QJEEM22Maf0c0: JEE Main 2022 Mathematics - Class 12 JEEM22Maf0c0 Solution: Let be the solution of the differential equation , and . If ,…

$\left( {1 - {x^2}} \right)dy = \left( {xy + \left( {{x^3} + 2} \right)\sqrt {1 - {x^2}} } \right)dx$

$\therefore$ ${{dy} \over {dx}} - {x \over {1 - {x^2}}}y = {{{x^3} + 3} \over {\sqrt {1 - {x^2}} }}$

$\therefore$ $I.F. = {e^{\int { - {x \over {1 - {x^2}}}dx} }} = \sqrt {1 - {x^2}}$

Solution is

$y.\,\sqrt {1 - {x^2}} = \int {\left( {{x^3} + 3} \right)dx}$

$y.\,\sqrt {1 - {x^2}} = {{{x^4}} \over 4} + 3x + c$

$\because$ $y(0) = 0 \Rightarrow c = 0$

$\therefore$ $y(x) = {{{x^4} + 12x} \over {4\sqrt {1 - {x^2}} }}$

$\therefore$ $\int_{{{ - 1} \over 2}}^{{1 \over 2}} {\sqrt {1 - {x^2}} y(x)dx = \int_{{{ - 1} \over 2}}^{{1 \over 2}} {\left( {{{{x^4} + 12x} \over 4}} \right)dx = \int_0^{{1 \over 2}} {{{{x^4}} \over 2}dx} } }$

$\therefore$ $k = {1 \over {320}}$

$\therefore$ $= {k^{ - 1}} = 320$