How to find the solution of ODE xy'+y=0. I got this so far... --> 10 points for best answer...

x*(dv/dx) = -2v, using substitution, v = y/x. Any ideas on how to work this? I'm confused. Thank you.

This is exact:
x dy/dx + y = (d/dx)(xy) = 0.

Integrating both sides yields xy = C.
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Alternately, using the substitution v = y/x:
y = xv ==> dy/dx = v + x dv/dx.

So, x dy/dx + y = 0 becomes
x(v + x dv/dx) + xv = 0
==> 2v + x dv/dx = 0

Separate variables:
dv/v = -2 dx/x

Integrate both sides:
ln v = -2 ln x + C'
==> ln (y/x) = -2 ln x + C'
==> ln (y/x) + 2 ln x = C'
==> ln((y/x) * x^2) = C'
==> xy = C, where C' = e^C.

I hope this helps!