Verified Solution • 2025–26 Session moderate

QJEEM20Ma391e: JEE Main 2020 Mathematics - Class 12 JEEM20Ma391e Solution: Let y = y(x) be the solution of the differential equation, xy'- …

Let y = y(x) be the solution of the differential equation,
xy'- y = x2(xcosx + sinx), x > 0. if y ( π\pi ) = π\pi then
y′′(π2)+y(π2)y''\left( {{\pi \over 2}} \right) + y\left( {{\pi \over 2}} \right) is equal to :

Multiple Choice Options

  • A 1+π21 + {\pi \over 2}
  • B 2+π2+π242 + {\pi \over 2} + {{{\pi ^2}} \over 4}
  • C 2+π22 + {\pi \over 2}
  • D 1+π2+π241 + {\pi \over 2} + {{{\pi ^2}} \over 4}

Step-by-Step Verified Solution

xy' - y = {x^2}(x{\mathop{\rm cosx}\nolimits} + sinx),\,x > 0,\,y(\pi ) = \pi

y′−1xy=x{xcos⁡x + sin⁡x}y' - {1 \over x}y = x\{ x\cos x\, + \,\sin x\}

I.F.=e−∫12dx=e−ln⁡x=1xI.F. = {e^{ - \int {{1 \over 2}dx} }} = {e^{ - \ln x}} = {1 \over x}

∴\therefore y.1x=∫1x.x(xcosx+sinx)dx

yx{{y \over x}} = ∫(x cos⁡ x+sin⁡ x)dx\int ( x\,\cos \,x + \sin \,x)dx

=x sin⁡ x+C= x\,\sin \,x + C

yx{{y \over x}} = x sin⁡ x+Cx\,\sin \,x + C

⇒y=x2=sin⁡ x+cx\Rightarrow y = {x^2} = \sin \,x + cx

x = π\pi , y = π\pi

π\pi = π\pi c ⇒\Rightarrow C = 1

y=x2sin⁡x+x⇒y(π2)=π24+π2y = {x^2}\sin x + x \Rightarrow y\left( {{\pi \over 2}} \right) = {{{\pi ^2}} \over 4} + {\pi \over 2}

y′=2xsin⁡x+x2cos⁡x+1y' = 2x\sin x + {x^2}\cos x + 1

y" = 2sin⁡x+2xcos⁡x+2xcos⁡x−x2sin⁡x2\sin x + 2x\cos x + 2x\cos x - {x^2}\sin x

y" (π2)=2−π24\left( {{\pi \over 2}} \right) = 2 - {{{\pi ^2}} \over 4}

⇒y(π2)\Rightarrow y\left( {{\pi \over 2}} \right) + y" (π2)=2+π2\left( {{\pi \over 2}} \right) = 2 + {\pi \over 2}

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