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Thread: Doubt on a demostration

  1. #1
    May 2010

    Doubt on a demonstration

    Hi there. I was seeing the proofs on my class on differential equations from last year. And I have some doubts about this demonstration.
    We demonstrate that given a solution f(x) for
    y''+Py'+Qy=0 (1)
    A linear independent solution is given by:
    g=f \int \frac{e^{\int Pdx}}{f^2}dx

    We used abel's theorem, and that for linear differential equations y'+Py=Q the solution is given by
    y=e^{-\int Pdx}\left [\int Q e^{\int Pdx}dx+k\right ] (2)
    The demonstration goes as follows:

    If g is solution of (1) because of abel's theorem:
    w(f,g,x)=fg'-f'g=ke^{-\int Pdx}
    g'-\frac{f'}{f}g=k\frac{e^{-\int Pdx}}{f}
    Because of (2) we have:
    g=e^{\int \frac{f'}{f}dx} \left [\int \frac{k_1 e^{-\int Pdx}}{f}e^{-\int \frac{f'}{f}dx}dx+k \right ]

    The problem comes with the next step. This is what follows in my notebook:

    g=f \left [\int \frac{k_1 e^{-\int Pdx}}{f} \frac{1}{f}dx +k \right ]

    I know that: e^{-\int \frac{f'}{f}dx} is a constant, and then it can goes out of the integral, but what bothers me is that f that is multiplied outside of the integrand, and divided inside of the inegrand, like multiplying by 1 (that's what I think the professor did, but I might be wrong). I think that can't be done by preserving the equality. Instead of it, considering k=0 and k1=1, we get the result that we were looking for:
    g=f \left [\int \frac{k_1 e^{-\int Pdx}}{f^2}\right ]

    I don't know if it's clear what I don't get. This is what I think that the professor did:

    \int \frac{k_1 e^{-\int Pdx}}{f}dx=f \left [\int \frac{k_1 e^{-\int Pdx}}{f} \frac{1}{f}dx \right ]
    And I think thats wrong, I don't think the equality is preserved by multiplying f outside the integrand and dividing inside of it. Perhaps I didn't copied something, I don't think my professor did a mistake really, perhaps there is some middle step that I've lost, I don't know. Anyway, I need to know the demonstration.
    Last edited by Ulysses; Mar 12th 2012 at 05:37 PM.
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  2. #2
    May 2010

    Re: Doubt on a demostration

    Ok, I was committing a silly mistake. Its solved.
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