Linear differential equations of the $n$th order $$ L(x)=x^{(n)}+a_1(t)x^{(n-1)}+\cdots +a_{n-1}(t)x'+a_n(t)x;\qquad a_1(t),a_2(t),\ldots\in C$$
$$U_j(\varphi)= \sum_{k=0}^{n-1}(M_{jk} \varphi^{k}(\alpha)-N_{jk} \varphi^{k}(\beta))= \gamma_j\quad \text{or} \quad U(\varphi)= \gamma$$ $$L(x)=b(t) \\ U(x)=0 \tag{1}$$
$(1)$ has to be solved. This is how the answer starts off, if anyone can pinpoint the detail, step that I am not seeing feel free to notify me.
Let $\varphi_{1}(t), \varphi_2(t),\ldots,\varphi_n(t)$ be the fundamental set of answers, the Vronski matrix : $W(t)=W(\varphi_1(t),\varphi_2(t),\ldots,\varphi_n(t))$ and the algebraic complement of the element $._{ni},\ \ $ $W_{ni}(t)$. We have the given formula: $$\varphi(t)= \sum_{i=1}^{n} \gamma_i \varphi_i(t)+ \sum_{i=1}^{n}\varphi_i(t) \int_{\alpha}^{t}\frac{W_{ni}(s)b(s)}{W(s)}ds $$ from here we have: $$\varphi(t)= \sum_{i=1}^{n} \gamma_i \varphi_i(t)+ \int_{\alpha}^{t}K^{*}(t,s)b(s)ds \tag{${*}{*}{*}{*}$}$$
where:
$$K^{*}(t,s)=\frac{1}{W(s)} \begin{vmatrix}
\varphi_1(s) & \varphi_2(s)& \cdots& \varphi_n(s)\\
\varphi_1'(s)& \varphi_2'(s)& \cdots & \varphi'_n(s)\\ \vdots& \vdots & \ddots & \vdots\\ 
\varphi^{(n-2)}_1(s)& \varphi^{(n-2)}_2(s)& \cdots & \varphi^{(n-2)}_n(s)\\ 
\varphi_1(s) & \varphi_2(s)& \cdots & \varphi_n(s)
\end{vmatrix}$$
Who understands: $({*}{*}{*}{*})$ ? Made an important update.
 A: I think I have got an idea how to decode the message, or at least its major part. 


*

*I start with a system of linear differential equations
$$
\dot z(t)=A(t)z(t)+f(t),\quad z(0)=z_0\in\mathbb{R}^n.\tag1
$$
If $\Phi(t)$ denotes the fundamental matrix of the corresponding homogeneous system, i.e.
$$
\dot\Phi(t)=A(t)\Phi(t),\qquad\Phi(0)=I,\tag2
$$
then the solution to (1) is given by the formula (take derivative to see it)
$$
z(t)=\Phi(t)z_0+\Phi(t)\int_0^t\Phi(s)^{-1}f(s)\,ds.\tag3
$$

*The linear differential equation of order $n$
$$
x^{(n)}(t)+a_1(t)x^{(n-1)}(t)+\ldots+a_n(t)x(t)=b(t)
$$
can be in the standard way converted to the form (1) if we denote
$$
z(t)=\left[\matrix{x(t)\\ x'(t)\\ \vdots\\x^{(n-2)}(t)\\x^{(n-1)}(t)}\right]\quad\Rightarrow
\quad
\dot z=\left[\matrix{0 & 1 & 0 &  \ldots & 0\\
0 & 0 & 1 &  \ldots & 0\\
\vdots & \vdots & \vdots & \ddots & \vdots \\
0 & 0 & 0  & \ldots & 1\\
-a_n & -a_{n-1} & -a_{n-2} & \ldots & -a_1}\right]z+
\left[\matrix{0\\ 0\\ \vdots\\ 0 \\b}\right].
$$
The fundamental matrix for the homogeneous part now will be the (scaled) Wronski matrix
$$
\Phi(t)=W(\phi_1(t),\ldots,\phi_n(t))W(\phi_1(0),\ldots,\phi_n(0))^{-1}=W(t)W(0)^{-1}.
$$
Now to get the general solution I am going to use (3), but I will a) move the initial time from zero to $\alpha$, and b) take only the first coordinate of $z$, which is $x=[1\ 0\ \ldots\ 0]z$, since the other coordinates are not interesting - just the derivatives of $x$.
$$
x(t)=[1\ 0\ \ldots\ 0]W(t)z_\alpha+[1\ 0\ \ldots\ 0]W(t)\int_\alpha^tW(s)^{-1}\left[\matrix{0\\ \vdots\\ 0 \\b(s)}\right]\,ds.
$$

