JEE Main 2022 (Online) 29th June Morning ShiftQuestion 1
Let $y=y(x)$ be the solution of the differential equation $\frac{dy}{dx}+\frac{\sqrt{2}y}{2\cos^{4}x-\cos2x}=xe^{\tan^{-1}(\sqrt{2}\cot2x)}$, $0<x<\frac{\pi}{2}$ with $y\left(\frac{\pi}{4}\right)=\frac{\pi^{2}}{32}$. If $y\left(\frac{\pi}{3}\right)=\frac{\pi^{2}}{18}e^{-\tan^{-1}(\alpha)}$, then the value of $3\alpha^{2}$ is equal to $\_\_\_\_\_\_$ .
Differential Equation
Answer$2$
Solution
Concept. The IF exponent is $-\tan^{-1}(\sqrt2\cot2x)$, which exactly cancels the exponential on the right. Formulas used. (1) $2\cos^4x-\cos2x=\cos^4x\left(\sec^4x-2\sec^2x+2\right)$(2) $\dfrac{1+t^2}{1+t^4}\,dt=\dfrac{du}{u^2+2}$ with $u=t-\frac1t$, and $\tan x-\cot x=-2\cot2x$Step 1 - The IF integral. By (1), with $t=\tan x$, $dx=\frac{dt}{1+t^2}$: $$\int\frac{\sqrt2\,dx}{2\cos^4x-\cos2x}=\int\frac{\sqrt2(1+t^2)}{1+t^4}\,dt=\int\frac{\sqrt2\,du}{u^2+2}=\tan^{-1}\frac{u}{\sqrt2}.$$ Step 2 - In terms of $x$. By (2), $u=-2\cot2x$, so the integral is $-\tan^{-1}\left(\sqrt2\cot2x\right)$ and $$\text{IF}=e^{-\tan^{-1}(\sqrt2\cot2x)}.$$ Step 3 - Solve. $Q\cdot\text{IF}=x$, so $y\,e^{-\tan^{-1}(\sqrt2\cot2x)}=\frac{x^2}{2}+C$. At $\frac\pi4$: $\cot\frac\pi2=0$, so $\frac{\pi^2}{32}=\frac{\pi^2}{32}+C$ and $C=0$. Step 4 - At $\frac\pi3$. $\sqrt2\cot\frac{2\pi}{3}=-\sqrt{\frac23}$: $$y=\frac{\pi^2}{18}\,e^{\tan^{-1}\left(-\sqrt{2/3}\right)}=\frac{\pi^2}{18}\,e^{-\tan^{-1}\sqrt{2/3}}.$$ So $\alpha^2=\frac23$ and $3\alpha^2=2$. Answer: 2