If `(x^2+y^2)dy=xydx` and y(1)=1 and `y(x_o)=e`, then `x_o=`

Options

A. `e sqrt(2)`
B. `e sqrt(3)`
C. `e sqrt(5)`
D. `e //sqrt(2)`

Answer

B

Explanation

Given, `(dy)/(dx)=(xy)/(x^(2)+y^(2))`
Put y=vx
`implies(dy)/(dx)=v+x(dv)/(dx)`
`:.v+x(dv)/(dx)=(x^(2)v)/(x^(2)(1+v^(2)))`
`impliesint(1+v^(2))/(v^(3))dv=-int(dx)/(x)`
`implies-(1)/(2v^(2))+logv=-logx+logc`
`implies-(1)/(2).(x^(2))/(y^(2))+logv=-log|y|=logc`
`:'y(1)=1,-(1)/(2)=logc`
`:.-(1)/(2).(x^(2))/(y^(2))+log|y|=-(1)/(2)`
`implieslog_(e)|y|+(1)/(2)=(x^(2))/(2y^(2))`
Again, when `x=x_(0),y=e`
`1+(1)/(2)=(x_(0)^(2))/(2e^(2))`
`impliesx_(0)=sqrt3e`