Verified Solution • 2025–26 Session moderate

QJEEM22Mad85d: JEE Main 2022 Mathematics - Class 12 JEEM22Mad85d Solution: Let y = y(x) be the solution of the differential equation , ,…

Let y = y(x) be the solution of the differential equation x(1−x2)dydx+(3x2y−y−4x3)=0x(1 - {x^2}){{dy} \over {dx}} + (3{x^2}y - y - 4{x^3}) = 0 , x > 1 , with y(2)=−2y(2) = - 2 . Then y(3) is equal to :

Multiple Choice Options

  • A −- 18
  • B −- 12
  • C −- 6
  • D −- 3

Step-by-Step Verified Solution

dydx+y(3x2−1)x(1−x2)=4x3x(1−x2){{dy} \over {dx}} + {{y(3{x^2} - 1)} \over {x(1 - {x^2})}} = {{4{x^3}} \over {x(1 - {x^2})}}

IF=e∫3x2−1x−x3dx=e−ln⁡∣x3−x∣=e−ln⁡(x3−x)=1x3−xIF = {e^{\int {{{3{x^2} - 1} \over {x - {x^3}}}dx} }} = {e^{ - \ln |{x^3} - x|}} = {e^{ - \ln ({x^3} - x)}} = {1 \over {{x^3} - x}}

Solution of D.E. can be given by

y. 1x3−x=∫4x3x(1−x2). 1x(x2−1)dxy.\,{1 \over {{x^3} - x}} = \int {{{4{x^3}} \over {x(1 - {x^2})}}.\,{1 \over {x({x^2} - 1)}}dx}

⇒yx3−x=∫−4x(x2−1)2dx\Rightarrow {y \over {{x^3} - x}} = \int {{{ - 4x} \over {{{({x^2} - 1)}^2}}}dx}

⇒yx3−x=2(x2−1)+c\Rightarrow {y \over {{x^3} - x}} = {2 \over {({x^2} - 1)}} + c

at x = 2, y = −- 2

−26=23+c⇒c=−1{{ - 2} \over 6} = {2 \over 3} + c \Rightarrow c = - 1

at x=3⇒y24=28−1⇒y=−18x = 3 \Rightarrow {y \over {24}} = {2 \over 8} - 1 \Rightarrow y = - 18

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