Verified Solution • 2025–26 Session moderate

QJEEM21Ma3b98: JEE Main 2021 Mathematics - Class 11 JEEM21Ma3b98 Solution: Let y = y(x) be solution of the following differential equation …

Let y = y(x) be solution of the following differential equation eydydx−2eysin⁡x+sin⁡xcos⁡2x=0,y(π2)=0{e^y}{{dy} \over {dx}} - 2{e^y}\sin x + \sin x{\cos ^2}x = 0,y\left( {{\pi \over 2}} \right) = 0 If y(0)=log⁡e(α+βe−2)y(0) = {\log _e}(\alpha + \beta {e^{ - 2}}) , then 4(α+β)4(\alpha + \beta ) is equal to ______________.

Step-by-Step Verified Solution

Let ey = t

⇒dtdx−(2sin⁡x)t=−sin⁡xcos⁡2x\Rightarrow {{dt} \over {dx}} - (2\sin x)t = - \sin x{\cos ^2}x

I.F.=e2cos⁡xI.F. = {e^{2\cos x}}

⇒t.e2cos⁡x=∫e2cos⁡x.(−sin⁡xcos⁡2x)dx\Rightarrow t.{e^{2\cos x}} = \int {{e^{2\cos x}}.( - \sin x{{\cos }^2}x)dx}

⇒ey.e2cos⁡x=∫e2z.z2dz,z=e2cos⁡x\Rightarrow {e^y}.{e^{2\cos x}} = \int {{e^{2z}}.{z^2}dz,z = {e^{2\cos x}}}

⇒ey.e2cos⁡x=12.cos⁡2x.e2cos⁡x−12cos⁡x.e2cos⁡x+e2cos⁡x4+C\Rightarrow {e^y}.{e^{2\cos x}} = {1 \over 2}.{\cos ^2}x.{e^{2\cos x}} - {1 \over 2}\cos x.{e^{2\cos x}} + {{{e^{2\cos x}}} \over 4} + C

at x=π2,y=0⇒C=34x = {\pi \over 2},y = 0 \Rightarrow C = {3 \over 4}

⇒ey=12cos⁡2x−12cos⁡x+14+34.e−2cos⁡x\Rightarrow {e^y} = {1 \over 2}{\cos ^2}x - {1 \over 2}\cos x + {1 \over 4} + {3 \over 4}.{e^{ - 2\cos x}}

⇒y=log⁡[cos⁡2x2−cos⁡x2+14+34e−2cos⁡x]\Rightarrow y = \log \left[ {{{{{\cos }^2}x} \over 2} - {{\cos x} \over 2} + {1 \over 4} + {3 \over 4}{e^{ - 2\cos x}}} \right]

Put x = 0

⇒y=log⁡[14+34e−2]⇒α=14,β=34\Rightarrow y = \log \left[ {{1 \over 4} + {3 \over 4}{e^{ - 2}}} \right] \Rightarrow \alpha = {1 \over 4},\beta = {3 \over 4}

Related Learning Resources & Exam Preparation

JEE Main PYQ Question Bank
Browse year-wise authentic previous years' questions with solutions
Authentic CBT Mock Tests
Simulate official exam environment with live timer and percentile scoring