# Doubt on a demostration

• Mar 12th 2012, 11:39 AM
Ulysses
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:
$w(f,g,x)=fg'-f'g=ke^{-\int Pdx}$
Then
$g'-\frac{f'}{f}g=k\frac{e^{-\int Pdx}}{f}$
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.
• Mar 13th 2012, 02:39 PM
Ulysses
Re: Doubt on a demostration
Ok, I was committing a silly mistake. Its solved.