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QJEEM23Ma695c: JEE Main 2023 Mathematics - Class 12 JEEM23Ma695c Solution: If is the solution curve of the differential equation , then is …

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

dydx+ytan⁡x=xsec⁡x,0≤x≤π3,y(0)=1\frac{d y}{d x}+y \tan x=x \sec x, 0 \leq x \leq \frac{\pi}{3}, y(0)=1 , then y(π6)y\left(\frac{\pi}{6}\right) is equal to

Multiple Choice Options

  • A π12−32log⁡e(23e)\frac{\pi}{12}-\frac{\sqrt{3}}{2} \log _{e}\left(\frac{2 \sqrt{3}}{e}\right)
  • B π12+32log⁡e(23e)\frac{\pi}{12}+\frac{\sqrt{3}}{2} \log _{e}\left(\frac{2 \sqrt{3}}{e}\right)
  • C π12+32log⁡e(2e3)\frac{\pi}{12}+\frac{\sqrt{3}}{2} \log _{e}\left(\frac{2}{e \sqrt{3}}\right)
  • D π12−32log⁡e(2e3)\frac{\pi}{12}-\frac{\sqrt{3}}{2} \log _{e}\left(\frac{2}{e \sqrt{3}}\right)

Step-by-Step Verified Solution

dydx+ytan⁡x=xsec⁡x\frac{d y}{d x}+y \tan x=x \sec x

∴\therefore I.F =e∫tan⁡xdx=sec⁡x=e^{\int \tan x d x}=\sec x

⇒ysec⁡x=∫xsec⁡2xdx\Rightarrow y \sec x=\int x \sec ^{2} x d x

⇒ysec⁡x=xtan⁡x−ln⁡∣sec⁡x∣+c\Rightarrow y \sec x=x \tan x-\ln |\sec x|+c

Given, y(0)=1y(0)=1

⇒1=c\Rightarrow 1=c

∴\therefore ysec⁡x=xtan⁡x−ln⁡∣sec⁡x∣+1y \sec x=x \tan x-\ln |\sec x|+1

⇒y=xsin⁡x−cos⁡xln⁡∣sec⁡x∣+cosx\Rightarrow y=x \sin x-\cos x \ln |\sec x|+cosx

Now, at x=π/6x=\pi / 6 , we have

amp;y=π12−32log⁡e23+32amp;=π12−32[log⁡e23−log⁡ee]amp;=π12−32log⁡e(2e3)\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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