- #1

fluidistic

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[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.