Question

solve the D.E. y(n)-y(n-2)+y(n-3)=delta (n)

Answer 1

$y[n]-y[n-2]+y[n-3]=\delta[n]$

For, $x[n]=u[n]$ ,

$D(z)=1-z^{-2}+z^{-3}=0\\ \Rightarrow z^3-z+1=0\\ \Rightarrow (z+1.32472)(z-0.66236 - 0.56228i)(z-0.66236 +0.56228i)=0\\ \Rightarrow z=0.66236 - 0.56228i=z_3 \;\;or,\;\; 0.66236 + 0.56228i=z_2\\\;\;or,\;\; 1.32472=z_1[let]$

$\therefore$ The solution :-

$$y[n]=\{c_1z_1^n+c_2z_2^n+c_3z_3^n+nc_1z_1^n+nc_2z_2^n+nc_3z_3^n+1)u[n]$$

Where,

$z_1=1.32472\\ z_2=z_3=0.66236+0.56228i\\z_3=0.66236-0.56228i$