1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Homogeneous ODE with variable coefficient

  1. Nov 2, 2008 #1
    Hello everyone!

    I'm trying to solve 2 questions for my assignment on homogeneous ODEs

    I can solve ODE's with variation of parameters and with the method of undetermined coefficients, but these 2 methods seem useless when the coefficients are not constant :

    (x^2)y" - 2xy' -54y = 0


    (x^2)y" - 3xy' + 4y = 0

    I'm not asking anyone to solve it for me, but just to tell me what method I have to use, and I'll find on my book or on the net.

    P.S : I know that you can use power series method, but I think there is another one since we are learning the power series method right now, with a brand new homework just on power series!

    Thanks guys
  2. jcsd
  3. Nov 2, 2008 #2


    User Avatar
    Science Advisor

    Those are "Euler type" or "equi-potential" equation. You can look for a solution of the form "y= xr" just as, with equations with constant coefficients, you try "y= erx". Of course, in either situation, there are other kinds of solutions.

    That works because making the substitution u= ln(x) converts a Euler-type equation to an equation with constant coefficients. For example, letting u= ln x, then dy/dx= (dy/du)(du/dx)= (1/x)dy/du. d2y/dx= (d/dx)((1/x) dy/du)= (-1/x2)dy/dx+ (1/x)((1/x)d2y/du2)= (1/x)2d2y/du2- (1/x2)dy/du.

    So x2y"- 3xy'+ 4y= 0 becomes x2(1/x2)d2/du2- (1/x2)dy/dx)- 3x((1/x)dy/du)+ 4y= d2y/du2- 4 dy/du+ 4y= 0. Solve that differential equation for y as a function of u and then replace u by ln(x).
  4. Nov 2, 2008 #3
    Thank you very much :D

    I'm trying it right now
  5. Nov 2, 2008 #4
    damn it!

    all terms cancel out when I got with

    y(u) = Ae2u+ Bue2u

    I'm going to try again tomorrow.

    thanks again!
  6. Nov 3, 2008 #5


    User Avatar
    Science Advisor

    What?? Yes, of course, "all terms cancel out"! They are supposed to- they make the equation equal to 0! That proves that y(u) satisfies y"- 4y'+4y= 0.

    And with u= ln x, y(x)= Ax2+ B x2 ln(x). Put that into the original equation. "All terms cancel out" is what you want.
  7. Nov 3, 2008 #6
    haha lol thanks :P

    amd another question : we always use u = ln (x) for Euler's type ODE ?
  8. Nov 3, 2008 #7


    User Avatar
    Science Advisor

    Yes, that gives a one to one relation between the set of all Euler-type equations and the set of all equations with constant coefficients. A solution of one automatically gives a solution of the other.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook