If y = y(x) is the solution curve of the differential equation...

Given:

$\dfrac{dy}{dx} + ytanx = xsecx$

This is a linear differential equation.

$I . F . = {e}^{∫ tanxdx} = \sec x$

Then solution of differential equation is

$y\left(\sec x\right) = ∫ x{sec}^{2}xdx$

$⇒ y\left(\sec x\right) = xtanx - ∫ tanxdx$

$⇒ y\left(\sec x\right) = xtanx - \ln \left(\sec x\right) + C$

Given:

$y\left(0\right) = 1 ⇒ c = 1$

$∴ y\left(\sec x\right) = xtanx - \ln \left(\sec x\right) + 1$

At $x = \dfrac{\pi}{6}$, we get

$y\left(\sec \left(\dfrac{\pi}{6}\right)\right) = \left(\dfrac{\pi}{6}\right)\tan \left(\dfrac{\pi}{6}\right) - \ln \left(\sec \left(\dfrac{\pi}{6}\right)\right) + 1$

$⇒ y\left(\dfrac{2}{\sqrt{3}}\right) = \left(\dfrac{\pi}{6}\right)\left(\dfrac{1}{\sqrt{3}}\right) - \ln \left(\dfrac{2}{\sqrt{3}}\right) + 1$

$⇒ y = \dfrac{\pi}{12} - \dfrac{\sqrt{3}}{2}\ln \left(\dfrac{2}{\sqrt{3}}\right) + \dfrac{\sqrt{3}}{2}$

$⇒ y = \dfrac{\pi}{12} - \dfrac{\sqrt{3}}{2}\left[\ln \left(\dfrac{2}{\sqrt{3}}\right) - lne\right]$

$⇒ y = \dfrac{\pi}{12} - \dfrac{\sqrt{3}}{2}\log_{e}\left(\dfrac{2}{e\sqrt{3}}\right)$