Gronwall-Bellman inequality I am trying to show what is asked in question 3.18. After using the hint I have 
$z(t)\leq k_1+k_2\int_{t_0}^t z(\tau)d\tau+k_3\int_{t_0}^t e^{\alpha (\tau-t_0)}d\tau \\
\quad \,= k_1+\dfrac{k_3}{\alpha}\big[e^{\alpha(t-t_0)}-1\big]+k_2\int_{t_0}^t z(\tau)d\tau\\$ 
then using the Gronwall-Bellman inequality I get 
$z(t)\leq k_1e^{k_2(t-t_0)}+\dfrac{k_3}{\alpha}\big[e^{(\alpha+k_2)(t-t_0)}-e^{k_2(t-t_0)}\big]$   
or equivalently 
$y(t)\leq k_1e^{-(\alpha-k_2)(t-t_0)}+\dfrac{k_3}{\alpha}\big[e^{k_2(t-t_0)}-e^{-(\alpha-k_2)(t-t_0)}\big]$, which is not quite what the book has: $y(t)\leq k_1e^{-(\alpha-k_2)(t-t_0)}+\dfrac{k_3}{\alpha-k_2}\big[1-e^{-(\alpha-k_2)(t-t_0)}\big]$.
I am not sure if I have made an error in application of Gronwalls inequality, or something else entirely.
If anyone can give me a hint as to which direction to go I would greatly appreciate it.

 A: 
As I do not know the specific form of the Gronall-Bellman inequality in your textbook, I provide a direct proof below.

Let 
\begin{align*}
v(t) = \int_{t_0}^t e^{-\alpha(t-\tau)}(k_2 y(\tau) + k_3) d\tau.
\end{align*}
Then \begin{align*}
y(t) \le v(t) + k_1e^{-\alpha(t-t_0)}.
\end{align*} Moreover,
\begin{align*}
\frac{d\left(v(t)e^{(\alpha-k_2)(t-t_0)}\right)}{dt} &= (\alpha-k_2) e^{(\alpha-k_2)(t-t_0)}v(t) + \frac{d\left(v(t)\right)}{dt}e^{(\alpha-k_2)(t-t_0)}\\
&=(\alpha-k_2) e^{(\alpha-k_2)(t-t_0)}v(t) \\
&\qquad + e^{(\alpha-k_2)(t-t_0)}\left(k_2y(t)+k_3 -\alpha \int_{t_0}^t e^{-\alpha(t-\tau)}(k_2 y(\tau) + k_3) d\tau\right) \\
&=(\alpha-k_2) e^{(\alpha-k_2)(t-t_0)}v(t) + e^{(\alpha-k_2)(t-t_0)}\left(k_2y(t)+k_3 -\alpha v(t)\right)\\
&\le (\alpha-k_2) e^{(\alpha-k_2)(t-t_0)}v(t) \\
&\qquad + e^{(\alpha-k_2)(t-t_0)}\left(k_2v(t)+k_3 -\alpha v(t) + k_2 k_1e^{-\alpha(t-t_0)}\right)\\
&= k_3 e^{(\alpha-k_2)(t-t_0)} + k_2 k_1 e^{-k_2(t-t_0)}.
\end{align*}
That is,
\begin{align*}
\frac{d\left(v(t)e^{(\alpha-k_2)(t-t_0)} - \frac{k_3}{\alpha-k_2}e^{(\alpha-k_2)(t-t_0)} + k_1 e^{-k_2(t-t_0)}\right)}{dt} \le 0.
\end{align*}
In other words, $$(v(t)e^{(\alpha-k_2)(t-t_0)} - \frac{k_3}{\alpha-k_2}e^{(\alpha-k_2)(t-t_0)} + k_1 e^{-k_2(t-t_0)}$$ is a decreasing function. Therefore,
\begin{align*}
v(t)e^{(\alpha-k_2)(t-t_0)} - \frac{k_3}{\alpha-k_2}e^{(\alpha-k_2)(t-t_0)} + k_1 e^{-k_2(t-t_0)} &\le - \frac{k_3}{\alpha-k_2} + k_1.
\end{align*}
Then,
\begin{align*}
v(t) \le k_1 e^{-(\alpha-k_2)(t-t_0)}+ \frac{k_3}{\alpha-k_2}\left( 1- e^{-(\alpha-k_2)(t-t_0)}\right) - k_1e^{-\alpha(t-t_0)}
\end{align*}
Finally,
\begin{align*}
y(t) &\le v(t) + k_1e^{-\alpha(t-t_0)}\\
&\le k_1 e^{-(\alpha-k_2)(t-t_0)}+ \frac{k_3}{\alpha-k_2}\left( 1- e^{-(\alpha-k_2)(t-t_0)}\right).
\end{align*}
