Find a particular solution of the differential equation
$(x - y)(dx + dy) = dx - dy$,
given that $y = - 1$, when $x = 0$ (Hint: put $x - y = t$)

$(x - y)(dx + dy) = dx - dy$
$\Rightarrow$ $(x - y + 1)dy = (1 - x + y)dx$
$\Rightarrow$ $\frac{{dy}}{{dx}} = \frac{{1 - x + y}}{{x - y + 1}}$

$\Rightarrow$ $\frac{{dy}}{{dx}} = \frac{{1 - (x - y)}}{{1 + (x - y)}}$

…..(1)
Let $x - y = t$
$\Rightarrow$ $\frac{d}{{dx}}(x - y) = \frac{{dt}}{{dx}}$

$\Rightarrow$ $1 - \frac{{dy}}{{dx}} = \frac{{dt}}{{dx}}$
$\Rightarrow$ $1 - \frac{{dt}}{{dx}} = \frac{{dy}}{{dx}}$
Substituting the values of $x - y$ and $\frac{{dy}}{{dx}}$in equation (1),

we get:
$1 - \frac{{dt}}{{dx}} = \frac{{1 - t}}{{1 + t}}$
$\Rightarrow$ $\frac{{dt}}{{dx}} = 1 - \left( {\frac{{1 - t}}{{1 + t}}} \right)$
$\Rightarrow$ $\frac{{dt}}{{dx}} = \frac{{(1 + t) - (1 - t)}}{{1 + t}}$

$\Rightarrow$ $\frac{{dt}}{{dx}} = \frac{{2t}}{{1 + t}}$
$\Rightarrow$ $\left( {\frac{{1 + t}}{t}} \right)dt = 2dx$

$\Rightarrow$ $\left( {1 + \frac{1}{t}} \right)dt = 2dx$

….(2)
Integrating both sides,

we get:
$t + \log |t| = 2x + C$
$\Rightarrow$ $(x - y) + \log |x - y| = 2x + C$
$\Rightarrow$ $\log |x - y| = x + y + C$ …(3)
Now, $y = - 1$ at $x = 0$

Therefore, equation (3) becomes: $\log 1 = 0 - 1 + {\rm{C}}$
$\Rightarrow$ $C = 1$

Substituting ${\rm{C}} = 1$ in equation (3)

we get:
$\log |x - y| = x + y + 1$
This is the required particular solution of the given differential equation.