$\text { The general solution of the differential equation }(1+\tan y)(d x-d y)+2 x d y=0 \text { is }$

The given differential equation is $(1+\tan y)(dx - dy) + 2x dy = 0$.

Rearranging the terms to form a linear differential equation in $x$:

$(1+\tan y) dx = (1+\tan y - 2x) dy$

$\dfrac{dx}{dy} = \dfrac{1+\tan y - 2x}{1+\tan y} = 1 - \dfrac{2x}{1+\tan y}$

$\dfrac{dx}{dy} + \left(\dfrac{2}{1+\tan y}\right) x = 1$

This is a linear differential equation of the form $\dfrac{dx}{dy} + P(y)x = Q(y)$, where $P(y) = \dfrac{2}{1+\tan y} = \dfrac{2\cos y}{\cos y + \sin y}$.

The integrating factor (IF) is $e^{\int P(y) dy} = e^{\int \dfrac{2\cos y}{\sin y + \cos y} dy}$.

Let $I = \int \dfrac{2\cos y}{\sin y + \cos y} dy$. Note that $2\cos y = (\cos y + \sin y) + (\cos y - \sin y)$.

$I = \int \left(1 + \dfrac{\cos y - \sin y}{\sin y + \cos y}\right) dy = y + \ln|\sin y + \cos y|$.

Thus, $IF = e^{y + \ln|\sin y + \cos y|} = e^y (\sin y + \cos y)$.

The solution is $x(IF) = \int Q(y)(IF) dy + c$.

$x e^y (\sin y + \cos y) = \int e^y (\sin y + \cos y) dy + c$.

Using integration by parts on $\int e^y (\sin y + \cos y) dy$:

Let $f(y) = \sin y + \cos y$, then $f'(y) = \cos y - \sin y$. This does not immediately simplify, but $\int e^y (\sin y + \cos y) dy = e^y \sin y + c$.

Therefore, $x e^y (\sin y + \cos y) = e^y \sin y + c$.

Dividing by $e^y$, we get $x(\sin y + \cos y) = \sin y + c e^{-y}$.

Answer: $x(\sin y+\cos y)=\sin y+c e^{-y}$