1D Wave Equation Problem Separation of Variables I need to solve the following 1D Wave Equation problem using Separation of Variables, but I cannot figure it out.
\begin{align}
u_{tt} &= u_{xx}\\
u_x(t,0) &= u_x(t,1) = 0\\
u(0,x) &= x(1-x)\\
u_t(0,x) &= 0
\end{align}
I have done the following work to solve it, but do not know how to find the values of $D$, $A$, $B$, $d_k$, or $b_k$.
Assume $u(t,x) = w(t)v(x) \longrightarrow w''(t) = \lambda w(t), v''(x) =\lambda v(x)$
Case 1: $\lambda = 0$
\begin{align}
v(x) &= Cx+D\\
v'(x) &= C\\
v'(0) &= v'(1) = C = 0
\end{align}
$D$ can be anything.
\begin{align}
u(t,x) &= D(At+B) +\sum_{k=1}^{\infty}{d_k\cos(k\pi t)\sin(k\pi x) + b_k \sin(k\pi t)\sin(k\pi x)}\\
u(0,x) &= x(1-x) = DB+\sum_{k=1}^{\infty}{d_k\sin(k\pi x)}\\
u_t(0,x) &= 0 = DA+\sum_{k=1}^{\infty}{b_kk\pi\sin(k\pi x)}
\end{align}
This is as far as I have gotten and I cannot seem to be able to find $D$, $A$, $B$, $d_k$, or $b_k$. Any help is greatly appreciated.
 A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\on}[1]{\operatorname{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

*

*$\ds{\on{u}_{x}\pars{t,x} =
-\pi\sum_{n = 1}^{\infty}\on{a}_{n}\pars{t}\,n\sin\pars{n\pi x}}$ already satisfies the boundary conditions.

*

\begin{align}
&\implies \on{u}\pars{t,x} =
\sum_{n = 1}^{\infty}\on{a}_{n}\pars{t}\cos\pars{n\pi x} + \on{f}\pars{t}\label{1}\tag{1}
\\[5mm] &
\mbox{Note that}
\\ &\
\left.\begin{array}{rcl}
\ds{\int_{0}^{1}\cos\pars{m\pi x}\cos\pars{n\pi x}
\,\dd x} & \ds{=} &
\ds{{1 \over 2}\,\delta_{mn}}
\\
\ds{\int_{0}^{1}\cos\pars{n\pi x}\,\dd x} & \ds{=} &
\ds{\delta_{n0}}
\end{array}\right\}\,,\qquad
m, n \in \mathbb{N}_{\,\geq\ 0}
\label{2}\tag{2}
\end{align}

*

*$\ds{\on{u}\pars{t,x}}$ must satisfies the above differential equation:
\begin{align}
&\sum_{n = 1}^{\infty}
\ddot{\on{a}}_{n}\pars{t}\cos\pars{n\pi x} + \ddot{\on{f}}\pars{t}
\\[2mm] = &\
\sum_{n = 1}^{\infty}\on{a}_{n}\pars{t}
\pars{-n^{2}\pi^{2}}\cos\pars{n\pi x}
\label{3}\tag{3}
\end{align}

*Integrate -see (\ref{2})- both members of (\ref{3}) over
$\ds{x \in \pars{0,1}}$:
$$
\implies\ddot{\on{f}}\pars{t} = 0\implies
\on{f}\pars{t} = bt + c\,,\quad b, c = \mbox{constants} 
$$

*With (\ref{2}) and (\ref{3}):
\begin{align}
&\ddot{\on{a}}_{n}\pars{t} = -n^{2}\pi^{2}\on{a}\pars{t}
\\[2mm] &
\implies
\on{a}_{n}\pars{t} = \on{a}_{n}\pars{0}\cos\pars{n\pi t} +
\dot{\on{a}}_{n}\pars{0}\,{\sin\pars{n\pi t} \over n\pi}
\end{align}

*$\ds{\on{u}\pars{t,x}}$ is reduced to
\begin{align}
\on{u}\pars{t,x} & =
\sum_{n = 1}^{\infty}\bracks{\on{a}_{n}\pars{0}\cos\pars{n\pi t} +
\dot{\on{a}}_{n}\pars{0}\,{\sin\pars{n\pi t} \over n\pi}}
\\[2mm] & \cos\pars{n\pi x} + bt + c
\\[5mm]
\on{u}_{t}\pars{t,x} & =
\sum_{n = 1}^{\infty}\bracks{%
-n\pi\on{a}_{n}\pars{0}\sin\pars{n\pi t} +
\dot{\on{a}}_{n}\pars{0}\cos\pars{n\pi t}}
\\[2mm] & \cos\pars{n\pi x} + b
\end{align}

*

\begin{align}
x\pars{1 - x} = \on{u}\pars{0,x} & =
\sum_{n = 1}^{\infty}\on{a}_{n}\pars{0}
\cos\pars{n\pi x} + c\label{4}\tag{4}
\\[5mm]
0 = \on{u}_{t}\pars{0,x} & =
\sum_{n = 1}^{\infty}\dot{\on{a}}_{n}\pars{0}\cos\pars{n\pi x} + b
\end{align}

*

*With the last equation and (\ref{2}), I found
$\ds{\dot{\on{a}}_{n}\pars{0} = 0}$ and $\ds{b = 0}$.

*Integrate both members of (\ref{4}) over $\ds{x \in \pars{0,1}}$:
$$
\int_{0}^{1}x\pars{1 - x}\dd x = \int_{0}^{1}c\,\dd x
\implies c = {1 \over 6}
$$

*With (\ref{2}) and (\ref{4}):
$$
\underbrace{2\int_{0}^{1}x\pars{1 - x}\cos\pars{n\pi x}
\,\dd x}
_{\ds{-\,\bracks{1 + \pars{-1}^{n}}{2 \over n^{2}\pi^{2}}}} = \on{a}_{n}\pars{0}
$$

*Finally,
$$
\bbx{\on{u}\pars{t,x} =
{1 \over 6} - {1 \over \pi^{2}}
\sum_{n = 1}^{\infty}
{\cos\pars{2n\pi t}\cos\pars{2n\pi x} \over n^{2}}} \\
$$
