Consider the following difference equation:
y[n] + 3y[n – 1] + 2y[n – 2] = 2x[n] – x[n – 1]
If y[-1] = 0, y[-2] = 1, x[n] = u[n], then y[n] can be represented as
1
\(\left[ { - \frac{1}{2}{{\left( 1 \right)}^{n - 1}} + \frac{4}{3}{{\left( { - 2} \right)}^{n - 1}} + \frac{1}{6}} \right]u\left( { - n - 1} \right)\)
2
\(\left[ { - \frac{1}{2}{{\left( -1 \right)}^{n + 1}} + \frac{4}{3}{{\left( { - 2} \right)}^{n +1}} + \frac{1}{6}} \right]u\left( {n + 1} \right)\)
3
\(\left[ { - \frac{1}{2}{{\left( 1 \right)}^{n - 1}} + \frac{4}{3}{{\left( { - 2} \right)}^{n - 1}}} \right]u\left( {n - 1} \right) + \frac{1}{6}u\left( n \right)\)
4
\(\left[ { - \frac{1}{2}{{\left( 1 \right)}^{n - 1}} + \frac{4}{3}{{\left( { - 2} \right)}^{n - 1}}} \right]u\left( { - n - 1} \right) + \frac{1}{6}u\left( n \right)\)