Step 1: Rewrite the differential equation in standard form.

$$x^2\, dy + (xy + y^2)\, dx = 0$$ $$x^2\, dy = -(xy + y^2)\, dx$$ $$\frac{dy}{dx} = -\frac{xy + y^2}{x^2} = -\frac{y}{x} - \left(\frac{y}{x}\right)^2$$

Step 2: Substitute $y = vx$ (since this is a homogeneous equation).

Let $y = vx$, so:

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

Substituting:

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

Step 3: Separate variables and integrate.

$$\frac{dv}{v^2 + 2v} = -\frac{dx}{x}$$ $$\int \frac{dv}{v(v+2)} = -\int \frac{dx}{x}$$

Using partial fractions:

$$\frac{1}{v(v+2)} = \frac{1/2}{v} - \frac{1/2}{v+2}$$ $$\frac{1}{2}\ln\left|\frac{v}{v+2}\right| = -\ln|x| + C$$ $$\ln\left|\frac{v}{v+2}\right| = -2\ln|x| + 2C$$ $$\frac{v}{v+2} = \frac{k}{x^2}$$

Step 4: Apply the initial condition $y = 1$ when $x = 1$.

When $x = 1$, $y = 1$, so:

$$v = \frac{y}{x} = 1$$ $$\frac{1}{1+2} = \frac{k}{1}$$ $$k = \frac{1}{3}$$

Step 5: Find the particular solution.

$$\frac{v}{v+2} = \frac{1}{3x^2}$$ $$3vx^2 = v + 2$$ $$v(3x^2 - 1) = 2$$ $$v = \frac{2}{3x^2 - 1}$$

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

$$y = \frac{2x}{3x^2 - 1}$$