Question

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

• A. $\frac{\pi}{12}-\frac{\sqrt{3}}{2} \log _e\left(\frac{2}{e \sqrt{3}}\right)$
• B. $\frac{\pi}{12}+\frac{\sqrt{3}}{2} \log _e\left(\frac{2}{e \sqrt{3}}\right)$
• C. $\frac{\pi}{12}+\frac{\sqrt{3}}{2} \log _e\left(\frac{2 \sqrt{3}}{e}\right)$
• D. $\frac{\pi}{12}-\frac{\sqrt{3}}{2} \log _e\left(\frac{2 \sqrt{3}}{e}\right)$

Answer 1

The Correct Option is A

The correct answer is (A) : $\frac{\pi}{12}-\frac{\sqrt{3}}{2} \log _e\left(\frac{2}{e \sqrt{3}}\right)$
Here I.F. $=\sec x$
Then solution of D.E :
$y(\sec x)=x\tan x-\ln (\sec x)+c$
Given $y(0)=1\Rightarrow c=1$
$\therefore y(\sec x)=x\tan x-\ln (\sec x)+1$
At $x=\frac{\pi }{6},y=\frac{\pi }{12}+\frac{\sqrt{3}}{2}\ln \frac{\sqrt{3}}{2}+\frac{\sqrt{3}}{2}$