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We have this theorem (Gronwall's inequality):

Let $f$, $g$ and $h$ be continuous nonnegative functions defined for $t\ge t_0$. If$$f(t)\le h(t)+\int_{t_0}^{t}g(s)f(s)\,ds\>,$$then$$f(t)\le h(t)+\int_{t_0}^{t}g(s)h(s)e^{\int_s^tg(u)\,du}\,ds$$

How do I prove another version of the theorem, letting $h(t)=k$ , where $k$ is any nonnegative constant, i.e If$$f(t)\le k+\int_{t_0}^{t}g(s)f(s)\,ds\>,$$then$$f(t)\le ke^{\int_{t_0}^tg(s)\,ds}\,ds$$

I worked out the following \begin{align} f(t)&\le k+\int_{t_0}^{t}g(s)ke^{\int_s^tg(u)\,du}\,ds\\\\&=k\left(1+\int_{t_0}^{t}g(s)e^{\int_s^tg(u)\,du}\,ds\right) \end{align} How should I proceed?

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HINT

Let $w(t)=v(t)\cdot\exp \left(- \int_{t_0}^{t}g(s)\mathrm{d}s\right)$

We have: $w'(t)<0, \forall t \ge t_0 \ge 0$, where $v(t)=k+\int_{t_0}^{t}g(s)f(s)\,\mathrm{d}s$.

Whence $w(t) \le w(t_0), \forall t \ge t_0 \ge 0$

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