Find the solution to $$a_n = 5a_{n−2} − 4a_{n−4}$$ with $$a_0 = 3$$
$$a_1 = 2$$ $$a_2 = 6$$ $$a_3 = 8$$
My answer: Observe that the degree of recurrence is 4. Hence, the characteristic equation is: $x^4 – 5x^2 +4 = 0$. Solving for $y=x^2$ we get $y_1=1$, $y_2 = 4$. Hence, the solutions are $x_1 = -1$, $x_2 = 1$, $x_3 = -2$, $x_4 = 2$. This implies that $a_n = \alpha_1(-1)^n + \alpha_2 + \alpha_3 (-2)^n + \alpha_4(2^n)$. It remains to solve the initial conditions for $\alpha_i-s : \alpha_1 = 1, \alpha_2 = 1, \alpha_3 = 0, \alpha_4 = 1$.
Where should I go from here to ensure the answer is fully answered and complete? Thank you!
Best Answer
Since the recurrence relation is linear, the general solution is a linear combination of 4 particular solution you found. Initial conditions determine unknown constants: $$ a_n = \alpha_1 + \alpha_2 (-1)^n + \alpha_3 2^n + \alpha_4 (-2)^n $$ Now substitute initial conditions for $n=0$, $n=1$, $n=2$ and $n=3$ to find $\alpha_1$, $\alpha_2$, $\alpha_3$ and $\alpha_4$.