \left|\frac{a y - 1}{y}\right| = -\ln|x + a| + C$$ 14. **Exponentiate both sides:** $$\left|\frac{a y - 1}{y}\right| = e^{C} \cdot \frac{1}{|x + a|}$$ Let $K = e^{C} > 0$: $$\left|\frac{a y - 1}{y}\right| = \frac{K}{|x + a|}$$ 15. **Write general implicit solution:** $$\frac{a y - 1}{y} = \pm \frac{K}{x + a}$$ Or equivalently, $$\frac{a y - 1}{y} (x + a) = \pm K$$ This is the implicit solution to the differential equation. --- **Final answer:** $$\frac{a y - 1}{y} (x + a) = C,$$ where $C$ is an arbitrary constant.