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):
\(
\begin{align}
\frac{d^2}{dt^2}y(t)+5\frac{d}{dt}y(t)+6y(t)&=\frac{d}{dt}x(t)+6x(t)\\
x(t)&=1-e^{-t}\\
y(0^-)&=1\\
\dot{y}(0^-)&=2
\end{align}
\)
Note - unilateral Laplace transforms and signals starting at time 0 are assumed; \(u(t)\) is assumed everywhere.
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
\(
\begin{align}
y(t)=y_h(t)+y_p(t)&=K_1e^{-2t}+K_2e^{-3t}+1-\frac{5}{2}e^{-t}\\
\dot{y}(t)&=-2K_1e^{-2t}-3K_2e^{-3t}+\frac{5}{2}e^{-t}
\end{align}
\)
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:
\(
\begin{align}
y_h(t)&=7e^{-2t}-\frac{9}{2}e^{-3t}\\
y_p(t)&=1-\frac{5}{2}e^{-t}\\
y(t)&=7e^{-2t}-\frac{9}{2}e^{-3t}+1-\frac{5}{2}e^{-t}
\end{align}
\)
Zero-input and Zero-state
The complete solution is also the sum of the zero-input and the zero-state solutions. The zero-input response is how the system responds solely to the initial state, neglecting any forcing function. The form of the zero-input response looks the same as the homogeneous response above, but the coefficients can be determined without knowledge of the forcing function (the "input"). That is,
\(
\begin{align}
y_{zi}(t)&=K_1e^{-2t}+K_2e^{-3t}\\
\dot{y}_{zi}(t)&=-2K_1e^{-2t}-3K_2e^{-3t}
\end{align}
\)
Given the state at time 0 - not as a result of any input, but rather as a result of the initial conditions on \(y\), we can find:
\(
\begin{align}
y_{zi}(0)&= K_1+K_2&=1\\
\dot{y}_{zi}(0)&=-2K_1-3K_2&=2
\end{align}
\)
from which we can determine
\(
\begin{align}
K_1= 5 & K_2= -4
\end{align}
\)
such that
\(
\begin{align}
y_{zi}(t)&=5e^{-2t}-4e^{-3t}
\end{align}
\)
The zero-state response is a bit more complicated, as it necessarily includes the particular solution from above as well as components of the homogeneous. Among the easier ways to solve for the zero-state response is to take the Laplace transform for the differential equation and set all the initial conditions to 0. This basically means using
\(
\begin{align}
\frac{d^n}{dt^n}y(t)&\rightarrow s^nY(s)
\end{align}
\)
which in this case yields:
\(
\begin{align}
s^2Y_{zs}(s)+5sY_{zs}(s)+6Y_{zs}(s)&=(s+6)\left(\frac{1}{s}-\frac{1}{s+1}\right)=\frac{s+6}{(s)(s+1)}\\
Y_{zs}(s)&=\frac{s+6}{(s)(s+1)(s+2)(s+3)}=\frac{1}{s}-\frac{5}{2}\frac{1}{s+1}+\frac{2}{s+2}-\frac{1}{2}\frac{1}{s+3}
\end{align}
\)
meaning
\(
\begin{align}
y_{zs}(t)&=1-\frac{5}{2}e^{-t}+2e^{-2t}-\frac{1}{2}e^{-3t}
\end{align}
\)
In summary:
\(
\begin{align}
y_{zi}(t)&=5e^{-2t}-4e^{-3t}\\
y_{zs}(t)&=1-\frac{5}{2}e^{-t}+2e^{-2t}-\frac{1}{2}e^{-3t}\\
y(t)&=7e^{-2t}-\frac{9}{2}e^{-3t}+1-\frac{5}{2}e^{-t}
\end{align}
\)
which...should look familiar.
