1. **Problem:** Find the general solution to the differential equation $$2x(y+1) - y y' = 0$$. **Step 1:** Rewrite the equation to isolate $$y'$$. $$2x(y+1) = y y' \implies y' = \frac{2x(y+1)}{y}$$ **Step 2:** Express $$y'$$ as $$\frac{dy}{dx}$$ and separate variables if possible. $$\frac{dy}{dx} = \frac{2x(y+1)}{y}$$ Rewrite as: $$\frac{y}{y+1} dy = 2x dx$$ **Step 3:** Integrate both sides. Left side: $$\int \frac{y}{y+1} dy = \int \left(1 - \frac{1}{y+1}\right) dy = \int 1 dy - \int \frac{1}{y+1} dy = y - \ln|y+1| + C_1$$ Right side: $$\int 2x dx = x^2 + C_2$$ **Step 4:** Combine constants and write the implicit solution. $$y - \ln|y+1| = x^2 + C$$ --- 2. **Problem:** Find the general solution to the differential equation $$x y' + 2y = 0$$. **Step 1:** Rewrite as: $$x \frac{dy}{dx} = -2y$$ **Step 2:** Separate variables: $$\frac{dy}{y} = -\frac{2}{x} dx$$ **Step 3:** Integrate both sides: $$\int \frac{1}{y} dy = \int -\frac{2}{x} dx$$ $$\ln|y| = -2 \ln|x| + C$$ **Step 4:** Simplify using logarithm properties: $$\ln|y| = \ln|x|^{-2} + C$$ Exponentiate both sides: $$y = Ae^{-2 \ln|x|} = A x^{-2}$$ where $$A = e^C$$. --- 3. **Problem:** Find the general solution to the differential equation $$y y' - 2 e^x = 0$$. **Step 1:** Rewrite as: $$y \frac{dy}{dx} = 2 e^x$$ **Step 2:** Separate variables: $$y dy = 2 e^x dx$$ **Step 3:** Integrate both sides: $$\int y dy = \int 2 e^x dx$$ $$\frac{y^2}{2} = 2 e^x + C$$ **Step 4:** Multiply both sides by 2: $$y^2 = 4 e^x + C'$$ where $$C' = 2C$$. --- 4. **Problem:** Find the particular solution to the differential equation $$x y y' - \ln x = 0$$ with initial condition $$y(1) = 0$$. **Step 1:** Rewrite as: $$x y \frac{dy}{dx} = \ln x$$ **Step 2:** Separate variables: $$y dy = \frac{\ln x}{x} dx$$ **Step 3:** Integrate both sides: Left side: $$\int y dy = \frac{y^2}{2} + C_1$$ Right side: $$\int \frac{\ln x}{x} dx$$ Use substitution: let $$t = \ln x$$, then $$dt = \frac{1}{x} dx$$, so $$\int t dt = \frac{t^2}{2} + C_2 = \frac{(\ln x)^2}{2} + C_2$$ **Step 4:** Equate and combine constants: $$\frac{y^2}{2} = \frac{(\ln x)^2}{2} + C$$ Multiply both sides by 2: $$y^2 = (\ln x)^2 + 2C$$ **Step 5:** Apply initial condition $$y(1) = 0$$: $$0^2 = (\ln 1)^2 + 2C \implies 0 = 0 + 2C \implies C = 0$$ **Step 6:** Final solution: $$y^2 = (\ln x)^2 \implies y = \pm \ln x$$ Since $$y(1) = 0$$, both signs satisfy initial condition, so solution is: $$y = \pm \ln x$$