Step 1: Rewrite the equation in standard form

Given:

$$2xy + y^2 - 2x^2 \frac{dy}{dx} = 0$$

Rearranging:

$$2x^2 \frac{dy}{dx} = 2xy + y^2$$ $$\frac{dy}{dx} = \frac{2xy + y^2}{2x^2} = \frac{y}{x} + \frac{y^2}{2x^2}$$

Step 2: Identify as homogeneous and substitute y = vx

This is a homogeneous differential equation.

Let $y = vx$, so

$$\frac{dy}{dx} = v + x\frac{dv}{dx}$$

Substituting:

$$v + x\frac{dv}{dx} = v + \frac{v^2}{2}$$

Step 3: Separate variables and integrate

$$x\frac{dv}{dx} = \frac{v^2}{2}$$ $$\frac{2}{v^2}dv = \frac{dx}{x}$$

Integrating:

$$-\frac{2}{v} = \ln|x| + C$$

Step 4: Substitute back and apply the condition

Substituting $v = \frac{y}{x}$:

$$-\frac{2x}{y} = \ln|x| + C$$

Given: $y = 2$ when $x = 1$

$$-\frac{2(1)}{2} = \ln(1) + C$$ $$-1 = 0 + C$$

∴ $C = -1$

Step 5: Write the particular solution

$$-\frac{2x}{y} = \ln x - 1$$ $$\frac{2x}{y} = 1 - \ln x$$

∴ $y = \frac{2x}{1 - \ln x}$