Solve `y((dy)/(dx))^(2)`

Explanation

`y((dy)/(dx))^(2)+2x(dy)/(dx)-y=0`
Solving quadratic in `(dy)/(dx)`, we get
`(dy)/(dx) = (-2x+-sqrt(4x^(2)+4y^(2)))/(2y)`
`=(-x+_sqrt(x^(2)+y^(2)))/(y)` which is homogenous
Put y=vx, i.e., `(dy)/(dx)=v+x(dv)/(dx)`
Then given equation transforms to
`v+x(dv)/(dx)=(-1+-sqrt(1+v^(2)))/(v)`
or `v^(2)+xv(dv)/(dx) -1+-sqrt(1+v^(2))`
or `(v^(2)+1)+sqrt(1+v^(2))=-xv(dv)/(dx)`
or `int(vdx)/((1+v^(2))+-sqrt(v^(2)+1))=-int(dx)/x`
`or int(vdv)/(sqrt(v^(2)+1)(sqrt(1+v^(2)+-1)))= -int(dx)/(x)`................(1)
Put `sqrt(1+v^(2))+-1` i.e., `v/sqrt(1+v^(2))dv=dt`
Then, equation (1) transforms to `int(dt)/(t) = -int(dx)/(x)`
or `"ln"t="ln"x+"ln"c`
or `tx=c`
or `(sqrt(1+v^(2))+-1)x=c`
Given when `x=0, y=sqrt(5)`
or `[sqrt(5)-0]=c` or `c=sqrt(5)`
`therefore x^(2)+y^(2)=5+x^(2)+-2sqrt(5)x`
or `y^(2)=5+-2sqrt(5)x`