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
$$y[n]=4(\frac{1}{2})^n-3(\frac{1}{3})^n-2(\frac{1}{2})^nu[-n-1]+2(\frac{1}{3})^nu[-n-1]$$
Is there a faster way to approach this problem (possibly using Fourier transform)? How would you solve this for complicated inputs, e.g. $x[n]=u[n]$?
