Solve recurrence relation $a(n) = -a(n - 2) + \cos({n} \cdot {\frac{\pi}{2}})$ Given recurrence equation
$$a(n) = -a(n - 2) + \cos({n} \cdot {\frac{\pi}{2}})$$
find the closed form solution.
Here is my attempt.
First solve the homogeneous equation:
$$a^{(0)}(n) = -a^{(0)}(n - 2)$$
My solution is:
$$a^{(0)}(n) = k_1 \cos({n} \cdot {\frac{\pi}{2}}) + k_2 \sin({n} \cdot {\frac{\pi}{2}})$$
Now the main concern is how should the particular solution look like.
I understand that if we have a recurrence equation $b(n) = c_1 b(n - 1) + c_2 b(n - 2) + 2^n$ and $f(n) = 2^n$ resonates with the solutions to $b^{(0)}(n) = c_1 b^{(0)}(n - 1) + c_2 b^{(0)}(n - 2)$, we just multiply it by $n$, so we look for $b(n) = k_3 n 2^n$.
If I repeat blindly the same multiplication, I get
$$a(n) = k_3 n \cos({n} \cdot {\frac{\pi}{2}}) + k_4 n \sin({n} \cdot {\frac{\pi}{2}})$$
Is it the correct way to look for particular solution?

If I proceed this way, I get the following general solution:
$$a(n) = k_1 \cos({n} \cdot {\frac{\pi}{2}}) + k_2 \sin({n} \cdot {\frac{\pi}{2}}) + \frac {n}{2} \cos({n} \cdot {\frac{\pi}{2}})$$
Is it the correct solution?
 A: $\newcommand{\angles}[1]{\left\langle\,{#1}\,\right\rangle}
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 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\ul}[1]{\underline{#1}}
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 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
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I'll consider $\ds{b_{n} = -b_{n - 2} + \ic^{n}}$ such that $\ds{\,\mathrm{a}\pars{n} = \Re\pars{b_{n}}}$.


With $\ds{\ul{\quad z \in \mathbb{R}\quad \mbox{and}\quad  0 < z < 1}}$:
\begin{align}
\sum_{n = 2}^{\infty}b_{n}\,z^{n} & =
-\sum_{n = 2}^{\infty}b_{n - 2}\,\,z^{n} + \sum_{n = 2}^{\infty}\ic^{n}\,z^{n}
\\[4mm]
\sum_{n = 0}^{\infty}b_{n}\,z^{n} - b_{0} - b_{1}\,z & =
-\sum_{n = 0}^{\infty}b_{n}\,\,z^{n + 2}\ -\ {z^{2} \over 1 - \ic z}
\\[4mm]
\pars{1 + z^{2}}\sum_{n = 0}^{\infty}b_{n}\,z^{n} & =
b_{0} + b_{1}\,z - {z^{2} \over 1 - \ic z}
\\[4mm]
\sum_{n = 0}^{\infty}b_{n}\,z^{n} & =
{b_{0} + b_{1}\,z \over 1 + z^{2}} -
{z^{2} \over \pars{1 + z^{2}}\pars{1 - \ic z}}
\end{align}

\begin{align}
\sum_{n = 0}^{\infty}\,\mathrm{a}\pars{n}\,z^{n} & =
{\,\mathrm{a}\pars{0} + \,\mathrm{a}\pars{1}\,z \over 1 + z^{2}} -
{z^{2} \over \pars{1 + z^{2}}^{2}}
\\[4mm] & =
\bracks{\,\mathrm{a}\pars{0} + \,\mathrm{a}\pars{1}\,z}
\sum_{n = 0}^{\infty}\pars{-1}^{n}\,z^{2n} -
z^{2}\sum_{n = 0}^{\infty}{-2 \choose n}z^{2n}
\end{align}

With
$\ds{{-2 \choose n} = {2 + n - 1 \choose n}\pars{-1}^{n} =
\pars{-1}^{n}\pars{n + 1}}$:
\begin{align}
&\sum_{n = 0}^{\infty}\,\mathrm{a}\pars{2n}\,z^{2n} +
\sum_{n = 0}^{\infty}\,\mathrm{a}\pars{2n + 1}\,z^{2n + 1}
\\[4mm] = &\
\,\mathrm{a}\pars{0}
\sum_{n = 0}^{\infty}\pars{-1}^{n}\,z^{2n} +
\sum_{n = 1}^{\infty}\pars{-1}^{n}\, n\,z^{2n} +
\,\mathrm{a}\pars{1}
\sum_{n = 0}^{\infty}\pars{-1}^{n}\,z^{2n + 1}
\\[4mm] = &\
\,\mathrm{a}\pars{0} +
\sum_{n = 1}^{\infty}\bracks{\,\mathrm{a}\pars{0} + n}\pars{-1}^{n}\,z^{2n} +
\sum_{n = 0}^{\infty}\,\mathrm{a}\pars{1}\pars{-1}^{n}\,z^{2n + 1}
\end{align}

$$
\color{#f00}{\,\mathrm{a}\pars{n}} =
\color{#f00}{\left\lbrace\begin{array}{lcl}
\ds{\pars{-1}^{\pars{n - 1}/2}\,\,\,\mathrm{a}\pars{1}} & \mbox{if} & \ds{n}\ \mbox{is}\  \ul{odd}
\\[2mm]
\ds{\pars{-1}^{n/2}\,\,\,\bracks{\mathrm{a}\pars{0} + {n \over 2}}} &
\mbox{if} & \ds{n}\ \mbox{is}\ \ul{even}
\end{array}\right.}
$$
A: The $\cos$ seems to add a lot of complexity. Why not consider instead separately the even and odd subsequences, which do not "interact" and — rewriting the recurrence relation — satisfy:
$$\begin{align}
a_{2n+2} &= -a_{2n} + \cos (n+1)\pi = -a_{2n} - (-1)^n\\
a_{2n+1} &= -a_{2n-1}
\end{align}$$
for all $n$. Solving this gives directly
$$\begin{align}
a_{2n} &= (-1)^n a_0 + (-1)^n n\\
a_{2n+1} &= (-1)^{n}a_1
\end{align}$$
(If the first one is not straightforward, you can compute the first few terms to get an idea, then show it by induction.)
