Consider the pure Poisson process defined by \begin{align} P_n'(t) &= -\lambda_n P_n(t) + \lambda_{n-1}P_{n-1}, \quad n \geq 1,\\ P'_0(t) &= -\lambda_0 P_0(t). \end{align}
with $P_0(0) = 1$. Let $\lambda_n > 0$ for all $n$, prove that for every fixed $n \geq 1$, the function $P_n(t)$ first increases, then decreases to $0$. If $t_n$ is the place of the maximum, then $t_1 < t_2< t_3<\dots$
The hint suggests to use induction and differentiate these set of equations, but I'm not sure how that gives arise to the answer.
It follows from the differential equation and initial condition for $\ P_0\ $ that $\ P_0(t)=e^{-\lambda_0t}\ $. Thus, $\ \lambda_0P_0\ $ is the density of an exponential distribution with parameter $\ \lambda_0\ $.
From the differential equation and initial condition for $\ P_n\ $ for $\ n\ge1\ $ (namely, $\ P_n(0)=0\ $), we get \begin{align} P_n(t)&=\lambda_{n-1}e^{-\lambda_nt}\int_0^te^{\lambda_ns}P_{n-1}(s)ds\\ &=\int_0^te^{-\lambda_n(t-s)}\lambda_{n-1}P_{n-1}(s)ds\\ &=\int_0^te^{-\lambda_n\sigma}\lambda_{n-1}P_{n-1}(t-\sigma)d\sigma \end{align} Thus, if $\ \lambda_{n-1}P_{n-1}\ $ is a probability density function, then $\ \lambda_nP_n\ $ is the convolution of this density function with that of an exponential distribution with parameter $\ \lambda_n\ $, and is hence also a probability density function—namely that of the sum of two independent random variables $\ T_n\sim\text{Exp}\big(\lambda_n\big)\ $ and $\ X_n\sim\lambda_{n-1}P_{n-1}\ $. Since $\ \lambda_0P_0 $ is the density function of a random variable $\ T_0\sim\text{Exp}\big(\lambda_0\big)\ $, it follows by induction that $\ \lambda_nP_n\ $ is the probability density function of a sum $$ \sum_{i=0}^nT_i $$ of independent random variables $\ T_0,T_1,\dots, T_n\ $ with $\ T_i\sim\text{Exp}\big(\lambda_i\big)\ $, and hence that $\ \lim_\limits{t\rightarrow\infty}P_n(t)=0\ $.
Evaluating the above integral for $\ P_1\ $ gives $$ P_1(t)=\cases{\frac{\lambda_0\big(e^{-\lambda_1t}-e^{-\lambda_0t}\big)}{\lambda_0-\lambda_1}&if $\ \lambda_1\ne\lambda_0\ $,\\ \lambda_0t e^{-\lambda_0t} &if $\ \lambda_1=\lambda_0\ $.} $$ In the first case, $\ P_1(t)\ $ is strictly increasing for $\ 0<t<t_1=\frac{\ln\lambda_0-\ln\lambda_1}{\lambda_0-\lambda_1}\ $, reaches a maximum at $\ t=t_1\ $ and is strictly decreasing for $\ t>t_1\ $. In the second case, $\ P_1(t)\ $ is strictly increasing for $\ 0<t<t_1'=\frac{1}{\lambda_0}\ $, reaches a maximum at $\ t=t_1'\ $ and is strictly decreasing for $\ t>t_1'\ $.
Suppose now that $\ n\ge2\ $ and $\ P_{n-1}(t)\ $ is strictly increasing for $\ 0<t<t_{n-1}\ $, reaches a maximum at $\ t=t_{n-1}\ $, and is strictly decreasing for $\ t>t_{n-1}\ $. Differentiating the last of the integral expressions above for $\ P_n(t)\ $ gives \begin{align} P_n'(t)&= \lambda_{n-1}\int_0^t e^{-\lambda_n\sigma}P_{n-1}'(t-\sigma)d\sigma+\lambda_{n-1} e^{-\lambda_nt}P_{n-1}(0)\\ &=\lambda_{n-1}\int_0^t e^{-\lambda_n\sigma}P_{n-1}'(t-\sigma)d\sigma\\ &=\lambda_{n-1}\int_0^t e^{-\lambda_n(t-s)}P_{n-1}'(s)ds\ . \end{align} For $\ t<t_{n-1}\ $, the integrand of this last integral is strictly positive over the interval $\ [0,t]\ $, and hence so is the integral itself. That is, $\ P_n'(t)>0\ $ for $\ t<t_{n-1}\ $, and so $\ P_n(t)\ $ is strictly increasing over the interval $\ \big[0,t_{n-1}\big]\ $. Since $\ P_n\ $ is continuously differentiable for $\ t>0\ $, and $\ \lim_\limits{t\rightarrow\infty}P_n(t)=0\ $, there must exist a point at which $\ P_n(t)\ $ reaches a maximum and $\ P_n'(t)\ $ vanishes. Let $$ t_n=\inf\big\{\,t\,\big|\,t>0\ \&\ P_n'(t)=0\,\big\}\ . $$ Then $\ t_n>t_{n-1}\ $, and at that point, we have $$ P_n'(t_n)=0= \lambda_{n-1}\int_0^{t_n} e^{-\lambda_n(t_n-s)}P_{n-1}'(s)ds\ , $$ and $\ P_n\ $ must be strictly increasing over the interval $\ \big[0,t_n\big]\ $. For $\ t>t_n\ $, we have
\begin{align} P_n'(t)&=\lambda_{n-1}\int_0^te^{-\lambda_n(t-s)}P_{n-1}'(s)ds\\ &=\lambda_{n-1}\left(\int_0^{t_n}e^{-\lambda_n(t-s)}P_{n-1}'(s)ds+\int_{t_n}^te^{-\lambda_n(t-s)}P_{n-1}'(s)ds\right)\\ &=\lambda_{n-1}\left(e^{\lambda(t_n-t)}\int_0^{t_n}e^{-\lambda_n(t_n-s)}P_{n-1}'(s)ds+\int_{t_n}^te^{-\lambda_n(t-s)}P_{n-1}'(s)ds\right)\\ &=\lambda_{n-1}\int_{t_n}^te^{-\lambda_n(t-s)}P_{n-1}'(s)ds\\ &<0\ , \end{align} because the integrand of the final integral is strictly negative over the interval $\ \big[t_n,t\big]\ $. It follows that $\ P_n(t)\ $ is strictly decreasing for $\ t>t_n\ $. This completes the proof that $\ P_n(t)\ $ is strictly increasing over an interval $\ \big[0,t_n\big]\ $, with $\ t_n>t_{n-1}\ $, strictly decreasing for $\ t>t_n\ $, and $\ \lim_\limits{t\rightarrow\infty}P_n(t)=0\ $. It follows by induction that these properties hold for all $\ n\ge1\ $.