How do I solve Kolmogorov differential equations for linear process of birth using generating function?
By linear process of birth I mean Poisson process for which
$$q_{i,i+1}=i\lambda\\ q_{i,i}=-i\lambda\\ q_{i,j}=0 $$
System of KDE is given as $$p_k'(t)=-k\lambda p_k(t)+(k-1)\lambda p_{k-1}(t)$$ with initial value $p_1(0)=1$.
Using the generating function $$G(x,t)=\sum_{k=0}^Np_kx^k$$
we get the time derivative
$$\frac{\partial G(x,t)}{\partial t}=-\sum_{k=0}^Nk\lambda p_kx^k+\sum_{k=1}^N(k-1)\lambda p_{k-1}x^k=\\=-x\lambda\sum_{k=0}^Nk p_kx^{k-1}+x^2\lambda \sum_{k=1}^N(k-1)p_{k-1}x^{k-2} =\lambda(x^2-x)\frac{\partial G(x,t)}{\partial x}$$
hence $$-\frac{dx}{\lambda(x^2-x)}=dt$$ $$x=\frac{e^{\lambda t}}{e^{\lambda t}-C}$$ $$C=e^{\lambda t}\left(1-\frac{1}{x}\right)$$
Now using the initial value
$$G(x,t)=f\left[e^{\lambda t}\left(1-\frac{1}{x}\right)\right]$$ $$G(x,0)=f\left(1-\frac{1}{x}\right)=x\qquad \Rightarrow \qquad f\left(y\right)=\frac{1}{1-y}$$
Which gives us
$$G(x,t)=\frac{1}{1-e^{\lambda t}\left(1-\frac{1}{x}\right)}=\frac{1}{1-e^{\lambda t}+\frac{e^{\lambda t}}{x}}$$
However this doesn't seem to come close to giving the correct solution for absolute probability which is
$$p_k(t)=e^{-\lambda t}\left(1-e^{-\lambda t}\right)^{k-1}$$.
Have I made a mistake in my calculations or is there actually a way to get to this solution with my result?
The answer you found is consistent with the solution, $$\frac{1}{\frac{e^{\lambda t}}{x}-e^{\lambda t}+1 }=\frac{xe^{-\lambda t}}{1-x(1-e^{-\lambda t})}=xe^{-\lambda t}\sum_{k=1}^\infty \bigl(x(1-e^{-\lambda t})\bigr)^{k-1}=\sum_{k=1}^\infty p_k(t)x^k \,.$$