Solve the following differential equation: `\ x(dy)/(dx)-y=2\ sqrt(y^2-x^2)`

Answer

`1/2log_(e)y+sqrt(y^(2)-x^(2))/(x)=log_(e)cx`

Explanation

Putting `y=vx` and `(dy)/(dx)=v+x(dv)/(dx)`, we get
`xv+x^(2)(dv)/(dx)=vx+2xsqrt(v^(2)-1)`
or `int(dv)/(2sqrt(v^(2)-1))=int(dx)/x`,
Integrating, we get
`1/2" ln "(v+sqrt(v^(2)-1))="ln "cx`
or `1/2"ln "(y+sqrt(y^(2)-x^(2)))/x="ln "cx`