QJEEM22Ma32a9: JEE Main 2022 Mathematics - Class 12 JEEM22Ma32a9 Solution: Let the solution curve of the differential equation , pass…

${{dy} \over {dx}} + {{xy} \over {{x^2} - 1}} = {{{x^4} + 2x} \over {\sqrt {1 - {x^2}} }}$

which is first order linear differential equation.

Integrating factor $(I.F.) = {e^{\int {{x \over {{x^2} - 1}}dx} }}$

$= {e^{{1 \over 2}\ln |{x^2} - 1|}} = \sqrt {|{x^2} - 1|}$

$= \sqrt {1 - {x^2}}$

$\because$ $x \in ( - 1,1)$

Solution of differential equation

$y\sqrt {1 - {x^2}} = \int {({x^4} + 2x)dx = {{{x^5}} \over 5} + {x^2} + c}$

Curve is passing through origin, $c = 0$

$y = {{{x^5} + 5{x^2}} \over {5\sqrt {1 - {x^2}} }}$

$\int\limits_{{{ - \sqrt 3 } \over 2}}^{{{\sqrt 3 } \over 2}} {{{{x^5} + 5{x^2}} \over {5\sqrt {1 - {x^2}} }}dx = 0 + 2\int\limits_0^{{{\sqrt 3 } \over 2}} {{{{x^2}} \over {\sqrt {1 - {x^2}} }}dx} }$

put $x = \sin \theta$

$dx = \cos \theta \,d\theta$

$I = 2\int\limits_0^{{\pi \over 3}} {{{{{\sin }^2}\theta \,.\,\cos \theta d\theta } \over {\cos \theta }}}$

$= \int\limits_0^{{\pi \over 3}} {(1 - \cos 2\theta )d\theta }$

$= \left. {\left( {\theta - {{\sin 2\theta } \over 2}} \right)} \right|_0^{{\pi \over 3}}$

$= {\pi \over 3} - {{\sqrt 3 } \over 4}$