pasmith said:
You haven't checked that -M(x,a,1) \ln x - 3\frac{\partial M}{\partial b} actually solves the ODE; it doesn't. If L(y) = xy'' + (b-x)y' - ay then differentiating partially with respect to b results in \frac{\partial}{\partial b}L(y) = L\left(\frac{\partial y}{\partial b}\right) + y' and you can also show that L(y \ln x) = L(y) \ln x + 2y' - y + \frac{(b-1)y}{x}. Since L(M) = 0 for all a and b you can see that L\left(M \ln x + 3 \frac{\partial M}{\partial b}\right) = -M' - M when b = 1.
The problem is that you haven't calculated the derivative of M(x,a-b+1, 2-b) with respect to b correctly. By \frac{\partial M}{\partial b} we mean the derivative of M(x,a,b) with respect to b with x and a constant. But you are looking at f(x,a,b) = M(x, g(a,b),h(a,b)) with g(a,b) = a-b+1 and h(a,b) = 2 - b rather than M(x,a,b). The multivariate chain rule then gives \frac{\partial f}{\partial b} = -\frac{\partial M}{\partial a} - \frac{\partial M}{\partial b}, and we must evaluate the derivatives of M at (x, a-b+1, 2-b) rather than at (x,a,b). In the limit, of course, these are both (x,a,1).
Thanks
@pasmith ! I found the error according to your correction, could you please check now? Especially if the argument on linear independence of the two solutions is valid?Expand the limit with l'Hopital's rule:
$$
\begin{aligned}
&\phantom{{}={}}\lim_{b\to 1}\frac{y_2-y_1}{b-1} \\
&= \biggl[\frac{\partial}{\partial b}(x^{1-b}M(x,a-b+1,2-b) - M(x,a,b))\biggr]_{b=1} \\
&= \biggl[-x^{1-b}\ln x M(x,a-b+1,2-b) + x^{1-b}\frac{\partial M(x,a-b+1,2-b)}{\partial b} - \frac{\partial M(x,a,b)}{\partial b}\biggr]_{b=1} \\
&= -M(x,a,1)\ln x + \biggl[\frac{\partial M(x,a-b+1,2-b)}{\partial b} - \frac{\partial M(x,a,b)}{\partial b}\biggr]_{b=1} \\
&= -M(x,a,1)\ln x + \biggl[-\frac{\partial M(x,a,b)}{\partial a} -\frac{\partial M(x,a,b)}{\partial b} - \frac{\partial M(x,a,b)}{\partial b}\biggr]_{b=1} \\
&= -M(x,a,1)\ln x - \frac{\partial M}{\partial a}(x,a,1) - 2\frac{\partial M}{\partial b}(x,a,1) \\
\end{aligned}
$$
Consider the differential operator ##L(y) = xy'' + (b-x)y' -ay##, then consider ##L(y\ln x)##, ##\frac{\partial}{\partial b}L(y)## and ##\frac{\partial}{\partial a}L(y)##:
$$
\begin{aligned}
&\phantom{{}={}} L(y\ln x) \\
&= x(y\ln x)'' + (b-x)(y\ln x)' - a(y\ln x) \\
&= xy''\ln x +2y' - y\frac{1}{x} + (b-x)y' \ln x + (b-x)y\frac{1}{x} - ay\ln x \\
&= \ln x(xy'' + (b-x)y' -ay) + 2y' -y + (b-1)y\frac{1}{x} \\
&= \ln x L(y) + 2y' -y + (b-1)y\frac{1}{x} \\
\end{aligned}
$$
$$
\begin{aligned}
&\phantom{{}={}} \frac{\partial}{\partial b}L(y) \\
&= \frac{\partial}{\partial b}(xy'' + (b-x)y' - ay) \\
&= x\frac{\partial }{\partial b}y'' + y' +(b-x)\frac{\partial}{\partial b}y' - a\frac{\partial}{\partial b}y \\
&= x(\frac{\partial y}{\partial b})'' + (b-x) (\frac{\partial y}{\partial b})' - a\frac{\partial y}{\partial b} + y' \\
&= L(\frac{\partial y}{\partial b}) + y'
\end{aligned}
$$
$$
\begin{aligned}
&\phantom{{}={}} \frac{\partial}{\partial a}L(y) \\
&= \frac{\partial}{\partial a}(xy'' + (b-x)y' - ay) \\
&= x\frac{\partial }{\partial a}y'' +(b-x)\frac{\partial}{\partial b}y' - y -a\frac{\partial}{\partial a}y \\
&= x(\frac{\partial y}{\partial a})'' + (b-x) (\frac{\partial y}{\partial a})' - a\frac{\partial y}{\partial a} -y \\
&= L(\frac{\partial y}{\partial a}) - y
\end{aligned}
$$
Check ##y_2 = -M(x,a,1)\ln x - \frac{\partial M}{\partial a}(x,a,1) - 2\frac{\partial M}{\partial b}(x,a,1)## is indeed a solution to the differential equation:
$$
\begin{aligned}
&\phantom{{}={}} L(y_2) \\
&= -L(M \ln x) - L(\frac{\partial M}{\partial a}) - 2L(\frac{\partial M}{\partial b}) \\
&= -\ln xL(M) - 2M' +M -\frac{\partial}{\partial b}L(M) - M -2\frac{\partial}{\partial a}L(M) + 2M' \\
&= 0
\end{aligned}
$$
Check the Wronskian of ##y_1 = M(x,a,1)## and ##y_2 = -M(x,a,1)\ln x - \frac{\partial M}{\partial a}(x,a,1) - 2\frac{\partial M}{\partial b}(x,a,1)##:
$$
\begin{aligned}
&\phantom{{}={}} y_1y_2'-y_2y_1' \\
&= M(-M'\ln x - M\frac{1}{x} - (\frac{\partial M}{\partial a})' -2(\frac{\partial M}{\partial b})') - M'(-M\ln x -\frac{\partial M}{\partial a} -2\frac{\partial M}{\partial b}) \\
&= -M^2 \frac{1}{x} -M(\frac{\partial M}{\partial a})' + M'\frac{\partial M}{\partial a} -2M(\frac{\partial M}{\partial a})' + 2M'\frac{\partial M}{\partial b}
\end{aligned}
$$
When ##x \to 0##, the Wronskian diverges because of the ##\frac{1}{x}## term, so it's non-zero. Hence ##y_1, y_2## are independent.