The solution of the equation `(dy)/(dx)=(x(2logx+1)/(siny+ycosy)` is (a) `( b ) (c) ysiny=( d ) x^(( e )2( f ))( g )logx+( h )(( i ) (j) x^(( k )2( l ))( m ))/( n )2( o ) (p)+c (q)` (r) (s) `( t ) (u) ycosy=( v ) x^(( w )2( x ))( y )(logx+1)+c (z)` (aa) (bb) `( c c ) (dd) ycosy=( e e ) x^(( f f )2( g g ))( h h )logx+( i i )(( j j ) (kk) x^(( l l )2( m m ))( n n ))/( o o )2( p p ) (qq)+c (rr)` (ss) (tt) `( u u ) (vv) ysiny=( w w ) x^(( x x )2( y y ))( z z )logx+c (aaa)` (bbb)

Options

A. `ysiny=x^(2)logx+x^(2)/2+c`
B. `ycosy=x^(2)(logx+1)+c`
C. `ycosy=x^(2)logx+x^(2)/2+c`
D. `ysiny=x^(2)logx+c`

Answer

D

Explanation

`(ycosy+siny)dy=(2xlogx+x)dx`
`ysiny-intsinydy+int(sinydx)`
`=x^(2)logx-intx^(2)1/xdx+intxdx+c`
`therefore ysiny=y^(2)logx+c`