Step 1: Rewrite the equation as $(x+y) \, dy = -(x-y) \, dx$, giving:

$$\frac{dy}{dx} = \frac{y-x}{x+y}$$

This is a homogeneous differential equation.

Step 2: Substitute $y = vx$, so $\frac{dy}{dx} = v + x\frac{dv}{dx}$. The equation becomes:

$$v + x\frac{dv}{dx} = \frac{vx - x}{x + vx} = \frac{v-1}{v+1}$$

Step 3: Simplify to get:

$$x\frac{dv}{dx} = \frac{v-1}{v+1} - v = \frac{-1-v^2}{v+1}$$

which gives:

$$\frac{v+1}{v^2+1} \, dv = -\frac{dx}{x}$$

Step 4: Integrate both sides:

$$\int\frac{v+1}{v^2+1} \, dv = -\int\frac{dx}{x}$$

This gives:

$$\tan^{-1}(v) + \frac{1}{2}\ln(v^2+1) = -\ln|x| + C$$

Step 5: Substitute back $v = \frac{y}{x}$:

$$\tan^{-1}\left(\frac{y}{x}\right) + \frac{1}{2}\ln\left(\frac{x^2+y^2}{x^2}\right) = -\ln|x| + C$$

Simplifying:

$$\tan^{-1}\left(\frac{y}{x}\right) + \frac{1}{2}\ln(x^2+y^2) = C$$

Step 6: Apply the condition $y = 1$ when $x = 1$:

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

giving:

$$C = \frac{\pi}{4} + \frac{1}{2}\ln 2$$