*The rest is pretty technical. With $z_\alpha=\gamma$ (not quite get the connections to $U(\phi)=\gamma$ here) the first term becomes
$$
\sum_{i=1}^n\phi_i(t)\gamma_i
$$
and in the second term an explicit calculation of $W^{-1}=\frac{\text{adj}\,W}{\det W}$ is done. Actually. we need only the last column of the inverse since others are multiplied with zeros anyway. This is how we obtain the algebraic complements $W_{ni}$ and (NB!) $\det W$ in the denominator.

*Finally, the function $K^*(t,s)$. We need only the last element of
$$
[\phi_1(t)\ \phi_2(t)\ \ldots\ \phi_n(t)]W(s)^{-1}=[*\ *\ \ldots\ K^*(t,s)]
$$
since the others are multiplied by zeros. Post-multiplying by $W(s)$ and taking transpose yields
$$
W(s)^T\left[\matrix{*\\*\\\vdots\\ K^*(t,s)}\right]=
\left[\matrix{\phi_1(t)\\\phi_2(t)\\\vdots\\ \phi_n(t)}\right].
$$
Now Cramer's rule gives
$$
K^*(t,s)=\frac{1}{\det(W(s))}\left|\matrix{\phi_1(s) & \phi_2(s) & \ldots &\phi_n(s)\\\phi_1'(s) & \phi_2'(s) & \ldots &\phi_n'(s)\\
\vdots & \vdots & \ddots & \vdots\\
\phi_1^{(n-2)}(s) & \phi_2^{(n-2)}(s) & \ldots &\phi_n^{(n-2)}(s)\\
\phi_1(t) & \phi_2(t) & \ldots &\phi_n(t)}\right|.
$$

