# Linear version of Gronwall's inequality, proof

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.

UPDATE

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.