JEE Main 2023 — Mathematics Calculus

Given:

$$\frac{dy}{dx}+y\mathrm{tan}x=x\mathrm{sec}x$$dxdy​+ytanx=xsecx

This is a linear differential equation.

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

Then solution of differential equation is

$$y(\mathrm{sec}x)=\int x{\mathrm{sec}}^{2}xdx$$y(secx)=∫xsec2xdx

$$\Rightarrow y(\mathrm{sec}x)=x\mathrm{tan}x-\int \mathrm{tan}xdx$$⇒y(secx)=xtanx−∫tanxdx

$$\Rightarrow y(\mathrm{sec}x)=x\mathrm{tan}x-\mathrm{ln}(\mathrm{sec}x)+C$$⇒y(secx)=xtanx−ln(secx)+C

Given:

$$y(0)=1\Rightarrow c=1$$y(0)=1⇒c=1

$$\therefore y(\mathrm{sec}x)=x\mathrm{tan}x-\mathrm{ln}(\mathrm{sec}x)+1$$∴y(secx)=xtanx−ln(secx)+1

At $x=\frac{\pi }{6}$x=6π​, we get

$$y(\mathrm{sec}(\frac{\pi }{6}))=(\frac{\pi }{6})\mathrm{tan}(\frac{\pi }{6})-\mathrm{ln}(\mathrm{sec}(\frac{\pi }{6}))+1$$y(sec(6π​))=(6π​)tan(6π​)−ln(sec(6π​))+1

$$\Rightarrow y(\frac{2}{\sqrt{3}})=(\frac{\pi }{6})(\frac{1}{\sqrt{3}})-\mathrm{ln}(\frac{2}{\sqrt{3}})+1$$⇒y(3​2​)=(6π​)(3​1​)−ln(3​2​)+1

$$\Rightarrow y=\frac{\pi }{12}-\frac{\sqrt{3}}{2}\mathrm{ln}(\frac{2}{\sqrt{3}})+\frac{\sqrt{3}}{2}$$⇒y=12π​−23​​ln(3​2​)+23​​

$$\Rightarrow y=\frac{\pi }{12}-\frac{\sqrt{3}}{2}[\mathrm{ln}(\frac{2}{\sqrt{3}})-\mathrm{ln}e]$$⇒y=12π​−23​​[ln(3​2​)−lne]

$$\Rightarrow y=\frac{\pi }{12}-\frac{\sqrt{3}}{2}{\mathrm{log}}_{e}(\frac{2}{e\sqrt{3}})$$⇒y=12π​−23​​loge​(e3​2​)