I am trying to solve the following PDE using the method of characteristics,
$$u_{t} = (1-s)\mu u_{s} + (s-1)\alpha u,$$ $$u(s,0) = s^i$$
We can reduce the PDE to the following set of ODEs along the characteristic curves,
$$\frac{dt}{1} = \frac{-ds}{\mu(1-s)} = \frac {du}{(s-1)\alpha u}.$$
On solving two ODEs we get,
$$t\mu - \ln|1-s| = A,$$ $$(s-1) \alpha t - \ln|u| = B .$$
So, the general solution is,
$$f(t\mu - \ln|1-s|, (s-1) \alpha t - \ln|u|) = 0.$$
I also know that after plugging in the initial condition we get $-\ln (s^{i}) = B$, $-\ln(1-s) = A$ which gets us,
$$-\ln[(1-e^{-A})^{i}] = B .$$
How do I get a specific solution from here?
Any help will be appreciated.
This is a first-order linear PDE with variable coefficients. The characteristic families are $$ \mu t - \ln (1-s) = A, \qquad\text{or}\qquad (1-s)e^{-\mu t} = e^{-A} = A' $$ $$ \tfrac\alpha\mu s - \ln u = B, \qquad\text{or}\qquad u\, e^{-\alpha s/\mu} = e^{-B} = B' $$ where composition with the exponential has been used twice (2nd integration in OP is incorrect). Introducing an explicit dependence $B' = F(A')$ yields the general solution $$ u = e^{\alpha s/\mu} F\big((1-s)e^{-\mu t}\big) \, , $$ where $F$ is an arbitrary function. At $t=0$, we have $u = e^{\alpha s/\mu} F(1-s)$ and $u = s^i$. Setting $r = 1-s$, we find $F(r) = e^{\alpha (r-1)/\mu} (1-r)^i$ by using the boundary condition. Thus, $$ u(s,t) = e^{\alpha (1-s)(e^{-\mu t}-1)/\mu} \big(1-(1-s)e^{-\mu t}\big)^i . $$ We can compute the partial derivatives to verify the solution (and it works!).