I'm studying how to solve linear constant coefficients difference equations but I have some troubles with a step of the procedure. Consider the system \begin{equation} \begin{cases} y[n]+y[n-1]-6 y[n-2]=x[n] \\ y[-1]=1 \\ y[-2]=-1 \end{cases} \end{equation} Firstable, I stated that $y[n]= y_p [n] + y_h[h]$, that is the sum of the particular and the homoogeneus solution. I found out that the homogeneus solution is $ y_h[n] = c _1 (-3 ) ^ n + c_2 2 ^ n $, whereas, supposing that $x$ has the particular form $x[n] = 8 u[n] $ (and so $y[n ] = \beta x[n]$), the particular solution turns out to be $ y_p [n] = - 2u [n]$. So I can write the whole solution as $y[n ] = y_{\text {h}}[n]+y_{\text {p}}[n]=c_{1}(-3)^{n}+c_{2}(2)^{n}-2 u[n] $. Evaluating the system for $n = 0,1 $, I obtained \begin{equation*} \begin{cases} y[1]+y[0]-6 y[-1]=8 u[1] \\ y[0]+y[-1]-6 y[-2]=8 u[0] \end{cases} \implies \begin{cases} c_1 = -\frac{9}{5} \\ c_2 = \frac{24}{5} \end{cases} \end{equation*} However, I tried to find these two coefficients in another way, namely, imposing the two conditions: \begin{equation} \begin{cases} y [-1 ] = 1\\ y[-2 ] =-1 \end{cases} \implies \begin{cases} c_{1}(-3)^{-1}+c_{2}(2)^{-1}-2u [-1] = -\frac{1}{3}c_1 + \frac{1}{2}c_2 = 1 \\ c_1 (-3 )^ {-2 } + c_2 (2) ^ {-2} -2u[-2] = \frac{1}{9}c_1 + \frac{1}{4} c_2 = -1 \end{cases} \end{equation} and from this system I got $c_1 = -\frac{27}{5}$, $c_2 = - \frac{8}{5} $, that is different from the previous values. Maybe it's a stupid mistake, but I can't understand why I obtain a different result. If someone would like to give me an explanation, it would be appreciated.


1 Answer 1


Your first solution is correct. The other approach you suggested for solving the problem is also correct, and it is even a bit less tedious than the first route you chose, but the result is wrong because your particular solution is wrong. Note that $y_p[n]=-2u[n]$ does not solve the given difference equation for all $n\ge 0$, simply because $y_p[-1]=y_p[-2]=0$ and, consequently, the difference equation is not satisfied for $n=0$ and $n=1$.

I would just get rid of the unit step in the particular solution and would write


This particular solution satisfies the given difference equation for all $n\ge 0$, and we don't care about other values of $n$.

With that particular solution, you can simply evaluate the complete solution


for $n=-1$ and $n=-2$ and obtain two equations which can be solved for $c_1$ and $c_2$:

$$\begin{align}-\frac13 c_1+\frac12 c_2-2&=1\\\frac19c_1+\frac14c_2-2&=-1\end{align}$$

This gives the same result as the one obtained by your first method in an even more straightforward way.


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