We are asked to solve the following differential equation: $(x-y)\frac{dy}{dx} + x + y = 0$

1. Problem Description

We are asked to solve the following differential equation:

$(x-y)\frac{dy}{dx} + x + y = 0$

2. Solution Steps

First, we rewrite the equation:

$(x-y)\frac{dy}{dx} = -x - y$

$\frac{dy}{dx} = \frac{-x-y}{x-y} = \frac{-(x+y)}{x-y} = \frac{x+y}{y-x}$

Let $y = vx$, then $\frac{dy}{dx} = v + x\frac{dv}{dx}$.

Substituting into the equation, we have:

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

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

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

Integrate both sides. Let $u = -v^2+2v+1$, then $du = (-2v+2)dv = -2(v-1)dv$.

So, $(v-1)dv = -\frac{1}{2}du$.

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

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

Therefore, we have

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

Multiply both sides by -2:

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

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

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

$-v^2+2v+1 = Cx^{-2}$

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

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

Multiply by $x^2$:

$-y^2 + 2xy + x^2 = C$

$x^2 + 2xy - y^2 = C$

3. Final Answer

$x^2 + 2xy - y^2 = C$