Question:
Suppose that $(X_t)_{t \geq 0}$ is a $(1,(\lambda_n)_{n \geq 0})$ simple birth process with $\lambda_n = (n+1)\lambda$. Show that
$$\phi(t) := \Bbb E[z^{X_t}]= ze^{-\lambda t}+ \int_0^t \lambda e^{-\lambda s} \phi(t-s)^2 \, ds$$
Hint: condition on the first arrival time.
Attempt:
Let $Z_0$ be the time of first arrival, so $Z_0 \sim \text{exp}(\lambda)$ with density function $f_{Z_0}(s)=\lambda e^{-\lambda s}$. Then
\begin{align} \phi(t) & = \int_0^\infty \Bbb E[z^{X_t} \, | \, Z_0=s] \, f_{Z_0}(s) \, ds \\ & = \int_t^\infty \Bbb E[z^{X_t} \, | \, Z_0=s] \, \lambda e^{-\lambda s} \, ds + \int_0^t \Bbb E[z^{X_t} \, | \, Z_0=s] \, \lambda e^{-\lambda s} \, ds \\ \end{align}
In the first integral, $X_t = 1$ because $t<s$ so that this is before the first arrival time $Z_0=s$. Thus we get
\begin{align} \dots & = \int_t^\infty \Bbb E[z \, | \, Z_0=s] \, \lambda e^{-\lambda s} \, ds + \int_0^t \Bbb E[z^{X_t} \, | \, Z_0=s] \, \lambda e^{-\lambda s} \, ds \\ & = \int_t^\infty z\lambda e^{-\lambda s} \, ds + \int_0^t \Bbb E[z^{X_t} \, | \, Z_0=s] \, \lambda e^{-\lambda s} \, ds \\ & = ze^{-\lambda t} + \int_0^t \lambda e^{-\lambda s} \, \Bbb E[z^{X_t} \, | \, Z_0=s] \, ds \end{align}
So far yet so close, I am having trouble dealing with the second integral.
Any hints?