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The given equation is $2x^{2}\frac{dy}{dx} = y^{2} + 2xy$. Rearranging it, we get $\frac{dy}{dx} = \frac{y^{2} + 2xy}{2x^{2}}$. Since the numerator and denominator are homogeneous functions of degree 2, this is a homogeneous differential equation.
Let $y = vx$. Then, differentiating with respect to $x$, we get $\frac{dy}{dx} = v + x\frac{dv}{dx}$. Substituting these into the equation:
$$v + x\frac{dv}{dx} = \frac{(vx)^{2} + 2x(vx)}{2x^{2}}$$ $$v + x\frac{dv}{dx} = \frac{v^{2} + 2v}{2}$$Subtract $v$ from both sides:
$$x\frac{dv}{dx} = \frac{v^{2} + 2v - 2v}{2} = \frac{v^{2}}{2}$$ $$\frac{2}{v^{2}} dv = \frac{1}{x} dx$$Integrating both sides:
$$\int 2v^{-2} dv = \int \frac{1}{x} dx$$ $$-2v^{-1} = \ln|x| + C$$ $$-\frac{2}{v} = \ln|x| + C$$ Since $v = \frac{y}{x}$, we have $-\frac{2x}{y} = \ln|x| + C$, or $y = \frac{-2x}{\ln|x| + C}$.Final Answer: y = \frac{-2x}{\ln|x| + C}
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