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← JEE Main 2022 · 29 Jun · Shift 1
MathematicsInteger+4 / -1Question 22 of 90

JEE Main 2022 · 29 Jun · Shift 1, Mathematics question 22

Let y=y(x)y=y(x) be the solution of the differential equation
dydx+2y2cos⁡4x−cos⁡2x=xetan⁡−1(2cot⁡2x),0<x<π2\frac{dy}{dx}+\frac{\sqrt2y}{2\cos^4x-\cos2x}=xe^{\tan^{-1}\left(\sqrt2\cot2x\right)},\quad0<x<\frac\pi2
with y(π4)=π232y\left(\dfrac\pi4\right)=\dfrac{\pi^2}{32}. If y(π3)=π218e−tan⁡−1(α)y\left(\dfrac\pi3\right)=\dfrac{\pi^2}{18}e^{-\tan^{-1}(\alpha)}, then the value of 3α23\alpha^2 is equal to ______.

Answer and worked solution

Official answer: 2

Integrating factor. With t=tan⁡xt=\tan x, ∫2 dx2cos⁡4x−cos⁡2x=∫2(1+t2)1+t4dt=tan⁡−1t2−12t=−tan⁡−1(2cot⁡2x)\displaystyle\int\frac{\sqrt2\,dx}{2\cos^4x-\cos2x}=\int\frac{\sqrt2(1+t^2)}{1+t^4}dt=\tan^{-1}\frac{t^2-1}{\sqrt2t}=-\tan^{-1}(\sqrt2\cot2x).

Solve. y e−tan⁡−1(2cot⁡2x)=x22+Cy\,e^{-\tan^{-1}(\sqrt2\cot2x)}=\frac{x^2}2+C; at π4\frac\pi4, C=0C=0.

At π3\frac\pi3. cot⁡2π3=−13\cot\frac{2\pi}3=-\frac1{\sqrt3}: y=π218e−tan⁡−12/3y=\frac{\pi^2}{18}e^{-\tan^{-1}\sqrt{2/3}}, so 3α2=23\alpha^2=2.

Answer: 2

Topics:Linear first-order equationsSubstitution
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