# Proving following theorem, Grönwall

Suppose $$y:[t_0,+\infty) \to [0,+\infty)$$, $$t_0 \in \mathbb{R}$$, is a non-negative continuous function and $$u:[t_0,+\infty)\to\mathbb{R}$$ is a non-decreasing continuous function. Let $$L\in\mathbb{R}^{+}$$ and $$0 \le y(t) \le u(t)+L\int_{t_0}^t y(s)ds, \, \forall t \ge t_0.\quad (1)$$ Show that $$0 \le y(t) \le u(t)\cdot e^{L(t-t_0)}, \, \forall t\ge t_0.$$

My attempt:

Define $$v(t):=\ln\left(u(t)+L\int_{t_0}^ty(s)ds\right),$$ then $$v'(t)=\frac{u'(t)+Ly(t)}{u(t)+L\int_{t_0}^t y(s)ds} \le \frac{u'(t)}{u(t)+L\int_{t_0}^ty(s)ds}+L.$$ Integrating both sides from $$t_0$$ to $$t$$: $$v(t)-v(t_0)\le\int_{t_0}^t \frac{u'(t)}{u(t)+L\int_{t_0}^t y(s)ds}dt+L(t-t_0).$$ Considering the fact that $$v(t_0)=\ln u(t)$$, substituting back the expression for $$v(t)$$ and exponentiating, we get $$u(t)+L\int_{t_0}^t y(s)ds\le u(t)\cdot e^{L(t-t_0)}\cdot e^{\int_{t_0}^t\frac{u'(t)}{u(t)+L\int_{t_0}^t y(s)ds}dt}.$$ I'm so close to finding the needed expression, but I don't know what to do with the last exponential.

Any ideas?

Thanks.