Let y = y ( x ) be a solution curve of the differential equation (y...

Given:

$\left(y + 1\right)\tan^{2}xdx + tanxdy + ydx = 0$

$⇒ \left[\left(y + 1\right)\tan^{2}x + y\right]dx + tanxdy = 0$

$⇒ \tan x\left(\dfrac{dy}{dx}\right) + \left(y + 1\right)\tan^{2}x + y = 0$

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

$⇒ \dfrac{dy}{dx} + y\left(\tan x + \cot x\right) = - \tan x$

This is a linear differential equation of the form $\dfrac{dy}{dx} + P\left(x\right)y = Q\left(x\right)$.

$\text{I.F} = {e}^{∫ ( \tan x + \cot x ) dx}$

$= {e}^{∫ \left(\dfrac{\tan^{2}x + 1}{\tan x}\right)dx}$

$= {e}^{∫ \left(\dfrac{\sec^{2}x}{\tan x}\right)dx}$

$= {e}^{\log_{e}\left|\tan x\right|}$

$= \tan x\left(∵ x \in \left(0, \dfrac{\pi}{2}\right)\right)$

Solution is

$ytanx = ∫ - \tan^{2}xdx + c$

$⇒ ytanx = ∫ \left(1 - \sec^{2}x\right)dx + c$

$⇒ ytanx = x - \tan x + c$

Now,

$\lim_{x \to {0}^{+}}xy = 1$

$⇒ \lim_{x \to {0}^{+}}\left(\dfrac{x}{\tan x}\right)\left(x - \tan x + c\right) = 1$

$⇒ 1 ( 0 - 0 + c ) = 1 ⇒ c = 1$

Then the function is

$ytanx = x - \tan x + 1$

Put $x = \dfrac{\pi}{4}$,

$y\left(\dfrac{\pi}{4}\right)\tan \left(\dfrac{\pi}{4}\right) = \left(\dfrac{\pi}{4}\right) - \tan \left(\dfrac{\pi}{4}\right) + 1$

$⇒ y\left(\dfrac{\pi}{4}\right) = \dfrac{\pi}{4}$