$\boldsymbol{\S}$ A. A special case : symmetric $\Omega^{\boldsymbol{-}1}$
Let the $2\times2$ real symmetric matrices
\begin{equation}
C^{\boldsymbol{-}1}\boldsymbol{=}
\begin{bmatrix}
\xi_1 & \xi \vphantom{\dfrac{a}{b}}\\
\xi &\xi_2 \vphantom{\dfrac{a}{b}}
\end{bmatrix}
\quad \text{and} \quad
L^{\boldsymbol{-}1}\boldsymbol{=}
\begin{bmatrix}
\eta_1 & \eta \vphantom{\dfrac{a}{b}}\\
\eta &\eta_2 \vphantom{\dfrac{a}{b}}
\end{bmatrix}
\tag{A-01}\label{A-01}
\end{equation}
Then
\begin{equation}
\Omega^{\boldsymbol{-}1}\boldsymbol{=}C^{\boldsymbol{-}1}L^{\boldsymbol{-}1}\boldsymbol{=}
\begin{bmatrix}
\xi_1\eta_1 \boldsymbol{+} \xi\eta & \xi_1\eta \boldsymbol{+} \xi\eta_2 \vphantom{\dfrac{a}{b}}\\
\hphantom{_1}\hphantom{_2}\xi\eta_1 \boldsymbol{+} \xi_2\eta & \hphantom{_1}\hphantom{_2}\xi\eta \boldsymbol{+}\xi_2\eta_2 \vphantom{\dfrac{a}{b}}
\end{bmatrix}
\tag{A-02}\label{A-02}
\end{equation}
With respect to the coordinates
\begin{equation}
\mathbf{V}
\boldsymbol{=}
\begin{bmatrix}
V_1\vphantom{\dfrac{a}{b}}\\
V_2\vphantom{\dfrac{a}{b}}
\end{bmatrix}
\tag{A-03}\label{A-03}
\end{equation}
the two coupled equations are
\begin{equation}
\dfrac{\mathrm d}{\mathrm dt}\left(\mathbf{\dot{V}}\right)\boldsymbol{-}\left(C^{\boldsymbol{-}1}\mathbf{J}\boldsymbol{+}\Omega^{\boldsymbol{-}1}\mathbf{K}\boldsymbol{-}\Omega^{\boldsymbol{-}1}\mathbf{V}\right)\boldsymbol{=}\boldsymbol{0}
\tag{A-04}\label{A-04}
\end{equation}
Now, if there exists a Lagrangian $\mathrm L\left(\mathbf{V},\mathbf{\dot{V}},t\right)$ for the problem then the Euler-Lagrange equations are
\begin{equation}
\dfrac{\mathrm d}{\mathrm dt}\left(\dfrac{\partial \mathrm L}{\partial \mathbf{\dot{V}}}\right)\boldsymbol{-}\dfrac{\partial \mathrm L}{\partial \mathbf{V}}\boldsymbol{=}\boldsymbol{0}
\tag{A-05}\label{A-05}
\end{equation}
where
\begin{equation}
\dfrac{\partial \mathrm L}{\partial \mathbf{V}}\boldsymbol{=}
\begin{bmatrix}
\dfrac{\partial \mathrm L}{\partial V_1} \vphantom{\dfrac{a}{\dfrac{a}{b}}}\\
\dfrac{\partial \mathrm L}{\partial V_2} \vphantom{\dfrac{a}{b}}
\end{bmatrix}
\quad \text{and} \quad
\dfrac{\partial \mathrm L}{\partial \mathbf{\dot{V}}}\boldsymbol{=}
\begin{bmatrix}
\dfrac{\partial \mathrm L}{\partial \dot{V}_1} \vphantom{\dfrac{a}{\dfrac{a}{b}}}\\
\dfrac{\partial \mathrm L}{\partial \dot{V}_2} \vphantom{\dfrac{a}{b}}
\end{bmatrix}
\tag{A-06}\label{A-06}
\end{equation}
Comparing equations \eqref{A-04} and \eqref{A-05} we note that the Lagrangian $\mathrm L\left(\mathbf{V},\mathbf{\dot{V}},t\right)$ must satisfy, except constants, the following two equations
\begin{align}
\dfrac{\partial \mathrm L}{\partial \mathbf{\dot{V}}} & \boldsymbol{=}\mathbf{\dot{V}}\vphantom{\dfrac{a}{\dfrac{a}{b}}}
\tag{A-07a}\label{A-07a}\\
\dfrac{\partial \mathrm L}{\partial \mathbf{V}} & \boldsymbol{=}C^{\boldsymbol{-}1}\mathbf{J}\boldsymbol{+}\Omega^{\boldsymbol{-}1}\mathbf{K}\boldsymbol{-}\Omega^{\boldsymbol{-}1}\mathbf{V}
\tag{A-07b}\label{A-07b}
