The solution of `(y+x+5)dy=(y-x+1)dx` is (a) `( b ) (c)log(( d ) (e) (f) (g)(( h ) (i) y+3( j ))^(( k )2( l ))( m )+( n ) (o)(( p ) (q) x+2( r ))^(( s )2( t ))( u ) (v))+( w ) (x)tan^(( y ) (z)-1( a a ))( b b ) (cc)(( d d ) y+3)/( e e )(( f f ) y+2)( g g ) (hh)+C (ii)` (jj) (kk) `( l l ) (mm)log(( n n ) (oo) (pp) (qq)(( r r ) (ss) y+3( t t ))^(( u u )2( v v ))( w w )+( x x ) (yy)(( z z ) (aaa) x-2( b b b ))^(( c c c )2( d d d ))( e e e ) (fff))+( g g g ) (hhh)tan^(( i i i ) (jjj)-1( k k k ))( l l l ) (mmm)(( n n n ) y-3)/( o o o )(( p p p ) y-2)( q q q ) (rrr)+C (sss)` (ttt) (uuu) `( v v v ) (www)log(( x x x ) (yyy) (zzz) (aaaa)(( b b b b ) (cccc) y+3( d d d d ))^(( e e e e )2( f f f f ))( g g g g )+( h h h h ) (iiii)(( j j j j ) (kkkk) x+2( l l l l ))^(( m m m m )2( n n n n ))( o o o o ) (pppp))+2( q q q q ) (rrrr)tan^(( s s s s ) (tttt)-1( u u u u ))( v v v v ) (wwww)(( x x x x ) y+3)/( y y y y )(( z z z z ) x+2)( a a a a a ) (bbbbb)+C (ccccc)` (ddddd) (eeeee) `( f f f f f ) (ggggg)log(( h h h h h ) (iiiii) (jjjjj) (kkkkk)(( l l l l l ) (mmmmm) y+3( n n n n n ))^(( o o o o o )2( p p p p p ))( q q q q q )+( r r r r r ) (sssss)(( t t t t t ) (uuuuu) x+2( v v v v v ))^(( w w w w w )2( x x x x x ))( y y y y y ) (zzzzz))-2( a a a a a a ) (bbbbbb)tan^(( c c c c c c ) (dddddd)-1( e e e e e e ))( f f f f f f ) (gggggg)(( h h h h h h ) y+3)/( i i i i i i )(( j j j j j j ) x+2)( k k k k k k ) (llllll)+C (mmmmmm)` (nnnnnn)

Options

A. `log((y+3)^(2)+(x+2)^(2))+tan^(-1)(y+3)/(y+2)+C`
B. `log((y+3)^(2)+(x-2)^(2))+tan^(-1)(y-3)/(x-2)=C`
C. `log((y+3)^(2)+(x+2)^(2))+2tan^(-1)(y+3)/(x+2)=C`
D. `log((y+3)^(2)+(x+2)^(2))-2tan^(-1)(y+3)/(x+2)=C`

Answer

A

Explanation

The intersection of `y-x+1=0` and `y+x+5=0` is `(-2.-3)`.
Put `x=X-2, y=Y-3`.
The given equation reduces to `(dY)/(dX) = (Y-X)/(Y+X)`
Putting `Y=vX`, we get
`X(dv)/(dX) = -(v^(2)+1)/(v+1)`
or `(-v/(v^(2)+1)-1/(v^(2)+1))dv=(dX)/X`
or `-1/2log(v^(2)+1)-tan^(-1)v=log|X|+` constant
or `log(Y^(2)+X^(2))+2tan^(-1)(y+3)/(x+2)=C`