Verified Solution • 2025–26 Session moderate

QJEEM20Maddfd: JEE Main 2020 Mathematics - Class 12 JEEM20Maddfd Solution: Let y = y(x) be the solution of the differential equation, , y…

Let y = y(x) be the solution of the differential equation,
2+sin⁡xy+1.dydx=−cos⁡x{{2 + \sin x} \over {y + 1}}.{{dy} \over {dx}} = - \cos x , y > 0,y(0) = 1.
If y( π\pi ) = a and dydx{{dy} \over {dx}} at x = π\pi is b, then the ordered pair (a, b) is equal to :

Multiple Choice Options

  • A (2, 1)
  • B (2,32)\left( {2,{3 \over 2}} \right)
  • C (1, -1)
  • D (1, 1)

Step-by-Step Verified Solution

2+sin⁡xy+1.dydx=−cos⁡x{{2 + \sin x} \over {y + 1}}.{{dy} \over {dx}} = - \cos x

⇒\Rightarrow ∫dyy+1=∫(−cos⁡x)dx2+sin⁡x\int {{{dy} \over {y + 1}}} = \int {{{\left( { - \cos x} \right)dx} \over {2 + \sin x}}}

⇒\Rightarrow ln⁡∣y+1∣=−ln⁡∣2+sin⁡x∣+ln⁡c\ln \left| {y + 1} \right| = - \ln \left| {2 + \sin x} \right| + \ln c

⇒\Rightarrow ln⁡∣(y+1)(2+sin⁡x)∣=ln⁡c\ln \left| {\left( {y + 1} \right)\left( {2 + \sin x} \right)} \right| = \ln c

As y(0) = 1

∴\therefore ln⁡∣(1+1)(2+0)∣=ln⁡c\ln \left| {\left( {1 + 1} \right)\left( {2 + 0} \right)} \right| = \ln c

⇒\Rightarrow ∴\therefore c = 4

∴\therefore (y+1)(2+sin⁡x){\left( {y + 1} \right)\left( {2 + \sin x} \right)} = 4

⇒\Rightarrow y = 2−sin⁡x2+sin⁡x{{{2 - \sin x} \over {2 + \sin x}}}

Given y( π\pi ) = a

∴\therefore y( π\pi ) = 2−sin⁡π2+sin⁡π{{{2 - \sin \pi } \over {2 + \sin \pi }}} = 1 = a

dydx=(2+sin⁡x)(−cos⁡x)−(2−sin⁡x).(cos⁡x)(2+sin⁡x)2{{dy} \over {dx}} = {{\left( {2 + \sin x} \right)\left( { - \cos x} \right) - \left( {2 - \sin x} \right).\left( {\cos x} \right)} \over {{{\left( {2 + \sin x} \right)}^2}}}

⇒\Rightarrow dydx∣x=π={\left. {{{dy} \over {dx}}} \right|_{x = \pi }} = 1 = b

So, (a, b) = (1, 1)

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