Need help solving the ODF $ f''(r) + \frac{1}{r}f'(r) = 0 $ I am currently taking complex analysis, and this homework question has a part that requires the solution to a differential equation. I took ODE over 4 years ago, so my skills are very rusty. The equation I derived is this:
$$ f''(r) + \frac{1}{r}f'(r) = 0 $$
I made this substitution: $ v(r) = f'(r) $ to get get:
$$ v'(r) + \frac{1}{r}v(r) = 0 \implies \frac{dv}{dr} = - \frac{v}{r} \implies \frac{dv}{v} = - \frac{dr}{r} $$
Unsure of what to do now. 
Edit: I forgot to add that the answer is $ a \log r + b $

You now have:
1/v dv = -1/r dr
We can integrate this:
ln(v) = -ln(r) + C1
v = e^(-ln(r) + c1)
v = 1/r + c1
Thus we have:
df/dr = 1/r + c1 (since v = f'(r) = df/dr)
df = (1/r + c1) dr
f = ln(r) + c1(r) + c2
And thats our answer!
