the general solution of $y(n+3)-\frac{2}{3}y(n+1)+\frac{1}{3}y(n) = 0$

Question

I have some trouble finding the correct solution for the difference equation

$$y(n+3)-\frac{2}{3}y(n+1)+\frac{1}{3}y(n) = 0$$

I've found that the characteristic equation of the difference equation is $\lambda^3-\frac{2}{3}\lambda+\frac{1}{3}$.

By computation I then have;

\begin{array}{lcl} \lambda^3-\dfrac{2}{3}\lambda+\dfrac{1}{3}\\ (\lambda+1)(\lambda^2-\lambda+\dfrac{1}{3})\\ \lambda_1 = -1, \lambda_2 = \dfrac{1-i\sqrt{\frac{1}{3}}}{2}, \lambda_3 = \dfrac{1+i\sqrt{\frac{1}{3}}}{2} \end{array}

I checked my answer using wolframalpha and it gave me the eigenvalues

$$\begin{array}{lcl} \lambda_1 = -1, \lambda_2 = \dfrac{1-i\sqrt{3}}{2}, \lambda_3 = \dfrac{1+i\sqrt{3}}{2} \end{array}$$

Could someone please tell me what I did wrong, or what I should do differently?

Answer 1

Wolfram Alpha gives the roots as $-1$ and $1/6 (3\pm i \sqrt(3))$. This last is $\frac{1\pm i\sqrt{1/3}}{2}$ which is your answer.

My guess is either you misread WA's answer or you entered the equation incorrectly.

Answer 2

How did you input the equation into Wolfram Alpha?

You have as solutions $$ \begin{align} a_n&=(-1)^n\tag{1}\\[9pt] b_n &=\frac12\left[\,\left(\frac{1+\frac i{\sqrt{3}}}{2}\right)^n+\left(\frac{1-\frac i{\sqrt{3}}}{2}\right)^n\,\right]\\ &=\mathrm{Re}\left[\,\left(\frac{1+\frac i{\sqrt{3}}}{2}\right)^n\,\right]\tag{2}\\[9pt] c_n &=\frac1{2i}\left[\,\left(\frac{1+\frac i{\sqrt{3}}}{2}\right)^n-\left(\frac{1-\frac i{\sqrt{3}}}{2}\right)^n\,\right]\\ &=\mathrm{Im}\left[\,\left(\frac{1+\frac i{\sqrt{3}}}{2}\right)^n\,\right]\tag{3} \end{align} $$ and any linear combination of $a_n$, $b_n$, and $c_n$.