QJEEM19Maaea8: JEE Main 2019 Mathematics - Class 12 JEEM19Maaea8 Solution: Let y = y(x) be the solution of the differential equation, , ,…

Question

Verified Solution • 2025–26 Session moderate

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

{{dy} \over {dx}} + y\tan x = 2x + {x^2}\tan x

,

x \in \left( { - {\pi \over 2},{\pi \over 2}} \right)

, such that y(0) = 1. Then :

A $y\left( {{\pi \over 4}} \right) - y\left( { - {\pi \over 4}} \right) = \sqrt 2$
B $y'\left( {{\pi \over 4}} \right) - y'\left( { - {\pi \over 4}} \right) = \pi - \sqrt 2$
C $y\left( {{\pi \over 4}} \right) + y\left( { - {\pi \over 4}} \right) = {{{\pi ^2}} \over 2} + 2$
D $y'\left( {{\pi \over 4}} \right) + y'\left( { - {\pi \over 4}} \right) = - \sqrt 2$

Accepted Answer

{{dy} \over {dx}} + y(\tan x) = 2x + {x^2}\tan x

I.F. =

{e^{\int {\tan x\,dx} }}

=

{e^{\ln \sec x}}

= sec x

y. sec x =

\int {(2x + {x^2}\tan x)\sec \,x\,dx}

\Rightarrow y\sec x = {x^2}\sec x + \lambda

\Rightarrow

y(0) = 0 +

\lambda

= 1

\Rightarrow

\lambda

= 1

\Rightarrow

y = x² + cos x

\Rightarrow

y' = 2x – sinx

\Rightarrow y'\left( {{\pi \over 4}} \right) = {\pi \over 2} - {1 \over {\sqrt 2 }}

\Rightarrow y'\left( { - {\pi \over 4}} \right) = - {\pi \over 2} + {1 \over {\sqrt 2 }}

\therefore

y'\left( {{\pi \over 4}} \right) - y'\left( { - {\pi \over 4}} \right) = \pi - \sqrt 2

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