A question on a step involving the constant of integration in solving this linear differential equation

Question

The following are some steps of solving a differential equation $$\begin{align} & (1-z)^2 g'(z) - 2(1-z) g(z) + 1 = - b g(z)\\ \iff & g'(z) + \left[ -\frac{2}{1-z} + \frac{b}{(1-z)^2}\right] g = - \frac{1}{(1-z)^2}\\ \iff & \frac{d}{dz}\left[ (1-z)^2 e^{bz/(1-z)} g(z) \right] = - e^{bz/(1-z)}\\ \implies & g(z) = \frac{1}{(1-z)^2}e^{-bz/(1-z)}\left[ 1 - \int_0^z e^{bt/(1-t)} dt \right] \\ & = \frac{1}{1-z} - \frac{b}{(1-z)^2} e^{-b/(1-z)}\left[{\mathrm {Ei}}\left(\frac{b}{1-z}\right) - {\mathrm {Ei}}(b)\right] \end{align}$$

This is a result given in one of my answers. I am curious how this came about because the user doesn't include a constant of integration. With the constant we would have $$\frac{1}{1-z} - \frac{b}{(1-z)^2} e^{-b/(1-z)}\left[{\mathrm {Ei}}\left(\frac{b}{1-z}\right) - \frac{c_1}{b}\right]$$ Why did he choose $c_1=b\mathrm{Ei}(b)$?

Answer 1

This was explained in a previous answer which has been deleted : https://math.stackexchange.com/questions/2218708/some-algebra-steps-confusion#comment4564303_2218708 .

In fact, the general solution of the ODE is : $$g(z)=\frac{1}{1-z} - \frac{b}{(1-z)^2} e^{-b/(1-z)}\left[{\mathrm {Ei}}\left(\frac{b}{1-z}\right) - \frac{c_1}{b}\right] \tag 1$$ To determine a unique $g(z)$, a condition must be added. The condition $g(0)=1$ implies $c_1=b\mathrm{Ei}(b)$ and the related particular solution is $$g(z)= \frac{1}{1-z} - \frac{b}{(1-z)^2} e^{-b/(1-z)}\left[{\mathrm {Ei}}\left(\frac{b}{1-z}\right) - {\mathrm {Ei}}(b)\right] \tag 2$$

In another answer Tough Recurrence Relation , in which the method of solving with series is used, it is assumed that : $$(g(z)-1) - 2zg(z) + (z + z^2g(z)) = (1-z)^2 g(z) + (z-1) = -b\int_0^z g(t)dt \tag 3 $$ One observe that this assumption is valid with $z=0$ and $g(0)=1$. Of course this determines a unique solution as explained above. But the counterpart is that the general solution isn't explicitly given.

This explains why a particular solution (2) is obtained instead of the general solution (1).

In fact, $c_1=b\mathrm{Ei}(b)$ was not voluntary and explicitly chosen. It comes from an implicit condition introduced in the calculus (at the line which copy is Eq.3 above).