Similarly we handle the Neumann conditions. Take the boundary value problem for \(0 < t < 1\text{,}\)
\begin{equation*}
x''(t) + 2 x(t) = f(t) ,
\end{equation*}
where again \(f(t) = t\) on \(0 < t < 1\text{,}\) but now satisfying the Neumann boundary conditions \(x'(0) = 0\text{,}\) \(x'(1)=0\text{.}\) We write \(f(t)\) as a cosine series
\begin{equation*}
f(t) = \frac{c_0}{2} + \sum_{n=1}^\infty c_n \cos (n \pi t) ,
\end{equation*}
where
\begin{equation*}
c_0 = 2 \int_0^1 t \,dt = 1 ,
\end{equation*}
and
\begin{equation*}
c_n = 2 \int_0^1 t \cos (n \pi t) \,dt =
\frac{2\bigl({(-1)}^n-1\bigr)}{\pi^2 n^2} =
\begin{cases}
\frac{-4}{\pi^2 n^2} & \text{if } n \text{ odd} , \\
0 & \text{if } n \text{ even}.
\end{cases}
\end{equation*}
We write \(x(t)\) as a cosine series
\begin{equation*}
x(t) = \frac{a_0}{2} + \sum_{n=1}^\infty a_n \cos (n \pi t) .
\end{equation*}
We plug in to obtain
\begin{equation*}
\begin{split}
x''(t) + 2 x(t) & =
\sum_{n=1}^\infty \Bigl[ - a_n n^2 \pi^2 \cos (n \pi t) \Bigr]
+
a_0 +
2
\sum_{n=1}^\infty \Bigl[ a_n \cos (n \pi t) \Bigr]
\\
& =
a_0 +
\sum_{n=1}^\infty a_n (2 - n^2 \pi^2 ) \cos (n \pi t)
\\
& = f(t)
=
\frac{1}{2} +
\sum_{\substack{n=1\\n~\text{odd}}}^\infty
\frac{-4}{\pi^2 n^2} \cos (n \pi t) .
\end{split}
\end{equation*}
Therefore, \(a_0 = \frac{1}{2}\text{,}\) \(a_n = 0\) for \(n\) even (\(n \geq 2\)) and for \(n\) odd we have
\begin{equation*}
a_n (2 - n^2 \pi^2)
=
\frac{-4}{\pi^2 n^2} ,
\end{equation*}
or
\begin{equation*}
a_n
=
\frac{-4}{n^2 \pi^2 (2 - n^2 \pi^2)} .
\end{equation*}
The Fourier series for the solution \(x(t)\) is
\begin{equation*}
x(t) =
\frac{1}{4} +
\sum_{\substack{n=1\\n~\text{odd}}}^\infty
\frac{-4}{n^2 \pi^2 (2 - n^2 \pi^2)}
\cos (n \pi t) .
\end{equation*}
