Homogeneous DE System

Consider solving a set of $n$ first order constant coefficient ordinary differential equations (ODE):

$$\frac{d}{dt}y_i(t)=\dot{y}_i(t)=\sum_{j=1}^n a_{ij} y_j(t)+x_i(t), \;\;\;\;\;\;\;(i=1,\cdots,n)$$

which can be written in matrix form:

$$\dot{\bf y}(t)={\bf A}{\bf y}(t)+{\bf x}(t)$$

where

$${\bf x}=\left[\begin{array}{c}x_1 \vdots x_n\end{array}\... ...& \ddots & \vdots a_{n1} & \cdots &a_{nn}\end{array}\right]$$

In particular, if ${\bf x}(t)={\bf0}$, we get a homogeneous ODE system:

$$\dot{\bf y}={\bf A}{\bf y}$$

We let the solution take the form of ${\bf y}(t)=e^{\lambda t}{\bf v}$, where scalor $\lambda$ and vector ${\bf v}$ are to be determined. We then have e $\dot{\bf y}(t)=\lambda e^{\lambda t}{\bf v}$. Substituting these into the DE system we get

$$\dot{\bf y}=\lambda e^{\lambda t}{\bf v}={\bf A}{\bf y} ={\bf A}e^{\lambda t}{\bf v}$$

Dividing both sides by $e^{\lambda t}\ne 0$, we get:

$${\bf Av}=\lambda{\bf v}$$

This happens to be the eigenequation of ${\bf A}$, and $\lambda$ and ${\bf v}$ can be found as the eigenvalue and the corresponding eigenvector of ${\bf A}$. In general, solving this equation we get $n$ eigenvalues $\lambda_1,\cdots,\lambda_n$ and their corresponding eigenvectors ${\bf v}_1,\cdots,{\bf v}_n$. We therefore get a set of $n$ fundamental solutions $e^{\lambda_it}{\bf v}_i\;(i=1,\cdots,n)$ of the ODE system, all satisfying

$${\bf Av}_i=\lambda_i{\bf v}_i,\;\;\;\;\;\;(i=1,\cdots,n)$$

These $n$ equations can be combined to become

$${\bf A}[{\bf v}_1,\cdots,{\bf v}_n]=[{\bf v}_1,\cdots,{\bf v... ...ight] \;\;\;\;\;\;\mbox{or}\;\;\;\;\;\;{\bf AV}={\bf V\Lambda}$$

where

$${\bf\Lambda}=\left[\begin{array}{ccc}\lambda_1&\cdots&0 \... ...ray}\right],\;\;\;\;\;\; {\bf V}=[{\bf v}_1,\cdots,{\bf v}_n]$$

are the eigenvalue and eigenvector matrices of ${\bf A}$.

The fundamental matrix of the ODE system $\dot{\bf y}={\bf Ay}$ is a matrix of which each column is a solution:

$$[e^{\lambda_1t}{\bf v}_1,\cdots,e^{\lambda_nt}{\bf v}_n] =[{\... ...\cdots&e^{\lambda_nt}\end{array}\right] ={\bf V}e^{{\Lambda}t}$$

The general solution of the ODE system is a linear combination of the $n$ fundamental solutions

$${\bf y}(t)=\sum_{i=1}^n c_i e^{\lambda_it}{\bf v}_i =[e^{\la... ...vdots c_n\end{array}\right] ={\bf V}e^{{\bf\Lambda}t}{\bf c}$$

where ${\bf c}=[c_1,\cdots,c_n]^T$ is a vector containing $n$ coefficients, which can be found based on the $n$ given initial conditions ${\bf y}(0)=[y_1(0),\cdots,y_n(0)]^T$. Specifically, evaluating the solution ${\bf y}(t)$ at $t=0$, we get

$${\bf y}(t)\bigg\vert _{t=0}=\sum_{i=1}^n c_i e^{\lambda_it}{... ..._1 \vdots c_n\end{array}\right] ={\bf V}{\bf c}={\bf y}(0)$$

Solving the equation ${\bf Vc}={\bf y}(0)$ we get the $n$ coefficients:

$${\bf c}={\bf V}^{-1}{\bf y}(0)$$

Substituting this ${\bf c}$ back into the general solution above, we finally get the homogeneous solution of the ODE system:

$${\bf y}(t)={\bf V}e^{{\bf\Lambda}t}{\bf c} ={\bf V}e^{{\bf\Lambda}t}{\bf V}^{-1}{\bf y}(0)$$

Example

Solve the following DE

$$\left\{\begin{array}{l}\dot{y}_1=2y_1+3y_2 \dot{y}_2=4y_1-2y_2\end{array}\right.$$

with initial conditions $y_1(0)=5$ and $y_2(0)=6$. The coefficient matrix is

$${\bf A}=\left[\begin{array}{rr}2 & 3 4 & -2\end{array}\right]$$

and its eigenvalues are $\lambda_1=4$ and $\lambda_2=-4$ and their corresponding eigenvectors are

$${\bf v}_1=c_1\left[\begin{array}{r}3 2\end{array}\right], ... ...; {\bf v}_2=c_2\left[\begin{array}{r}-1 2\end{array}\right],$$

The solution is

$${\bf y}=c_1e^{4t}\left[\begin{array}{r}3 2\end{array}\right] +c_2e^{-4t}\left[\begin{array}{r}-1 2\end{array}\right],$$

The coefficients can be found by

$${\bf c}={\bf V}^{-1}{\bf y}(0)=\left[\begin{array}{rr}3 & -1... ...d{array}\right] =\left[\begin{array}{r}2 1\end{array}\right]$$

The solution is

$${\bf y}=2e^{4t}\left[\begin{array}{r}3 2\end{array}\right] +e^{-4t}\left[\begin{array}{r}-1 2\end{array}\right],$$

i.e.,

$$\left\{\begin{array}{l}y_1(t)=6e^{4t}-e^{-4t} y_2(t)=4e^{4t}+2e^{-4t}\end{array}\right.$$