Solve this


Let $\mathrm{y}=\mathrm{y}(\mathrm{x})$ be a solution curve of the differential equation $(y+1) \tan ^{2} x d x+\tan x d y+y d x=0$,\

$x \in\left(0, \frac{\pi}{2}\right) .$ If $\lim _{x \rightarrow 0+} x y(x)=1$, then the value of

$y\left(\frac{\pi}{4}\right)$ is:

  1. $-\frac{\pi}{4}$

  2. $\frac{\pi}{4}-1$

  3. $\frac{\pi}{4}+1$

  4. $\frac{\pi}{4}$

Correct Option: 4,


$(y+1) \tan ^{2} x d x+\tan x d y+y d x=0$

or $\frac{d y}{d x}+\frac{\sec ^{2} x}{\tan x} \cdot y=-\tan x$

$\mathrm{IF}=\mathrm{e}^{\int \frac{\sec ^{2} x}{\tan x} d x}=\mathrm{e}^{\ln \tan x}=\tan x$

$\therefore \mathrm{y} \tan \mathrm{x}=-\int \tan ^{2} \mathrm{xdx}$

or $y \tan x=-\tan x+x+C$

or $y=-1+\frac{x}{\tan x}+\frac{C}{\tan x}$

or $\lim _{x \rightarrow 0} x y=-x+\frac{x^{2}}{\tan x}+\frac{C x}{\tan x}=1$

or $\mathrm{C}=1$

$y(x)=\cot x+x \cot x-1$


$y(x)=\cot x+x \cot x-1$

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