Suppose you have a system whose kinematics are given by a list of $N$"configuration" variables $q = \left(q^a: 0 ≤ a < N\right)$, for some number $N = 1, 2, 3, ⋯$ of "degrees of freedom" of the system, that satisfy a$$\frac{dq^a}{dt} = v^a \hspace 1em (0 ≤ a < N). \label{Kinematic Law}\tag{Kinematic Law}$$for a list $v = \left(v^a: 0 ≤ a < N\right)$ describing the respective components of the "velocities".
Suppose its dynamics is described by a list of corresponding components of momentum $p = \left(p_a: 0 ≤ a < N\right)$ that satisfy a$$\frac{dp_a}{dt} = f_a \hspace 1em (0 ≤ a < N), \label{Dynamic Law}\tag{Dynamic Law}$$for a list $f = \left(f_a: 0 ≤ a < N\right)$ describing the respective components of the force that governs the dynamics of the momentum.
Up to this point, the description is an empty shell whose only material assertion is that the system in question actually does have such attributes that describe it. What fills in the shell is a further set of relations that connect $q$, $v$, $p$ and $f$ to one another - the constitutive laws that describe the system, in question.
A cardinal example is where the $p$'s are functions of the $v$'s, alone, that satisfy a "principle of reciprocity":$$\frac{∂p_a}{∂v^b} = m_{ab} = m_{ba} = \frac{∂p_b}{∂v^a} \hspace 1em (0 ≤ a,b < N), \label{1a}\tag{1a}$$with the derivatives $m_{ab}$ defining coefficients that would then be called its "moments of inertia" - i.e. its "mass matrix". The simplest cases in point are for $n$ bodies, with $N = 3n$ being the total number of coordinates, where $m$ is a constant diagonal matrix - actually a $3×n$-block diagonal containing 3 copies, each, of the $n$ masses of the bodies as its diagonal components.
We can also consider cases where the $p$'s may also be functions of the $q$'s, continuing to maintain reciprocity even in this more general context. In such cases, we can express the $p$'s themselves in terms of a generating function$\bar{T}(q,v)$ by$$p_a = \frac{∂\bar{T}}{∂v^a} \hspace 1em (0 ≤ a < N). \label{1b}\tag{1b}$$One example may be where each body has, in addition to a "kinetic momentum" $P_a = ∂T(v)/∂v^a$, also a "potential momentum" $A_a(q)$, so that we could write $p_a = P_a(v) + A_a(q)$. In this case, the coefficients of inertia arise solely from $T(v)$, while $\bar{T}(q,v)$ may be expressed as:$$\bar{T}(q,v) = T(v) + V(q,v), \hspace 1em V(q,v) = \sum_{0 ≤ a < N} v^a A_a(q). \label{Ex1}\tag{Ex1}$$
Suppose the components of the force $f$ are functions of those of the configuration $q$ that satisfy their own reciprocity principle:$$\frac{∂f_a}{∂q^b} = -k_{ab} = -k_{ba} = \frac{∂f_b}{∂q^a} \hspace 1em (0 ≤ a,b < N). \label{2a}\tag{2a}$$This case is generic for forces derivable from a potential $U(q)$, the simplest case in point being that for harmonic oscillators:$$U(q) = \frac{1}{2} \sum_{0≤a,b<N} {k_{ab} q^a q^b},$$where the matrix $k$ of "spring coefficients" is positive definite, with $f_a = -{∂U}/{∂q^a}$.
More generally, we can conceive of forces that may also be velocity-dependent, that may be derivable from a generating function $\bar{U}(q,v)$ that depends on both the configurations $q$and their velocities $v$:$$f_a = -\frac{∂\bar{U}}{∂q^a} \hspace 1em (0 ≤ a < N). \label{2b}\tag{2b}$$
The generating function plays the role of potential energy. An example of a velocity-dependent potential, continuing on from our example ($\ref{Ex1}$) above, would be:$$\bar{U}(q,v) = U(q) - \sum_{0≤b<N} v^b A_b(q) = U(q) - V(q,v). \label{Ex2}\tag{Ex2}$$This, of course, generalizes from the fairly well-known example of electromagnetism, where we can write the force law in the form:$$\frac{d}{dt}(m𝐯+ e𝐀(𝐫)) = -\frac{∂\bar{U}(t,𝐫,𝐯)}{∂𝐫}, \hspace 1em \bar{U}(t,𝐫,𝐯) = φ(t,𝐫) - 𝐯·𝐀(t,𝐫).$$
