The solution of `(x d d+y dy)/(x dy-y dx)=sqrt((1-x^2-y^2)/(x^2+y^2))` is (a) `( b ) (c)sqrt(( d ) (e) (f) x^(( g )2( h ))( i )+( j ) y^(( k )2( l ))( m ) (n))( o )=sin{( p ) (q)tan^(( r ) (s)-1( t ))( u )(( v ) (w) (x) y/( y ) x (z) (aa) (bb))+C}( c c )` (dd) (ee) `( f f ) (gg)sqrt(( h h ) (ii) (jj) x^(( k k )2( l l ))( m m )+( n n ) y^(( o o )2( p p ))( q q ) (rr))( s s )=cos{( t t ) (uu)tan^(( v v ) (ww)-1( x x ))( y y )(( z z ) (aaa) (bbb) y/( c c c ) x (ddd) (eee) (fff))+C}( g g g )` (hhh) (iii) `( j j j ) (kkk)sqrt(( 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 )=(tan{s i (xxx) n^(( y y y ) (zzz)-1( a a a a ))( b b b b )(( c c c c ) (dddd) (eeee) y/( f f f f ) x (gggg) (hhhh) (iiii))+C}( j j j j )` (kkkk) (llll) `( m m m m ) (nnnn) y+xtan(( o o o o ) (pppp) c+( q q q q ) (rrrr)sin^(( s s s s ) (tttt)-1( u u u u ))( v v v v )sqrt(( w w w w ) (xxxx) (yyyy) x^(( z z z z )2( a a a a a ))( b b b b b )+( c c c c c ) y^(( d d d d d )2( e e e e e ))( f f f f f ) (ggggg))( h h h h h ) (iiiii))( j j j j j )` (kkkkk)

Options

A. `sqrt(x^(2)+y^(2))= sin{(tan^(-1)y//x)+C}`
B. `sqrt(x^(2)+y^(2))=cos{(tan^(-1)y//x)+C}`
C. `sqrt(x^(2)+y^(2))=(tan(sin^(-1)y//x)+C)`
D. `y=xtan(c+sin^(-1)sqrt(x^(2)+y))`

Answer

A, D

Explanation

The differential equation can be rewritten as `(xdx+ydy)/sqrt(1-(x^(2)+y^(2)))=(xdy-ydx)/sqrt(x^(2)+y^(2))`
Since `dtan^(-1)(y//x)=(xdy-ydx)/(x^(2)+y^(2))`
and `d(x^(2)+y^(2))=2(xdx+ydy)`
We have `(1/2d(x^(2)+y^(2))/(sqrt(x^(2)+y^(2))sqrt(1-(x^(2)+y^(2)))))=(xdy-ydx)/(x^(2)+y^(2))=d{(tan^(-1)(y//x))}`
Put `x^(2)+y^(2)=r^(2)` in the LHS. Then
`(tdt)/(tsqrt(1-t^(2)))=d{tan^(-1)(y//x)}`
Integrating both sides, we get
`sin^(-1)t=tan^(-1)(y//x)+c`
i.e., `sin^(-1)sqrt((x^(2)+y^(2)))=tan^(-1)(y//x)+c`
i.e., `sin^(-1)sqrt((x^(2)+y^(2)))=tan^(-1)(y//x)+c`