Question about Gronwall Inequality So, for this question, (as it is for practice so I wish for a hint or connection),
I am trying to find a comparison to Gronwall's inequality and the finalized form of the integral equation as (Int-Eqn)'

MY REMARKS:
So, I understand Gronwall's inequality is used to solve ODEs but with inequalities to get a bound, but what am I missing? Why is the (Int-Eqn)' supposed to be different different? This appears like the same form and a direct application. Such as where I am told to apply where $g(t)$ is a constant, then this would imply that
$$f(t) \leq x_0e^{\int_{t_0}^t 5 ds}$$
other than this would be in the form of a norma exponential and then there is no bound since $e^{t}$ is exponentially growing?
My text only has 2 pages on this theorem, so, if one could give a few bullet points of what I am missing, that would be great. I am not posting requesting for a solution, more as **why are these two supposed to be different? What am I missing here?**Thanks community!
 A: HINT. The lemma of Gronwall says exactly the following. Suppose that $y$ is a sub-solution to the integral equation
$$\tag{1}x(t)=y_0+ \int_{t_0}^t g(s) x(s)\, ds,\quad t\ge t_0, $$
which means that $y$ solves (1) with $\le$ in place of $=$. Then $y(t)\le x(t)$ for all $t\ge t_0$, where $x$ is the unique solution to (1). In words, “sub-solutions are dominated by solutions”.
A: One typical application is that you have a vector system $\dot x(t)=f(t,x(t))$ where you get an upper estimate for the ODE function $\|f(t,x)\|\le g(t)\|x\|$.
Then the integral for of the initial value problem
$$
x(t)=x_0+\int_0^t\dot x(s)\,ds=x_0+\int_0^tf(s,x(s))\,ds
$$
leads to an integral inequality for $\phi(t)=\|x(t)\|$
$$
\phi(t)\le K+\int_0^t g(s)\phi(s)\,ds
$$
with additionally $K=\|x_0\|$.
Then you can use the comparison interpretation of the theorem to say that $\|x(t)\|=\phi(t)\le u(t)$ where $u\ge 0$ solves the equation version of the inequality
$$
u(t)=K+\int_0^tg(s)u(s)\,ds,
$$
which in reverse is equivalent to the IVP
$$
u(0)=K,~~~\dot u(t)=g(t)u(t).
$$

As a direct proof consider
$$
\frac{d}{dt}\ln\left(K+\int_0^tg(s)\phi(s)\,ds\right)
=\frac{g(t)\phi(t)}{K+\int_0^tg(s)\phi(s)\,ds}\le g(t)
$$
which directly implies
$$
\phi(t)\le K+\int_0^tg(s)\phi(s)\,ds\le K\exp\left(\int_0^tg(s)\,ds\right).
$$
