JEE Main 2022 (Online) 28th July Morning ShiftQuestion 1
If $y=y(x)$, $x\in\left(0,\frac\pi2\right)$ be the solution curve of the differential equation $(\sin^22x)\frac{dy}{dx}+(8\sin^22x+2\sin4x)y=2e^{-4x}(2\sin2x+\cos2x)$, with $y\left(\frac\pi4\right)=e^{-\pi}$, then $y\left(\frac\pi6\right)$ is equal to :
Differential Equation
Solution
Concept. Divide by $\sin^22x$; the IF is $e^{8x}\sin^22x$, and the right side becomes $(e^{4x}\sin2x)'$. Formulas used. (1) $\dfrac{2\sin4x}{\sin^22x}=4\cot2x$ and $\int4\cot2x\,dx=2\ln\sin2x$(2) $\left(e^{4x}\sin2x\right)'=2e^{4x}(2\sin2x+\cos2x)$Step 1 - Linear form. By (1), $$y'+(8+4\cot2x)\,y=\frac{2e^{-4x}(2\sin2x+\cos2x)}{\sin^22x},\qquad\text{IF}=e^{8x}\sin^22x.$$ Step 2 - Solve. By (2), $$\left(ye^{8x}\sin^22x\right)'=2e^{4x}(2\sin2x+\cos2x)\ \Longrightarrow\ ye^{8x}\sin^22x=e^{4x}\sin2x+C.$$ Step 3 - Constant. At $\frac\pi4$: $e^{-\pi}e^{2\pi}=e^{\pi}+C$, so $C=0$ and $y=\dfrac{e^{-4x}}{\sin2x}$. Step 4 - At $\frac\pi6$. $y=\dfrac{e^{-2\pi/3}}{\sqrt3/2}=\dfrac{2}{\sqrt3}e^{-2\pi/3}$. Answer: (A)