Partial differential equation with initial condition for time derivative I am trying to solve the PDE
\begin{align}
u_{tt} - u_{xx} + 2u &= 0
\end{align}
where  $0 \leq x \leq \pi, \ t \geq 0$. The initial conditions are
$$u(x,0) = 0, \quad u_t(x,0)  = \frac{x}{\pi}$$
and the boundary conditions are
$$ u_x(0,t) = 0, \quad u_x(\pi,t) = 0$$
I am trying to solve using separation of variables. Making the ansatz
$$u(x,t) = F(x)G(t)$$
after substitution I get
$$\frac{F''(x)-2F(x)}{F(x)} = \frac{\ddot{G}(t)}{G(t)}=-n^2-2$$
I get finally a solution of the form
$$ u(x,t)=\sum_{n=1}^{\infty} A_n \cos(nx) \sin(\sqrt{n^2+2}t)  $$
Now I still have to comply with the initial condition for the time derivative
$$u_t(x,0)  = \frac{x}{\pi}$$
This implies
$$ u(x,t)=\sum_{n=1}^{\infty} A_n \cos(nx) \sqrt{n^2+2} = \frac{x}{\pi}  $$
But $x/\pi$ is an odd function, so it can not be represented by a sum of cosine functions. What did I do wrong?
 A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
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 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
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\begin{align}
&\mbox{Lets}\ \mrm{u}_{x}\pars{x,t} =
\sum_{n = 1}^{\infty}a_{n}\pars{t}\sin\pars{nx}\
\mbox{which already satisfies}\
\mrm{u}_{x}\pars{0,t} = \mrm{u}_{x}\pars{\pi,t} = 0.
\\[5mm] &\
\mbox{Then,}\
\mrm{u}\pars{x,t} =
-\sum_{n = 1}^{\infty}a_{n}\pars{t}\,{\cos\pars{nx} \over n} + \mrm{f}\pars{t}.
\\ &\ \mrm{f}\pars{t}\ \mbox{is a time dependent }\ arbitrary\ \mbox{function ( for the time being ).}
\end{align}

$\ds{\mrm{u}\pars{x,t}}$ must satisfy the differential equation. Namely,
\begin{align}
&0 = \bracks{-\sum_{n = 1}^{\infty}\ddot{a}_{n}\pars{t}\,{\cos\pars{nx} \over n} + \ddot{\mrm{f}}\pars{t}} -
\bracks{\sum_{n = 1}^{\infty}a_{n}\pars{t}n\cos\pars{nx}}
\\[5mm] &\ +
2\bracks{-\sum_{n = 1}^{\infty}a_{n}\pars{t}\,{\cos\pars{nx} \over n} + \mrm{f}\pars{t}}
\end{align}
Integrating both sides over $\ds{\pars{0,\pi}} \implies
\ddot{\mrm{f}}\pars{t} + 2\,\mrm{f}\pars{t} = 0 \implies
\mrm{f}\pars{t} = A\sin\pars{\root{2}t} + B\cos\pars{\root{2}t}$.
$\ds{A\ \mbox{and}\ B}$ are constants.
Similarly, integrate after multiplying both sides for a factor
$\ds{\cos\pars{nx}}$ to get
\begin{align}
&\ddot{a}_{n}\pars{t} + \pars{n^{2} + 2}a_{n}\pars{t} = 0
\\ &\ \implies
a_{n}\pars{t} =
a_{n}\pars{0}\cos\pars{\root{n^{2} + 2}t} +
\dot{a}_{n}\pars{0}\,{\sin\pars{\root{n^{2} + 2}t} \over \root{n^{2} + 2}}
\end{align}
The general solution becomes:
\begin{align}
\mrm{u}\pars{x,t} & =
-\sum_{n = 1}^{\infty}\bracks{a_{n}\pars{0}\cos\pars{\root{n^{2} + 2}t} +
\dot{a}_{n}\pars{0}\,{\sin\pars{\root{n^{2} + 2}t} \over \root{n^{2} + 2}}}\,{\cos\pars{nx} \over n}
\\ & +
A\sin\pars{\root{2}t} + B\cos\pars{\root{2}t}
\end{align}
Also,
$$
0 = \mrm{u}\pars{x,0} =
-\sum_{n = 1}^{\infty}a_{n}\pars{0}\,{\cos\pars{nx} \over n} + B
\implies a_{n}\pars{0} = B = 0
$$
The general solution is reduced to
\begin{align}
\mrm{u}\pars{x,t} & =
-\sum_{n = 1}^{\infty}
\dot{a}_{n}\pars{0}\,{\sin\pars{\root{n^{2} + 2}t} \over \root{n^{2} + 2}}\,{\cos\pars{nx} \over n}
+
A\sin\pars{\root{2}t}
\end{align}
In addition,
\begin{align}
\mrm{u}_{t}\pars{x,0} & = {x \over \pi} =
-\sum_{n = 1}^{\infty}
\dot{a}_{n}\pars{0}\,{\cos\pars{nx} \over n}
+
\root{2}A
\end{align}
Integrating both sides over $\ds{\pars{0,\pi} \implies
{\pi \over 2} = \root{2}A\pi \implies A = {\root{2} \over 4}}$. Also,
\begin{align}
&\int_{0}^{\pi}{x \over \pi}\,\cos\pars{nx}\,\dd x =
-\,{\pi \over 2n}\dot{a}_{n}\pars{0} \implies
{\pars{-1}^{n} - 1 \over n^{2}\pi} = -\,{\pi \over 2n}\dot{a}_{n}\pars{0}
\\[5mm] &\
\implies \dot{a}_{n}\pars{0} =
{2 \over \pi^{2}}\,{1 - \pars{-1}^{n} \over n}
\end{align}
Finally,
\begin{align}
\mrm{u}\pars{x,t} & =
-\sum_{n = 1}^{\infty}
{2 \over \pi^{2}}\,{1 - \pars{-1}^{n} \over n}\,{\sin\pars{\root{n^{2} + 2}t} \over \root{n^{2} + 2}}\,{\cos\pars{nx} \over n}
+
{\root{2} \over 4}\,\sin\pars{\root{2}t}
\\[5mm] & =
\color{red}{-\,{4 \over \pi^{2}}\sum_{n = 0}^{\infty}
{1 \over \pars{2n + 1}^{2}}\,{\sin\pars{\root{\bracks{2n + 1}^{2} + 2}t} \over \root{\bracks{2n + 1}^{2} + 2}}\cos\pars{\bracks{2n + 1}x}}
\\[2mm] &\
\color{red}{+ {\root{2} \over 4}\,\sin\pars{\root{2}t}}
\end{align}
