The solution of the primitive integral equation `(x^2+y^2)dy=x ydx` is `y=y(x)dot` If `y(1)=1` and `y(x_0)=e ,` then `x_0` is (a) `( b ) (c)2sqrt(( d ) (e)(( f ) (g) (h) e^(( i )2( j ))( k )-1( l ))( m ))( n ) (o)` (p) (b) `( q ) (r)2sqrt(( s ) (t)(( u ) (v) (w) e^(( x )2( y ))( z )+1( a a ))( b b ))( c c ) (dd)` (ee) (c) `( d ) (e)sqrt(( f )3( g ))( h )e (i)` (j) (d) `( k ) (l)sqrt(( m ) (n) (o)(( p ) (q) e^(( r )2( s ))( t )+1)/( u )2( v ) (w) (x))( y ) (z)` (aa)

Options

A. `sqrt(2 (e^(2) - 1))`
B. `sqrt(2(e^(2) + 1))`
C. `sqrt3 e`
D. `sqrt((1)/(2) (e^(2) + 1))`

Answer

C

Explanation

We have
`(dy)/(dx) = (xy)/(x^(2) + y^(2))`
put y = vx
`implies (dy)/(dx) = v .1 + x (dv)/(dx)`
`v + x (dv)/(dx) = (v)/(1 +v^(2))`
`x(dv)/(dx) = (v)/(1+ v^(2))-v = (v -v-v^(3))/(1+v^(2))`
`(1+v^ (2))/(v^(3))dv = -(dx)/(x)`
`int (v^(-3) + (1)/(v))dv - int(dx)/(x)`
`(v^(-2))/(-2) + "log v " = - "log" x + C `
`-(1)/(2((y)/(x))^(2))+ "log" ((y)/(x)) = - "log"x + C `
`-(x^(2))/(2y^(2)) + "log y" = C " " ....(1)`
y(1) = 1 at x = 1 , y = 1
C - 1/2
put in (1)
`-(x^(2))/(2y^(2)) + "log y" = -(1)/(2) `
`y(x_(0)) = e`,
at x = `x_(0) , y = e`
`-(x_(0)^(2))/(2e^(2)) + 1= (-1)/(2)`
`x_(0) = sqrt3 e `