# Is non-causal, non-LTI solution to difference equation correct?

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 ($$y_h[0]=y_h[1]=1$$):

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

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 $$y_h[-1]=y_h[0]=1$$:

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

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

$$y[n]=4(\frac{1}{2})^n-3(\frac{1}{3})^n-2(\frac{1}{2})^nu[-n-1]+2(\frac{1}{3})^nu[-n-1]$$

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

• In what way is that nonlinear? In what way is it time varying? Mar 30, 2020 at 15:51
• @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. Mar 30, 2020 at 16:08
• @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. Mar 30, 2020 at 16:10
• It's not causal, but that doesn't mean it's not LTI. Mar 30, 2020 at 16:11
• 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. Jun 25, 2020 at 22:19