I've got this question on special functions take-home exam. I will add some literature in the nearest time.
Let us have an equation $xy''+(c-x)y'-ay=0$ which is a confluent hypergeometric function equation. It's an ODE
$f = F_{1,1}(a;c;x)$ and $g = x^{1-c}F_{1,1}(a+1-c;2-c;x)$ are the solutions of this equation on a subset of a complex plane. This analytical decomposition is constructed around 0.
$F_{1,1}(a;c;x) = \sum_{n=0}^\infty \frac{(a)_n}{(b)_n n!}x^n$
I try to calculate $W = fg'-f'g$?
Assuming $W \neq 0$
$\partial_x(W) = f'g'+fg''-f'g'-f''g = f \cdot \frac{ ag -(c-x)g'}{x} - \frac{af -(c-x)f'}{x} \cdot{g} = W \cdot \frac{x-c}{x}$
Now I need to solve $\frac{W'}{W} = 1- \frac{c}{x}$
$(\ln W)'= -c\int_1^x \frac{1}{s}ds + x + C_1$
$W = C_2 \cdot x^{-c} e^x$
Could you help me with finding $C_2$?
UPD: As written at 4.2 of Aske, Roy, Andrews, special functions. Applying Mellin transform and its inverse we get an integral
$\Large{F_{1,1}(a;c;x) = \frac{\Gamma(c)}{\Gamma(a)2\pi i} \int_{-i \infty}^{i \infty} \frac{\Gamma(a+s)\Gamma(-s)}{\Gamma(c+s)} (-x)^s ds }$
Which is well-defined for all complex numbers $Re z < 0$ except $0$ and negative integers and contour divides poles of enumerator and denominator.
I can's substitute $x$ for $-1$ so I'm stuck there.
UPD 2: DLMF https://dlmf.nist.gov/13.2.E33 states that
$\Large{W(f,g) =\frac{sin(\pi c)\Gamma(c)\Gamma(2-c) e^x x^{-c}}{\pi}}$ and by sine product formula
$\Large{W(f,g) = \frac{\Gamma(c)\Gamma(2-c) e^x x^{-c}}{\Gamma(c) \Gamma(1-c)} = (1-c) x^{-c} e^x} $
Let's make an asymptotic expansion at 0: $f = 1 + \frac{a}{c}x+ \frac{a(a+1)}{2c(c+1)}x^2 + o(x^2)$, $g = x^{1-c}(1 + \frac{a+ 1-c }{2-c}x + o(x))$
$f'g = x^{1-c}(\frac{a}{c}+ \frac{a(a+1)}{c(c+1)}x + o(x) )(1 + \frac{a+ 1-c }{2-c}x+ o(x))$,
$fg'= (1-c)x^{-c}(1 + \frac{a}{c}x + o(x))(1 + \frac{a+ 1-c }{2-c}x + o(x)) + \\ x^{1-c}(1 + \frac{a}{c}x + o(x))(\frac{a+1-c}{2-c} + \frac{a+1-c(a+2-c)}{2-c(1-c)}x + o(x))$
But I don't see how this can help. That would be great to know that the least powers of $x$ win near $0$, and the least one is $x^{-c}$ and the answer is $1-c$ but why is it legal to do that?
So, we see the Taylor series, and we need its part at $x^{-c}$ because our solution is of the form $C x^{-c} e^x$. Its $1-c$.
Answer: $(1-c) x^{-c} e^x$