\end{align}
From equation \eqref{A-07a} and partly because of the first two terms in the rhs of equation \eqref{A-07b} we note that one part $\mathrm L_1\left(\mathbf{V},\mathbf{\dot{V}},t\right)$ of the Lagrangian would be
\begin{equation}
\mathrm L_1\left(\mathbf{V},\mathbf{\dot{V}},t\right)\boldsymbol{=}\frac12\left(\mathbf{\dot{V}}\boldsymbol{\cdot}\mathbf{\dot{V}}\right)\boldsymbol{+}\left[\left(C^{\boldsymbol{-}1}\mathbf{J}\right)\boldsymbol{\cdot}\mathbf{V}\right]\boldsymbol{+}\left[\left(\Omega^{\boldsymbol{-}1}\mathbf{K}\right)\boldsymbol{\cdot}\mathbf{V}\right]
\tag{A-08}\label{A-08}
\end{equation}
while a second part $\mathrm L_2\left(\mathbf{V},\mathbf{\dot{V}},t\right)$ of the Lagrangian must satisfy the equation
\begin{equation}
\dfrac{\partial \mathrm L_2}{\partial \mathbf{V}} \boldsymbol{=}\boldsymbol{-}\Omega^{\boldsymbol{-}1}\mathbf{V}
\tag{A-09}\label{A-09}
\end{equation}
If the matrix $\Omega^{\boldsymbol{-}1}$ of equation \eqref{A-02} is symmetric, that is if the elements of the matrices $C^{\boldsymbol{-}1}$ and $L^{\boldsymbol{-}1}$ satisfy the condition
\begin{equation}
\left(\xi_1\boldsymbol{-}\xi_2\right)\eta\boldsymbol{=}\left(\eta_1\boldsymbol{-}\eta_2\right)\xi
\tag{A-10}\label{A-10}
\end{equation}
then
\begin{equation}
\mathrm L_2\left(\mathbf{V},\mathbf{\dot{V}},t\right) \boldsymbol{=}\boldsymbol{-}\frac12\left[\left(\Omega^{\boldsymbol{-}1}\mathbf{V}\right)\boldsymbol{\cdot}\mathbf{V}\right]
\tag{A-11}\label{A-11}
\end{equation}
and so
\begin{align}
&\mathrm L\left(\mathbf{V},\mathbf{\dot{V}},t\right) \boldsymbol{=}\mathrm L_1\left(\mathbf{V},\mathbf{\dot{V}},t\right)\boldsymbol{+}\mathrm L_2\left(\mathbf{V},\mathbf{\dot{V}},t\right) \qquad \textbf{for symmetric } \Omega^{\boldsymbol{-}1}
\nonumber\\
& \boldsymbol{=}\frac12\left(\mathbf{\dot{V}}\boldsymbol{\cdot}\mathbf{\dot{V}}\right)\boldsymbol{-}\frac12\left[\left(\Omega^{\boldsymbol{-}1}\mathbf{V}\right)\boldsymbol{\cdot}\mathbf{V}\right]\boldsymbol{+}\left[\left(C^{\boldsymbol{-}1}\mathbf{J}\right)\boldsymbol{\cdot}\mathbf{V}\right]\boldsymbol{+}\left[\left(\Omega^{\boldsymbol{-}1}\mathbf{K}\right)\boldsymbol{\cdot}\mathbf{V}\right]
\tag{A-12}\label{A-12}
\end{align}
$\boldsymbol{=\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!=}$
$\boldsymbol{\S}$ B. The general case : A systematic way to find the Lagrangian for two coupled second order linear differential equations
The efforts to find a Lagrangian for two coupled second order linear differential equations (as in the question) would be unsuccessful because of the so called $^{\prime\prime}$cross terms$^{\prime\prime}$ that appear at an intermediate step , for example terms like $V_1 V_2, \dot{V}_1 \dot{V}_2, \dot{V}_1 V_2$ etc. These terms "couple" the two equations. So we must find a method to eliminate terms of this kind. This will give us at first two uncoupled second order linear differential equations and next a well-defined Lagrangian.
