Solve the recurrence relation :

$ \begin{aligned} &a_{n}+6 a_{n-1}+12 a_{n-2}+8 a_{n-3}=2^{n}, n \geq 3 \\\\ &a_{0}=0, a_{1}=0, a_{2}=2 \end{aligned} $

Solution:

The characteristic equation is,

$ \begin{aligned} \Rightarrow \quad \lambda^{3}+6 \lambda^{2}+12 \lambda+8 &=0 \\\\(\lambda+2)^{3} &=0 \\\\ \quad \lambda =-2,-2,-2 \end{aligned} \\ $

Hence the homogeneous solution is,

$ a_{n}^{(h)}=\left(A_{1} n^{2}+A_{2} n+A_{3}\right)(-2)^{n} \\ $

The particular solution will be of the form $P\left(2^{n}\right)$. Substituting the given recurrence relation, we get,

$ \begin{aligned} \mathrm{P}\left(2^{n}\right)+6 P\left(2^{n-1}\right)+12 P\left(2^{n-2}\right)+8 P\left(2^{n-3}\right) &=2^{n} \\\\ \Rightarrow 2^{n} P\left[1+\frac{6}{2}+\frac{12}{4}+\frac{8}{8}\right] =2^{n} \\\\ \text { Or } P[1+3+3+1] =1 \Rightarrow \quad P=\frac{1}{8} \\\\ \therefore \quad a_{n}^{(P)} =\frac{1}{8}\left(2^{n}\right)=2^{n-3} \\ \end{aligned} $

The total solution is,

$ a_{n}=a_{n}^{(h)}+a_{n}^{(P)}=\left(A_{1} n^{2}+A_{2} n+A_{3}\right)(-2)^{n}+2^{n-3} \\ $

Using the initial conditions,

$ \begin{aligned} a_{0} &=0, a_{1}=0, a_{2}=2 \\\\ 0 &=A_{3}+\frac{1}{8} \\ \end{aligned} $

$ \Rightarrow \quad \mathrm{A}_{3}=-\frac{1}{8}....(1) \\ $

$ \begin{aligned} A_{1}+A_{2}+A_{3} =\frac{1}{8}......(2) \\\\ 2 =\left(A_{1} 4+A_{2} \cdot 2+A_{3}\right)(-2)^{2}+2^{2-3} \\ \end{aligned} $

$ \begin{aligned} &2=\left(4 \mathrm{~A}_{1}+2 \mathrm{~A}_{2}+\mathrm{A}_{3}\right)(4)+\frac{1}{2} \\\\ &\frac{3}{2}=4\left(4 \mathrm{~A}_{1}+2 \mathrm{~A}_{2}+\mathrm{A}_{3}\right) \\ \end{aligned} $

$ \Rightarrow \quad 4 A_{1}+2 A_{2}+A_{3}=\frac{3}{8}....(3) \\ $

From equations (1), (2), and (3),

$ \mathrm{A}_{1}=0, \mathrm{~A}_{2}=\frac{1}{4}, \mathrm{~A}_{3}=-\frac{1}{8} \\ $

$ a_{n}=\left(\frac{1}{4} n-\frac{1}{8}\right)(-2)^{n}+2^{n-3} \\ $