Find the general solution of $(1 + \tan y)(dx - dy) + 2xdy = 0$.

Given differential equation is $(1 + \tan y)(dx - dy) + 2xdy = 0$

on dividing throughout by dy,

we get
$(1 + \tan y)\left( {\frac{{dx}}{{dy}} - 1} \right) + 2x = 0$

$\Rightarrow$ $(1 + \tan y)\frac{{dx}}{{dy}} - (1 + \tan y) + 2x = 0$

$\Rightarrow$ $(1 + \tan y)\frac{{dx}}{{dy}} + 2x = (1 + \tan y)$

$\Rightarrow$ $\frac{{dx}}{{dy}} + \frac{{2x}}{{1 + \tan y}} = 1$

which is a linear differential equation.
On comparing it with

$\frac{{dx}}{{dy}} + Px = Q,$

we get
$P = \frac{2}{{1 + \tan y}},Q = 1$

$IF = {e^{\int {\frac{2}{{1 + \tan y}}} dy}} = {e^{\int {\frac{{2\cos y}}{{\cos y + \sin y}}} dy}}$

$= {e^{\int {\frac{{\cos y + \sin y + \cos y - \sin y}}{{\cos y + \sin y}}} dy}}$

$= {e^{\int {\left( {1 + \frac{{\cos y - \sin y}}{{\cos y + \sin y}}} \right)} dy}} = {e^{y + \log (\cos y + \sin y)}}$

The general solution is
$x \cdot {e^y}(\cos y + \sin y) = \int 1 \cdot {e^y}(\cos y + \sin y)dy + C$

$\Rightarrow$ $x \cdot {e^y}(\cos y + \sin y) = \int {{e^y}} (\sin y + \cos y)dy + C$

$\Rightarrow$ $x \cdot {e^y}(\cos y + \sin y) = {e^y}\sin y + C$

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