I have a differential equation $$\frac{d^2y(t)}{dt^2}+y(t) = \frac{dx(t)}{dt} + x^2(t)$$ and I need to see if this system with input $x(t)$ and output $y(t)$ is time invariant and linear. I tried to break up the equation by substituting a linear operator D as the derivative.

So I get: $$y(t) = \frac{Dx(t)+x^2(t)}{D^2 + 1}$$

But the problem is, this equation is meaningless since you are dividing by the derivative operator. However, mathematically, it works out, so can I get $y(t)$ in this form and test for linearity or time invariance using this method?

  • $\begingroup$ I think you should play with some inputs and ouputs and shift them etc to see if you can also show the shift in the new output through the differential equation... Put some material on your work so that you may get better help. $\endgroup$ – Fat32 Feb 8 '17 at 0:00
  • $\begingroup$ @Fat32 Perhaps there's a quick proof you know of? $\endgroup$ – JobHunter69 Feb 8 '17 at 0:23

Let $y(t)=\mathcal{H}(x(t))$ describe the system by relating its input $x(t)$ to the output $y(t)$. Then, to check linearity simply check with $x(t)=1$ (i.e. a constant). You will see, that

$$\mathcal{H}(x(t)=1)+\mathcal{H}(x(t)=1)\neq \mathcal{H}(x(t)=1+1)$$

which is one requirement for the linearity. So, the system is non-linear. Actually, you can immediately see this from the fact that $x(t)$ counts quadratically, which is a non-linear relation.

Regarding time-invariance, you see that the time-argument is the same for all functions and the time-value is not used in the equation itself (such as $t\cdot x(t)$ would ne not time-invariant). So, the system is time-invariant.

  • $\begingroup$ Are you saying that I can treat y(t) like what I wrote above? $\endgroup$ – JobHunter69 Feb 21 '17 at 6:31

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