The solution of the differential equation `x(x^2+1)((dy)/(dx))=y(1-x^2)+x^3logx` is (a) `( b ) (c) (d)(( e ) y(( f ) (g) (h) x^(( i )2( j ))( k )+1( l )))/( m ) x (n) (o)=( p )1/( q )4( r ) (s) (t) x^(( u )2( v ))( w )logx+( x )1/( y )2( z ) (aa) (bb) x^(( c c )2( d d ))( e e )+c (ff)` (gg) (hh) `( i i ) (jj) (kk)(( l l ) y(( m m ) (nn) (oo) x^(( p p )2( q q ))( r r )-1( s s )))/( t t ) x (uu) (vv)=( w w )1/( x x )2( y y ) (zz) (aaa) x^(( b b b )2( c c c ))( d d d )logx-( e e e )1/( f f f )4( g g g ) (hhh) (iii) x^(( j j j )2( k k k ))( l l l )+c (mmm)` (nnn) (ooo) `( p p p ) (qqq) (rrr)(( s s s ) y(( t t t ) (uuu) (vvv) x^(( w w w )2( x x x ))( y y y )+1( z z z )))/( a a a a ) x (bbbb) (cccc)=( d d d d )1/( e e e e )2( f f f f ) (gggg) (hhhh) x^(( i i i i )2( j j j j ))( k k k k )logx-( l l l l )1/( m m m m )4( n n n n ) (oooo) (pppp) x^(( q q q q )2( r r r r ))( s s s s )+c (tttt)` (uuuu) (vvvv) None of these

Options

A. `y(x^(2)+1)//x=1/4x^(2)logx+1/2x^(2)+c`
B. `y^(2)(x^(2)-1)//x=1/2x^(2)logx-1/4x^(2)+c`
C. `y(x^(2)+1)x=1/2x^(2)logx-1/4x^(2)+c`
D. None of these

Answer

C

Explanation

The given equation can be rewritten as
`(dy)/(dx) +(x^(2)-1)/(x(x^(2)-1)y)=(x^(2)logx)/(x^(2)+1)`…………..(1)
which is linear, Also,
`P=(x^(2)-1)/(x(x^(2)+1))` and `Q=(x^(2)logx)/(x^(2)+1)`
`intPdx=int[(2x)/(x^(2)+1)-1/x]dx` [resolving into partial fractions]
`log(x^(2)+1)-logx`
`therefore` I.F. `=[e^(log[x^(2)+1)//x)]=(x^(2)+1)/(x)`
Hence, the required solution of equation (1) is
`y(x^(2)+1)/(x) = int((x^(2)+1)/(x)(x^(2)logx)/(x^(2)+1))dx+c`
`=intxlogxdx+c`
`=1/2x^(2)logx-int1/x^(2)/2dx+c`
`therefore y(x^(2)+1)//x=1/2x^(2)logx-1/4x^(2)+c`