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Ordinary Differential Equation--System with Constant Coefficients

To solve the system of differential equations

{d{\bf x}\over dt} = {\hbox{\sf A}}{\bf x}(t) + {\bf p}(t),
\end{displaymath} (1)

where ${\hbox{\sf A}}$ is a Matrix and ${\bf x}$ and ${\bf p}$ are Vectors, first consider the homogeneous case with ${\bf p}={\bf0}$. Then the solutions to
{d{\bf x}\over dt} = {\hbox{\sf A}}{\bf x}(t)
\end{displaymath} (2)

are given by
{\bf x}(t)=e^{\hbox{{\sf A}}t}{\bf x(t)}.
\end{displaymath} (3)

But, by the Matrix Decomposition Theorem, the Matrix Exponential can be written as
e^{\hbox{{\sf A}}t}={\hbox{\sf u}}{\hbox{\sf D}}{\hbox{\sf u}}^{-1},
\end{displaymath} (4)

where the Eigenvector Matrix is
{\hbox{\sf u}} = \left[{\matrix{{\bf u}_1 & \cdots & {\bf u}_n\cr}}\right]
\end{displaymath} (5)

and the Eigenvalue Matrix is
{\hbox{\sf D}} = \left[{\matrix{e^{\lambda_1t} & 0 &\cdots &...
... & \ddots & 0\cr
0 & 0 & \cdots & e^{\lambda_nt}\cr}}\right].
\end{displaymath} (6)

Now consider

$\displaystyle e^{\hbox{{\sf A}}t}{\hbox{\sf u}}$ $\textstyle =$ $\displaystyle {\hbox{\sf u}}{\hbox{\sf D}}{\hbox{\sf u}}^{-1}{\hbox{\sf u}} = {\hbox{\sf u}}{\hbox{\sf D}}$  
  $\textstyle =$ $\displaystyle \left[\begin{array}{cccc} u_{11} & u_{21} & \cdots & u_{n1}\\  u_...
...dots & \vdots & \ddots & 0\\  0 & 0 & \cdots & e^{\lambda_nt}\end{array}\right]$  
  $\textstyle =$ $\displaystyle \left[\begin{array}{ccc} u_{11}e^{\lambda_1t} & \cdots & u_{n1}e^...
...dots\\  u_{n1}e^{\lambda_1t} & \cdots & u_{n2}e^{\lambda_nt}\end{array}\right].$ (7)

The individual solutions are then
{\bf x}_i = (e^{\hbox{{\sf A}}t}{\hbox{\sf u}})\cdot \hat {\bf e}_i = {\bf u}_i e^{\lambda_i t},
\end{displaymath} (8)

so the homogeneous solution is
{\bf x}=\sum_{i=1}^n c_i {\bf u}_i e^{\lambda_it},
\end{displaymath} (9)

where the $c_i$s are arbitrary constants.

The general procedure is therefore

1. Find the Eigenvalues of the Matrix ${\hbox{\sf A}}$ ($\lambda_1$, ..., $\lambda_n$) by solving the Characteristic Equation.

2. Determine the corresponding Eigenvectors ${\bf u}_1$, ..., ${\bf u}_n$.

3. Compute
{\bf x}_i \equiv e^{\lambda_it}{\bf u}_i
\end{displaymath} (10)

for $i = 1$, ..., $n$. Then the Vectors ${\bf x}_i$ which are Real are solutions to the homogeneous equation. If ${\hbox{\sf A}}$ is a $2\times 2$ matrix, the Complex vectors ${\bf x}_j$ correspond to Real solutions to the homogeneous equation given by $\Re({\bf x}_j)$ and $\Im({\bf x}_j)$.

4. If the equation is nonhomogeneous, find the particular solution given by
{\bf x}^*(t) = {\hbox{\sf X}}(t) \int {\hbox{\sf X}}^{-1}(t){\bf p}(t)\,dt,
\end{displaymath} (11)

where the Matrix ${\hbox{\sf X}}$ is defined by
{\hbox{\sf X}}(t) \equiv \left[{\matrix{{\bf x}_1 & \cdots & {\bf x}_n\cr}}\right].
\end{displaymath} (12)

If the equation is homogeneous so that ${\bf p}(t)={\bf0}$, then look for a solution of the form
{\bf x}={\boldsymbol{\xi}}e^{\lambda t}.
\end{displaymath} (13)

This leads to an equation
({\hbox{\sf A}}-\lambda {\hbox{\sf I}})\boldsymbol{\xi}={\bf0},
\end{displaymath} (14)

so ${\bf\xi}$ is an Eigenvector and $\lambda$ an Eigenvalue.

5. The general solution is
{\bf x}(t) = {\bf x}^*(t) + \sum_{i=1}^n c_i {\bf x}_i.
\end{displaymath} (15)

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© 1996-9 Eric W. Weisstein