1. **Problem:** Find the general solution to the differential equation $$2x(y + 1) - y y' = 0.$$ 2. **Step 1: Rewrite the equation** We have $$2x(y + 1) - y \frac{dy}{dx} = 0.$$ Rearranging to isolate $$\frac{dy}{dx}$$ gives: $$-y \frac{dy}{dx} = -2x(y + 1)$$ $$\Rightarrow y \frac{dy}{dx} = 2x(y + 1).$$ 3. **Step 2: Separate variables if possible** Rewrite as: $$y \frac{dy}{dx} = 2x(y + 1)$$ Divide both sides by $$y(y+1)$$ (assuming $$y \neq 0$$ and $$y \neq -1$$): $$\frac{\cancel{y}}{\cancel{y}(y+1)} \frac{dy}{dx} = \frac{2x(y+1)}{y(y+1)}$$ which simplifies to: $$\frac{1}{y+1} \frac{dy}{dx} = \frac{2x}{y}.$$ 4. **Step 3: Rearrange to separate variables** Multiply both sides by $$\frac{y+1}{2x}$$: $$\frac{dy}{dx} = \frac{2x(y+1)}{y}$$ We want to separate $$x$$ and $$y$$ terms: Rewrite original as: $$y \frac{dy}{dx} = 2x(y+1)$$ Divide both sides by $$y+1$$: $$y \frac{dy}{dx} \frac{1}{y+1} = 2x$$ Rewrite $$\frac{dy}{dx}$$ as $$\frac{dy}{dx} = \frac{2x(y+1)}{y}$$ This is not straightforward to separate, so try substitution. 5. **Step 4: Use substitution** Let $$u = y + 1$$, so $$y = u - 1$$ and $$\frac{dy}{dx} = \frac{du}{dx}$$. Rewrite the equation: $$2x u - (u - 1) \frac{du}{dx} = 0$$ Rearranged: $$(u - 1) \frac{du}{dx} = 2x u$$ 6. **Step 5: Separate variables** $$\frac{du}{dx} = \frac{2x u}{u - 1}$$ Rewrite as: $$\frac{u - 1}{u} du = 2x dx$$ Simplify left side: $$\left(1 - \frac{1}{u}\right) du = 2x dx$$ 7. **Step 6: Integrate both sides** $$\int \left(1 - \frac{1}{u}\right) du = \int 2x dx$$ $$\int 1 du - \int \frac{1}{u} du = \int 2x dx$$ $$u - \ln|u| = x^2 + C$$ 8. **Step 7: Substitute back for $$u$$** Recall $$u = y + 1$$, so: $$y + 1 - \ln|y + 1| = x^2 + C$$ **Final answer:** $$\boxed{y + 1 - \ln|y + 1| = x^2 + C}$$