Verified Solution • 2025–26 Session moderate

QJEEM21Maf5f2: JEE Main 2021 Mathematics - Class 12 JEEM21Maf5f2 Solution: Let y = y(x) be the solution of the differential equation such…

Let y = y(x) be the solution of the differential equation

dydx=2(y+2sin⁡x−5)x−2cos⁡x{{dy} \over {dx}} = 2(y + 2\sin x - 5)x - 2\cos x such that y(0) = 7. Then y( π\pi ) is equal to :

Multiple Choice Options

  • A 2eπ2+52{e^{{\pi ^2}}} + 5
  • B eπ2+5{e^{{\pi ^2}}} + 5
  • C 3eπ2+53{e^{{\pi ^2}}} + 5
  • D 7eπ2+57{e^{{\pi ^2}}} + 5

Step-by-Step Verified Solution

dydx−2xy=2(2sin⁡x−5)x−2cos⁡x{{dy} \over {dx}} - 2xy = 2(2\sin x - 5)x - 2\cos x

IF = e−x2{e^{ - {x^2}}}

So, y.e−x2=∫e−x2(2x(2sin⁡x−5)−2cos⁡x)dxy.{e^{ - {x^2}}} = \int {{e^{ - {x^2}}}(2x(2\sin x - 5) - 2\cos x)dx}

⇒y.e−x2=e−x2(5−2sin⁡x)+c\Rightarrow y.{e^{ - {x^2}}} = {e^{ - {x^2}}}(5 - 2\sin x) + c

⇒y=5−2sin⁡x+c.ex2\Rightarrow y = 5 - 2\sin x + c.{e^{{x^2}}}

Given at x = 0, y = 7

⇒\Rightarrow 7 = 5 + c ⇒\Rightarrow c = 2

So, y=5−2sin⁡x+2ex2y = 5 - 2\sin x + 2{e^{{x^2}}}

Now, at x = π\pi ,

y = 5 + 2 ex2{e^{{x^2}}}

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