# Finding steady state response of LTI, dissipative systems

The input, $x(t)$, and output, $y(t)$ of a linear time invariant, dissipative system are related via the differential equation

$$d^4(y(t)) + 6d^3(y(t)) + 14d^2(y(t))+14d(y(t)) + 5y(t) = - d^2(x(t)) - 9x(t)$$

If $x(t) = \cos(t)-\sin(t)$, determine the steady state response.

Now, I know

$$H(iw) = \frac{-(iw)^2 - 9}{(iw)^4 + 6(iw)^3 + 14(iw)^2 + 14(iw) + 5}$$

But I'm having a hard time understanding how this transfer function will help find the steady state response. Especially since there is no circuit given.

If a sinusoidal signal is applied to an LTI system, the output will be having same frequency as input but will have different phase and amplitude. If the input is $\cos(wt)$, then output will be: $$|H(jw)|\cos\left(wt+\angle H(jw)\right)$$ Where $H(jw)$ is the frequency response. $\angle H(jw)$ is the argument of $H(jw)$.
In your case, the input is sum of two sinusoids with $w=1$. Find the output for each one after evaluating magnitude and phase of $H(jw)$ at $w=1$ and add them to get the result.
Use $\small j$ instead of $\small i$. $\small j$ is used in engineering to avoid confusion with current.
Express the TF as a complex number: $\small A+jB$, then the modulus and argument give the steady-state frequency domain gain and phase angle as functions of frequency, $\omega\: \small rad \:s^{-1}$
To find the response to $\small cos(t)-sin(t)$, you could use the double angle representation: $\small cos(t)-sin(t)=\sqrt2\:sin(t+3\pi/4)$. Hence, multiply the gain, calculated from the above analysis, by $\small \sqrt2$ and add $\small 135^o$ to the phase angle.