Find a solution to the difference equation $y[n]-\frac{5}{6}y[n-1]+\frac{1}{6}y[n-2]=\frac{1}{3}x[n-1]$ that is neither casual nor LTI, where $y[0]=y[1]=1$ and $x[n]=\delta[n]$

The homogenous solution for the difference equation is

$$y_h[n] =A_1(\frac{1}{2})^n + A_2(\frac{1}{3})^n$$

Solving for specific solution ($y_h[0]=y_h[1]=1$):

\begin{equation} y_1[n]=4(\frac{1}{2})^n-3(\frac{1}{3})^n \tag{1} \end{equation}

We want to "modify" $y_1[1]$ or $y_1[0]$ or $y_1[-1]$ because the right hand side of the difference equation is non-zero at $n=1$.

Since $y[1]$ and $y[0]$ are specifically set to $1$ in the question, we must change $y[-1]$.

$$y[1]-\frac{5}{6}y[0]+\frac{1}{6}y[-1]=\frac{1}{3}x[0]$$ $$1-\frac{5}{6}+\frac{1}{6}y[-1]=\frac{1}{3}$$ $$y[-1]=1$$

Since $y_1[-1]=-1$, $y_1$ will be the correct output only for $n \geq 0$. This means that we need to find another "piece" $y_2$ for $n < 0$ such that $y_2[0]=y_1[0]$.

Solving another specific solution with $y_h[-1]=y_h[0]=1$:

\begin{equation} y_2[n]=2(\frac{1}{2})^n-(\frac{1}{3})^n \tag 2 \end{equation}

Hence, our solution is:

$$ y[n] = \left\{ \begin{array}{ll} 4(\frac{1}{2})^n-3(\frac{1}{3})^n & n\geq 0 \\ 2(\frac{1}{2})^n-(\frac{1}{3})^n & n \lt 0 \\ \end{array} \right. $$

which can be written as


Is there a faster way to approach this problem (possibly using Fourier transform)? How would you solve this for complicated inputs, e.g. $x[n]=u[n]$?

  • $\begingroup$ In what way is that nonlinear? In what way is it time varying? $\endgroup$
    – TimWescott
    Mar 30, 2020 at 15:51
  • $\begingroup$ @TimWescott The input $x[n] = \delta[n] = 0$ for $n < 0$; however, $y[n] \ne 0$ for $n < 0$. That means the solution is not linear. $\endgroup$ Mar 30, 2020 at 16:08
  • $\begingroup$ @TimWescott I think the use of $u[-n-1]$ makes it time varying. If input is delayed by 1, then the coefficient $u[-n-1]$ will not be delayed. I'm not sure if I'm right on this one, so correct me if I'm wrong. $\endgroup$ Mar 30, 2020 at 16:10
  • $\begingroup$ It's not causal, but that doesn't mean it's not LTI. $\endgroup$
    – Hilmar
    Mar 30, 2020 at 16:11
  • 1
    $\begingroup$ Yes, that's a good point. It's kind of a gray area -- is the system forcing the constraints $y[0] = y[1] = 1$, or is that part of the problem statement? If the system is somehow magically forcing that, then it is indeed time-varying and nonlinear. Normally when you see a problem in a book with constraints like that, it's because someone is trying to determine the system behavior under a certain set of circumstances, and the system itself is linear. $\endgroup$
    – TimWescott
    Jun 25, 2020 at 22:19


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