Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Cauchy-Euler's equation

  1. Nov 11, 2012 #1

    fluidistic

    User Avatar
    Gold Member

    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.
     
  2. jcsd
  3. Nov 11, 2012 #2

    LeonhardEuler

    User Avatar
    Gold Member

    I put ln(r) into that equation and come up with
    [tex]\frac{2r}{r} -\frac{r^2}{r^2} = 1 \neq 0[/tex]
     
  4. Nov 11, 2012 #3

    fluidistic

    User Avatar
    Gold Member

    Whoops. :blushing:
    Nevermind then... I made some algebra mistake.
    Problem solved.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Cauchy-Euler's equation
  1. Cauchy-Euler Equation (Replies: 9)

Loading...