What is the general solution to Cauchy-Euler's equation?

[fluidistic]

I had to solve the DE:
2rT'+r^2T''=0 where T(r). I noticed it's a Cauchy-Euler's equation so I proposed a solution of the form T(r)=r^k. This gave me k=0 or k=1.
Thus, I thought, the general solution to that homogeneous DE is under the form T(r)=\frac{c_1}{r}+c_2. Wolfram alpha also agrees on this.
However I noticed that T(r)=c_3 \ln r (or even c_3 \ln r + c_4) also satisfies the DE!
I don't understand:
1)How is that possible?!
2)What is the general way to find such a solution?
3)Isn't the general solution then under the form T(r)=\frac{c_1}{r}+c_2+ c_3 \ln r. I guess not, because some initial conditions would not be enough to solve for the 3 constants?
I don't understand what's going on. Any help is appreciated.

 

[LeonhardEuler]

I put ln(r) into that equation and come up with
\frac{2r}{r} -\frac{r^2}{r^2} = 1 \neq 0

 

[fluidistic]

I put ln(r) into that equation and come up with
\frac{2r}{r} -\frac{r^2}{r^2} = 1 \neq 0

Whoops.
Nevermind then... I made some algebra mistake.
Problem solved.