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Cauchy-Euler's equation

by fluidistic
Tags: cauchyeuler, equation
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fluidistic
#1
Nov11-12, 07:58 PM
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I had to solve the DE:
[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.
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LeonhardEuler
#2
Nov11-12, 08:13 PM
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I put ln(r) into that equation and come up with
[tex]\frac{2r}{r} -\frac{r^2}{r^2} = 1 \neq 0[/tex]
fluidistic
#3
Nov11-12, 08:30 PM
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Quote Quote by LeonhardEuler View Post
I put ln(r) into that equation and come up with
[tex]\frac{2r}{r} -\frac{r^2}{r^2} = 1 \neq 0[/tex]
Whoops.
Nevermind then... I made some algebra mistake.
Problem solved.


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