If integrating factor of `x(1-x^2)dy+(2x^2y-y-a x^3)dx=0` is `e^(intp dx ,)` then `P` is equal to (a) `( b ) (c) (d)(( e )2( f ) x^(( g )2( h ))( i )-a (j) x^(( k )3( l ))( m ))/( n )(( o ) x(( p ) (q)1-( r ) x^(( s )2( t ))( u ) (v)))( w ) (x) (y)` (z) (b) `( a a ) (bb)2( c c ) x^(( d d )3( e e ))( f f )-1( g g )` (hh) (c) `( d ) (e) (f)(( g )2( h ) x^(( i )2( j ))( k )-a)/( l )(( m ) a (n) x^(( o )3( p ))( q ))( r ) (s) (t)` (u) (d) `( v ) (w) (x)(( y )2( z ) x^(( a a )2( b b ))( c c )-1)/( d d )(( e e ) x(( f f ) (gg)1-( h h ) x^(( i i )2( j j ))( k k ) (ll)))( m m ) (nn) (oo)` (pp)

Options

A. `(2x^(2)-ax^(2))/(x(1-x^(2))`
B. `2x^(3)-1`
C. `(2x^(2)-a)/(ax^(3))`
D. `(2x^(2)-1)/(x(1-x^(2))`

Answer

D

Explanation

`x(1-x^(2))dy+(2x^(2)-y-ax^(3))dx=0`
or `x(1-x^(2))(dy)/(dx)+2x^(2)y-y-ax^(3)=0`
or `x(1-x^(2))(dy)/(dx) + y(2x^(2)-1)=ax^(3)`
or `(dy)/(dx) + (2x^(2)-1)/(x(1-x^(2))y)=(ax^(3))/(x(1-x^(2))`
Which is of the form `(dy)/(dx) + Py=Q`
Its integrating factor is `e^(intPdx)`
Here, `P=(2x^(2)-1)/(x(1-x^(2))`