I'm trying to find the solution to the following difference equation:
$$y[n] - \frac{1}{4} y[n-1] - \frac{1}{8} y[n-2] =3x[n] $$
with $x[n]=(\frac{1}{2})^nu[n]$ where $u[n]$ is a step signal defined as :
$$ u[n]= 1 \forall n\ge 0$$ $$= 0 \forall n<0 $$
but somehow my solution does not match the given solution.
The complementary solution or homogeneous solution is : $k_1(\frac{1}{2})^n+k_2 (\frac{-1}{4})^n]u[n]$
And after that I take the particular solution of the form $$y_p[n]=kn(\frac{1}{2})^nu[n]$$ since the input is of the same form as the roots of the characteristic equation.
and get k as equal to 2.
and evaluate the coefficients of complementary solution as $$y[n]=\frac{8}{3} (\frac{1}{2})^n u[n]+\frac{1}{3} (\frac{-1}{4})^n u[n]+2n (\frac{1}{2})^n u[n]$$
but the solution given is $$\frac{1}{3} (\frac{1}{4})^n u[n] + 4(n+1)(\frac{1}{2})^{(n+1)} u[n+1] + \frac{2}{3} (\frac{1}{2})^n u[n]$$
I don't where the anomaly in my evaluation is? Or is the given solution wrong? I'm new to the concept of difference equations.
I wish to solve this difference equation without using z transform.
A resolution method not based on the $z$ transform.
Making $y_1(k) = y(k), y_2(k) = y_1(k+1)$ and $Y = (y_1, y_2)^{\dagger}$ we have
$$ Y_{k+1}= M Y_k + B x_k $$
considering now $M = T\Lambda T^{-1}$ we have
$$ T^{-1}Y_{k+1}=\Lambda T^{-1} Y_k + T^{-1}B x_k $$
making now $Z_k = T^{-1} Y_k,\ \ \ V = T^{-1}B $ we follow with
$$ Z_{k+1} = \Lambda Z_k + V x_k $$
the homogeneous solution is direct
$$ Z_k^h = \Lambda^k C_0 $$
now assuming a particular solution $Z_k^p = \Lambda^k C_k$ and substituting into the complete recurrence
$$ \Lambda^{k+1}C_{k+1} = \Lambda^{k+1}C_k+V x_k $$
gives the recurrence
$$ C_{k+1}-C_k = \Lambda^{-k-1}V x_k $$
then
$$ C_k = \sum_{j=0}^{k-1}\Lambda^{-j-1}V x_j $$
and finally
$$ Z_k=Z_k^h + Z_k^p = \Lambda^k C_0+\Lambda^k\sum_{j=0}^{k-1}\Lambda^{-j-1}V x_j $$
hence
$$ Y_k = T Z_k $$
NOTE
$$ M = \left( \begin{array}{cc} 0 & 1 \\ \frac{1}{8} & \frac{1}{4} \\ \end{array} \right),\ \ T = \left( \begin{array}{cc} 2 & -4 \\ 1 & 1 \\ \end{array} \right),\ \ \Lambda = \left( \begin{array}{cc} \frac{1}{2} & 0 \\ 0 & -\frac{1}{4} \\ \end{array} \right),\ \ B = \left( \begin{array}{c} 0 \\ 1 \\ \end{array} \right),\ \ V = \left( \begin{array}{c} \frac{2}{3} \\ \frac{1}{3} \\ \end{array} \right) $$