Solve the following recurrence relation :

(i) $a_{r}-7 a_{r-1}+10 a_{r-2}=0$ given that $a_{0}=0$, $a_{1}=3$

(ii) $a_{r}-4 a_{r-1}+4 a_{r}-2=0$ given that $a_{0}=1$ and $a_{1}=6$

Solution:

(i) The characteristics equation is,

$ \begin{aligned} \lambda^{2}-7 \lambda+10 &=0 \\\\ (\lambda-5)(\lambda-2) &=0 \\\\ \lambda =5, \lambda=2 \\\\ \end{aligned} $

Therefore the solution of the given recurrence relation is

$ { A_{r}} =\mathrm{A}\left({5^{r}}\right)+\mathrm{B}\left(2^{\mathrm{r}}\right)....(1) \\ $

To where A and B are constants.

To find A and B putting, $r=0$ in equation (1)

$ a_{0}=A+B \\\\ $

$ A + B = 0....(2) \\\\ $

Also putting $r=1$ in equation (1)

$ 5 \mathrm{~A}+2 \mathrm{~B}=3....(3) \\\\ $

From equation (2) and equation (3)

$ \begin{aligned} &A=1 \\\\ &B=-1 \\ \end{aligned} $

Hence the homogeneous solution of the given recurrence relation is,

$ a_{r}=5^{r}-2^{r} \\\\ $

(ii) The characteristics equation is,

$ \begin{aligned} \lambda^{2}-4 \lambda+4 &=0 \\\\ (\lambda-2)^{2} &=0 \\\\ \lambda =2,2 \\\\ \end{aligned} $

Therefore the solution of the given recurrence relation is,

$ a_{r}=(A+B r) 2^{r}.....(4) \\\\ $

Where A and B are constants.

To find A and B putting $r=0$ in equation (4)

$ \begin{aligned} a_{0} &=A \\\\ A &=1 \\\\ \end{aligned} $

Also putting $r=1$ in equation (4)

$ \begin{aligned} a_{1} &=(A+B \cdot 1) 2^{1} \\\\ 6 &=(A+B) 2 \\\\ \Rightarrow & A+B =3 \\\\ \text { or } & 1+B =3 \\\\ \text { or } & B =2 \\\\ \end{aligned} $

Hence the homogeneous solution of the given recurrence relation is

$ a_{r}=(1+2 r) 2^{r} \\\\ $