`(x-y)dy-(x+y)dx=0`

Explanation

To solve the differential equation \((x - y) dy - (x + y) dx = 0\), we will follow these steps:

### Step 1: Rearranging the Equation
We start with the given equation:
\[ (x - y) dy - (x + y) dx = 0 \]
We can rearrange this as:
\[ (x - y) dy = (x + y) dx \]
Dividing both sides by \((x - y)(x + y)\), we get:
\[ \frac{dy}{dx} = \frac{x + y}{x - y} \]

### Step 2: Identifying Homogeneity
The right-hand side \(\frac{x + y}{x - y}\) is a homogeneous function of degree 0 because if we replace \(x\) and \(y\) with \(kx\) and \(ky\), we have:
\[ \frac{kx + ky}{kx - ky} = \frac{x + y}{x - y} \]
Thus, this is a homogeneous differential equation.

### Step 3: Substitution
To solve this, we use the substitution \(y = vx\), where \(v\) is a function of \(x\). Therefore, we have:
\[ dy = v dx + x dv \]
Substituting \(y\) and \(dy\) into the equation gives:
\[ (x - vx)(v dx + x dv) = (x + vx)dx \]
Simplifying this leads to:
\[ (x(1 - v))(v dx + x dv) = x(1 + v)dx \]

### Step 4: Simplifying the Equation
Expanding both sides:
\[ x(1 - v)v dx + x^2(1 - v) dv = x(1 + v)dx \]
Rearranging gives:
\[ x^2(1 - v) dv = x(1 + v - v^2) dx \]
Dividing both sides by \(x(1 - v)\):
\[ \frac{dv}{dx} = \frac{1 + v}{x(1 - v)} \]

### Step 5: Separation of Variables
Now we separate the variables:
\[ (1 - v) dv = \frac{(1 + v)}{x} dx \]
Integrating both sides:
\[ \int (1 - v) dv = \int \frac{(1 + v)}{x} dx \]

### Step 6: Performing the Integrals
The left side integrates to:
\[ v - \frac{v^2}{2} + C_1 \]
The right side integrates to:
\[ \ln |x| + v \]
Thus, we have:
\[ v - \frac{v^2}{2} = \ln |x| + C \]

### Step 7: Substituting Back
Recall that \(v = \frac{y}{x}\). Substituting back gives:
\[ \frac{y}{x} - \frac{1}{2} \left(\frac{y}{x}\right)^2 = \ln |x| + C \]
Multiplying through by \(2x^2\) results in:
\[ 2y - y^2 = 2x \ln |x| + 2Cx^2 \]

### Final Step: Rearranging the Solution
Rearranging gives the final implicit solution:
\[ y^2 - 2y + 2Cx^2 + 2x \ln |x| = 0 \]