Adjoint differential equations Consider the vector differential equations
\begin{equation}
\mathbf{x}^{\prime}=\mathbf{A}(t)\cdot\mathbf{x}\tag{1}
\end{equation}
and
\begin{equation}
\mathbf{y}^{\prime}=-\mathbf{A}^{\ast}(t)\cdot\mathbf{y},\tag{2}
\end{equation}
where $\mathbf{A}^{\ast}$ is the complex conjugate transpose of $\mathbf{A}$ and $\mathbf{x},\mathbf{y}$ are column vectors. It is well-known that (1) and (2) are said to be adjoint to one another.
Further, we know that if $\mathbf{x}$ and $\mathbf{y}$ are solutions of (1) and (2), respectively, then
\begin{equation}
\mathbf{y}^{\ast}\cdot\mathbf{x}=\text{constant}.\notag
\end{equation}
Now, consider the higher-order (scalar) differential equations
\begin{equation}
\sum_{i=1}^{n}p_{i}(t)x^{(i)}(t)=0,\tag{3}
\end{equation}
where $p_{n}(t)\neq0$, and
\begin{equation}
\sum_{i=1}^{n}(-1)^{(i)}[p_{i}y]^{(i)}(t)=0.\tag{4}
\end{equation}
Also, (3) and (4) are said to be adjoint to one another.
Further, if $x$ and $y$ are solutions of (3) and (4), respectively, then (see [1, (8.17) on pp. 67])
\begin{equation}
\sum_{i=0}^{n}\sum_{j=0}^{i-1}(-1)^{j}x^{(i-j-1)}(t)[p_{i}z]^{(j)}(t)=\text{constant}.\label{hmfeq}\tag{*}
\end{equation}
The inner sum in \eqref{hmfeq} resembles the matrix multiplication formula.
So, recognizing the similarities between systems and scalar equations, is it possible to obtain the result for higher-order equations by transforming them into vector equations? I could not establish any bridge here. I am experiencing problems in transforming (4) into a useful matrix representation.
References
[1]. P. Hartman, Ordinary Differential Equations, SIAM, 2002.
 A: Okay, I believe, I have constructed the bridge between the adjoint of a vector equation and the adjoint of a scalar equation. I will explain it step by step here for those who might be interested.
Higher-Order Two-Term Scalar Equations
Now, consider the following higher-order two-term scalar differential equation
\begin{equation}
x^{(n)}(t)+p(t)x(t)=0,\label{hottceeq1}\tag{1}
\end{equation}
where $p$ is a complex-valued continuous function.
Then, the matrix representation for \eqref{hottceeq1} is
\begin{equation}
\left(
  \begin{array}{c}
    x\\
    x^{\prime}\\
    \vdots\\
    x^{(n-1)}
  \end{array}
\right)^{\prime}
=
\left(
  \begin{array}{cccc}
    &1&&\\
    &&\ddots&\\
    &&&1\\
    -p(t)&&&
  \end{array}
\right)
\cdot
\left(
  \begin{array}{c}
    x\\
    x^{\prime}\\
    \vdots\\
    x^{(n-1)}
  \end{array}
\right).\label{hottceeq2}\tag{2}
\end{equation}
Thus, we obtain
\begin{equation}
(-1)
\left(
  \begin{array}{cccc}
    &1&&\\
    &&\ddots&\\
    &&&1\\
    -p(t)&&&
  \end{array}
\right)^{\ast}
=
\left(
  \begin{array}{cccc}
    &&&\overline{p}(t)\\
    (-1)&&&\\
    &\ddots&&\\
    &&(-1)&
  \end{array}
\right)\notag
\end{equation}
and
\begin{equation}
\left(
  \begin{array}{c}
    -y^{(n-1)}\\
    y^{(n-2)}\\
    \vdots\\
    (-1)^{n}y
  \end{array}
\right)^{\prime}
=
\left(
  \begin{array}{cccc}
    &&&\overline{p}(t)\\
    (-1)&&&\\
    &\ddots&&\\
    &&(-1)&
  \end{array}
\right)
\cdot
\left(
  \begin{array}{c}
    -y^{(n-1)}\\
    y^{(n-2)}\\
    \vdots\\
    (-1)^{n}y
  \end{array}
\right),\label{hottceeq3}\tag{3}
\end{equation}
where we have constructed the unknown matrix from bottom to the top.
