# Differential equations and LTI systems [duplicate]

I've seen in many textbooks on Signals and Systems that an LTI (Linear Time-Invatiant) system can be described as a constant-coefficient linear differential equation, such as

$$\sum_{k=1}^N a_k \frac{d^k}{dt^k}y(t) = \sum_{l=1}^M b_l \frac{d^l}{dt^l}x(t)$$

Later on, it is stated that the response of any LTI system can be decomposed into two terms: a zero-input response, and a zero-state response: $y(t) = y_{zi}(t) + y_{zs}(t)$ (among other decompositions) which are independent.

Zero-input (ZI) response $y_{zi}(t)$ is the output of the system when the input is zero, that is, the output to the initial conditions of the system. This means that ZI-response is the same as long as the initial conditions are the same and no matter what the input is.

Zero-state (ZS) response $y_{zs}(t)$ is the output of the system when the input in nonzero and the initial condition are zero. Thus, ZS-response depends only on the input (via convolution integral etc.)

Now, the question: if the system is LTI, it means it satisfies the linearity and the time-invariance property. That is:

Linearity: for input-output pairs $x_1(t) \to y_1(t)$, $x_2(t) \to y_2(t)$, we must have $$ax_1(t) + bx_2(t) \to ay_1(t) + by_2(t)$$

However, this is not true for nonzero ZI-response $y_{zi}(t)$. If $y_{zi}(t) = e^{-t}u(t)$, for example, then this response is not scaled by $a$ or $b$ when an input is applied, because it does not depend on the input! Thus, any system with nonzero ZI-response cannot be linear!

Time-invariance: if the input-output pairs are $x(t) \to y(t)$, and $x(t-t_0) \to y(t-t_0)$, then the system is time-invariant.

However, since again the ZI-response is the same as long as the initial conditions do not change, there is no shifting of the ZI-response when the input is shifted! Thus, a system cannot be time-invariant!

This makes me assume that a system described by a constant-coefficient differential equation is LTI only when the zi-response is zero, that is, when the initial conditions are zero! The last one makes the system causal as well.

Where do I make the mistake??? It seems to me that linearity and time-invariance (and causality) is only valid when initial conditions are zero but this is rarely noted in textbooks...

• You're right: non-zero initial conditions make the system non-linear and time-varying, simply because part of the output does not depend on the input signal. For the system to be LTI you need a linear constant coefficients differential equation AND require the initial conditions to be zero. – Matt L. Feb 19 '17 at 18:10

Your "mistake" if it can be called that is to assume that your input signal was zero for $t\le0$.