Solving infinite ladder of differential equations using generating functions. I am interested in solving the following infinite ladder of coupled differential equations. For any integer $k \geq 0$, we have a real-valued function of a single real variable, $p_k (t)$, which satisfies
$$\dot{p}_k(t) = (k+1)p_{k+1}(t) - kp_k(t)$$
Here, $t \geq 0$ ("time"), and the dot denotes a derivative. The choice of notation $p_k$ is intentional, as these form a set of probabilities. That is,
$$\forall t\geq 0, \quad  p_k(t) \geq 0 \, \,\text{and}\,\, \sum_{k=0}^\infty p_k(t)=1$$
(One can show that the differential equations conserve this sum.) To solve this problem, I attempted to introduce a generating function of the form
$$g(z, t) \equiv \sum_{k=0}^\infty z^k p_k(t).$$
This function has the property that $g(0, t) = 0$ and $g(1,t) = 1$. Moreover, by differentiating the equation with respect to $t$, I found that it satisfies the following first-order, linear partial differential equation.
$$ \partial_t g(z,t) + (z-1)\partial_z g(z,t) = 0$$
This seems promising to me, as I seem to have a well defined boundary value problem. Namely, letting $t \in [0,\infty)$ and $z \in [0,1]$, I set the values of $g$ at the boundaries $z = 0, 1$, and with the corresponding initial condition $g(z,0)$. This seems like a well-posed problem. However, I'm having trouble finding the solution. I believe the general solution to the differential equation is
$$ g(z,t) = f(e^{-t}(1-z)) $$
where $f$ is any differentiable function of a single variable. But when I try to satisfy the boundary conditions, I hit a snag. The $z = 1$ condition implies $f(0) = 1$, but the $z = 0$ condition implies $f(e^{-t}) = 0$. I'm pretty sure this breaks the camel's back: it seems to be saying $f = 0$ for all values!
Am I missing something? Are there modifications to this process that can lead me to a solution? Thanks in advance!
 A: Thanks to some feedback in the comments to my original post, I was able to come to a solution to the problem. As pointed out by Robert Israel, my error was assuming that $g(0, t) = 0$, when, in fact, there is still a contribution from the $k = 0$ term in the defining sum. That is, $g(0, t) = p_0(t)$. My general solution to the PDE above was correct, so now we have $f(1-z) = p_0(t)$.
The form of the function $f$ is going to be determined from our initial condition: $g(z, 0) = f(1-z) = \sum_{k=0}^\infty z^k p_k(0)$. Let's consider a power series expansion of $f(x)$ about $x =0$, with coefficients $f_k$.
$$f(x) = \sum_{k=0}^\infty f_k x^k$$
Plugging in $x = 1-z$ and reexpanding in terms of powers of $z$, we can express $f_k$ in terms of the various $p_j(0)$. I found the result to be
$$f_k = (-1)^k \sum_{n=k}^\infty \binom{n}{k} p_n(0)$$.
Now we may return to our solution for the generating function. I found
\begin{aligned}
g(z, t) &= f(e^{-t} (1-z)) = \sum_{k=0}^\infty f_k e^{-kt}(1-z)^k \\
&= \sum_{k=0}^\infty (-1)^k \left(\sum_{n=k}^\infty \binom{n}{k} p_n(0)\right) e^{-kt}(1-z)^k
\end{aligned}
Finally, I want to relate this to the functions $p_k(t)$, which should now solve the original ladder of differential equations. I can do this by, once more, expanding the $(1-z)^k$ in each term and regrouping into powers $z^j$. Then I can match the coefficients. Going through this process, I find that
$$\boxed{p_k(t) = (-1)^k \sum_{n=k}^\infty \binom{n}{k}e^{-nt}(-1)^n \sum_{m=n}^\infty \binom{m}{n} p_m(0)}$$.
I should double check that this answer indeed satisfies the original equation, and that it satisfies the basic properties of probabilities for all $t$. Perhaps it can be simplified. Nevertheless, I tested this solution for a few simple initial probability distributions $\{p_m(0)\}$ and they all satisfied the requirements. So I'm pretty convinced and I'm going to call that good.
I decided to finish the solution using the method I began with, but as Cesareo suggested in the comments, this problem might be easier solved by taking a Laplace transform from the beginning and solving the resulting recurrence relation.
