Understanding solution to first-order differential equation written in integral form, $y(x)=\int_0^xy(t)dt + x + 1$

Question

I have a differential equation and I'm trying to understand the solution printed in the back of the book. I will specify what part I'm struggling to understand:

Problem statement: Solve the following equation. $y(x)=\int_0^xy(t)dt + x + 1$.

Verbatim solution from back of the book that I'm trying to understand:
$y(x)=\int_0^xy(t)dt + x + 1 ⇒ \frac{dy}{dx} = y(x) + 1 ⇒ \frac{dy}{dx}- y = 1 ⇒ u(x) = e^{\int-1dx} = e^{-x} ⇒ \frac{d}{dx}[e^{-x}y] = e^{-x} ⇒ e^{-x}y = \int e^{-x}dx = -e^{-x} + C ⇒ y = Ce^{x} - 1.$

But there is an implied initial condition here: $y(0)=1.$ Therefore, $1=y(0)=Ce^{0}-1⇒C=2$, so that the solution is $y=2e^{x}-1.$ [Try any other value of C and see that the function doesn't satisfy the original equation.]

I have 2 questions:

1. Why is $y(x)=\int_0^xy(t)dt = y(x)$, and not $y(x)-y(0)$?

2. Why is there an implied initial condition when there is no implied condition stated in the problem?

Answer 1

1) This is the direct result of the first part of the Fundamental Theorem of Calculus.

2) You can see in the original equation, when $x=0$ the integral vanishes, as does $x$, leaving you the condition $y(0)=1$. This is certainly true, so this condition has to be satisfied.