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

$x(\cos y + \sin y) = \sin y + Ce^{-y}$

The correct answer is Option (1) → $x(\cos y + \sin y) = \sin y + Ce^{-y}$ ##

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

On dividing throughout by $dy$, we get

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

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

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

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

which is a linear differential equation.

On comparing it with $\frac{dx}{dy} + Px = Q$, we get

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

$\text{I.F.} = e^{\int \frac{2}{1 + \tan y} , dy} = e^{\int \frac{2 \cos y}{\cos y + \sin y} , dy} \quad [∵\text{I.F.} = e^{\int P 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)} $

$= e^y \cdot e^{\log (\cos y + \sin y)} $

$= e^y \cdot (\cos y + \sin y) \quad [∵e^{\log x} = x]$

The general solution is $x \cdot \text{I.F.} = \int Q \cdot \text{I.F.} \, dy + C$

$x \cdot e^y (\cos y + \sin y) = \int 1 \cdot e^y (\cos y + \sin y) \, dy + C$

$⇒x \cdot e^y (\cos y + \sin y) =\int e^y(\sin y+\cos y) \, dy + C$

$⇒x \cdot e^y (\cos y + \sin y) =e^y\sin y+C$ $[∵\int e^x\{f(x)+f'(x)\}dx=e^xf(x)+C]$

$⇒x(\sin y+\cos y)=\sin y + Ce^{-y}$