JEE Main 2022 — Mathematics Calculus

Given $\frac{dy}{dx}+\frac{\sqrt{2}}{2{\mathrm{cos}}^{4}x-\mathrm{cos}2x}y=x{e}^{{\mathrm{tan}}^{-1}(\sqrt{2}\mathrm{cot}2x)}$dxdy​+2cos4x−cos2x2​​y=xetan−1(2​cot2x)

This is a linear differential equation.

Here $\displaystyle I=\int \frac{dx}{2{\mathrm{cos}}^{4}x-\mathrm{cos}2x}=\int \frac{dx}{(\frac{1}{2}+\frac{{\mathrm{cos}}^{2}2x}{2}+\mathrm{cos}2x)-\mathrm{cos}2x}$I=∫2cos4x−cos2xdx​=∫(21​+2cos22x​+cos2x)−cos2xdx​

$$=2\sqrt{2}\int \frac{{\mathrm{sec}}^{2}2xdx}{2+{\mathrm{tan}}^{2}2x}$$=22​∫2+tan22xsec22xdx​

Put $\mathrm{tan}2x=t$tan2x=t

$$I={\mathrm{tan}}^{-1}\frac{\mathrm{tan}2x}{\sqrt{2}}$$I=tan−12​tan2x​

$$\therefore IF={e}^{{\mathrm{tan}}^{-1}(\frac{\mathrm{tan}2x}{\sqrt{2}})}={e}^{{\mathrm{cot}}^{-1}(\sqrt{2}\mathrm{cot}2x)}$$∴IF=etan−1(2​tan2x​)=ecot−1(2​cot2x)

So general solution will be

$$y{e}^{{\mathrm{cot}}^{-1}(\sqrt{2}\mathrm{cot}2x)}=\int x{e}^{{\mathrm{tan}}^{-1}(\sqrt{2}\mathrm{cot}2x)}{e}^{{\mathrm{cot}}^{-1}(\sqrt{2}\mathrm{cot}2x)}dx$$yecot−1(2​cot2x)=∫xetan−1(2​cot2x)ecot−1(2​cot2x)dx

$$y{e}^{{\mathrm{cot}}^{-1}(\sqrt{2}\mathrm{cot}2x)}={e}^{\frac{\pi }{2}}(\frac{{x}^{2}}{2})+c$$yecot−1(2​cot2x)=e2π​(2x2​)+c

$$y(\frac{\pi }{4})=\frac{{\pi }^{2}}{32}\Rightarrow c=0$$y(4π​)=32π2​⇒c=0

i.e. $y=\frac{{x}^{2}}{2}{e}^{{\mathrm{tan}}^{-1}(\sqrt{2}\mathrm{cot}2x)}$y=2x2​etan−1(2​cot2x)

Given $y(\frac{\pi }{3})=\frac{{\pi }^{2}}{18}{e}^{-{\mathrm{tan}}^{-1}\alpha }$y(3π​)=18π2​e−tan−1α

$\Rightarrow \frac{{\pi }^{2}}{18}{e}^{-{\mathrm{tan}}^{-1}\alpha }=\frac{{\pi }^{2}}{18}{e}^{{\mathrm{tan}}^{-1}(\sqrt{2}\mathrm{cot}\frac{2\pi }{3})}$⇒18π2​e−tan−1α=18π2​etan−1(2​cot32π​)$=\frac{{\pi }^{2}}{18}{e}^{-{\mathrm{tan}}^{-1}(\sqrt{\frac{2}{3}})}$=18π2​e−tan−1(32​​)

Hence $\alpha =\sqrt{\frac{2}{3}}\Rightarrow 3{\alpha }^{2}=2$α=32​​⇒3α2=2