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Math Help - Solving a DE

  1. #1
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    Solving a DE

    \psi'(y)+y^2\psi'(y)-\lambda\psi(y)=0

    How can I approach this?

    The Cauchy Euler method didn't work.
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  2. #2
    A Plied Mathematician
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    It's separable.
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  3. #3
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    Quote Originally Posted by dwsmith View Post
    \psi'(y)+y^2\psi'(y)-\lambda\psi(y)=0

    How can I approach this?

    The Cauchy Euler method didn't work.
    What you have posted can be re-written as \psi'(y) (1+y^2) - \lambda\psi(y)=0 and is separable.
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  4. #4
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    Quote Originally Posted by mr fantastic View Post
    What you have posted can be re-written as \psi'(y) (1+y^2) - \lambda\psi(y)=0 and is separable.
    It's also first order linear if written as

    \displaystyle \psi'(y) - \frac{\lambda}{1+y^2}\psi(y) = 0.
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  5. #5
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by dwsmith View Post
    [snip]

    The Cauchy Euler method didn't work.

    [/snip]
    FYI: Cauchy Euler Equations are second order equations, not first order equations. Thus, the technique involved in solving C-E equations is not applicable.
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  6. #6
    A Plied Mathematician
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    ...Cauchy Euler Equations are second order equations...
    Really? I thought you could have a CE equation of any order. You just needed to have the appropriate power of x multiplying the corresponding order of differentiation. The wiki certainly seems to indicate that. There's nothing preventing a CE equation from being third order, or fourth. Why not first-order?
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  7. #7
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by Ackbeet View Post
    Really? I thought you could have a CE equation of any order. You just needed to have the appropriate power of x multiplying the corresponding order of differentiation. The wiki certainly seems to indicate that. There's nothing preventing a CE equation from being third order, or fourth. Why not first-order?
    Ah..I never really thought of that. I was only familiar with them as second order equations.
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