First order partial differential equation - method of characteristics I am trying to solve following PDE:

\begin{equation}
\frac{\partial g}{\partial t} = (1 - x) \mu \frac{\partial g}{\partial x} - \tau (1 - x) g, \quad g = g(x,t)
\end{equation}
  where 
  \begin{equation}
g(x,0) = x^{n_0}, \quad x(1,t) = 1
\end{equation}

I wanted to solve this by method of characteristics. First, I have parameterized  my variables with parameter $s$ with initial conditions
\begin{equation}
t(0) = 0, \quad x(0) = x_{0}
\end{equation}
from which
\begin{align}
\frac{\partial g}{\partial t} &= \tau (x - 1) g \\ 
\frac{\partial t}{ \partial s} &= 1, \quad \implies t = s \\
\frac{\partial x}{\partial s} &= \mu (x - 1) \implies x - 1 = (x_{0} - 1) e^{\mu s}
\end{align}
From first differential equation (using third)
\begin{equation}
\ln \frac{g(x,t)}{g_{0}} = \frac{\tau}{\mu} (x_{0} - 1) e^{\mu t} \implies g(x,t) = g_{0} \exp \left\{ \frac{\tau}{\mu}(x_{0} - 1) e^{\mu t} \right\}
\end{equation}
From boundary condition $ g(x,0) = x^{n_{0}} $ we have
\begin{equation}
g_{0} = x^{n_{0}} \exp \left\{ - \frac{\tau}{\mu} (x_{0} - 1) \right\}
\end{equation}
Using third differential equation again
\begin{equation}
g_{0} = x^{n_{0}} \exp \left\{ - \frac{\tau}{\mu} (x - 1) e^{- \mu t}  \right\}
\end{equation}
and so I got
\begin{equation}
g(x,t) = x^{n_{0}} \exp \left\{ \frac{\tau}{\mu}(x-1)(1 - e^{-\mu t}) \right\}
\end{equation}
I would be generally satisfied with this solution... but mathematica gives:
\begin{equation}
\left\{\left\{g(x,t)\to \left(1 + (x-1)e^{-\mu t} \right)^{n_0} \exp \left(\frac{\tau}{\mu} (x-1)(1-e^{-\mu t}) \right) \right\}\right\}
\end{equation}
And I have no idea what I did wrong.... Any ideas?
 A: I have found what was wrong with it.
I forgot to consider, that also $ g_0 $ can be $ x_0 $ depended.... Corrected solution is following:
\begin{align}
\frac{\partial g}{\partial t} &= \tau (x - 1) g \\
\frac{\partial t}{ \partial s} &= 1, \quad \implies t = s \\
\frac{\partial x}{\partial s} &= \mu (x - 1) \implies x - 1 = (x_{0} - 1) e^{\mu s}
\end{align}
Using third equation on first one $ (s = t) $
\begin{equation}
\ln \frac{g(x,t)}{g_{0}(x_{0})} = \frac{\tau}{\mu} (x_{0} - 1) e^{\mu t} \implies g(x,t) = g_{0}\left[ (x-1)e^{-\mu t} + 1 \right] \exp \left\{ \frac{\tau}{\mu} (x - 1) \right\}
\end{equation}
where we have used, that constant $ g_{0} $ can be also $ x_{0} $ depended. From boundary condition $ g(x,0) = x^{n_{0}} $ we have
\begin{equation}
g_{0}[x] = x^{n_{0}} \exp \left\{ - \frac{\tau}{\mu} (x - 1) \right\} \implies g_{0}\left[ (x-1)e^{-\mu t} + 1 \right] = ((x-1)e^{-\mu t} + 1)^{n_{0}} \exp \left\{ - \frac{\tau}{\mu} (x-1)e^{-\mu t} \right\}
\end{equation}
and so we have
\begin{equation}
g(x,t) = \left(1 + (x-1)e^{-\mu t} \right)^{n_0} \exp \left\{ \frac{\tau}{\mu} (x-1)(1-e^{-\mu t}) \right\}
\end{equation}