This system gives us the adjoint equation
\begin{equation}
(-1)^{n}y^{(n)}(t)+\overline{p}(t)y(t)=0.\notag
\end{equation}
Note that, if $n=\text{even}$, then both \eqref{hottceeq2} and \eqref{hottceeq3} represent \eqref{hottceeq1},
i.e., \eqref{hottceeq1} is self-adjoint.
Further, using the definition of the inner product in the first post, we get
\begin{equation}
\left\langle\left(
  \begin{array}{c}
    x\\
    x^{\prime}\\
    \vdots\\
    x^{(n-1)}
  \end{array}
\right),
\left(
  \begin{array}{c}
    -y^{(n-1)}\\
    y^{(n-2)}\\
    \vdots\\
    (-1)^{n}y
  \end{array}
\right)\right\rangle
=\sum_{j=0}^{n-1}(-1)^{n-j}x^{(n-1-j)}\overline{y}^{(j)}=\text{constant}.\notag
\end{equation}
Higher-Order Autonomous Scalar Equations
Next, consider the
\begin{equation}
x^{(n)}(t)+p_{1}x^{(n-1)}(t)+\cdots+p_{n}x(t)=0,\label{hoaseeq1}\tag{4}
\end{equation}
where $p_{1},p_{2},\cdots,p_{n}$ are complex numbers.
Then, the matrix representation for \eqref{hoaseeq1} is
\begin{equation}
\left(
  \begin{array}{c}
    x\\
    x^{\prime}\\
    \vdots\\
    x^{(n-1)}
  \end{array}
\right)^{\prime}
=
\left(
  \begin{array}{cccc}
    &1&&\\
    &&\ddots&\\
    &&&1\\
    -p_{n}&-p_{n-1}&\cdots&-p_{1}
  \end{array}
\right)
\cdot
\left(
  \begin{array}{c}
    x\\
    x^{\prime}\\
    \vdots\\
    x^{(n-1)}
  \end{array}
\right).\notag
\end{equation}
On the other hand, by using the adjoint coefficient matrix, we form the differential system
\begin{equation}
\begin{aligned}[]
&\left(
  \begin{array}{c}
    (-1)^{n-1}y^{(n-1)}+(-1)^{n-2}\overline{p}_{1}y^{(n-2)}+\cdots+\overline{p}_{n-1}y\\
    (-1)^{n-2}y^{(n-2)}+(-1)^{n-3}\overline{p}_{2}y^{(n-3)}+\cdots+\overline{p}_{n-2}y\\
    \vdots\\
    y
  \end{array}
\right)^{\prime}\\
=&
\left(
  \begin{array}{cccc}
    &&&\overline{p}_{n}\\
    (-1)&&&\overline{p}_{n-1}\\
    &\ddots&&\vdots\\
    &&(-1)&\overline{p}_{1}
  \end{array}
\right)
\cdot
\left(
  \begin{array}{c}
    (-1)^{n-1}y^{(n-1)}+(-1)^{n-2}\overline{p}_{1}y^{(n-2)}+\cdots+\overline{p}_{n-1}y\\
    (-1)^{n-2}y^{(n-2)}+(-1)^{n-3}\overline{p}_{2}y^{(n-3)}+\cdots+\overline{p}_{n-2}y\\
    \vdots\\
    y
  \end{array}
\right),
\end{aligned}
\notag
\end{equation}
which gives the scalar equation
\begin{equation}
(-1)^{n}y^{(n)}(t)+(-1)^{n-1}\overline{p}_{1}y^{(n-1)}(t)+\cdots+\overline{p}_{n}y(t)=0.\notag
\end{equation}
Further, putting $p_{0}:=1$ for simplicity, we have
\begin{equation}
\left\langle\left(
  \begin{array}{c}
    x\\
    x^{\prime}\\
    \vdots\\
    x^{(n-1)}
  \end{array}
\right),
\left(
  \begin{array}{c}
  \begin{array}{c}
    \sum_{j=0}^{n-1}(-1)^{j}\overline{p}_{n-1-j}y^{(j)}\\
    \sum_{j=0}^{n-2}(-1)^{j}\overline{p}_{n-2-j}y^{(j)}\\
    \vdots\\
    y
  \end{array}
  \end{array}
\right)\right\rangle
=\sum_{k=0}^{n-1}\sum_{j=0}^{k}(-1)^{j}p_{k-j}x^{(n-1-k)}\overline{y}^{(j)}=\text{constant}.\notag
\end{equation}
Higher-Order Scalar Equations with Variable Coefficients
Finally, we consider
\begin{equation}
x^{(n)}(t)+p_{1}(t)x^{(n-1)}(t)+\cdots+p_{n}(t)x(t)=0,\notag
\end{equation}
where $p_{i}(t)$ ($i=1,2,\cdots,n$) is complex-valued and is $i$ times continuously differentiable function.
