Verified Solution • 2025–26 Session moderate

QJEEM21Ma1bd6: JEE Main 2021 Mathematics - Class 12 JEEM21Ma1bd6 Solution: Let y = y(x) be the solution of the differential equation xdy = …

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

equation xdy = (y + x3 cosx)dx with y( π\pi ) = 0, then y(π2)y\left( {{\pi \over 2}} \right) is equal to :

Multiple Choice Options

  • A π24+π2{{{\pi ^2}} \over 4} + {\pi \over 2}
  • B π22+π4{{{\pi ^2}} \over 2} + {\pi \over 4}
  • C π22−π4{{{\pi ^2}} \over 2} - {\pi \over 4}
  • D π44−π2{{{\pi ^4}} \over 4} - {\pi \over 2}

Step-by-Step Verified Solution

xdy=(y+x3cos⁡x)dxxdy = (y + {x^3}\cos x)dx

⇒\Rightarrow xdy=ydx+x3cos⁡xdxxdy = ydx + {x^3}\cos xdx

⇒\Rightarrow xdy−ydxx2=x3coxdxx2{{xdy - ydx} \over {{x^2}}} = {{{x^3}coxdx} \over {{x^2}}}

⇒\Rightarrow ddx(yx)=∫xcos⁡xdx{d \over {dx}}\left( {{y \over x}} \right) = \int {x\cos xdx}

⇒yx=xsin⁡x−∫1.sin⁡xdx\Rightarrow {y \over x} = x\sin x - \int {1.\sin xdx}

⇒\Rightarrow yx=xsin⁡x+cos⁡x+C{y \over x} = x\sin x + \cos x + C

⇒0=−1+C⇒C=1,x=π,y=0\Rightarrow 0 = - 1 + C \Rightarrow C = 1,x = \pi ,y = 0

so, yx=xsin⁡x+cos⁡x+1{y \over x} = x\sin x + \cos x + 1

y=x2sin⁡x+xcos⁡x+xy = {x^2}\sin x + x\cos x + x

x=π2x = {\pi \over 2}

y(π2)=π24+π2y\left( {{\pi \over 2}} \right) = {{{\pi ^2}} \over 4} + {\pi \over 2}

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