Question
Solve the following differential equation: $\cos^{2}\text{x}\frac{\text{dy}}{\text{dx}}+\text{y}=\tan\text{x}.$

Answer

Given differential equation can be written as $\frac{\text{dy}}{\text{dx}}+\text{sec}^{2}\text{x}\cdot\text{y}=\tan\text{x}\cdot\sec^{2}\text{x}$
$\text{I.F.}=\text{e}^{\int{\text{sec}^{2}\text{x dx}}}=\text{e}^{\tan\text{x}}$
$\therefore$ The Solution is $\text{y}\cdot\text{e}^{\text{tan x}}=\int\tan\text{x }\cdot\text{e}^{\text{tan x}}\text{sec}^{2}\text{ x}\text{ dx}$
$=\int\text{t.e}^\text{t}\text{dt},$ 
where tan $x = 1.$
$\Rightarrow\text{y}\cdot\text{e}^{\text{tan x}}= (t - 1) e^{t }+ c$
$\Rightarrow\text{y}\cdot\text{e}^{\text{tan x}}= (\tan x - 1) e^{\tan x} + c.$
Alternate Answer
$y = (\tan x - 1) + \text{c}\cdot\text{c}^{\text{-tan x}}$.

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