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ODE: find general solution

  1. Mar 8, 2009 #1
    1. The problem statement, all variables and given/known data

    Find the general solution of the following ODE:
    x[tex]^{2}[/tex]Y''(x) + xY'(x) - CY(x)=0

    where C is a constant.

    2. Relevant equations



    3. The attempt at a solution

    First I want to do this in the case where C=0; this gives:
    x[tex]^{2}[/tex]Y''(x) + xR'(x)=0

    How do i solve this ODE? Any hints? Thank you.
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Mar 8, 2009 #2

    HallsofIvy

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    Please show some work! Do you know anything about "Euler type" equations? (Also called "equipotential" equations.)
     
  4. Mar 8, 2009 #3
    I didn't show any working because i got stuck at that point :)
    I know how to find the general solution of ODEs but not when there are functions of x infront of the Y'' and Y'.
    I don't know about Euler type Equations.

    Thank you
     
  5. Mar 8, 2009 #4

    gabbagabbahey

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    Have you learned the method of power series solutions yet?
     
  6. Mar 8, 2009 #5
    no.
    This question is in the section of partial differential equation.
    I have the two equations:
    xY''(x) + xY'(x) - CY(x)=0
    and
    ZZ''(t)+CZ(t)=0
    I can do the second one since there's no function of t infron of the Z. But the first one...I havent come across something like that.
     
  7. Mar 8, 2009 #6

    gabbagabbahey

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    You MUST have been taught at least one method of solving such ODE's....what methods have you learned?
     
  8. Mar 8, 2009 #7
    the lecturer gave us an example. He found the general seperable solution of Laplace's Equation: V[tex]_{xx}[/tex]+V[tex]_{yy}[/tex]=0

    Where the PDE was converted into the ODEs:
    X''(x)-[tex]\lambda[/tex]X(x)=0
    Y''(y)+[tex]\lambda[/tex]Y(y)=0

    This is simple to solve.

    My original question was: find the general solution of:

    V[tex]_{xx}[/tex]+(1/x)V[tex]_{x}[/tex]+(1/x^2)V[tex]_{tt}[/tex]
    then i said: let V(x,t)=Y(x)Z(t)
    so:
    V[tex]_{xx}[/tex]=Z(t)Y''(x)
    V[tex]_{tt}[/tex]=Y(x)Z''(t)
    I substituted these into the original:
    V[tex]_{xx}[/tex]+(1/x)V[tex]_{x}[/tex]+(1/x^2)V[tex]_{tt}[/tex]
    and divided by Z(t)Y(x) and then multiplied by x^2; to get:
    x^2Y''(x)+xY'(x)-CY(x)=0
    Z''(t)+CZ(t)=0
     
  9. Mar 8, 2009 #8
    sorry, the powers are supposed to be subscripts.
     
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