I am reading a proof of the following theorem:

Assume $\phi$ is a continuous function in $[0,T]$ that satisfies $$\phi(t) = \alpha + \int_0^t (\beta \phi(s) + \gamma)ds, \hspace{0.5mm} t\in [0,T], $$ where $\alpha, \gamma \in \mathbb{R}$ and $\beta > 0$. Then $$\phi(t) \leq \alpha e^{\beta t} + \frac{\gamma}{\beta}(e^{\beta t} - 1), \hspace{0.5mm} t\in [0,T]. $$

The proof begins by setting $$\psi(t) = \alpha + \int_0^t (\beta \phi(s) + \gamma)ds, $$ in order to obtain $$\psi'(t) \leq \beta \psi(t) + \gamma \implies \psi'(t) - \beta \psi(t) \leq \gamma \implies \left(e^{-\beta t} \psi(t) \right)' \leq \gamma e^{-\beta t}. $$

After this, it says that this implies that $e^{-\beta t} \psi(t) - \psi(0) \leq \frac{\gamma}{\beta}(1-e^{-\beta t}), $ and this is what confuses me.

I think that $$\left(e^{-\beta t} \psi(t) \right)' \leq \gamma e^{-\beta t} \implies e^{-\beta t} \psi(t) \leq -\frac{\gamma}{\beta}e^{-\beta t}, $$ but does subtracting $\psi(0) = \alpha$ on the left-hand side somehow correspond to adding $\gamma / \beta$ on the right-hand side? I don't see how it would.


Could it be that I have to integrate $\gamma e^{-\beta t} $ from $0$ to $t$? Because then that would be $\frac{\gamma}{\beta}(1-e^{-\beta t})$. But then I am still confused about what allows us to subtract $\psi(0)$ from the left-hand side.


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