The problem asks to solve the differential equation: $y' = \frac{x-y}{x+y}$

1. Problem Description

The problem asks to solve the differential equation:

$y' = \frac{x-y}{x+y}$

2. Solution Steps

This is a homogeneous differential equation. We can solve this by substituting $y = vx$. Then $y' = v + x \frac{dv}{dx}$.

Substituting into the differential equation:

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

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

Now we separate variables:

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

Integrate both sides:

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

Let $u = 1-2v-v^2$, then $du = (-2-2v)dv = -2(1+v)dv$, so $(1+v)dv = -\frac{1}{2}du$

$\int \frac{1+v}{1-2v-v^2} dv = \int \frac{-\frac{1}{2}du}{u} = -\frac{1}{2} \int \frac{1}{u} du = -\frac{1}{2} \ln|u| + C_1 = -\frac{1}{2} \ln|1-2v-v^2| + C_1$

$\int \frac{1}{x} dx = \ln|x| + C_2$

Therefore,

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

$\ln|1-2v-v^2| = -2 \ln|x| - 2C$

$\ln|1-2v-v^2| = \ln|x^{-2}| + C'$

$1-2v-v^2 = e^{\ln|x^{-2}| + C'} = e^{C'} x^{-2} = Ax^{-2}$ where $A = e^{C'}$

$1-2v-v^2 = \frac{A}{x^2}$

Substituting back $v = \frac{y}{x}$,

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

$1 - \frac{2y}{x} - \frac{y^2}{x^2} = \frac{A}{x^2}$

$x^2 - 2xy - y^2 = A$

3. Final Answer

$x^2 - 2xy - y^2 = A$