
#1
Nov1112, 07:58 PM

PF Gold
P: 3,173

I had to solve the DE:
[tex]2rT'+r^2T''=0[/tex] where [itex]T(r)[/itex]. I noticed it's a CauchyEuler'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. 



#2
Nov1112, 08:13 PM

PF Gold
P: 864

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


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