Solve $\frac{\mathrm{d}y}{\mathrm{d}x} = (x-y)/(x+y)$ Solve

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

It is homogeneous, thus let $y = vx$. From this, $\frac{\mathrm{d}y}{\mathrm{d}x} = x\frac{\mathrm{d}v}{\mathrm{d}x} + v$
Thus,
$v'x + v = (1-v)/(1+v)$ thus, 
$\frac{2}{1-v} + \ln(v - 1) = \ln(x) + C$. 
Which by substituion,
$\frac{2x}{x-y} + \ln(y - x) = 2\ln(x) + C$
But I cant get it any further
 A: $$\frac { dy }{ dx } =\frac { x-y }{ x+y } \\ \frac { dy }{ dx } =\frac { 1-\frac { y }{ x }  }{ 1+\frac { y }{ x }  } \\ y=zx\\ z^{ \prime  }x+z=\frac { 1-z }{ 1+z } \\ z^{ \prime  }x=\frac { 1-z }{ 1+z } -z=\frac { 1-2z-z^{ 2 } }{ 1+z } \\ \int { \frac { 1+z }{ 1-2z-z^{ 2 } } dz } =\int { \frac { dx }{ x }  } \\ -\int { \frac { 1+z }{ { z }^{ 2 }+2z-1 } dz } =\int  \frac { dx }{ x } \\ -\frac { 1 }{ 2 } \int  \frac { d\left( z^{ 2 }+2z-1 \right)  }{ z^{ 2 }+2z-1 } =\int { \frac { dx }{ x }  } \\ \\ \ln { \left| z^{ 2 }+2z-1 \right|  } =\ln { \frac { C }{ x^{ 2 } }  } \\ z^{ 2 }+2z-1=\frac { C }{ x^{ 2 } } \\ \frac { y^{ 2 } }{ x^{ 2 } } +2\frac { y }{ x } -1=\frac { C }{ x^{ 2 } }  $$
so the answer is :

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

A: Your step from
$$v'x + v = (1-v)/(1+v)$$
to
$$
\frac{2}{1-v} + \ln(v - 1) = \ln(x) + C
$$
is wrong. Separating $v$ and $x$ yields
$$
v'\cdot\frac{1+v}{1-2v-v^2}=\frac1x\;,
$$
and then integrating leads to
$$
-\frac12\log\left(1-2v-v^2\right)=\log x+C\;,
\\
1-2v-v^2=cx^{-2}\;,
$$
and thus
$$
x^2-2xy-y^2=c\;,
$$
in agreement with Battani's answer.
A: $$
\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{x-y}{x+y}
$$
The equation
$$
(y-x)\,\mathrm{d}x+(y+x)\,\mathrm{d}y=0
$$
is exact, so we just need to integrate along any path to get
$$
y^2+2xy-x^2=C
$$
