particular solution of a difference equation

Question

I am unable to find a particular solution of the following difference equation $$ y[k-1]-5y[k]+6y[k+1]=-u[k-1]+4u[k] $$ with $u[k]=\big(\frac{1}{2}\big)^k$.

This is what I tried so far. Because $u[k]=\big(\frac{1}{2}\big)^k$ we try $y_p[k]=\alpha \big(\frac{1}{2}\big)^k$. Substituting this is in the difference equation gives us $$ \begin{align} \alpha \bigg(\frac{1}{2}\bigg)^{k-1}-5\alpha \bigg(\frac{1}{2}\bigg)^k + 6\alpha \bigg(\frac{1}{2}\bigg)^{k+1} &= -\bigg(\frac{1}{2}\bigg)^{k-1} + 4 \bigg(\frac{1}{2}\bigg)^k\\ \bigg(\frac{1}{2}\bigg)^{k-1}\bigg(\alpha -\frac{5\alpha}{2}+\frac{6\alpha}{4}\bigg)&= \bigg(\frac{1}{2}\bigg)^{k-1} \end{align} $$ Thus $\Big(\alpha -\frac{5\alpha}{2}+\frac{6\alpha}{4}\Big)=1$ which results in $0\alpha=2$ which has no possible solution.

The problem is that I have no idea which particular solution I could try that solves this problem, namely the fact that $1-5/2+6/4=0$.

Any help would be greatly appreciated.

Answer 1

We change notation slightly. Note that $-u_{k-1}+4u_k=\frac{4}{2^k}-\frac{2}{2^k}=\frac{2}{2^k}$. We look for a solution of the shape $y_i=\alpha\frac{i}{2^i}$.

Substitute in $6y_{k+1}-5y_k+y_{k-1}$, forgetting about the $\alpha$ for a while. We get $$\frac{6(k+1)}{2^{k+1}}-\frac{5k}{2^k}+\frac{k-1}{2^{k-1}}.$$ This simplifies to $$\frac{1}{2^{k+1}}\left(6k+6 -10k+4k+4k-4\right).$$ I am sure you can finish from here.

Answer 2

When $y[k] = c \lambda^k$ is a solution of the homogeneous equation $L(y) = 0$, you should look for a particular solution of $L(y) = a \lambda^k$ of the form $y[k] = c k \lambda^k$.