The Integrating Factor of the differential equation $(1 - {y^2})\cfrac{{dx}}{{dy}} + yx = ay( - 1 < y < 1)$ is

• $\cfrac{1}{{{y^2} - 1}}$

• $\cfrac{1}{{\sqrt {{y^2} - 1} }}$

• $\cfrac{1}{{1 - {y^2}}}$

• $\cfrac{1}{{\sqrt {1 - {y^2}} }}$

Option d is Correct

The given equation can be written as :
$\cfrac{{dx}}{{dy}} + \cfrac{y}{{1 - {y^2}}}x = \cfrac{{ay}}{{1 - {y^2}}}$

which is linear equation of type $\cfrac{{dx}}{{dy}} + Px = Q$

Where $P = \cfrac{y}{{1 - {y^2}}},Q = \cfrac{{ay}}{{1 - {y^2}}}$

$\therefore$ ${\rm{I}}{\rm{.F}}{\rm{.}} = {e^{\int {Pdy} }} = {e^{\int {\left( {\cfrac{y}{{1 - {y^2}}}} \right)dy} }} = {e^{\left( { - \cfrac{1}{2}\int {\cfrac{{ - 2y}}{{1 - {y^2}}}dy} } \right)}}$

Let $I = - \cfrac{1}{2}\int {\cfrac{{ - 2y}}{{1 - {y^2}}}dy}$

$\Rightarrow$ $I = - \cfrac{1}{2}\log \left( {1 - {y^2}} \right) = \log {\left( {1 - {y^2}} \right)^{ - 1/2}} = \log \left( {\cfrac{1}{{\sqrt {1 - {y^2}} }}} \right)$

$\therefore$ ${\rm{I}}{\rm{.F}}{\rm{.}} = {e^{\int {Pdx} }} = {e^{\log \left( {\cfrac{1}{{\sqrt {1 - {y^2}} }}} \right)}} = \cfrac{1}{{\sqrt {1 - {y^2}} }}$

MISCELLANEOUS