$\left( {x + y} \right)dy + (x - y)dx = 0;y = 1$ when $x = 1$

: $(x + y)dy + (x - y)dx = 0 \Rightarrow \cfrac{{dy}}{{dx}} = \cfrac{{y - x}}{{x + y}}$

$\Rightarrow \cfrac{{dy}}{{dx}} = \cfrac{{\cfrac{y}{x} - 1}}{{\cfrac{y}{x} + 1}}$

…(1)
Since R.H.S. is of the form $g(y/x)$, so it is homogenous function of degree zero.

Therefore, equation (1) is a homogeneous differential equation. To solve this equation, we put

$y = vx \Rightarrow \cfrac{{dy}}{{dx}} = v + x\cfrac{{dv}}{{dx}}$

So eq. (1) becomes
$v + x\cfrac{{dv}}{{dx}} = \cfrac{{v - 1}}{{v + 1}} \Rightarrow x\cfrac{{dv}}{{dx}} = \cfrac{{v - 1}}{{v + 1}} - v$

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

… (2)
Integrating (2) both sides,

we get
$\int {\cfrac{{v + 1}}{{{v^2} + 1}}} d\nu = - \int {\cfrac{{dx}}{x}}$

$\Rightarrow \cfrac{1}{2}\int {\cfrac{{2v}}{{{v^2} + 1}}} dv + \int {\cfrac{{dv}}{{{v^2} + 1}}} = - \log |x| + {C_1}$

$\Rightarrow \cfrac{1}{2}\log |{v^2} + 1| + {\tan ^{ - 1}}v + \log |x| = {C_1}$

$\Rightarrow \cfrac{1}{2}\log \left( {\cfrac{{{y^2} + {x^2}}}{{{x^2}}}} \right) + \log |x|( + {\tan ^{ - 1}}(y/x) = {C_1}$

$\Rightarrow \log \left( {\cfrac{{{x^2} + {v^2}}}{{{x^2}}}} \right) + 2\log |x| + 2{\tan ^{ - 1}}\left( {\cfrac{y}{x}} \right) = 2{C_1}$

$\Rightarrow \log ({x^2} + {y^2}) + 2{\tan ^{ - 1}}(y/x) = C$

…(3)
We are given that when $x = 1,y = 1$
$\therefore \log ({1^2} + {1^2}) + 2{\tan ^{ - 1}}(1/1) = C$

$\Rightarrow \log 2 + 2x\cfrac{\pi }{4} = C \Rightarrow C = \log 2 + \cfrac{\pi }{2}$

Substituting the value of $C$ in (3),

we get the required particular solution as
$\log ({x^2} + {y^2}) + 2{\tan ^{ - 1}}\left( {\cfrac{y}{x}} \right) = \log 2 + \cfrac{\pi }{2}$