A function `y=f(x)` satisfies `(x+1)f^(prime)(x)-2(x^2+x)f(x)=(e^x^2)/((x+1)),AAx >-1.` If `f(0)=5,` then `f(x)` is (a) `( b ) (c)(( d ) (e) (f)(( g )3x+5)/( h )(( i ) x+1)( j ) (k) (l))( m ) e^( n ) (o) (p) x^((( q )2( r ))( s ) (t))( u ) (v)` (w) (b) `( x ) (y)(( z ) (aa) (bb)(( c c )6x+5)/( d d )(( e e ) x+1)( f f ) (gg) (hh))( i i ) e^( j j ) (kk) (ll) x^((( m m )2( n n ))( o o ) (pp))( q q ) (rr)` (ss) (c) `( d ) (e)(( f ) (g) (h)(( i )6x+5)/( j )(( k ) x+1)( l ) (m) (n))( o ) e^( p ) (q) (r) x^((( s )2( t ))( u ) (v))( w ) (x)` (y) (d) `( z ) (aa)(( b b ) (cc) (dd)(( e e )5-6x)/( f f )(( g g ) x+1)( h h ) (ii) (jj))( k k ) e^( l l ) (mm) (nn) x^((( o o )2( p p ))( q q ) (rr))( s s ) (tt)` (uu)

Options

A. `((3x+5)/(x+1))e^(x^(2))`
B. `((6x+5)/(x+1))e^(x^(2))`
C. `((6x+5)/(x+1)^(2))e^(x^(2))`
D. `((5-6x)/(x+1))e^(x^(2))`

Answer

B

Explanation

`f^(')(x) - (2x(x+1))/(x+1)f(x)=e^(x^(2))/(x+1)^(2)`
I.F. `=e^(int-2xdx)=e^(-x^(2))`
Thus, the solution is `f(x)e^(-x^(2))=int(dx)/(x+1)^(2)+C`
or `f(x)e^(-x^(2))=int(dx)/(x+1)^(2)+C`
Given, f(0) =5. Thus, C=6
`therefore f(x) = ((6x+5)/(x+1))e^(x^(2))`