Solve `(x-y)(dx+dy)=dx-dy`, given that `y=-1`, where `x=0`.

Explanation

To solve the differential equation \((x-y)(dx+dy) = dx - dy\) with the condition \(y = -1\) when \(x = 0\), we will follow these steps:

### Step 1: Rearranging the Equation
We start with the given equation:
\[ (x-y)(dx + dy) = dx - dy \]
We can expand the left-hand side:
\[ (x-y)dx + (x-y)dy = dx - dy \]
Now, we can rearrange the equation to isolate the terms involving \(dx\) and \(dy\):
\[ (x-y)dx + (x-y)dy - dx + dy = 0 \]
This simplifies to:
\[ ((x-y) - 1)dx + ((x-y) + 1)dy = 0 \]

### Step 2: Separating Variables
We can separate the variables \(dx\) and \(dy\):
\[ \frac{dy}{dx} = \frac{1 - (x-y)}{(x-y) + 1} \]
This can be rewritten as:
\[ \frac{dy}{dx} = \frac{1 + y - x}{x - y + 1} \]

### Step 3: Substitution
Let \(t = x - y\). Then, we have:
\[ y = x - t \quad \text{and} \quad dy = dx - dt \]
Substituting these into the equation gives:
\[ \frac{dx - dt}{dx} = \frac{1 + (x - t) - x}{t + 1} \]
This simplifies to:
\[ 1 - \frac{dt}{dx} = \frac{-t}{t + 1} \]
Rearranging gives:
\[ \frac{dt}{dx} = 1 + \frac{t}{t + 1} \]

### Step 4: Further Simplification
We can simplify the right-hand side:
\[ \frac{dt}{dx} = \frac{(t + 1) + t}{t + 1} = \frac{2t + 1}{t + 1} \]

### Step 5: Integrating Both Sides
Now we separate variables again:
\[ \frac{t + 1}{2t + 1} dt = dx \]
Integrating both sides:
\[ \int \frac{t + 1}{2t + 1} dt = \int dx \]
The left side can be integrated using substitution or partial fractions.

### Step 6: Finding the Constant of Integration
After integrating, we will have:
\[ \text{(integrated result)} = x + C \]
Now, we will use the initial condition \(y = -1\) when \(x = 0\) to find \(C\).

### Step 7: Final Solution
Substituting back \(t = x - y\) and solving for \(y\) gives us the final solution in terms of \(x\) and \(y\).