Solve $(x + y)(dx - dy) = dx + dy$.

Given differential equation is

$(x + y)(dx - dy) = dx + dy$

$\Rightarrow$ $(x + y)\left( {1 - \frac{{dy}}{{dx}}} \right) = 1 + \frac{{dy}}{{dx}}$
……(i)
Put $x + y = z$

$\Rightarrow$ $1 + \frac{{dy}}{{dx}} = \frac{{dz}}{{dx}}$

On substıutıng these values in Eq. (i),

we get
$z\left( {1 - \frac{{dz}}{{dx}} + 1} \right) = \frac{{dz}}{{dx}}$

$\Rightarrow$ $z\left( {2 - \frac{{dz}}{{dx}}} \right) = \frac{{dz}}{{dx}}$

$\Rightarrow$ $2z - z\frac{{dz}}{{dx}} - \frac{{dz}}{{dx}} = 0$

$\Rightarrow$ $2z - (z + 1)\frac{{dz}}{{dx}} = 0$

$\Rightarrow$ $\frac{{dz}}{{dx}} = \frac{{2z}}{{z + 1}}$

$\Rightarrow$ $\left( {\frac{{z + 1}}{z}} \right)dz = 2dx$
On integrating both sides,

we get
$\int {\left( {1 + \frac{1}{z}} \right)} dz = 2\int d x$

$\Rightarrow$ $z + \log z = 2x - \log C$

$\Rightarrow$ $(x + y) + \log (x + y) = 2x - \log C$

$\Rightarrow$ $2x - x - y = \log C + \log (x + y)$

$\Rightarrow$ $x - y = \log |C(x + y)|$

$\Rightarrow$ ${e^{x - y}} = C(x + y)$

$\Rightarrow$ $(x + y) = \frac{1}{C}{e^{x - y}}$

$\Rightarrow$ $x + y = K{e^{x - y}}$