You mean y'= v+ xv'

First, I would have written y'= v+ xv'= (v^2- 1)/(2v)= v/2- 1/(2v) so that xv'= -1/(2v)- v/2.1/x dx = 1/(f(v)-v) dv

Okay, yes, that is what I got.f(v)-v= (v^2-1)/2v - v

= (-2v^2+v^2-1)/2v

= -(v^2+1)/2v

You mean "ln|v^1+ 1|" don't you?integrating:

int(dx/x) = int(2v/-(v^2+1))

ln|x| + c = -ln|v^2-1|

This is wrong or two reasons. First, it should be "-(v^2+ 1)" on the right side.x + c = -(v^2-1)

More importantly, e^(ln(x)+ C)= e^C e^(ln(x))= cx where c= e^C.

That is, cx= -(v^2+ 1)

Fromx + v^2 -1 = c

v = +-(c-x+1)

y = xv

y = +-x(c-x+1)

..

fairly sure i am correct up until i take the logs out.

The answer book says the answer is

x^2 + y^2 = cx

I am unsure how to get this.