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Trying to find the general solution to this homogeneous difference equation:

$y_k - 2\cos\theta y_{k-1} + y_{k-2} = 0.$

The characteristic equation is

$\lambda^2 - 2\cos\theta \lambda + 1 = 0.$

Not sure how to factor this, but tried

$(\lambda - \cos\theta)(\lambda - \cos\theta) = 0$

but I am stuck as to how to get $\cos^2\theta = 1$ using trigonometric identities.

By using the quadratic formula I get a discriminant of

$4(\cos^2\theta - 1)$

and I am stuck on how to simplify this to get the general solution.

Any help is appreciated. This is not for homework, but self study.

1 Answers 1

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To solve the difference equation $y(k) - [2\cos(θ)]y(k-1) + y(k-2) = 0$, assume the solution has the form $ y(k)=\lambda^k $ , then you substitute back in the difference equation which gives the equation

$ λ^2 - 2\cos(θ)λ + 1=0 $ $\Rightarrow \lambda_1 = \frac{2\cos(\theta)+\sqrt{4\cos(\theta)^2-4}}{2} \,, \lambda_2 = \frac{2\cos(\theta)-\sqrt{4\cos(\theta)^2-4}}{2} $

Note that,

$ \sqrt{4\cos(\theta)^2-4} = 2\sqrt{\cos^2(\theta)-1}=2 \sqrt{-1}\sqrt{1-\cos^2(\theta)}= 2 i\sin(\theta)\,.$ So, the roots are

$ \lambda_1= \cos(\theta)+i\sin(\theta)\,, \lambda_1= \cos(\theta)-i\sin(\theta)\,.$

The general solution can be written as

$ y(k)=c_1{\lambda_1}^k + c_2 {\lambda_2}^k \,. $