JEE Main 2022 — Differential Equations Question with Solution

Given $\frac{dy}{dx}+\frac{\sqrt{2}}{2{\cos }^{4}x-\cos 2x}y=x{e}^{{\tan }^{-1}\left(\sqrt{2}\cot 2x\right)}$

This is a linear differential equation.

Here $I=\int \frac{dx}{2{\cos }^{4}x-\cos 2x}=\int \frac{dx}{\left({\displaystyle \frac{1}{2}}+{\displaystyle \frac{{\cos }^{2}2x}{2}}+\cos 2x\right)-\cos 2x}$

$=2\sqrt{2}\int \frac{{\sec }^{2}2xdx}{2+{\tan }^{2}2x}$

Put $\tan 2x=t$

$I={\tan }^{-1}\frac{\tan 2x}{\sqrt{2}}$

$\therefore IF=e^{{\tan }^{-1}\left(\frac{\tan 2\mathrm{x}}{\sqrt{2}}\right)}=e^{{\cot }^{-1}\left(\sqrt{2}\cot 2x\right)}$

So general solution will be

$y{e}^{{\cot }^{-1}\left(\sqrt{2}\cot 2x\right)}=\int x{e}^{{\tan }^{-1}\left(\sqrt{2}\cot 2x\right)}{e}^{{\cot }^{-1}\left(\sqrt{2}\cot 2x\right)}dx$

$y{e}^{{\cot }^{-1}\left(\sqrt{2}\cot 2x\right)}=e^{\frac{\mathrm{\pi }}{2}}\left(\frac{x^2}{2}\right)+c$

$y\left(\frac{\pi }{4}\right)=\frac{{\pi }^2}{32}\Rightarrow c=0$

i.e. $y=\frac{x^2}{2}{e}^{{\tan }^{-1}\left(\sqrt{2}\cot 2x\right)}$

Given $y\left(\frac{\pi }{3}\right)=\frac{{\pi }^2}{18}{e}^{-{\tan }^{-1}\alpha }$

$\Rightarrow \frac{{\pi }^2}{18}{e}^{-{\tan }^{-1}\alpha }=\frac{{\pi }^2}{18}{e}^{{\tan }^{-1}\left(\sqrt{2}\cot \frac{2\pi }{3}\right)}$$=\frac{{\pi }^2}{18}{e}^{-{\tan }^{-1}\left(\sqrt{\frac{2}{3}}\right)}$

Hence $\alpha =\sqrt{\frac{2}{3}}\Rightarrow 3{\alpha }^2=2$