The following is a brief description of the different ways a complete solution to a differential equation might be broken up. The three possibilities are:
- Homogeneous and Particular (aka natural and forced)
- Zero-input and Zero-state
- Transient and Steady-state
To give an example, consider the following equation (from Haykin/Van Veen, p. 503, with a different input):
\(
\begin{align}
\frac{d^2}{dt^2}y(t)+5\frac{d}{dt}y(t)+6y(t)&=\frac{d}{dt}x(t)+6x(t)\\
x(t)&=u(t)-e^{-t}u(t)\\
y(0^-)&=1\\
\dot{y}(0^-)&=2
\end{align}
\)
Homogeneous and Particular
The homogeneous solution is the solution to the differential equation when the forcing function \(x(t)=0\). In other words,
\(
\begin{align}
\frac{d^2}{dt^2}y_h(t)+5\frac{d}{dt}y_h(t)+6y_h(t)&=0\\
\end{align}
\)
Using a general complex exponential as an example:
\(
\begin{align}
y_h(t)&=Ke^{st}
\end{align}
\)
yields
\(
\begin{align}
s^2Ke^{st}+5sKe^{st}+6Ke^st=0\\
\end{align}
\)
which is only true if \(K=0\) (trivial) or
\(
\begin{align}
s^2+5s+6&=0
s&=-2, -3
\end{align}
\)
meaning
\(
\begin{align}
y_h(t)=K_1e^{-2t}+K_2e^{-3t}
\end{align}
\)
It is impossible to find the constants without also knowing what the particular solution is.
The particular solution is going to resemble the forcing function - in this case, a constant plus an exponential with a decay of -1. That is,
\(
\begin{align}
y_p(t)&=C_1+C_2e^{-t}\\
\dot{y}_p(t)&=-C_2e^{-t}\\
\ddot{y}_p(t)&=C_2e^{-t}
\end{align}
\)
and
\(
\begin{align}
\frac{d^2}{dt^2}y_p(t)+5\frac{d}{dt}y_p(t)+6y_p(t)&=\frac{d}{dt}x(t)+6x(t)\\
\left(C_2e^{-t} \right) +
5\left(-C_2e^{-t} \right) +
6\left(C_1+C_2e^{-t} \right)
&=0+e^{-t}+6-6e^{-t}
\end{align}
\)
from which, using harmonic balance, the following equations may be found:
\(
\begin{align}
6C_1=6 &\longrightarrow C_1=1\\
2C_2=-5 & \longrightarrow C_2=-\frac{5}{2}
\end{align}
\)
such that
\(
\begin{align}
y(t)=y_h(t)+y_p(t)&=K_1e^{-2t}+K_2e^{-3t}+1-\frac{5}{2}e^{-t}\\
\dot{y}&=-2K_1e^{-2t}-3K_2e^{-3t}+\frac{5}{2}e^{-t}
\end{align}
\)
Since the values for \(y\) and \(\dot{y}\) are known at time 0, solve:
\(
\begin{align}
y(0)=K_1+K_2+1-\frac{5}{2}=1&\longrightarrow K_1+K_2=\frac{5}{2}\\
\dot{y})(0)=-2K_1-3K_2+\frac{5}{2}=2&\longrightarrow 2K_1+3K_2=\frac{1}{2}
\end{align}
\)
such that
\(
\begin{align}
K_1&=7 & K_2&=-\frac{9}{2}
\end{align}
\)
meaning:
\(
\begin{align}
y_h(t)&=7e^{-2t}-\frac{9}{2}e^{-3t}\\
y_p(t)&=1-\frac{5}{2}e^{-t}\\
y(t)&=7e^{-2t}-\frac{9}{2}e^{-3t}+1-\frac{5}{2}e^{-t}
\end{align}
\)
