A differential-algebraic equation is, loosely speaking, an equation of the form $$E\mathbf x'(t)=Ax(t)+f(t)$$ where $E,A\in\mathbb C^{n\times n}$, $t\in[0,\infty)$ and $f:[0,\infty)\to\mathbb C^n$, perhaps subject to some initial condition $\mathbf x(t_0)=\mathbf c\in\mathbb C^n$.
Based on the response here On an example for consistent initial values in differential-algebraic equations, I see that if the matrix $E$ were invertible, then one could recover a 'usual' linear system of first order differential equations, and could try to apply methods from there to derive a solution.
On the other hand, if $E$ is not invertible, in particular when the kernel of $E$ is non-trivial, then one needs to use both the differential and algebraic part to determine which initial values are admissable and even determine the solution.
As the examples below highlight, a differential-algebraic equation can be viewed as a system involving a differential equation and an algebraic equation.
Example 1:
\begin{align*} x_1'(t)+x_2'(t)+x_1(t) & = 0\quad \text{when $t\in(0,\infty)$} \\ x_2(t)&= 0\quad \text{when $t\in(0,\infty)$}\\ \mathbf x(0)&= \mathbf c, \end{align*} where the last equation corresponds to the initial condition.
Example 2: \begin{align*} x_2'(t) & = x_1(t)\\ 0&= x_2(t)-t^3\quad \\ 0&= x_3(t)-t, \end{align*} for $t\in[0,\infty)$, which is independent of an initial condition.
One thing I have read, which I don't completely understand, says that a differential-algebraic equation $E\mathbf x'(t)=Ax(t)+f(t)$ involves a differential equation on $\ker(E)^\perp$ and an algebraic equation on $\ker(E)$.
My questions are on this point. How does one see that we should work with the differential part on $\ker(E)^\perp$ and the algebraic part on $\ker(E)$? How does this 'distinction' arise? Namely, how does the fact that we should make a distinction between where we consider the differential and algebraic part in their relation to $\ker(E)$ arise?
I see two things. First, you are aware that because of finite dimension of $\mathbb{C}^n$,
$$ \mathbb{C}^n = \ker(E)\oplus \ker(E)^{\perp}$$
Secondly, given a function $y$, if
\begin{align} y' &\in \ker(E) \\ \implies 0 &= Ay + f \tag 1 \end{align}
which gives an algebraic equation on $\ker(E)$. Next, given a function $z$ (for clarity), if
\begin{align} z' &\in \ker(E)^{\perp} \\ \implies Ez' &= Az + f \tag 2 \end{align}
which is a differential equation on $\ker(E)^{\perp}$. This now gives a solution
$$x = y \in \ker(E) + z \in \ker(E)^{\perp}$$
You've a part given by $(1)$ and $\ker(E)$ and a part given by $(2)$ and $\ker(E)^{\perp}$.