What we find, with the choice ($\ref{Ex2}$) for $\bar{U}$, is that the force and momentum satisfy a third principle of reciprocity:$$\frac{∂f_a}{∂v^b} = \frac{∂}{∂v^b}\left(-\frac{∂U}{∂q^a} + \sum_{0≤b<N} v^b \frac{∂A_b}{∂q^a}\right) = \frac{∂A_b}{∂q^a} = \frac{∂p_b}{∂q^a}.$$
So, let's suppose in the general case that in addition to ($\ref{1a}$) and ($\ref{2a}$), we have a third principle of reciprocity:$$\frac{∂f_a}{∂v^b} = l_{ab} = \frac{∂p_b}{∂q^a} \hspace 1em (0 ≤ a,b < N). \label{3a}\tag{3a}$$
From ($\ref{1a}$), we infer that there is a generating function $\bar{T}(q,v)$ for the momentum $p$ that satisfies ($\ref{1b}$). From ($\ref{2a}$), we also infer that a second generating function $\bar{U}(q,v)$ exists for the force $f$ that satisfies ($\ref{2b}$). Applying ($\ref{1b}$) and ($\ref{2b}$) to ($\ref{3a}$), we then find that:$$\frac{∂^2\bar{T}}{∂q^a∂v^b} = \frac{∂p_b}{∂q^a} = \frac{∂f_a}{∂v^b} = -\frac{∂^2\bar{U}}{∂q^a∂v^b} \hspace1em⇒\hspace1em \frac{∂^2}{∂q^a∂v^b}\left(\bar{T} + \bar{U}\right) = 0 \hspace 1em (0 ≤ a,b < N). \label{3b}\tag{3b}$$
Therefore, even though $\bar{T}(q,v)$ may not be a function of $v$ alone, nor $\bar{U}(q,v)$ a function of $q$ alone, their sum does separate into sums of such functions:$$\bar{T}(q,v) + \bar{U}(q,v) = T(v) + U(q).$$Therefore, defining$$V(q,v) ≡½ ((T(q,v) - T(v)) - (U(q,v) - U(q))),$$we may write:$$\bar{T}(q,v) = T(v) + V(q,v), \hspace 1em \bar{U}(q,v) = U(v) - V(q,v),$$just like in ($\ref{Ex1}$) and ($\ref{Ex2}$).
From this, it follows, for $(0 ≤ a < N)$, that:$$p_a = \frac{∂}{∂v^a}(T(v) + V(q,v)) = \frac{∂}{∂v^a}(T(v) - U(q) + V(q,v)),\\f_a = -\frac{∂}{∂q^a}(U(q) - V(q,v)) = \frac{∂}{∂q^a}(T(v) - U(q) + V(q,v)).$$Applying this to the $\ref{Dynamic Law}$ and adding in the $\ref{Kinematic Law}$, we find that:$$\frac{d}{dt}\left(\frac{∂L}{∂v^a}\right) = \frac{∂L}{∂q^a}, \hspace 1em \frac{dq^a}{dt} = v^a \hspace 1em (0 ≤ a < N),$$where$$L(q,v) = T(v) - U(q) + V(q,v),$$defines the "Lagrangian" $L(q,v)$ for kinematics and dynamics.
A Lagrangian is a function for generating the constitutive relations that fill in the empty shell provided by the $\ref{Kinematic Law}$ and $\ref{Dynamic Law}$. The reciprocity conditions ($\ref{1a})$, ($\ref{2a}$) and ($\ref{3a}$) determine where this is possible.
Example:
The simplest example, apart from the $L(q,v) = T(v) - U(q)$ cases is where the $k$, $l$ and $m$ coefficients are all constant, with $l_{ab} = -l_{ba}$:$$L = ½ \sum_{0≤a,b<N} \left(m_{ab} v^a v^b + l_{ab} \left(q^a v^b - v^a q^b\right) - k_{ab} q^a q^b\right).$$Then, for $(0 ≤ a < N)$, we have the following constitutive relations:$$p_a = \sum_{0≤b<N} \left(m_{ab} v^b - l_{ab} q^b\right) = +\sum_{0≤b<N} \left(m_{ba} v^b + l_{ba} q^b\right), \\f_a = \sum_{0≤b<N} \left(l_{ab} v^b - k_{ab} q^b\right) = -\sum_{0≤b<N} \left(l_{ba} v^b + k_{ba} q^b\right).$$Substituting this into the ($\ref{Dynamic Law}$) and applying the ($\ref{Kinematic Law}$), we get the following system of second order equations:$$\sum_{0≤b<N} \left(m_{ba}\frac{d^2q^b}{dt^2} + 2l_{ba}\frac{dq^b}{dt} + k_{ba}q^b\right) = 0 \hspace 1em (0 ≤ a < N).$$As a special case of this example, for $N = 2$, with $(x,y) = \left(q^0,q^1\right)$, if the $m$ and $k$ matrices are diagonal, their diagonal components written respectively as just $m$ and $k$, and if $l_{01} = -l_{10}$ is written as just $l$, then we have:$$m \frac{d^2x}{dt^2} - 2l \frac{dy}{dt} + k x = 0, \hspace 1emm \frac{d^2y}{dt^2} + 2l \frac{dx}{dt} + k y = 0.$$Writing the velocity components are $(\dot{x},\dot{y}) = \left(v^0,v^1\right)$, we can find two constants of motion from this:$$H = m\frac{\dot{x}^2 + \dot{y}^2}{2} + k\frac{x^2 + y^2}{2}, \hspace 1em J = m\left(x\dot{y} - y\dot{x}\right) + l \left(x^2 + y^2\right).$$