JEE Main 2022 · 28 Jul · Shift 1 · Mathematics Q10

← JEE Main 2022 · 28 Jul · Shift 1

MathematicsSingle correct+4 / -1Question 10 of 90

If $y=y(x)$ , $x\in(0,\pi/2)$ be the solution curve of the differential equation $\left(\sin^22x\right)\dfrac{dy}{dx}+\left(8\sin^22x+2\sin4x\right)y=2e^{-4x}(2\sin2x+\cos2x)$ , with $y(\pi/4)=e^{-\pi}$ , then $y(\pi/6)$ is equal to :

\dfrac2{\sqrt3}e^{-2\pi/3}

\dfrac2{\sqrt3}e^{2\pi/3}

\dfrac1{\sqrt3}e^{-2\pi/3}

\dfrac1{\sqrt3}e^{2\pi/3}

Answer and worked solution

Official answer: A

Divide by $\sin^22x$ : $y'+(8+4\cot2x)y=\frac{2e^{-4x}(2\sin2x+\cos2x)}{\sin^22x}$ .

Integrating factor $e^{8x}\sin^22x$ ; the right side becomes $\frac d{dx}\left(e^{4x}\sin2x\right)$ .

So $ye^{8x}\sin^22x=e^{4x}\sin2x+C$ , and $y(\pi/4)=e^{-\pi}$ gives $C=0$ . Hence $y=\frac{e^{-4x}}{\sin2x}$ and $y(\pi/6)=\frac2{\sqrt3}e^{-2\pi/3}$ .

(A)

Topics:Linear first-order equations

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JEE Main 2022 · 28 Jul · Shift 1, Mathematics question 10