Verified Solution • 2025–26 Session moderate

QJEEM19Maae7e: JEE Main 2019 Mathematics - Class 12 JEEM19Maae7e Solution: If , (0 < x < ) and = 0 then is equal to :-

If cos⁡xdydx−ysin⁡x=6x\cos x{{dy} \over {dx}} - y\sin x = 6x , (0 < x < π2{\pi \over 2} )
and y(π3)y\left( {{\pi \over 3}} \right) = 0 then y(π6)y\left( {{\pi \over 6}} \right) is equal to :-

Multiple Choice Options

  • A −π22- {{{\pi ^2}} \over {2 }}
  • B −π243- {{{\pi ^2}} \over {4\sqrt 3 }}
  • C π223{{{\pi ^2}} \over {2\sqrt 3 }}
  • D −π223- {{{\pi ^2}} \over {2\sqrt 3 }}

Step-by-Step Verified Solution

cos⁡xdydx−ysin⁡x=6x\cos x{{dy} \over {dx}} - y\sin x = 6x

⇒\Rightarrow dydx−ytan⁡x=6x.sec⁡x{{dy} \over {dx}} - y\tan x = 6x.\sec x

This is a linear differential equation.

∴\therefore I.F = e∫−tan⁡xdx{e^{\int { - \tan xdx} }} = ∣cos⁡x∣\left| {\cos x} \right|

As 0 < x < π2{\pi \over 2} , so I.F = cos⁡x\cos x

So solution is

y(I.F) = ∫6x.sec⁡x(I.F)dx\int {6x.\sec x\left( {I.F} \right)} dx

⇒\Rightarrow y cosx = ∫6x.sec⁡xcos⁡xdx\int {6x.\sec x\cos x} dx

⇒\Rightarrow y cosx = 6∫xdx6\int x dx

⇒\Rightarrow y cosx = 3x2 + C

As y(π3)=0y\left( {{\pi \over 3}} \right) = 0

∴\therefore (0)×(12)=3π29+C\left( 0 \right) \times \left( {{1 \over 2}} \right) = {{3{\pi ^2}} \over 9} + C

⇒C=−π23\Rightarrow C = {{ - {\pi ^2}} \over 3}

⇒ycos⁡x=3x2−π23\Rightarrow y\cos x = 3{x^2} - {{{\pi ^2}} \over 3}

For y(π6)y\left( {{\pi \over 6}} \right)

y32=3π236−π23y{{\sqrt 3 } \over 2} = {{3{\pi ^2}} \over {36}} - {{{\pi ^2}} \over 3}

⇒\Rightarrow 3y2=−3π212{{\sqrt 3 y} \over 2} = {{ - 3{\pi ^2}} \over {12}}

⇒y=−π223\Rightarrow y = {{ - {\pi ^2}} \over {2\sqrt 3 }}

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