> [!NOTE] Theorem > Let $y_{k}+a y_{k-1}+by_{k-2}=0$for all $k\in \mathbb{N}_{\geq 2},$ be a homogenous second order linear [[Recurrence Relation|recurrence relation]] with $a,b\in \mathbb{R}.$ > > Let $\lambda_{1,2}$ denote the solution to the equation $0=\lambda^{2}+a\lambda+b.$ > > Then any solution to the equation has the form $y_{k}= \begin{cases} c_{1}\lambda_{1}^{k}+c_{2}\lambda_{2}^{k}, &a^{2}-4b>0 \\ c_{1}\lambda^{k} + c_{2}k\lambda^{k}, & a^{2}-4b =0 \\ r^{k} (c_{1}\cos(k\theta)+c_{2}\sin(k\theta)), &a^{2}-4b<0 \text{ and } \lambda_{1}= \bar{\lambda}_{2} =r e^{i\theta} \end{cases}$for some $c_{1},c_{2}\in \mathbb{R}.$ **Note**: that in the third case $r$ and $\theta$ are the argument and modulus of $\lambda_{1}.$ **Proof**: Suppose $y_{k}=\lambda^{k}$ is a solution for some $\lambda\in \mathbb{R}.$ This is satisfied if $\lambda$ satisfies the auxiliary equation $\lambda^{2}+a\lambda+b = 0.$Note that trying the ansatz $y= \lambda^{k}$ yields this same equation. Solving gives $\lambda_{1,2}= -\frac{a}{2} \pm \frac{1}{2} \sqrt{ a^{2} -4b }$We distinguish the following possibilities: 1. For distinct real solutions ( $a^{2}-4b>0$), we obtain solutions of the form $y_{k} = c_{1} \lambda_{1}^{k} + c_{2} \lambda_{2}^{k}$ 2. For distinct complex solutions ($a^{2}-4b<0$), $\lambda_{1}= \bar{\lambda_{2}}$ are complex conjugates. We can write them in the form $\lambda_{1} = r e^{i \theta}$, $\lambda_{2} = r e^{-i \theta}$ where $r = \sqrt{ b }$ and $\theta = \arctan\left( - \frac{\sqrt{ 4b-a^{2} }}{a} \right) + \begin{cases} \pi & \text{if } a>0,\ \\ 0 & \text{otherwise} \end{cases}$are the modulus and argument of $\lambda_{1}$. However, for any $c_{1},c_{2} \in \mathbb{R}$, $c_{1}\lambda_{1}^{k} + c_{2} \lambda_{2}^{k} = c_{1} r^{k} e^{i k \theta} + c_{2} r^{k} e^{-i k \theta}$ is complex valued, in general. Note that $r^{k} \cos(k \theta ) = r^{k} \frac{e^{ik\theta} + e^{-ik\theta}}{2} = \frac{\lambda_{1}^{k}+ \lambda_{2}^{k}}{2}$and similarly $r^{k} \sin(k \theta ) = r^{k} \frac{e^{ik\theta} - e^{-ik\theta}}{2} = \frac{\lambda_{1}^{k} - \lambda_{2}^{k}}{2}$So we obtain real-valued solutions of the form $y_{k} = c_{1} r^{k}\cos(k \theta) + c_{2} r^{k} \sin(k\theta)$where $c_{1},c_{2} \in \mathbb{R}$ are arbitrary numbers. 3. For repeated real solutions ($a^{2}-4b=0$), $\lambda=-\frac{a}{2}$ is the only root. Then $y_{k}= k\lambda^{k}$ is a solution. We obtain solutions of the form $y_{k} = c_{1} \lambda^{k} + c_{2} k \lambda^{k}$where $c_{1},c_{2}\in \mathbb{R}$ are arbitrary numbers. **Proof**: Follows from [[Linear Recurrence Relation with Constant Coefficients]].