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Let $F(z,t)=\sum_{n=r}^k z^n p_{r,n}(t)$ with $|z|<1$ a generating function of probability. $F$ satisfies the following PDE: $$ \frac{\partial }{\partial t}F(z,t) + (z-1)(az+b)\frac{\partial }{\partial z}F(z,t) = (z-1)(ak+\frac{br}{z})F(z,t). $$ Solved with the initial condition $F(z,0)=z^n$, the solution of the PDE is $$ F(z,t)=z^r \left \{ \frac{az(1-e^{-(a+b)t)})+ae^{-(a+b)t}+b}{a+b} \right \}^{k-r} $$ I tried to solve using the method of characteristics, obtaining $$ \frac{dt}{1} = \frac{dz}{(z-1)(az+b)}=\frac{dF(z,t)}{(z-1)(ak+\frac{br}{z})F(z,t)} $$ but I don't really know how the solution above comes up. Any idea?

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