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

Transformation of a Cauchy-Euler equation

  1. Mar 26, 2013 #1
    Can anyone explain to me how I would go about transforming a Cauchy-Euler equation for an equation such as:
    x2y'' - xy' = ln x

    I know you have to start with x = et or t = ln x however I'm not sure what to do next...
     
  2. jcsd
  3. Mar 26, 2013 #2
    Well making the substitution [tex]x=e^t[/tex] does indeed get the solution, you should get via chain rule:

    [tex]\frac{d}{dx}=e^{-t}\frac{d}{dt}, \frac{d^2}{dx^2}=e^{-2t}(\frac{d^2}{dt^2}-\frac{d}{dt})[/tex]

    which when substituted into your original equation should yield an inhomogeneous 2nd order ODE with constant coefficients which we can do via the standard method of finding complementary solutions and then a particular solution.
     
  4. Mar 26, 2013 #3
    With z(x)=y'(x), the ODE is of the first order :
    x²z'+xz=ln(x)
    Solving x²z'-xz=0 leads to z=c*x where c=constant
    Then bing back z=u(x)*x into x²z'-xz=ln(x)
    which leads to u' =ln(x)/(x^3)
    u(x)=-(2*ln(x)+1)/(4*x²)+C where C=constant
    y'=z(x)=u(x)*x=-(2ln(x)+1)/(4*x)+C*x
    y= -(ln(x)+1)*ln(x)/4 +A*x² +B where A, B are constants.
     
  5. Mar 26, 2013 #4

    Mark44

    Staff: Mentor

Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Transformation of a Cauchy-Euler equation
  1. Cauchy-Euler Equation (Replies: 9)

Loading...