The problem is to solve the following differential equation: $2xy \frac{dy}{dx} + x^2 - y^2 = 0$.

1. Problem Description

The problem is to solve the following differential equation:

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

2. Solution Steps

First, we rewrite the equation as

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

Then, we can write it as

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

This is a homogeneous differential equation. Let $y = vx$, then $\frac{dy}{dx} = v + x \frac{dv}{dx}$.

Substituting into the equation, we have

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

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

Separating the variables, we have

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

Integrating both sides, we have

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

$\ln(v^2 + 1) = -\ln|x| + C$.

$\ln(v^2 + 1) + \ln|x| = C$.

$\ln(x(v^2 + 1)) = C$.

$x(v^2 + 1) = e^C = A$, where $A$ is a constant.

Substituting $v = \frac{y}{x}$ back into the equation, we get

$x(\frac{y^2}{x^2} + 1) = A$.

$x(\frac{y^2 + x^2}{x^2}) = A$.

$\frac{y^2 + x^2}{x} = A$.

$y^2 + x^2 = Ax$.

$x^2 - Ax + y^2 = 0$.

Completing the square for $x$,

$(x - \frac{A}{2})^2 - \frac{A^2}{4} + y^2 = 0$.

$(x - \frac{A}{2})^2 + y^2 = \frac{A^2}{4}$.

This represents a circle centered at $(\frac{A}{2}, 0)$ with radius $\frac{|A|}{2}$.

Let $A = 2C$.

Then the equation is

$y^2 + x^2 = 2Cx$.

3. Final Answer

$x^2 + y^2 = Ax$, where $A$ is an arbitrary constant.

Alternatively, $x^2 + y^2 = 2Cx$, where $C$ is an arbitrary constant.

Also, $(x - C)^2 + y^2 = C^2$.