The differential equation for the family of curve `x^2+y^2-2a y=0,` where `a` is an arbitrary constant, is (a) `( b ) (c)2(( d ) (e) (f) x^(( g )2( h ))( i )-( j ) y^(( k )2( l ))( m ) (n))( o ) y^(( p )prime( q ))( r )=x y (s)` (t) (u) `( v ) (w)2(( x ) (y) (z) x^(( a a )2( b b ))( c c )+( d d ) y^(( e e )2( f f ))( g g ) (hh))( i i ) y^(( j j )prime( k k ))( l l )=x y (mm)` (nn) (oo) `( p p ) (qq)(( r r ) (ss) (tt) x^(( u u )2( v v ))( w w )-( x x ) y^(( y y )2( z z ))( a a a ) (bbb))( c c c ) y^(( d d d )prime( e e e ))( f f f )=2x y (ggg)` (hhh) (iii) `( j j j ) (kkk)(( l l l ) (mmm) (nnn) x^(( o o o )2( p p p ))( q q q )+( r r r ) y^(( s s s )2( t t t ))( u u u ) (vvv))( w w w ) y^(( x x x )prime( y y y ))( z z z )=2x y (aaaa)` (bbbb)

Options

A. `2(x^(2)-y^(2))y^(')=xy`
B. `2(x^(2)+y^(2))y^(')=xy`
C. `(x^(2)-y^(2))y^(')=2xy`
D. `(x^(2)+y^(2))y^(')=2xy`

Answer

C

Explanation

The given family of curves is `x^(2)+y^(2)-2ay=0` ………….(1)
Differentiating w.r.t x, we get `2x+2y(dy)/(dx)-2a(dy)/(dx)=0`
or `2x+2y(dy)/(dx) -(x^(2)+y^(2))/(y)(dy)/(dx)=0` [Using equation (1)]
or `2xy+(2y^(2)-x^(2)-y^(2))y^(')=0`
or `(x^(2)-y^(2))y^(')=2xy`