JEE MAIN - Mathematics (2023 - 1st February Morning Shift - No. 13)
If $$y=y(x)$$ is the solution curve of the differential equation
$$\frac{d y}{d x}+y \tan x=x \sec x, 0 \leq x \leq \frac{\pi}{3}, y(0)=1$$, then $$y\left(\frac{\pi}{6}\right)$$ is equal to
$$\frac{d y}{d x}+y \tan x=x \sec x, 0 \leq x \leq \frac{\pi}{3}, y(0)=1$$, then $$y\left(\frac{\pi}{6}\right)$$ is equal to
$$\frac{\pi}{12}-\frac{\sqrt{3}}{2} \log _{e}\left(\frac{2 \sqrt{3}}{e}\right)$$
$$\frac{\pi}{12}+\frac{\sqrt{3}}{2} \log _{e}\left(\frac{2 \sqrt{3}}{e}\right)$$
$$\frac{\pi}{12}+\frac{\sqrt{3}}{2} \log _{e}\left(\frac{2}{e \sqrt{3}}\right)$$
$$\frac{\pi}{12}-\frac{\sqrt{3}}{2} \log _{e}\left(\frac{2}{e \sqrt{3}}\right)$$
Explanation
$\frac{d y}{d x}+y \tan x=x \sec x$
$\therefore $ I.F $=e^{\int \tan x d x}=\sec x$
$\Rightarrow y \sec x=\int x \sec ^{2} x d x$
$\Rightarrow y \sec x=x \tan x-\ln |\sec x|+c$
Given, $$ y(0)=1 $$
$\Rightarrow 1=c$
$$ \therefore $$ $ y \sec x=x \tan x-\ln |\sec x|+1$
$\Rightarrow y=x \sin x-\cos x \ln |\sec x|+cosx$
Now, at $x=\pi / 6$, we have
$$ \begin{aligned} & y=\frac{\pi}{12}-\frac{\sqrt{3}}{2} \log _e \frac{2}{\sqrt{3}}+\frac{\sqrt{3}}{2} \\\\ & =\frac{\pi}{12}-\frac{\sqrt{3}}{2}\left[\log _e \frac{2}{\sqrt{3}}-\log _e e\right]\\\\ &=\frac{\pi}{12}-\frac{\sqrt{3}}{2} \log _e\left(\frac{2}{e \sqrt{3}}\right) \end{aligned} $$
$\therefore $ I.F $=e^{\int \tan x d x}=\sec x$
$\Rightarrow y \sec x=\int x \sec ^{2} x d x$
$\Rightarrow y \sec x=x \tan x-\ln |\sec x|+c$
Given, $$ y(0)=1 $$
$\Rightarrow 1=c$
$$ \therefore $$ $ y \sec x=x \tan x-\ln |\sec x|+1$
$\Rightarrow y=x \sin x-\cos x \ln |\sec x|+cosx$
Now, at $x=\pi / 6$, we have
$$ \begin{aligned} & y=\frac{\pi}{12}-\frac{\sqrt{3}}{2} \log _e \frac{2}{\sqrt{3}}+\frac{\sqrt{3}}{2} \\\\ & =\frac{\pi}{12}-\frac{\sqrt{3}}{2}\left[\log _e \frac{2}{\sqrt{3}}-\log _e e\right]\\\\ &=\frac{\pi}{12}-\frac{\sqrt{3}}{2} \log _e\left(\frac{2}{e \sqrt{3}}\right) \end{aligned} $$
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