The two-stage MVK model is a continuous time Markov model of cancer formation that describes the occurrence and growth of intermediate cells and malignant cells arising from a population of normla cells. Let $N(t), I(t)$ and $M(t)$ denote the number of normal, intermediate and malignant cells at time $t$.
Let $\mu_1(t) = $ rate at which a normal cell gives rise to an intermediate cell, $\alpha(t) = $ rate at which an intermediate cell divides into two intermediate cells, $\beta(t) = $ rate at which an intermediate cell dies and $\mu(t)$ = rate at which an intermediate cell divides into one intermediate cell and one malignant cell. Finally, let $\nu(t) = \mu_1(t)N(t)$.
For $t \geq \tau$ we have the following generating functions
$$\Psi(y,z;\tau,t) = \mathbb{E}[y^{I(t)}z^{J(t)} \mid I(\tau) = 0, M(\tau) = 0]$$
with
$$J(t) = \begin{cases} 1 &\mbox{if } M(\tau) > 0, \tau \leq t \\ 0 &\mbox{otherwise} \end{cases}$$
and
$$\Phi(y,z;\tau,t) = \mathbb{E}[y^{I(t)}z^{J(t)} \mid I(\tau) = 1, M(\tau) = 0]$$
The article I am reading (A Numerical Solution to the Nonohomogeneous Two-Stage MVK Model of Cancer) now states that these generating functions satisfy the Kolmogorov backward equations such that
$$\frac{d\Phi(y,z;\tau,t)}{d\tau} = -\alpha(\tau)\Phi^2(y,z;\tau,t) - \beta(\tau) - z\mu(\tau)\Phi(y,z;\tau,t) + [\alpha(\tau) + \beta(\tau) + \mu(\tau)]\Phi(y,z;\tau,t)$$
and
$$\frac{d\Psi(y,z;\tau,t)}{d\tau} = -\nu(\tau)\Psi(y,z;\tau,t)[\Phi(y,z;\tau,t) - 1]$$
but I do cannot deduce this myself. Any help would be much appreciated.
Given the desired backward equations, the definitions of these generating functions appear to be slightly incorrect. Given a certain intermediate cell at time $\tau$, there will be, at any future time $t\ge \tau$, a certain number $I_1(t,\tau)$ of intermediate cells and $M_1(t,\tau)$ of malignant cells that descend from it via the intermediate cell dividing into progeny intermediate cells and malignant cells. Also, at time $t$, independent of how many intermediate or malignant cells there are at time $\tau$, there will be a certain number $I_0(t,\tau)$ of intermediate cells and $M_0(t,\tau)$ of malignant cells that do not descend from the intermediate or malignant cells alive at time $\tau$ but descend from normal cells transformed between times $\tau$ and $t$. Then it appears that the intended definitions are \begin{eqnarray*} \Phi(y,z;\tau,t)&=&\mathbb{E}[y^{I_1(t,\tau)}z^{M_1(t,\tau)} ],\\ \Psi(y,z;\tau,t)&=&\mathbb{E}[y^{I_0(t,\tau)}z^{M_0(t,\tau)} ]\\ &=&\mathbb{E}[y^{I(t)}z^{M(t)} \mid I(\tau) = 0, M(\tau) = 0]. \end{eqnarray*} Notice that we now have $M(t)$ in the exponent of $z$ instead of $J(t)$, and that $\Phi(y,z;\tau,t)$ now only includes the descendants of the intermediate cell alive at time $\tau$.
Given these changes, the backward equations can be derived by looking at what happens in the interval ${\cal I}_\tau$ between times $\tau$ and $\tau + d\tau$, and then taking the limit as $d\tau\to 0$. For example, suppose that we start with zero intermediate or malignant cells at time $\tau$. Then, assuming that $\nu(t)$ is differentiable, the probability that one intermediate cell appears during ${\cal I}_\tau$ is $\nu(\tau) d\tau + O(d\tau^2)$, and the probability that nothing happens is $1-\nu(\tau) d\tau + O(d\tau^2)$. All other events have probability $O(d\tau^2)$. Therefore \begin{eqnarray*} \Psi(y,z;\tau,t)&=&\mathbb{E}[y^{I(t)}z^{J(t)} \mid I(\tau) = 0, M(\tau) = 0]\\ &=&\mathbb{E}[y^{I(t)}z^{J(t)} \mid I(\tau+d\tau) = 1, M(\tau) = 0]\ \nu(\tau) d\tau\\ &\ & \ \ \ + \mathbb{E}[y^{I(t)}z^{J(t)} \mid I(\tau+d\tau) = 0, M(\tau) = 0]\ (1-\nu(\tau) d\tau)\\ &\ &\ \ \ + O(d\tau^2)\\ &=&\mathbb{E}[y^{I_0(t,\tau+d\tau)+I_1(t,\tau+d\tau)} z^{M_0(t,\tau+d\tau)+M_1(t,\tau+d\tau)} ]\ \nu(\tau) d\tau\\ &\ & \ \ \ + \mathbb{E}[y^{I_0(t,\tau+d\tau)} z^{M_0(t,\tau+d\tau)} ]\ (1-\nu(\tau) d\tau)\\ &\ &\ \ \ + O(d\tau^2)\\ &=&\Psi(y,z;\tau+d\tau,t)\Phi(y,z;\tau+d\tau,t)\nu(\tau) d\tau\\ &\ &\ \ +\Psi(y,z;\tau+d\tau,t)(1-\nu(\tau) d\tau)+O(d\tau^2). \end{eqnarray*} Rearranging and dividing by $d\tau$ gives $$ \frac{\Psi(y,z;\tau+d\tau,t)-\Psi(y,z;\tau,t)}{d\tau}= $$ $$ \Psi(y,z;\tau+d\tau,t)\nu(\tau) -\Psi(y,z;\tau+d\tau,t)\Phi(y,z;\tau+d\tau,t)\nu(\tau) +O(d\tau). $$ Taking the limit as $d\tau\to 0$ then gives the desired equation.
The differential equation for $\Phi(y,z;\tau,t)$ is derived in the same way. In this case the likely events during ${\cal I}_\tau$ are that (1) the intermediate cell divides into two intermediate cells with probability $\alpha(\tau) d\tau+O(d\tau^2)$, (2) that the intermediate cell dies with probability $\beta(\tau) d\tau+O(d\tau^2)$, (3) that the intermediate cell produces a malignant cell with probability $\mu(\tau) d\tau+O(d\tau^2)$, and (4) that nothing happens with probability $1-(\alpha(\tau)+\beta(\tau)+\mu(\tau))d\tau+O(d\tau^2)$. All other events have probability $O(d\tau^2)$. The events (1)-(4) give rise to the four terms on the right-hand side of the equation for $d\Phi(y,z;\tau,t)/d\tau$.