Textbook Section2.1 Example5

[RunboZhang]

Hi guys, I have been stuck at example 5 of section2.1(page30) for quite a while. In particular, I do not understand why the lower bound of the integral is the initial point t=0. Why can't it be the upper bound?
(example screenshot is attached below)

[Jinqiu Liang]

In my opinion, firstly this question is an initial value problem, which means the number that the equation (y(0)=1) gives you is the lower limit of the integral, which is 0. You can also use the upper bound, however, if t is infinity, then we can not substitute t as a number into the equation and then solve it. In this way, choosing the lower limit number 0 is the easiest and the fastest method.

[Victor Ivrii]

You can select any lower limit you wish, the difference goes to the constant. However, as Ella correctly observed, it makes sense to select $t=0$ since the $t_0=0$ in the initial problem