The general solution of the differential equation `(dy)/(dx)+sin(x+y)/2=sin(x-y)/2` is (a) `( b ) (c)logtan(( d ) (e) (f) y/( g )2( h ) (i) (j))=c-2sinx (k)` (l) (m) `( n ) (o)logtan(( p ) (q) (r) y/( s )4( t ) (u) (v))=c-2sin(( w ) (x) (y) x/( z )2( a a ) (bb) (cc))( d d )` (ee) (ff) `( g g ) (hh)logtan(( i i ) (jj) (kk) y/( l l )2( m m ) (nn)+( o o )pi/( p p )4( q q ) (rr) (ss))=c-2sinx (tt)` (uu) (vv) `( w w ) (xx)logtan(( y y ) (zz) (aaa) y/( b b b )4( c c c ) (ddd)+( e e e )pi/( f f f )4( g g g ) (hhh) (iii))=c-2sin(( j j j ) (kkk) (lll) x/( m m m )2( n n n ) (ooo) (ppp))( q q q )` (rrr)

Options

A. `log tan(y/2)=c-2sinx`
B. `log tan(y/2)=c-2sin(x/2)`
C. `log tan(y/2+pi/4)=c-2sinx`
D. `log tan(y/4+pi/4)=c-2sin(x/2)`

Answer

B

Explanation

We have `(dy)/(dx) = sin(x-y)/2-sin(x+y)/2`
`=-2cosx/2siny/2`
or `1/2" cosec "y/2dy=-cosx/2dx`
or `logtany/4-(sinx/2)/(1/2)+c`
or `logtan(y/4)=c-2sinx/2`