The solution to a linear constant coefficient differential equation of the form $$\sum_{k = 0}^{N} a_k y^{(k)} (t) = \sum_{k = 0}^{M} b_k x^{(k)} (t)$$ can be written as $y(t) = y_{ZI} (t) + y_{ZS} (t)$ (sum of the zero-input response and the zero-state response) or $y(t) = y_h (t) + y_p (t)$ (the sum of the homogeneous solution and the particular solution). The zero-input response is found by solving the homogeneous equation with initial conditions applied to the solution to the homogeneous equation alone, whereas the zero-state response is found by solving the differential equation with zero initial conditions applied to the solution to the full equation (the sum of the homogeneous solution and the particular solution). However, if we split the solution up to a homogeneous solution and a particular solution (the second decomposition above) we apply initial conditions to the sum of the homogeneous and particular solutions, but not to the homogeneous solution alone first, before adding to it the particular solution. What is the reason for that?

I also wanted to know how this relates to AC steady state analysis in circuit theory. When we do steady state analysis, are we obtaining $y_p (t)$ or $y_{ZS} (t)$? Can you explain why? Is the steady-state response the particular solution or the zero-state response? Is it not one or the other in general?

I would be very grateful if you could explain this in terms of the time-domain description of systems, without resorting to s-plane concepts, poles, etc.

  • $\begingroup$ Have you read this answer? I think it should clarify at least some of your doubts. $\endgroup$
    – Matt L.
    Feb 7, 2018 at 13:35
  • $\begingroup$ I have read it and I found it very helpful. However, I am still struggling to understand why it works, and also the AC steady state analysis point at the end of my question. $\endgroup$
    – user33568
    Feb 7, 2018 at 13:41

3 Answers 3


The two approaches should return the same solution. They are just two different ways to get to the same place.

In the ZSR/ZIR method, you are solving two different IVPs - they have the same expression on the left-side of the equality, but different initial conditions and different right-hand sides (one has a $0$, the another one has some function of $t$). In the more mathematical non-engineer like approach (homogeneous/particular), you solve just one IVP with both things: the right-hand side of the equation not set to $0$ and the original initial conditions. Thus one can intuitively see the second approach as some kind of superposition of the two IVPs proposed in the first one.

Just in case you don't believe this yet, let me just write down an example (without doing it in detail as that's not the point of this post) to show that both methods work. Let's solve the following differential equation:

$$\left\{ \begin{array}{ll} y''(t)+4y'(t)+3y(t)=e^{-2t} \\ y(0)=1 \ , \ y'(0)=1 \end{array} \right.$$

The mathematical approach ($y_p+y_h$)

I propose:


Check that it satisfies the differential equation and the initial conditions:

$$y_p''(t)+4y_p'(t)+3y_p(t)=-4e^{-2t}+8e^{-2t}-3e^{-2t}=e^{-2t} \ \implies\mathrm{OK}$$ $$y_p(0)=-e^0=1\ \implies\mathrm{OK}$$ $$y_p'(0)=-2e^0=1\ \implies\mathrm{OK}$$

Therefore we conclude that $y_p(t)$ is a valid particular solution. Then we should solve the homogeneous equation. I'm not going to do it step by step, but you can check thay you get something of the form $$y_h(t)=Ae^{-t}+Be^{-3t}$$

If we impose the initial conditions to $y_p(t)+y_h(t)$, the constants turn out to be $A=5/2$ and $B=-1/2$. So the solution can be written as:

$$y(t)=y_p(t)+y_h(t)=\frac52 e^{-t}-e^{-2t}-\frac12 e^{-3t}$$

The engineer approach ($y_{zs}+y_{zi}$)

As stated at the beginning, this approach requires us to solve two different IVPs. The first one is:

$$\left\{ \begin{array}{ll} y_{zi}''(t)+4y_{zi}'(t)+3y_{zi}(t)=0 \\ y_{zi}(0)=1 \ , \ y_{zi}'(0)=1 \end{array} \right.$$

