i hope you can guide me in this mess i have.
Gived this signal
$y[n+2]-y[n+1]-30y[n]=(1/5)^n+1; \qquad (n \in \Bbb N, n>= 2) $
$y[0]=0,$
$y[1]=1.$
get the general y[n] expression.
Applying Z transform to all the members of the first expression I get:
$$(z^2y(z)-z)-(z*y(z))-30y(z)=\frac{z}{(5z-1)}$$
$$y(z)(z^2-z-30)-z=\frac{z}{(5z-1)}$$
$$y(z)(z^2-z-30)=\frac{5z^2}{(5z-1)}$$
$$y(z)= \frac{5z^2}{(z+5)(z-6)(5z-1)}$$
After applying partial fraction expansion I get
$$y(z)= \frac{125}{286(z+5)} + \frac{180}{319(z-6)} - \frac{5}{754\color{red}{(5z-1)}}$$
Applying Laurent to the all three roots (1/5,6,-5)
$Rmax = max \{|1/5|,|6|,|-5|\} = 6$
$A(0,6,\infty )$
-5 => $$\frac{1}{z+5} = \frac{1}{z} * \frac{1}{1+\frac{5}{z}} =\sum_{z=1}^\infty (-1)^{n-1} \frac{5^{n-1}}{Z^n}$$
6 => $$\frac{1}{z-6} = \frac{1}{z} * \frac{1}{1-\frac{6}{z}} =\sum_{z=1}^\infty \frac{6^{n-1}}{Z^n}$$
1/5 => $$\frac{1}{z-(1/5)} = \frac{1}{z} * \frac{1}{1-\frac{1/5}{z}} =\sum_{z=1}^\infty \frac{(1/5)^{n-1}}{Z^n}$$
Now I can get the general expression
$$Y(z) = \sum_{z=1}^\infty \frac{1}{Z^n} [ -\frac{5}{754}\left(\frac{1}{5}\right)^{n-1} \color{red}{\frac{1}{5}} + \frac{125}{286}*(-1)^{n-1}*5^{n-1} + \frac{180}{319}*6^{n-1} ]$$
and using
$$y(n) = -\frac{5}{754}\left(\frac{1}{5}\right)^{n-1} \color{red}{\frac{1}{5}} + \frac{125}{286}*(-1)^{n-1}*5^{n-1} + \frac{180}{319}*6^{n-1}$$
to prove $y[0]=0, n=0$;
$$y(n) = -\frac{5}{754}\left(\frac{1}{5}\right)^{-1} \color{red}{\frac{1}{5}} + \frac{125}{286}*(-1)^{-1}*5^{-1} + \frac{180}{319}*6^{-1}$$
$$y(n) = -\frac{5}{754} - \frac{125}{286*5} + \frac{180}{319*6}$$
$$y[0] = 0$$
problem solved!
My bad, was a mistake with the (1/5) i've put in red.
The problem was with the (5z-1) I've put in red. I forgot to put (1/5) after Laurent's expansion in the general expression.
peace