If $y=y(x)$ is the solution curve of the differential equation $\frac{\mathrm{dy}}{\mathrm{dx}}+\mathrm{y} \tan \mathrm{x}=\mathrm{x} \sec \mathrm{x}, 0 \leq \mathrm{x} \leq \frac{\pi}{3}$, $y(0)=1$, then $y\left(\frac{\pi}{6}\right)$ is equal to

(A) $\frac{\pi}{12}-\frac{\sqrt{3}}{2} \log _{\mathrm{e}}\left(\frac{2}{\mathrm{e} \sqrt{3}}\right)$

(B) $\frac{\pi}{12}+\frac{\sqrt{3}}{2} \log _{\mathrm{e}}\left(\frac{2 \sqrt{3}}{\mathrm{e}}\right)$

(C) $\frac{\pi}{12}-\frac{\sqrt{3}}{2} \log _{\mathrm{e}}\left(\frac{2 \sqrt{3}}{\mathrm{e}}\right)$

(D) $\frac{\pi}{12}+\frac{\sqrt{3}}{2} \log _{\mathrm{e}}\left(\frac{2}{\mathrm{e} \sqrt{3}}\right)$

JEE Main · 2023 · 01 Feb 2023 · Shift 1 · Differential Equations

y = y (x)

is the solution curve of the differential equation

\frac{dy}{dx} + y tan x = x sec x, 0 \leq x \leq \frac{π}{3}

y (0) = 1

, then

y (\frac{π}{6})

is equal to

Answer: (A)

The integrating factor is

e^{\int t a n x d x} = sec x

, and multiplying gives

\frac{d}{d x} (y sec x) = x sec^{2} x

. Integrating by parts,

y sec x = x tan x - lo g_{e} (sec x) + C

, and

y (0) = 1

gives

C = 1

. At

x = \frac{π}{6}

y \cdot \frac{2}{3} = \frac{π}{6 3} - lo g_{e} \frac{2}{3} + 1

, so

y = \frac{π}{12} + \frac{3}{2} (1 - lo g_{e} \frac{2}{3}) = \frac{π}{12} - \frac{3}{2} lo g_{e} (\frac{2}{e 3})

JEE Main · 2023 · 01 Feb 2023 · Shift 1 · Single correct option