If `y(x)` satisfies the differential equation `y^(prime)-ytanx=2xs e c x` and `y(0)=0` , then (a) `( b ) (c) y(( d ) (e) (f)pi/( g )4( h ) (i) (j))=( k )(( l ) (m)pi^(( n )2( o ))( p ))/( q )(( r )8sqrt(( s )2( t ))( u ))( v ) (w) (x)` (y) (b) `( z ) (aa) (bb) y^(( c c )prime( d d ))( e e )(( f f ) (gg) (hh)pi/( i i )4( j j ) (kk) (ll))=( m m )(( n n ) (oo)pi^(( p p )2( q q ))( r r ))/( s s )(( t t ) 18)( u u ) (vv) (ww)` (xx) (c) `( d ) (e) y(( f ) (g) (h)pi/( i )3( j ) (k) (l))=( m )(( n ) (o)pi^(( p )2( q ))( r ))/( s )9( t ) (u) (v)` (w) (d) `( x ) (y) (z) y^(( a a )prime( b b ))( c c )(( d d ) (ee) (ff)pi/( g g )3( h h ) (ii) (jj))=( k k )(( l l )4pi)/( m m )3( n n ) (oo)+( p p )(( q q )2( r r )pi^(( s s )2( t t ))( u u ))/( v v )(( w w )3sqrt(( x x )3( y y ))( z z ))( a a a ) (bbb) (ccc)` (ddd)

Options

A. `y(pi/4)=pi^(2)/(8sqrt(2))`
B. `y^(')(pi/4)=pi^(2)/18`
C. `y(pi/3)=pi^(2)/9`
D. `y^(')(pi/3)=(4pi)/3+(2pi^(2))/(3sqrt(3))`

Answer

A, D

Explanation

`(dy)/(dx) -ytanx=2xsecx`
`therefore cosx(dy)/(dx) + (-sinx)y=2x`
`therefore d/(dx)(ycosx)=2x`
Integrating, we get
`y(x)cosx=x^(2)+c`, where c=0 since y(0)=0
When `x=pi/4, y(pi/4)= pi^(2)/(8sqrt(2))`
When `x=pi/3, y(pi/3)=(2pi^(2))/9`
When `x=pi/4, y^(')(pi/4)=pi^(2)/(8sqrt(2))+pi/sqrt(2)`
When `x=pi/3, y^(')(pi/3)=(2pi^(2))/(3sqrt(3))+(4pi)/3`