Recently came across to solve system of linear differential equation with the matrix method, I have learnt that the eigenvectors forms part of the complementary function.
A matrix method such as :
$$ dy/dt = 5x+7y$$ $$ dx/dt =9x +4y $$
which then we replace a vector x $=(x,y)$ and a matrix M with entries (5 7,9 4)
Is there any reason for this (such as the one in ordinary Differential equation is due to the fact that the exponential is the eigenfunction if the differential which also exist as a matrix method, but why the eigenvectors in the front of it?)
Make a variable substitution. If you have a homogeneous linear system of ODEs $\dot{x} = Ax$, where $A$ has a basis of eigenvectors $v_1, \ldots, v_n$ and corresponding eigenvalues $\lambda_1, \ldots, \lambda_n$, then make a substitution in terms of a new vector variable $y$, the coordinate vector of $x$ with respect to $v_1, \ldots, v_n$. Then, $$x = y_1 v_1 + \ldots + y_n v_n \implies \dot{x} = \dot{y}_1 v_1 + \ldots + \dot{y}_n v_n.$$ Hence, the system becomes, \begin{align*} \dot{x} &= Ax \\ \dot{y}_1 v_1 + \ldots + \dot{y}_n v_n &= A(y_1 v_1 + \ldots + y_n v_n) \\ &= y_1 Av_1 + \ldots + y_n A v_n \\ &= \lambda_1 y_1v_1 + \ldots + \lambda_ny_nv_n. \end{align*} By the linear independence of $v_1, \ldots, v_n$, we may equate the (non-constant, but still scalar) coefficients of $v_i$ to obtain $$\dot{y}_i = \lambda_iy_i,$$ for each $i$. In other words, we have now decoupled the system into $n$ totally independent first order ODEs, in terms of the coordinates of $y$. These have the usual solutions: $$y_i = A_ie^{\lambda_i t},$$ for some constant $A_i$. Thus, changing back, $$x = y_1 v_1 + \ldots + y_n v_n = A_1 e^{\lambda_1 t} v_1 + \ldots + A_n e^{\lambda_n t} v_n.$$