If y = y x is the solution curve of the differential equation d y d x + y tan x = x sec x ,    0…

Given:

$\frac{dy}{dx}+y\tan x=x\sec x$

This is a linear differential equation.

$\mathrm{I}.\mathrm{F}.=e^{\int \tan xdx}=\sec x$

Then solution of differential equation is

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

$\Rightarrow y\left(\sec x\right)=x\tan x-\int \tan xdx$

$\Rightarrow y\left(\sec x\right)=x\tan x-\ln \left(\sec x\right)+C$

Given:

$y\left(0\right)=1\Rightarrow c=1$

$\therefore y\left(\sec x\right)=x\tan x-\ln \left(\sec x\right)+1$

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

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

$\Rightarrow y\left(\frac{2}{\sqrt{3}}\right)=\left(\frac{\mathrm{\pi }}{6}\right)\left(\frac{1}{\sqrt{3}}\right)-\ln \left(\frac{2}{\sqrt{3}}\right)+1$

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

$\Rightarrow y=\frac{\pi }{12}-\frac{\sqrt{3}}{2}\left[\ln \left(\frac{2}{\sqrt{3}}\right)-\ln e\right]$

$\Rightarrow y=\frac{\pi }{12}-\frac{\sqrt{3}}{2}{\log }_e\left(\frac{2}{e\sqrt{3}}\right)$