$${{dy} \over {dx}} + {{y(3{x^2} - 1)} \over {x(1 - {x^2})}} = {{4{x^3}} \over {x(1 - {x^2})}}$$

$$IF = {e^{\int {{{3{x^2} - 1} \over {x - {x^3}}}dx} }} = {e^{ - \ln |{x^3} - x|}} = {e^{ - \ln ({x^3} - x)}} = {1 \over {{x^3} - x}}$$

Solution of D.E. can be given by

$$y.\,{1 \over {{x^3} - x}} = \int {{{4{x^3}} \over {x(1 - {x^2})}}.\,{1 \over {x({x^2} - 1)}}dx} $$

$$ \Rightarrow {y \over {{x^3} - x}} = \int {{{ - 4x} \over {{{({x^2} - 1)}^2}}}dx} $$

$$ \Rightarrow {y \over {{x^3} - x}} = {2 \over {({x^2} - 1)}} + c$$

at x = 2, y = $$-$$2

$${{ - 2} \over 6} = {2 \over 3} + c \Rightarrow c = - 1$$

at $$x = 3 \Rightarrow {y \over {24}} = {2 \over 8} - 1 \Rightarrow y = - 18$$