Solve the following differential equation: x² dy + (xy + y²) dx = 0

$x^2dy = -(xy + y^2)dx \Rightarrow \frac{dy}{dx} = -\frac{xy+y^2}{x^2}$, a homogeneous equation. Put $y = vx$: $v + x\frac{dv}{dx} = -v - v^2 \Rightarrow x\frac{dv}{dx} = -v(v+2)$. Separating: $\frac{dv}{v(v+2)} = -\frac{dx}{x} \Rightarrow \frac{1}{2}\left(\frac{1}{v} - \frac{1}{v+2}\right)dv = -\frac{dx}{x}$. Integrating: $\frac{1}{2}\ln\left|\frac{v}{v+2}\right| = -\ln|x| + \ln c_1 \Rightarrow \frac{v}{v+2} = \frac{C}{x^2}$. Putting $v = \frac{y}{x}$: $\frac{y}{y+2x} = \frac{C}{x^2}$. Hence $x^2y = C(y + 2x)$.