Because of linearity we make a change of the variables from old $V_1, V_2$ to new $q_1, q_2$ via a linear transformation
\begin{align}
V_1 & \boldsymbol{=}a_{11}q_1\boldsymbol{+}a_{12}q_2
\tag{B-01a}\label{B-01a}\\
V_2 & \boldsymbol{=}a_{21}q_1\boldsymbol{+}a_{22}q_2
\tag{B-01b}\label{B-01b}
\end{align}
or
\begin{equation}
\mathbf{V}\boldsymbol{=}
\begin{bmatrix}
V_1\vphantom{\dfrac{a}{b}}\\
V_2\vphantom{\dfrac{a}{b}}
\end{bmatrix}
\boldsymbol{=}
\begin{bmatrix}
a_{11} & a_{12}\vphantom{\dfrac{a}{b}}\\
a_{21} & a_{22}\vphantom{\dfrac{a}{b}}
\end{bmatrix}
\begin{bmatrix}
p_1\vphantom{\dfrac{a}{b}}\\
p_2\vphantom{\dfrac{a}{b}}
\end{bmatrix}
\boldsymbol{=}A\mathbf{q}
\tag{B-02}\label{B-02}
\end{equation}
that is
\begin{equation}
\mathbf{V}\boldsymbol{=}A\mathbf{q}
\,,\qquad
A\boldsymbol{=}
\begin{bmatrix}
a_{11} & a_{12}\vphantom{\dfrac{a}{b}}\\
a_{21} & a_{22}\vphantom{\dfrac{a}{b}}
\end{bmatrix}
\tag{B-03}\label{B-03}
\end{equation}
and we'll try to find, if there exists, an invertible transformation $\:A\:$ that eliminates the cross terms so uncoupling the two equations.
If on our initial equation
\begin{equation}
\mathbf{\ddot{V}}\boldsymbol{+}\Omega^{\boldsymbol{-}1}\mathbf{V}\boldsymbol{=}C^{\boldsymbol{-}1}\mathbf{J}\boldsymbol{+}\Omega^{\boldsymbol{-}1}\mathbf{K}
\tag{B-04}\label{B-04}
\end{equation}
we apply from the left the transformation $\:A^{\boldsymbol{-}1}\:$ we have
\begin{equation}
A^{\boldsymbol{-}1}\mathbf{\ddot{V}}\boldsymbol{+}A^{\boldsymbol{-}1}\Omega^{\boldsymbol{-}1}\mathbf{V}\boldsymbol{=}A^{\boldsymbol{-}1}C^{\boldsymbol{-}1}\mathbf{J}\boldsymbol{+}A^{\boldsymbol{-}1}\Omega^{\boldsymbol{-}1}\mathbf{K}
\tag{B-05}\label{B-05}
\end{equation}
Making use of \eqref{B-03} we replace $\:\mathbf{V}\:$ by $\:A\mathbf{q}\:$ so
\begin{equation}
A^{\boldsymbol{-}1}\left(A\mathbf{\ddot{q}}\right)\boldsymbol{+}A^{\boldsymbol{-}1}\Omega^{\boldsymbol{-}1}\left(A\mathbf{q}\right)\boldsymbol{=}A^{\boldsymbol{-}1}C^{\boldsymbol{-}1}\mathbf{J}\boldsymbol{+}A^{\boldsymbol{-}1}\Omega^{\boldsymbol{-}1}\mathbf{K}
\nonumber
\end{equation}
that is
\begin{equation}
\mathbf{\ddot{q}}\boldsymbol{+}\left(A^{\boldsymbol{-}1}\Omega^{\boldsymbol{-}1} A\right)\mathbf{q}\boldsymbol{=}\left(A^{\boldsymbol{-}1}C^{\boldsymbol{-}1 }A\right)\mathbf{j}\boldsymbol{+}\left(A^{\boldsymbol{-}1}\Omega^{\boldsymbol{-}1 }A\right)\mathbf{k}
\tag{B-06}\label{B-06}
\end{equation}
or
\begin{align}
&\mathbf{\ddot{q}}\boldsymbol{+}W\,\mathbf{q} \boldsymbol{=}U\,\mathbf{j}\boldsymbol{+}W\,\mathbf{k}
\tag{B-07a}\label{B-07a}\\
&\text{where} \nonumber\\
&W\boldsymbol{=}A^{\boldsymbol{-}1}\Omega^{\boldsymbol{-}1}A\,, \quad U\boldsymbol{=}A^{\boldsymbol{-}1}C^{\boldsymbol{-}1}A\,, \quad \mathbf{j}\boldsymbol{=}A^{\boldsymbol{-}1}\mathbf{J}\,,\quad \mathbf{k}\boldsymbol{=}A^{\boldsymbol{-}1}\mathbf{K}
\tag{B-07b}\label{B-07b}
\end{align}
Now, the two second order linear differential equations \eqref{B-07a} would be uncoupled if the matrix $\:W\:$ could be diagonal
\begin{equation}
W\boldsymbol{=}A^{\boldsymbol{-}1}\Omega^{\boldsymbol{-}1 }A\boldsymbol{=}
\begin{bmatrix}
\mathrm w_1 & 0 \vphantom{\dfrac{a}{b}}\\
0 & \mathrm w_2\vphantom{\dfrac{a}{b}}
\end{bmatrix}
\tag{B-08}\label{B-08}
\end{equation}
This uncoupling is shown explicitly below
\begin{align}
\ddot{q}_1\boldsymbol{+}\mathrm w_1 p_1 &\boldsymbol{=}\left(U\,\mathbf{j}\right)_1 \boldsymbol{+}\left(W\,\mathbf{k}\right)_1
\tag{B-09a}\label{B-09a}\\
\ddot{q}_2\boldsymbol{+}\mathrm w_2 p_2 &\boldsymbol{=}\left(U\,\mathbf{j}\right)_2 \boldsymbol{+}\left(W\,\mathbf{k}\right)_2
\tag{B-09b}\label{B-09b}
\end{align}
These two independent $^{\prime\prime}$motions$^{\prime\prime}$ are called normal modes and the variables $q_1,q_2$ normal coordinates.
