Question about a solution of $y'(t) \le Ky(t)$ for $t \ge 0$ implies $y(t) \le y(0)e^{Kt}$ There is a step in the following solution that I am not sure it is correct, but maybe I am missing something. The problem is:

Let $K$ be a real constant. Suppose that $y(t)$ is a positive differentiable function satisfying $y'(t) \le Ky(t)$ for $t \ge 0$. Prove that $y(t) \le e^{Kt}y(0)$ for $t \ge 0$.

And the solution is:

From the given inequality we get $0 \ge e^{-Kt}y'(t)-Ke^{-Kt}y(t)=\frac{d}{dt}\left(e^{-Kt}y(t)\right)$ for $t \ge 0$; integrating from $0$ to $t$ we find that, for $t \ge 0$, $e^{-Kt}y(t)-y(0) \le 0$ from which the desired inequality follows.

I am not sure about the step where the author integrates from $0$ to $t$: in the hypotheses we have, we know that $y$ is differentiable and, since $e^{-Kt}$ is indefinitely differentiable, so $y(t)e^{-Kt}$ is at most differentiable in this context. So $\frac{d}{dt}y(t)e^{-Kt}$ is not  continuous, and $\frac{d}{dt}y(t)e^{-Kt}$ is not monotonic as far as we know, hence there is no theorem (as far as I know) that guarantees the integrability of $\frac{d}{dt}y(t)e^{-Kt}$ in the interval $[0,t]$.
Moreover, where the hypothesis that $y$ is positive is used?
My solution is similar, but after getting to $\frac{d}{dt}\left(e^{-Kt}y(t)\right) \le 0$ I proceed by saying that so $y(t)e^{-Kt}$ is decreasing for $t \ge 0$ and so $y(t)e^{-Kt} \le y(0)e^{-K \cdot 0}=y(0)$, so $y(t) \le e^{Kt}y(0)$. Is my solution correct? I am not using that $y$ is positive as well.

Correct. The author has a step that requires a justification that is not given. And you have remedied this.
Another approach: If $z(t)$ is differentiable and $z'(t)\le K$ for all $t$ then $t\ge 0\implies z(t)\le z(0)+ Kt.$ Proof: If $z>0$ and $z(t)>z(0)+Kt$ then by the MVT there exists $t^*\in (0,t)$ with $z'(t^*)=\frac {z(t)-z(0)}{t-0}>K$, a contradiction. So let $z(t)=\ln y(t)$ & the rest is easy.
