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Homework Help: Differential equation x y' +x^2 y'' = k^2 y

  1. Feb 17, 2008 #1
    [SOLVED] differential equation

    1. The problem statement, all variables and given/known data
    x y' +x^2 y'' = k^2 y

    where y=y(x), k is constant.

    How do you prove that x^r, where r is a real number form a basis for that differential equation? They are obviously linearly independent. But how do you prove that they span the solution space?

    2. Relevant equations

    3. The attempt at a solution
  2. jcsd
  3. Feb 17, 2008 #2
    By solving it! :smile:
    Try the substitution [itex]y(x)=u(x)\,x^k[/itex]
  4. Feb 17, 2008 #3
    i didn't get what exactly you want to prove! do you want to solve it for x?
  5. Feb 17, 2008 #4
    When I do that I get (2k+1)u'(x)+x u''(x)=0. Is that a contradiction?
  6. Feb 17, 2008 #5
    No it isn't! Now let [itex]u'(x)=a(x),\,u''(x)=a'(x)[/itex] which yields a separable 1st order ODE.
  7. Feb 17, 2008 #6
    I want to find out if the set of solutions {x^r}, [tex]r \in \mathbb{R}[/tex] spans the entire solution space of that equation.
  8. Feb 17, 2008 #7
    Then I get [tex]a(x)=\frac{C}{x^{2k+1}}[/tex], where C is a constant. Very nice Rainbow Child!
  9. Feb 17, 2008 #8
    oh... i see, thank you
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