I had to solve the DE:(adsbygoogle = window.adsbygoogle || []).push({});

[tex]2rT'+r^2T''=0[/tex] where [itex]T(r)[/itex]. I noticed it's a Cauchy-Euler's equation so I proposed a solution of the form [itex]T(r)=r^k[/itex]. This gave me k=0 or k=1.

Thus, I thought, the general solution to that homogeneous DE is under the form [itex]T(r)=\frac{c_1}{r}+c_2[/itex]. Wolfram alpha also agrees on this.

However I noticed that [itex]T(r)=c_3 \ln r[/itex] (or even [itex]c_3 \ln r + c_4[/itex]) 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 [itex]T(r)=\frac{c_1}{r}+c_2+ c_3 \ln r[/itex]. 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.

**Physics Forums | Science Articles, Homework Help, Discussion**

Dismiss Notice

Join Physics Forums Today!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Cauchy-Euler's equation

**Physics Forums | Science Articles, Homework Help, Discussion**