Isn't it wrong to write that Indefinite Integral = Definite Integral with a variable in its Upper Limit?
$\int f(t) \; dt = \int_{t_0}^t f(s) \; ds \quad \text{ where $t_0$ is some convenient lower limit of integration.}$
Isn't this wrong? Because LHS $\neq$ RHS in general!
Rather, LHS $\ni$ RHS, because LHS = RHS only if $C = -g(t_0)$.
By the Fundamental Theorem of Calculus (FTC), LHS = $\int{f(t) \; dt} = \{ g(t) + C : C\in \mathbb{R}\}$. It's wrong to write $g(t) + C$ because this means THE antiderivative of $f$. But the LHS is the SET of antiderivatives of $f$ that differ from $g$ by $C$.
By the FTC again, RHS = $\int_{t_0}^t f(s)\ ds = g(t) - g(t_0)$. Then LHS = RHS $\implies C = -g(t_0)$.
Boyce, Elementary Differential Equations (2017 11 edn), pp 28–9. The top-most equation hails from equating (33), but I defined $f(t) :=u(t)g(t)$ and equated (33) to the RHS of (32).
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Your observation RHS $\in$ LHS is correct. LHS is not a unique function but a set of functions differing by a constant.
I take issue with your usage of $C$. There is not really a meaningful definition of $C$ for an arbitrary antiderivative. For example, what is the constant term of $\cos(x) + 3$? You could say it's 3 but what if we rewrite it as $u(x) + 4$ where $u(x) = \cos(x) + 1$?
Note that if $g \in \int f$ then $\int_{t_0}^t f = g(t) - g(t_0)$. Whatever constant is included in $g$ cancels in $g(t) - g(t_0)$ so $g$ and $t_0$ don't have anything to do with each other.
If instead we write $F = \int_{t_0}^t f$ then we have $F(t_0) = \int_{t_0}^{t_0} t = 0$. This equation forces a certain constant term of $F$. A different choice of $t_0$ can (but doesn't necessarily) lead to a different constant term of $F$.
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${\int{f(t) \; dt} = \int_{t_0}^t f(s) \; ds \quad \text{ where $t_0$ is some convenient lower limit of integration.}}$
isn't actually in the source text at all. Unpacking some of the surrounding text to more formal notation, it goes from equation (32) $$\exists c: \mu(t) y = \int \mu(t) g(t) \,dt + c$$ so equation (33) $$\exists t_0, c: y = \frac{1}{\mu(t)} \left( \int_{t_0}^t \mu(s) g(s) \,ds + c \right)$$
You could argue that it would have been slightly clearer to use $c'$ instead of $c$ in (33), but equally you could argue that it's valid to eliminate $c$ entirely and say that the general solution (my emphasis) is $$y = \frac{1}{\mu(t)} \left( \int_{t_0}^t \mu(s) g(s) \,ds\right)$$ because the arbitrary choice of $t_0$ induces an arbitrary constant of integration. (Actually that would be a bit iffy, because the entire range of possible $t_0$ might not produce a sufficiently wide choice of the constant of integration, but the main point I'm trying to make is that you're not giving enough weight to the context).
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