Verified Solution • 2025–26 Session moderate

QJEEM21Ma7745: JEE Main 2021 Mathematics - Class 12 JEEM21Ma7745 Solution: Let y = y(x) be the solution of the differential equation ,…

Let y = y(x) be the solution of the differential equation cos⁡ec2xdy+2dx=(1+ycos⁡2x)cos⁡ec2xdx\cos e{c^2}xdy + 2dx = (1 + y\cos 2x)\cos e{c^2}xdx , with y(π4)=0y\left( {{\pi \over 4}} \right) = 0 . Then, the value of (y(0)+1)2{(y(0) + 1)^2} is equal to :

Multiple Choice Options

  • A e1/2
  • B e −- 1/2
  • C e −- 1
  • D e

Step-by-Step Verified Solution

dydx+2sin⁡2x=1+ycos⁡2x{{dy} \over {dx}} + 2{\sin ^2}x = 1 + y\cos 2x

⇒dydx+(−cos⁡2x)y=cos⁡2x\Rightarrow {{dy} \over {dx}} + ( - \cos 2x)y = \cos 2x

I.F.=e∫−cos⁡2xdx=e−sin⁡2x2I.F. = {e^{\int { - \cos 2xdx} }} = {e^{ - {{\sin 2x} \over 2}}}

Solution of D.E.

y(e−sin⁡2x2)=∫(cos⁡2x)(e−sin⁡2x2)dx+cy\left( {{e^{ - {{\sin 2x} \over 2}}}} \right) = \int {(\cos 2x)\left( {{e^{ - {{\sin 2x} \over 2}}}} \right)} dx + c

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

Given

y(π4)=0y\left( {{\pi \over 4}} \right) = 0

⇒0=−e−12+c⇒c=e−12\Rightarrow 0 = - {e^{{{ - 1} \over 2}}} + c \Rightarrow c = {e^{{{ - 1} \over 2}}}

⇒y(e−sin⁡2x2)=−e−sin⁡2x2+e−12\Rightarrow y\left( {{e^{ - {{\sin 2x} \over 2}}}} \right) = - {e^{ - {{\sin 2x} \over 2}}} + {e^{{{ - 1} \over 2}}}

at x = 0

y=−1+e−12y = - 1 + {e^{{{ - 1} \over 2}}}

⇒y(0)=−1+e−12⇒(y(0)+1)2=e−1\Rightarrow y(0) = - 1 + {e^{{{ - 1} \over 2}}} \Rightarrow {(y(0) + 1)^2} = {e^{ - 1}}

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