Step 1: Rewrite the given differential equation in the form $\frac{dy}{dx} = F(x,y)$.

Given:

$$x \, dy - y \, dx = \sqrt{x^2 + y^2} \, dx$$

Rearranging:

$$x \, dy = y \, dx + \sqrt{x^2 + y^2} \, dx = \left(y + \sqrt{x^2 + y^2}\right) dx$$

Thus,

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

Step 2: Show that the equation is homogeneous.

A differential equation is homogeneous if it can be written as $\frac{dy}{dx} = f\left(\frac{y}{x}\right)$. Here,

$$\frac{dy}{dx} = \frac{y}{x} + \frac{\sqrt{x^2 + y^2}}{x} = \frac{y}{x} + \sqrt{1 + \left(\frac{y}{x}\right)^2}$$

which is a function of $\frac{y}{x}$ alone. Hence, the given differential equation is homogeneous.

Step 3: Solve using substitution $y = vx$, where $v$ is a function of $x$.

Then

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

Substituting in the equation:

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

Cancelling $v$ on both sides, we get:

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

Step 4: Separate variables and integrate.

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

Integrating both sides:

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

We know

$$\int \frac{dv}{\sqrt{1+v^2}} = \ln\left|v + \sqrt{1+v^2}\right| + C$$

Thus,

$$\ln\left|v + \sqrt{1+v^2}\right| = \ln|x| + \ln C_1$$

where $C_1$ is an arbitrary constant.

Taking exponentials:

$$v + \sqrt{1+v^2} = C_1 x$$

(since $v + \sqrt{1+v^2} > 0$ and we can absorb sign in constant).

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

$$\frac{y}{x} + \sqrt{1 + \frac{y^2}{x^2}} = C_1 x$$

Multiply both sides by $x$:

$$y + \sqrt{x^2 + y^2} = C_1 x^2$$

Let $C = C_1$ be the constant. Hence, the general solution is:

$$\boxed{y + \sqrt{x^2 + y^2} = C x^2}$$