# Cauchy-Euler's equation

by fluidistic
Tags: cauchyeuler, equation
 PF Patron P: 3,143 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.
 PF Patron P: 864 I put ln(r) into that equation and come up with $$\frac{2r}{r} -\frac{r^2}{r^2} = 1 \neq 0$$
PF Patron
P: 3,143
 Quote by LeonhardEuler 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.

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