1. The problem asks to solve two differential equations: a) $$\frac{dy}{dx} = \frac{1+x}{1+y}$$ b) $$\frac{dy}{dx} = \frac{x+y}{x}$$ 2. Solve a) $$\frac{dy}{dx} = \frac{1+x}{1+y}$$: Step 1: Rearrange to separate variables: $$ (1+y) dy = (1+x) dx $$ Step 2: Integrate both sides: $$ \int (1+y) dy = \int (1+x) dx $$ This gives: $$ y + \frac{y^2}{2} = x + \frac{x^2}{2} + C $$ Step 3: The implicit general solution is: $$ y + \frac{y^2}{2} = x + \frac{x^2}{2} + C $$ 3. Solve b) $$\frac{dy}{dx} = \frac{x+y}{x}$$: Step 1: Rewrite equation: $$ \frac{dy}{dx} = 1 + \frac{y}{x} $$ Step 2: Substitute $$ v = \frac{y}{x} \Rightarrow y = vx $$ Then: $$ \frac{dy}{dx} = v + x\frac{dv}{dx} $$ Step 3: Substitute into differential equation: $$ v + x\frac{dv}{dx} = 1 + v $$ Simplify: $$ x\frac{dv}{dx} = 1 $$ Step 4: Separate variables: $$ dv = \frac{1}{x} dx $$ Step 5: Integrate both sides: $$ \int dv = \int \frac{1}{x} dx $$ Gives: $$ v = \ln|x| + C $$ Step 6: Recall $$ v=\frac{y}{x} $$ so: $$ \frac{y}{x} = \ln|x| + C $$ Multiply both sides by $$ x $$: $$ y = x\ln|x| + Cx $$ 4. Final answers: a) $$ y + \frac{y^2}{2} = x + \frac{x^2}{2} + C $$ b) $$ y = x\ln|x| + Cx $$ These solutions describe the implicit and explicit solutions to the given differential equations.