proof of Gronwall’s lemma

The inequality

$$\varphi (t)\le K+L{\int }_{t_0}^t\psi (s)\varphi (s)𝑑s$$(1)

is equivalent to

$$\frac{\varphi (t)}{K+L{\int }_{t_0}^t\psi (s)\varphi (s)𝑑s}\le 1$$

Multiply by $L\psi (t)$ and integrate, giving

$${\int }_{t_0}^t\frac{L\psi (s)\varphi (s)ds}{K+L{\int }_{t_0}^s\psi (\tau )\varphi (\tau )𝑑\tau }\le L{\int }_{t_0}^t\psi (s)𝑑s$$

Thus

$$\ln \left(K+L{\int }_{t_0}^t\psi (s)\varphi (s)𝑑s\right)-\ln K\le L{\int }_{t_0}^t\psi (s)𝑑s$$

and finally

$$K+L{\int }_{t_0}^t\psi (s)\varphi (s)𝑑s\le K\exp \left(L{\int }_{t_0}^t\psi (s)𝑑s\right)$$

Using (1) in the left hand side of this inequality gives the result.