Use Gronwall's lemma to prove that the IVP $$\frac{dy}{dt}=e^{\sin t}y(t),~y(0)=y_{0}~~~(*),$$ with $y_0$ being given, has an infinite interval of existence for its solutions.
My approach: Gronwall's lemma states that
Suppose that $g(t)$ is a continuous real-valued function that satisfies $g(t) \geq 0$ and $$g(t) \leq C+K \int_{0}^{t} g(s)~ds$$ for all $t \in [0,a],$ where $C$ and $K$ are positive constants. It then follows that $$g(t) \leq C e^{Kt},~~\textrm{for all}~t \in [0,a].$$
Solving $(*)$ we get $$y(t)=y_0 \cdot e^{ \int_{0}^{t} e^{\sin s}~ds}.$$ How can I set up an inequality for this and apply Gronwall's to conclude what's being asked ? What I'm understanding is existence of an infinite interval means that the solution exists for all values of $t.$ Any help is much appreciated.
Partial answer:
$$\frac{dy}{dt}=e^{\sin t}y(t),~y(0)=y_{0}$$
$$\implies y(t)=y(0)+\int_0^t\exp(\sin(t))y(s)\,ds$$
$$\implies |y(t)|\le |y_0|+e\int_0^t|y(s)|\,ds$$
Applying Gronwall:
$$|y(t)|\le |y_0|\cdot\exp(et)$$
So, $y$ cannot blow up to $\infty$ in finite time, as it is bounded above by an exponential function.