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

$\displaystyle y''+Py'+Qy=0$ (1)

A linear independent solution is given by:

$\displaystyle g=f \int \frac{e^{\int Pdx}}{f^2}dx$

We used abel's theorem, and that for linear differential equations $\displaystyle y'+Py=Q$ the solution is given by

$\displaystyle 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:

$\displaystyle w(f,g,x)=fg'-f'g=ke^{-\int Pdx}$

Then

$\displaystyle g'-\frac{f'}{f}g=k\frac{e^{-\int Pdx}}{f}$

Because of (2) we have:

$\displaystyle 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:

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

I know that: $\displaystyle e^{-\int \frac{f'}{f}dx}$ is a constant, and then it can goes out of the integral, but what bothers me is thatfthat 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:

$\displaystyle 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:

$\displaystyle \int \frac{k_1 e^{-\int Pdx}}{f}dx=f \left [\int \frac{k_1 e^{-\int Pdx}}{f} \frac{1}{f}dx \right ]$

And I think thats wrong, I don't think the equality is preserved by multiplyingfoutside 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.