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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

    and

    (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

    HallsofIvy

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    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

    HallsofIvy

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    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

    HallsofIvy

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    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.
     
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