The solution of the differential equation `(x+(x^3)/(3!)+(x^5)/(5!)+)/(1+(x^2)/(2!)+(x^4)/(4!)+)=(dx-dy)/(dx+dy)i s` (a) `( b ) (c)2y( d ) e^(( e ) (f)2x (g))( h )=C( i ) e^(( j ) (k)2x (l))( m )+1( n )` (o) (p) `( q ) (r)2y( s ) e^(( t ) (u)2x (v))( w )=C( x ) e^(( y ) (z)2x (aa))( b b )-1( c c )` (dd) (ee) `( f f ) (gg) y( h h ) e^(( i i ) (jj)2x (kk))( l l )=C( m m ) e^(( n n ) (oo)2x (pp))( q q )+2( r r )` (ss) (d) None of these

Options

A. `2ye^(2x)=Ce^(2x)+1`
B. `2ye^(2x)=Ce^(2x)-1`
C. `ye^(2x)=Ce^(2x)+2`
D. None of these

Answer

B

Explanation

Applying componendo and dividendo, we get
`(dy)/(dx) = e^(-x)/e^(x)=e^(-2x)`
or `2y=-e^(-2x)+C`
or `2ye^(2x)=Ce^(2x)-1`