The solution of the differential equation `(dy)/(dx)=1/(x y[x^2siny^2+1])` is (a) `( b ) (c) (d) x^(( e )2( f ))( g )(cos( h ) y^(( i )2( j ))( k )-sin( l ) y^(( m )2( n ))( o )-2C (p) e^( q ) (r)-( s ) y^((( t )2( u ))( v ) (w))( x ))=2( y )` (z) (aa) `( b b ) (cc) (dd) y^(( e e )2( f f ))( g g )(cos( h h ) y^(( i i )2( j j ))( k k )-sin( l l ) y^(( m m )2( n n ))( o o )-2C (pp) e^( q q ) (rr)-( s s ) y^((( t t )2( u u ))( v v ) (ww))( x x ))=2( y y )` (zz) (aaa) `( b b b ) (ccc) (ddd) x^(( e e e )2( f f f ))( g g g )(cos( h h h ) y^(( i i i )2( j j j ))( k k k )-sin( l l l ) y^(( m m m )2( n n n ))( o o o )-( p p p ) e^( q q q ) (rrr)-( s s s ) y^((( t t t )2( u u u ))( v v v ) (www))( x x x ))=4C (yyy)` (zzz) (aaaa) None of these

Options

A. `x^(2)(cosy^(2)-siny^(2)-2Ce^(-y^(2)))=2`
B. `y^(2)(cosx^(2)-siny^(2)-2Ce^(-y^(2)))=4C`
C. None of these
D. a system of circles

Answer

A

Explanation

`(dy)/(dx) = 1/(xy[x^(2)siny^(2)+1])`
or `1/x^(3)(dx)/(dy) -1/x^(2)y=ysiny^(2)`
Putting `-1//x^(2)=u`, we get
`(du)/(dy)+2uy=2ysiny^(2)`.
I.F. `=e^(y^(2))`
Thus, solution is `ue^(y^(3))=int2ysiny^(2)e^(y^(2))dy+C`
`=int(sint)e^(t)dt+C`
`=1/2e^(y^(3))(siny^(2)-cosy^(2))+c`
or `2u=(siny^(2)-cosy^(2))+2Ce^(-y^(2))`
or `2=x^(2)[cosy^(2)-siny^(2)-2Ce^(-y^(2))]`