# Cauchy-Euler's equation

1. Nov 11, 2012

### 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.

2. Nov 11, 2012

### LeonhardEuler

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

3. Nov 11, 2012

### fluidistic

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