The general solution of the differential equation, `y^(prime)+yvarphi^(prime)(x)=0` , where `varphi(x)` is a known function, is (a) `( b ) (c) y=c (d) e^(( e ) (f)-varphi(( g ) x (h))( i ))( j )+varphi(( k ) x (l))-1( m )` (n) (o) `( p ) (q) y=c (r) e^(( s ) (t)+varphi(( u ) x (v))( w ))( x )+varphi(( y ) x (z))-1( a a )` (bb) (cc) `( d d ) (ee) y=c (ff) e^(( g g ) (hh)-varphi(( i i ) x (jj))( k k ))( l l )-varphi(( m m ) x (nn))+1( o o )` (pp) (d) `( q q ) (rr) y=c (ss) e^(( t t ) (uu)-varphi(( v v ) x (ww))( x x ))( y y )+varphi(( z z ) x (aaa))+1( b b b )` (ccc)

Options

A. `y=ce^(-phi(x))+phi(x)-1`
B. `y=ce^(+phi(x))+phi(x)-1`
C. `y=ce^(-phi(x))-phi(x)+1`
D. `y=ce^(-phi(x))+phi(x)+1`

Answer

A

Explanation

`therefore (dy)/(dx) +y/(xlog_(e)x)=2/x`
`therefore` I.F. `=e^(phi(x))phi(x)phi^(')(x)dx`
`=inte^(t)tdt`, where `phi(x)=t`
`=te^(t)-e^(t)+c`
`=phi(x)e^(phi(x))+phi(x)-1`