Recurrence relation of degree 2

Question

i have got this recurrence relation : $$ g(n)+5g(n-1)-6g(n-2) = 14-2^n\cdot2 $$ so far I have figured out the characteristic polynomial: $ x^2+5x-6 = 0 $ this factors to $ (x+6)(x-1)=0$ so my homogenous solution is $$g(n)= C_1+ (-6)^nC_2$$

Now, how do I deal with this error function? $$f(h) = -2^n\cdot 2 +14$$ I tried plugging in $$An(-2)^n\cdot 2+Bn$$ to obtain the particular solution, but this led to nonsense. What did I do wrong?

Answer 1

Hint:

Let $g(m)=h(m)+Am+B+C2^m$

$$14-2\cdot2^n$$ $$=h(n)+An+B+C2^n+5(h(n-1)+An-A+B+C2^{n-1})-6\{h(n-2)+An-2A+B+C2^{n-2}\}$$

$$=h(n)+5h(n-1)-6h(n-2)+7A+C2^{n-2}(4+10-6)$$

Set $7A=14\implies A=?$

and $C2^{n-2}(4+10-6)=-2\cdot2^n\implies C=-1$