The solution of the differential equation `x^2(dy)/(dx)cos1/x-ysin1/x=-1,` where `vec-1` as`xvecoo` is (a) `( b ) (c) y=sin( d )1/( e ) x (f) (g)-cos( h )1/( i ) x (j) (k) (l)` (m) (b) `( n ) (o) y=( p )(( q ) x+1)/( r )(( s ) xsin( t )1/( u ) x (v) (w))( x ) (y) (z)` (aa) (c) `( d ) (e) y=cos( f )1/( g ) x (h) (i)+sin( j )1/( k ) x (l) (m) (n)` (o) (d) `( p ) (q) y=( r )(( s ) x+1)/( t )(( u ) xcos( v )1/( w ) x (x) (y))( z ) (aa) (bb)` (cc)

Options

A. `y=sin(1/x)-cos(1/x)`
B. `y=(x+1)/(xsin(1/x))`
C. `y=cos(1/x)+sin(1/x)`
D. `y=(x+1)/(xcos(1//x))`

Answer

A

Explanation

`x^(2)(dy)/(dx)cos(1/x)-ysin(1/x)=-1`
or `(dy)/(dx) -y/x^(2)tan(1/x)=-sec(1/x)1/x^(2)`(linear)
I.F. `=e^((-int(1/x)^(2)tan(1/x)dx))=sec(1/x)`
Thus, solution is y sec `1/x = -intsec^(2)(1/x)1/x^(2)dx=tan(1/x)+c`
Given `yto -1, x to infty`. Thus, `c=-1`
Hence, equation of curve is `y=sin(1/x)-cos(1/x)`