Now, from \eqref{B-08} the constants $\:\mathrm w_1,\mathrm w_2\:$ are the eigenvalues of the matrix $\:\Omega^{\boldsymbol{-}1}\:$ while the columns of the matrix $\:A\:$ are the eigenvectors respectively
\begin{align}
\mathbf{a}_1 & \boldsymbol{=}
\begin{bmatrix}
a_{11} \vphantom{\dfrac{a}{b}}\\
a_{21} \vphantom{\dfrac{a}{b}}
\end{bmatrix}\boldsymbol{=}\text{eigenvector of eigenvalue } \mathrm w_1
\tag{B-10a}\label{B-10a}\\
\mathbf{a}_2 & \boldsymbol{=}
\begin{bmatrix}
a_{12} \vphantom{\dfrac{a}{b}}\\
a_{22} \vphantom{\dfrac{a}{b}}
\end{bmatrix}\boldsymbol{=}\text{eigenvector of eigenvalue } \mathrm w_2
\tag{B-10b}\label{B-10b}
\end{align}
Note that depending on the matrix $\:\Omega^{\boldsymbol{-}1}\:$ the eigenvalues $\:\mathrm w_1,\mathrm w_2\:$ could be either both real or both complex conjugates.
Now, since the diagonal matrix $\:W\:$ is symmetric we make use of the results of $\boldsymbol{\S}$ A and we build the Lagrangian for the Euler-Lagrange equations \eqref{B-09a},\eqref{B-09b} according to equation \eqref{A-12}
\begin{equation}
\mathrm L\left(\mathbf{q},\mathbf{\dot{q}},t\right) \boldsymbol{=}
\tfrac12\left(\mathbf{\dot{q}}\boldsymbol{\cdot}\mathbf{\dot{q}}\right)\boldsymbol{-}\tfrac12\left[\left(W\mathbf{q}\right)\boldsymbol{\cdot}\mathbf{q}\vphantom{\dfrac{a}{b}}\right]\boldsymbol{+}\left[\left(U\mathbf{j}\right)\boldsymbol{\cdot}\mathbf{q}\vphantom{\dfrac{a}{b}}\right]\boldsymbol{+}\left[\left(W\mathbf{k}\right)\boldsymbol{\cdot}\mathbf{q}\vphantom{\dfrac{a}{b}}\right]
\tag{B-11}\label{B-11}
\end{equation}
Explicitly
\begin{align}
\mathrm L\left(\mathbf{q},\mathbf{\dot{q}},t\right) & \boldsymbol{=}
\tfrac12\left(\dot{q}^2_1\boldsymbol{+}\dot{q}^2_2\right)\boldsymbol{-}\tfrac12\left(\mathrm w_1 q^2_1\boldsymbol{+}\mathrm w_2 q^2_2\right)
\tag{B-12}\label{B-12}\\
&\boldsymbol{+} \left[\left(U\mathbf{j}\right)_1\boldsymbol{+}\left(W\mathbf{k}\right)_1\vphantom{\dfrac{a}{b}}\right]q_1\boldsymbol{+} \left[\left(U\mathbf{j}\right)_2\boldsymbol{+}\left(W\mathbf{k}\right)_2\vphantom{\dfrac{a}{b}}\right]q_2
\nonumber
\end{align}
Note that the above Lagrangian doesn't contain $^{\prime\prime}$cross terms$^{\prime\prime}$ like $q_1 q_2, \dot{q}_1 \dot{q}_2, \dot{q}_1 q_2$ etc.
