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Answer:  $x=(\tan y+C) e^{-y}$

Hint: To solve this equation we use  where  are constants.

Give:  $d x+x d y=e^{-y} \sec ^{2} y d y$

Solution:  \begin{aligned} & d x=e^{-y} \sec ^{2} y d y-x d y \\ & \end{aligned}

$=\frac{d x}{d y}+x=e^{-y} \sec ^{2} y$

First order linear differential equation form

\begin{aligned} &=\frac{d x}{d y}=P x-Q \\ &P=1 \text { and } e^{-y} \sec ^{2} y \end{aligned}

If of differential equation is

\begin{aligned} &I f=e^{\int R d y} \\ & \end{aligned}

$=e^{\int 1 d y} \\$

$\int d y=y+C \\$

$=I f=e^{y}$                                        $\quad\left[e^{\log x}=x\right]$

\begin{aligned} &=x(I f)=\int Q I F d y+C \\ & \end{aligned}

$=x e^{y}=\int e^{-y} \sec ^{2} y e^{y} d y+C \\$

$=x e^{y}=\int \sec ^{2} y d y+C \\$

$=x e^{y} \times \frac{1}{e^{y}}=(\tan y+C) \times \frac{1}{e^{y}}$                              $\quad\left[\int \sec ^{2} y d y=\tan x+C\right] \\$

$=x=(\tan y+C) e^{-y}$

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