The solution (the generic one, with the constants $A$ and $B$) to this problem is the same as in the previous case, but with different constants (now we only have the homogeneous solution term). In this case, $A=2$ and $B=-1$. Therefore:


Now we have the other IVP:

$$\left\{ \begin{array}{ll} y_{zs}''(t)+4y_{zs}'(t)+3y_{zs}(t)=e^{-2t} \\ y_{zs}(0)=0 \ , \ y_{zs}'(0)=0 \end{array} \right.$$

Note that his is exactly the same IVP as the one stated at the beginning, but with different initial conditions. So we can use the homogeneous/particular approach to solve this IVP as well, and find the constants $A$ and $B$ that suit the initial conditions in this case. We find that

$$y_{zs}(t)=\frac12 e^{-t}-2e^{-2t}+\frac12 e^{-3t}$$

If we sum the two solutions:

$$y(t)=y_{zs}(t)+y_{zi}(t)=\frac52 e^{-t}-e^{-2t}-\frac12 e^{-3t}$$

We got the same result.


As can be seen from the example, both methods arrive to the same solution because they are just two different ways of decomposing the problem into two easier ones, but they are equivalent.

The approach using ZSR and ZIR is more practical in real life. Take for example a circuit. We usually know the initial conditions with no input. When we turn the input on (i.e. we make a sudden change in the circuit), voltages and currents could change abruptly. The ZIR uses these intitial conditions with no input, and the ZSR uses the input when we know all initial conditions are zero (for example, we discharge all capacitors before turning the circuit on). Performing the mathematical approach would not be simple, as we would have to know the initial conditions just when the input appeared, and that can be rather hard to measure or observe. With ZSR/ZIR, we can measure the two things independently and then add them up to get the total solution.

Regarding the steady state question, note that when you solve the differential equation you get the signal $y(t)$. Namely, you can evaluate the output at any instant $t$ you want. Thus, you don't have just the transient response or the steady state one: you have both. You can observe whatever you want as you have the signal as a function of time. The steady state will be reached at $t\to\infty$ (in real life, after some time, when all components in the circuit are acting as the input had been turned on for a long time).

  • $\begingroup$ Did you make a mistake in the homogeneous solution + particular solution approach? The particular solution does not seem to satisfy the initial conditions. $\endgroup$
    – user33568
    Feb 14, 2018 at 5:36
  • $\begingroup$ @0MW Oops, sorry about that. Yes, there is a mistake indeed. Note that the final result is correct, though. I'll fix it when I have the time, thanks for pointing it out. $\endgroup$
    – Tendero
    Feb 14, 2018 at 12:19

In addition to Tendero’s explanation, the zero state response, zero input response lends itself directly to the use of the one sided Laplace Transform.

My undergraduate linear systems text book, Signals, Systems, and Control by Lathi, defined a linear system to be, linear only if it was both zero input linear and zero state linear.

In circuit theory, circuits are causal. For DSP, we can process data that is recorded so there are interesting things that can be done that are not causal. Because circuits are causal, we need to concern ourselves at time t=0 to time approaching infinity. The choice of t=0 is actually arbitrary because the circuit may have been used prior to our t=0 time and keeping with causality, we use initial conditions to account for any prior signals. On a deeper level, linear circuits, when in addition are feed random input are continuous time Markov processes in their state variables. You only need to know the current state and input forward from the current time to predict their outputs. All the history prior to that point in time is included in the state variables. The homogeneous and particular solution approach isn’t cast as a causal relationship. You could run the solutions back in time and t=0 has no special physical interpretation


In system's engineering, the zero input response of a system described by a linear differential equation refers to the homogeneous solution of the differential equation. True False

  • $\begingroup$ Is this a question or an answer? What is the meaning of “True False” at the end of the post? $\endgroup$
    – ZaellixA
    Apr 2 at 11:31

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.