Use of this Lagrangian in the equations below
\begin{align}
\dfrac{\mathrm d}{\mathrm dt}\left(\dfrac{\partial \mathrm L}{\partial \dot{q}_1}\right)\boldsymbol{-}\dfrac{\partial \mathrm L}{\partial q_1}\boldsymbol{=}0
\tag{B-13a}\label{B-13a}\\
\dfrac{\mathrm d}{\mathrm dt}\left(\dfrac{\partial \mathrm L}{\partial \dot{q}_2}\right)\boldsymbol{-}\dfrac{\partial \mathrm L}{\partial q_2}\boldsymbol{=}0
\tag{B-13b}\label{B-13b}
\end{align}
yields equations \eqref{B-09a} and \eqref{B-09b} as expected.
Now, based on \eqref{B-11} we can build the Lagrangian $\:\mathrm L\left(\mathbf{V},\mathbf{\dot{V}},t\right)\:$ for the initial coordinates $\:V_1,V_2\:$ from $\:\mathrm L\left(\mathbf{q},\mathbf{\dot{q}},t\right)$. We simply replace $\:\mathbf{q}\:$ by $\:A^{\boldsymbol{-}1}\mathbf{V}\:$ in \eqref{B-11} and we have
\begin{align}
&\mathrm L\left(\mathbf{V},\mathbf{\dot{V}},t\right)\boldsymbol{=}
\tag{B-14}\label{B-14}\\
&\tfrac12\left[\left(A^{\boldsymbol{-}1}\mathbf{\dot{V}}\right)\boldsymbol{\cdot}\left(A^{\boldsymbol{-}1}\mathbf{\dot{V}}\right)\vphantom{\dfrac{a}{b}}\right]\boldsymbol{-}\tfrac12\left[\left(A^{\boldsymbol{-}1}\Omega^{\boldsymbol{-}1}\mathbf{V}\right)\boldsymbol{\cdot}\left(A^{\boldsymbol{-}1}\mathbf{V}\right)\vphantom{\dfrac{a}{b}}\right]
\nonumber\\
&\boldsymbol{+}\left[\left(A^{\boldsymbol{-}1}C^{\boldsymbol{-}1}\mathbf{J}\right)\boldsymbol{\cdot}\left(A^{\boldsymbol{-}1}\mathbf{V}\right)\vphantom{\dfrac{a}{b}}\right]\boldsymbol{+}\left[\left(A^{\boldsymbol{-}1}\Omega^{\boldsymbol{-}1}\mathbf{K}\right)\boldsymbol{\cdot}\left(A^{\boldsymbol{-}1}\mathbf{V}\right)\vphantom{\dfrac{a}{b}}\right]
\nonumber
\end{align}
If $\:\Omega^{\boldsymbol{-}1}\:$ is (real) symmetric then the Lagrangian of \eqref{B-14} must yield that of \eqref{A-12}. But these two expressions are very different and it seems that we have a contradiction here. But there is no contradiction : in case of symmetric matrix $\:\Omega^{\boldsymbol{-}1}\:$ the eigenvalues $\:\mathrm w_1,\mathrm w_2\:$ are both real, the eigenvectors $\:\mathbf{a}_1,\mathbf{a}_2 $ of equations \eqref{B-10a},\eqref{B-10b} are orthogonal and the matrix $\:A\:$ of equations \eqref{B-02},\eqref{B-03} is orthogonal . For this matrix we have $\:A^{\boldsymbol{-}1}\boldsymbol{=}A^{\boldsymbol{\top}}\boldsymbol{=}\text{transpose of }A$. Replacing $\:A^{\boldsymbol{-}1}\:$ by $\:A^{\boldsymbol{\top}}\:$ the expression \eqref{B-14} becomes identical to \eqref{A-12}.In other words, since $\:A^{\boldsymbol{-}1}\:$ is also orthogonal it leaves the inner product of two vectors invariant, so in \eqref{B-14} we could replace any inner product $\:\left(A^{\boldsymbol{-}1}\mathbf{x}\right)\boldsymbol{\cdot}\left(A^{\boldsymbol{-}1}\mathbf{y}\right)\vphantom{\dfrac{a}{b}}\:$ by
$\:\left(\mathbf{x}\boldsymbol{\cdot}\mathbf{y}\right)\vphantom{\dfrac{a}{b}}$.
$\boldsymbol{=\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!=}$
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