A: I will post a different solution to this question, a solution based on the Green's function method. Now, again, the question reads as follows . Given the fundamental set of solutions $\phi_j(t)$ to the homogeneous equation prove that the particular solution to the in-homogeneous equation can be constructed by integrating a kernel (or a Green's function) with the right hand side. 
In other words if:
\begin{equation}
(1) \quad \hat{L}(t) \phi_j(t) = 0
\end{equation}
for $j=1,\cdots,n$ then the particular solution to the equation below:
\begin{equation}
(2) \quad \hat{L}(t) y(t) = b(t)
\end{equation}
is given as follows:
\begin{equation}
y(t) = \int\limits_0^t \frac{1}{\det(W(\xi))} \left|
\begin{array}{rrrr}
\phi_1(\xi) & \phi_2(\xi) & \cdots & \phi_n(\xi)\\
\phi_1^{'}(\xi) & \phi_2^{'}(\xi) & \cdots & \phi_n^{'}(\xi)\\
\vdots\\
\phi_1^{(n-2)}(\xi) & \phi_2^{(n-2)}(\xi) & \cdots & \phi_n^{(n-2)}(\xi)\\
\phi_1(t) & \phi_2(t) & \cdots & \phi_n(t)\\
\end{array}
 \right| b(\xi) d\xi
\end{equation}
Proof: Clearly the solution to equation (2) has the form:
\begin{equation}
(3) \quad y(t) = \int\limits_0^t G(t,\xi) b(\xi) d\xi
\end{equation}
where 
\begin{equation}
(4) \quad \hat{L}(t) G(t,\xi) = \delta(t-\xi)
\end{equation}
This follows from applying the operator $\hat{L}(t)$ to both sides of equation (3) and using the linearity of the operator.
Now, fix $\xi$ and treat $t$ as a variable. Clearly equation (4) is very similar to equation (1) . In fact the two equations are exactly the same except at $t=\xi$. It is therefore reasonable to assume that the sought for Green's function $G(t,\xi)$ can be written as some sort of linear combination of the fundamental solutions . In other words we assume that:
\begin{eqnarray}
(5) \quad G(t,\xi)= \left\{
\begin{array}{rr}
\sum\limits_{j=1}^n C^{+}_j \cdot \phi_j(t) & \mbox{if $t\ge \xi$}\\
\sum\limits_{j=1}^n C^{-}_j \cdot \phi_j(t) & \mbox{if $t< \xi$}\\
\end{array}
\right.
\end{eqnarray} 
We determine the unknown constants $C^{\pm}$ using smoothness conditions. From the definition of the Dirac delta function it follows that, at $t=\xi$, the function itself along with all its derivatives up to order $(n-2)$ must be continuous whereas the $(n-1)$st derivative must have a jump of magnitude one. Therefore we write:
\begin{equation}
\left(
\begin{array}{rrrr}
\phi_1(\xi) & \phi_2(\xi) & \cdots & \phi_n(\xi)\\
\phi_1^{'}(\xi) & \phi_2^{'}(\xi) & \cdots & \phi_n^{'}(\xi)\\
\vdots\\
\phi_1^{(n-1)}(\xi) & \phi_2^{(n-1)}(\xi) & \cdots & \phi_n^{(n-1)}(\xi)\\
\phi_1^{(n-0)}(\xi) & \phi_2^{(n-0)}(\xi) & \cdots & \phi_n^{(n-0)}(\xi)
\end{array}
\right)
\left(
\begin{array}{r}
C_1^{+}-C_1^{-}\\
C_2^{+}-C_2^{-}\\
\vdots\\
C_{n-1}^{+}-C_{n-1}^{-}\\
C_{n-0}^{+}-C_{n-0}^{-}
\end{array}
\right)=
\left(
\begin{array}{r}
0\\
0\\
\vdots\\
0\\
1
\end{array}
\right)
\end{equation}
We solve the above system of equations using the Kramer rule and we have:
\begin{equation}
\left(
\begin{array}{r}
C_1^{+}-C_1^{-}\\
C_2^{+}-C_2^{-}\\
\vdots\\
C_{n-1}^{+}-C_{n-1}^{-}\\
C_{n-0}^{+}-C_{n-0}^{-}
\end{array}
\right)=
\left(
\begin{array}{r}
(W^{-1})_{1,n}\\
(W^{-1})_{2,n}\\
\vdots\\
(W^{-1})_{n-1,n}\\
(W^{-1})_{n,n}
\end{array}
\right)=
\frac{1}{\det(W)}
\left(
\begin{array}{r}
(M)_{n,1}\\
(M)_{n,2}\\
\vdots\\
(M)_{n,n-1}\\
(M)_{n,n}
\end{array}
\right)
\end{equation}
Here $W$ is the matrix of the fundamental system and $M_{i,j}$ is the algebraic complement of $W_{i,j}$. 
Now, since we are only interested in constructing a particular solution to the in-homogeneous system we can assume that  $C^{-}_j=0$ for $j=1,\cdots,n$. Therefore from (5) the Greens function reads:
\begin{equation}
G(t,\xi)= \frac{1}{\det(W)} \cdot \sum\limits_{j=1}^n (-1)^{n+j} \cdot M_{n,j} \cdot \phi_j(t)=
\frac{1}{\det(W)}
\left|
\begin{array}{rrrr}
\phi_1(\xi) & \phi_2(\xi) & \cdots & \phi_n(\xi)\\
\phi_1^{'}(\xi) & \phi_2^{'}(\xi) & \cdots & \phi_n^{'}(\xi)\\
\vdots\\
\phi_1^{(n-2)}(\xi) & \phi_2^{(n-2)}(\xi) & \cdots & \phi_n^{(n-2)}(\xi)\\
\phi_1(t) & \phi_2(t) & \cdots & \phi_n(t)\\
\end{array}
 \right|
\end{equation}
as expected.
