If $z$ satisfies; $z'\le az+b$, $\ z(0)=z_0>0$ with constants $a,b$ why is true that $z(t)\le z_0+bt$, if $a=0$
It is clear that it can't be justified only by integrating. We had only Gronwall lemma in the lecture, the other case, where $a\neq 0$ I have already proved (To show was: $z(t)\le z_0e^{at}+\frac ba(e^{at}-1)$), but didn't use the lemma, now I'm stuck, do you have an idea, Thanks.
On
If $$ z'\le az+b,\quad\text{then}\quad \mathrm{e}^{-at}\big(z'(t)-az(t)\big)\le b\,\mathrm{e}^{-at}, $$ or $$ \big(\mathrm{e}^{-at}z(t)\big)'\le -\frac{b}{a}\big(\mathrm{e}^{-at}\big)' $$ and integrating in $[0,t]$ we obtain $$ \left(\mathrm{e}^{-at}z(t)+\frac{b}{a}\mathrm{e}^{-at}\right)- \left(z(0)+\frac{b}{a}\right)\le 0, $$ or $$ z(t)\le \left(z(0)+\frac{b}{a}\right)\mathrm{e}^{at}-\frac{b}{a}. $$ If $a=0$, then see Pedro's answer.
If you have $a=0$ then $z'\leq b$. So you just have to integrate the inequality
\begin{equation} \int_{0}^{t}z'dt \leq \int_{0}^{t}bdt, \end{equation}
and obtain
\begin{equation} z(t)\leq z_0 +bt. \end{equation}