A question on a step involving the constant of integration in solving this linear differential equation 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 $c_1=b\mathrm{Ei}(b)$?
 A: 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).
