The function y=f(x) is the solution o...

The function `y=f(x)` is the solution of the differential equation `(dy)/(dx)+(x y)/(x^2-1)=(x^4+2x)/(sqrt(1-x^2))` in `(-1,1)` satisfying `f(0)=0.` Then `int_((sqrt(3))/2)^((sqrt(3))/2)f(x)dx` is (a) `( b ) (c) (d)pi/( e )3( f ) (g)-( h )(( i )sqrt(( j )3( k ))( l ))/( m )2( n ) (o) (p)` (q) (b) `( r ) (s) (t)pi/( u )3( v ) (w)-( x )(( y )sqrt(( z )3( a a ))( b b ))/( c c )4( d d ) (ee) (ff)` (gg) (c) `( d ) (e) (f)pi/( g )6( h ) (i)-( j )(( k )sqrt(( l )3( m ))( n ))/( o )4( p ) (q) (r)` (s) (d) `( t ) (u) (v)pi/( w )6( x ) (y)-( z )(( a a )sqrt(( b b )3( c c ))( d d ))/( e e )2( f f ) (gg) (hh)` (ii)

`(pi)/(3)-(sqrt3)/(2)`

`(pi)/(3)-(sqrt3)/(4)`

`(pi)/(6)-(sqrt3)/(4)`

`(pi)/(6)-(sqrt3)/(2)`

Text Solution

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The correct Answer is:

We have,
`(dy)/(dx)+(xy)/(x^(2)-1)=(x^(4)+2x)/(sqrt(1-x^(2)))" ...(i)"`
This is a linear differential equation with
`"I.F."=e^(int(x)/(x^(2)-1)dx)=e^((1)/(2)log|x^(2)-1|)=e^((1)/(2)log(1-x^(2)))=sqrt(1-x^(2))`
Multiplying both sides of (i) by I.F. and integrating with respect to x, we obtain
`ysqrt(1-x^(2))=int(x^(4)+2x)dx+C`
or, `ysqrt(1-x^(2))=(x^(5))/(5)+x^(2)+C" ...(ii)"`
Putting x = 0 and y = f (0) = 0 in (ii), we get C = 0
Putting C = 0 in (ii), we get
`ysqrt(1-x^(2))=(x^(5))/(5)+x^(2)`
`rArr" "f(x)-(x^(5))/(5sqrt(1-x^(2)))+(x^(2))/(sqrt(1-x^(2)))`
`therefore" "int_(-sqrt3//2)^(sqrt3//2)f(x)dx=int_(-sqrt3//2)^(sqrt3//2)(x^(5))/(sqrt(1-x^(2)))dx+int_(-sqrt3//2)^(sqrt3//2)(x^(2))/(sqrt(1-x^(2)))dx`
`=2int_(0)^(sqrt3//2)(x^(2))/(sqrt(1-x^(2)))dx`
`=2int_(0)^(pi//3)sin^(2)d theta," where x "=sin theta`
`=int_(0)^(pi//3)(1-cos2 theta)d theta`
`[theta-(1)/(2)sin2 theta]_(0)^(pi//3)=(pi)/(3)-(sqrt3)/(4)`

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