ode; zero input response: what does q(D)y = 0 mean?

Question

my textbook on ordinary differential equations has a chapter on system modeling, and I 'm not sure if I understand what is meant by

The zero-input response is the solution to $q(\mathbf{D})y=0$, $y(0)=a_0$, $y'(0)=a_1$, ...

One of the examples given of how to determine the zero-input respnse from a system modeled by q(D) is

$q(\mathbf{D}) = (\mathbf{D} + 1)^2 + 4$, $y(0)=1$, $y'(0)=3$

I can see how the answer

$y=e^{-t}(\cos(2t) + 2\sin(2t))$

is derived by searching for the characteristic modes from the standard basis, but then I recalculated the linear equation with the derivative back from this answer, and I expected the result to be a constant 0.

I tried this in matlab:

syms x 

y = exp(-x) * (cos(2*x) + 2*sin(2*x));

qD = (diff(y) + y)^2 + 4*y;

diffy = diff(y);

hold on

fplot(y, [0 5])
fplot(qD, [0 5])
fplot(diffy, [0 5])

%fplot(hn)

legend ('y', 'qD', 'diffy') 

but I got anything but a constant 0 for $q(\mathbf D)$.

so is my matlab formula for $q(\mathbf D)$ wrong? or what else is meant by $q(\mathbf{D})=0$?

regards, Danny.

Answer 1

$$q(\mathbf{D}) = (\mathbf{D} + 1)^2 + 4$$ $q(D)y=0$ (and not $q(D)=0)$ means: $$( (\mathbf{D} + 1)^2 + 4)y=0$$ $$(D^2+2D+5)y=0$$ $$y''+2y'+5y=0$$ Characteristic polynomial is: $$(r+1)^2-4i^2=0 \implies r=-1 \pm 2i$$ $$\implies y(t)=e^{-t}(c_1 \cos(2t)+c_2 \sin (2t)$$

Answer 2

Apparently you are using $\mathbf{D}$ as the derivative operator. So an ODE of a dynamic system can be written as: $$q(\mathbf{D})y(t)=u(t)$$ being $q(\mathbf{D})$ a polynomial of $\mathbf{D}$ and $u(t)$ the input of the system.

The zero input response of a system is the response taking into account only its internal dynamics. So $y(t)$ is the solution of the equation $q(\mathbf{D})y(t)=0$.

The function you calculated in matlab is already the application of the operator $\mathbf{D}$, as you are using the diff functions, so it is expected to result 0.