Post History
#2: Post edited
by
Peter Taylor
·
2021-02-01T14:54:58Z (almost 4 years ago)
- > ${\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 $$y = \frac{1}{\mu(t)} \left( \int_{t_0}^t \mu(s) g(s) \\,ds ight)$$ because the arbitrary choice of $t_0$ induces an arbitrary constant of integration.
- > ${\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 $$y = \frac{1}{\mu(t)} \left( \int_{t_0}^t \mu(s) g(s) \\,ds ight)$$ 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).
#1: Initial revision
by
Peter Taylor
·
2021-02-01T14:52:00Z (almost 4 years ago)
> ${\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.