A first order system tends to be in the form
\[ \frac{dy}{dt}+k y=0 \]where $k$ is a constant
The solutions are of the form $y=e^{at}$, so it is possible to solve by comparing coefficients.
Substitute the solution into the first order equation we get
\[ ae^{at}+k e^{at}=0 \]so $a=-k$ and the solution is $y=e^{-kt}$
It is also possible to get a solution with the Laplace transform in which case the first order equation has the Laplace transform
\[ (sy-y_0)+k y=0 \]with $y_0$ is the value of $y$ at $t=0$
A first order system can have an input driving the response. These are of the form
\[ \frac{dy}{dt}+k y=u \]where $u$ is an input variable. The solution is slightly more complicated and for an arbitrary input the solution would be solved by numerical integration. Some specific inputs can be calculated so for a 'step change' the solution is of the form
\[ y=U_0\left(1-e^{-kt}\right) \]where $U_0$ is the size of the step.
Second order systems are widely used for modelling e.g.
A general form is
\[ \frac{d^2y}{ dt^2}+\beta\frac{dy}{dt}+\omega^2 y=0 \]where $\beta$ (beta) and $\omega$ are constant values (damping and natural frequency). Sometimes it will be convenient to write $\beta=2\zeta\omega$ i.e. use A state space form can be expressed by letting $x_1 \equiv y$ and $x_2 \equiv dy/dy$ so
\[ \begin{bmatrix}\dot{x_1}\\\dot{x_2}\\\end{bmatrix} = \begin{bmatrix}0 & 1\\-\omega^2&-\beta\end{bmatrix} \begin{bmatrix}{x_1}\\{x_2}\\\end{bmatrix} \]i.e. $\vec{\dot{x}}=A\vec{x}$ We can check this by substituting back
\[ \begin{bmatrix}\frac{dy}{dt}\\\frac{d^2y}{dt^2}\\\end{bmatrix} = \begin{bmatrix}\dot{x_1}\\\dot{x_2}\\\end{bmatrix} = \begin{bmatrix}0 & 1\\-\omega^2&-\beta\end{bmatrix} \begin{bmatrix}{y}\\\frac{dy}{dt}\\\end{bmatrix} \]A mechanical example is a mass connected to a spring and damper
By equating forces on the mass $m$ it is possible to derive the free running system as
\[ m\ddot{x}+c\dot{x}+k x=0 \]or the force driven system as
\begin{equation} m\ddot{x}+c\dot{x}+k x=f \label{eq:massspring} \end{equation}Equation \eqref{eq:massspring} can also be written as
\[ \ddot{x}+\frac{c}{m}\dot{x}+\frac{k}m x=\frac{f}{m} \]so it is then possible to see that $\omega^2=\frac{k}{m}$ and $\beta=\frac{c}{m}$
See (http:w8interactive.html)Recall that for a square matrix $A$ the Eigenvectors $v$ are special vectors that when real have the same direction when transformed by A.
So if
\[ v'=A v \]then $v$ is an Eigenvector if $v'=\lambda v$. i.e.
\begin{equation} Av=\lambda v \label{eq:eigval} \end{equation}The same is true if the Eigenvectors are complex, but it is may be harder to conceptulise a complex scaling value $\lambda$ and complex direction of $v$. One additional cavet is that if the Eigenvalue is real and negative then $v'$ will point in exactly the oposite direction to $v$, that is to say it now as negative magnitude.
From the equation \eqref{eq:eigval} above we can compute a solution by noting that
\[ (A-\lambda I)v=0 \]and by realsing that for any non zero vector $v$ the matrix $A-\lambda I$ must be singular and in which case the determinant must be zero hence
\[ \begin{vmatrix}A-\lambda I\end{vmatrix} = 0 \]The stability of the system is dictated by the Eigenvalues of the $A$ matrix, more specifically whether the real parts of the Eigenvalues are greater than zero. For a second order system (where we use the substitution $\beta=2\zeta\omega$) the the Eigenvalues are
\[ \lambda=-\omega(\zeta \pm\sqrt{\zeta^2 - 1}) \]Now omega ($\omega$) only effects the magnitude of the Eigenvalues, so we can map out the behaviour of the system based on the value of zeta ($\zeta$).
| zeta | lambda ($\lambda$) | behaviour |
| $\zeta<-1$ | real positive numbers | exponentially increasing (unstable) |
| $-1<\zeta< 0$ | complex with positive real part | increasing oscillations (unstable) |
| $\zeta=0$ | imaginary (real part is zero) | stable oscillations |
| $0<\zeta< 1$ | complex with negative real part | decreasing oscillations (stable) |
| $\zeta>1$ | real negative numbers | exponentially decreasing (stable) |