Thus, the matrix representation is
\begin{equation}
\left(
  \begin{array}{c}
    x\\
    x^{\prime}\\
    \vdots\\
    x^{(n-1)}
  \end{array}
\right)^{\prime}
=
\left(
  \begin{array}{cccc}
    &1&&\\
    &&\ddots&\\
    &&&1\\
    -p_{n}(t)&-p_{n-1}(t)&\cdots&-p_{1}(t)
  \end{array}
\right)
\cdot
\left(
  \begin{array}{c}
    x\\
    x^{\prime}\\
    \vdots\\
    x^{(n-1)}
  \end{array}
\right).\notag
\end{equation}
Then, the associated adjoint matrix is
\begin{equation}
-
\left(
  \begin{array}{cccc}
    &1&&\\
    &&\ddots&\\
    &&&1\\
    -p_{n}(t)&-p_{n-1}(t)&\cdots&-p_{1}(t)
  \end{array}
\right)^{\ast}
=
\left(
  \begin{array}{cccc}
    &&&\overline{p}_{n}(t)\\
    (-1)&&&\overline{p}_{n-1}(t)\\
    &\ddots&&\vdots\\
    &&(-1)&\overline{p}_{1}(t)
  \end{array}
\right),\notag
\end{equation}
which yields the system
\begin{equation}
\begin{aligned}[]
&\left(
  \begin{array}{c}
    \sum_{j=0}^{n-1}(-1)^{j}[p_{n-1-j}\overline{y}]^{(j)}\\
    \sum_{j=0}^{n-2}(-1)^{j}[p_{n-2-j}\overline{y}]^{(j)}\\
    \vdots\\
    \overline{y}
  \end{array}
\right)^{\prime}\\
&=
\left(
  \begin{array}{cccc}
    &&&\overline{p}_{n}(t)\\
    (-1)&&&\overline{p}_{n-1}(t)\\
    &\ddots&&\vdots\\
    &&(-1)&\overline{p}_{1}(t)
  \end{array}
\right)
\cdot
\left(
  \begin{array}{c}
    \sum_{j=0}^{n-1}(-1)^{j}[p_{n-1-j}\overline{y}]^{(j)}\\
    \sum_{j=0}^{n-2}(-1)^{j}[p_{n-2-j}\overline{y}]^{(j)}\\
    \vdots\\
    \overline{y}
  \end{array}
\right),
\end{aligned}\notag
\end{equation}
where we put $p_{0}(t):\equiv1$ for simplicity.
Transforming this into the differential equation, we get
\begin{equation}
(-1)^{n}y^{(n)}(t)+(-1)^{n-1}[\overline{p}_{1}y]^{(n-1)}(t)+\cdots+[\overline{p}_{n-1}y]^{\prime}(t)+\overline{p}_{n}(t)y(t)=0.\notag
\end{equation}
Using the inner product in the first post gives us
\begin{equation}
\left\langle\left(
  \begin{array}{c}
    x\\
    x^{\prime}\\
    \vdots\\
    x^{(n-1)}
  \end{array}
\right),
\left(
  \begin{array}{c}
  \begin{array}{c}
    \sum_{j=0}^{n-1}(-1)^{j}[p_{n-1-j}\overline{y}]^{(j)}\\
    \sum_{j=0}^{n-2}(-1)^{j}[p_{n-2-j}\overline{y}]^{(j)}\\
    \vdots\\
    \overline{y}
  \end{array}
  \end{array}
\right)\right\rangle
=\sum_{k=0}^{n-1}\sum_{j=0}^{k}(-1)^{j}x^{(n-1-k)}[p_{k-j}\overline{y}]^{(j)}=\text{constant},\label{finaleq}\tag{#}
\end{equation}
which is the desired identity.
I believe that \eqref{finaleq} and (*) in the first post are equivalent.
