PDE Initial-Boundary value problem question The problem: The ends of a stretched string are fixed at the origin and at the point,  $ x=\pi$ on the horizontal x-axis. The string is initially at rest along the x-axis, and then drops under it's own weight. The veritcal displacement $y(x,t)$ satisfies the PDE: 
$Y_{tt} = v^{2}Y_{xx} - g$ 
$ \big( 0 < x <\pi, t>0 \big)$ 
where $g$ is acceleration due to gravity. Write the above as an initial boundary-value problem, and show that the solution is given by 
$y(x,t) = \frac{4g}{\pi v^{2}} \big [ \sum_{n=1}^{\infty} \frac{\sin(2n-1)x}{(2n-1)^{3}} \cos(2n-1)vt - \frac{\pi}{8}x(\pi - x) \big]$ 
I honestly have no idea how to set this up because the wording of this problem is confusing me. 
Much help is appreciated. 
Thank you for all of your help. 
 A: From "the string is initially at rest", we know that $\partial_{t}y(x,0) = 0$. From "along the $x$-axis", we have $y(x,0) = 0$. So our initial-boundary-value problem is
\begin{align}
y_{tt}  &= \nu^{2}y_{xx} -g, \qquad 
y(x,0) = 0, \qquad
\partial_{t}y(x,0) =  0, \qquad
y(0,t) = y(\pi, t) = 0
\end{align}
You can solve the inhomogeneous part of the equation by taking $y_{p}(x,t)$ to be a second order polynomial in $x$ and $t$, (which will differentiate down to a constant under the action of $\partial_{t}^{2} - \nu^{2}\partial_{x}^{2}$, i.e., take $y_{p}(x,t) = a_{20}t^{2}+a_{10}t + a_{11}tx + a_{01}x + a_{02}x^2 + a_{00}$. The conditions $y(0,t) = y(\pi,t) = 0$ imply that $y_{p}(x,t) = cx(x-\pi)$. Hitting this with $\partial_{t}^{2} - \nu^{2}\partial_{x}^{2}$ and setting the result $=-g$ shows $c = \frac{g}{2\nu^{2}}$, i.e.,
\begin{align}
y(x,t) = y_{h}(x,t) - \frac{g}{2\nu^{2}}x(\pi-x),
\end{align} 
where $y_{h}$ solves
\begin{align}
\partial_{t}^{2}y_{h}  &= \nu^{2}\partial_{x}^{2}y_{h}, \qquad 
y_{h}(x,0) = 0, \qquad
\partial_{t}y_{h}(x,0) =  0, \qquad
y_{h}(0,t) = y_{h}(\pi, t) = 0
\end{align}
By separation of variables and application of boundary conditions, we see that the eigenfunctions are of the form $\cos(\nu mt)\sin(mx)$, so
\begin{align}
y(x,t) = \sum_{m=1}^{\infty} a_{m}\cos(\nu mt)\sin(mx)  - \frac{g}{2\nu^{2}}x(\pi-x).
\end{align}
Since $y(x,0) = 0$, then
\begin{align}
 \sum_{m=1}^{\infty} a_{m}\sin(mx) =  \frac{g}{2\nu^{2}}x(\pi-x)
\end{align}
Multiplication by $\sin(\ell x)$ and integration over $[0,\pi]$ gives
\begin{align}
a_{\ell} &= \frac{g}{\pi\nu^{2}}\int_{0}^{\pi} x(\pi -x) \sin(\ell x), \mathrm{d}x \\
&=\frac{2g(1-(-1)^{\ell})}{\pi \nu^2 \ell^3}
\end{align}
so that $a_{2n} = 0$ and $a_{2n-1} = \frac{4g}{\pi\nu^{2}(2n-1)^{3}}$, giving
\begin{align}
y(x,t) = \sum_{n=1}^{\infty} \frac{4g}{\pi\nu^{2}(2n-1)^{3}}\cos((2n-1)\nu t)\sin((2n-1)x)  - \frac{g}{2\nu^{2}}x(\pi-x).
\end{align}
A: The vertical displacement $y(x,t)$ satisfies the PDE
$$y_{tt}=v^2y_{xx}-g$$ for $x \in (0,\pi)$ and $t>0$, together with the boundary conditions $$y(0,t)=y(\pi,t)=0$$ (since the ends are fixed), and the initial conditions $$y(x,0)=0, \quad y_t(x,0)=0 \qquad\text{for }x \in [0,\pi].$$ Notice that $y(x,0)=0$ since the string is initially at rest along the $x$-axis. Similarly, $y_t(x,0)=0$ since the string has no initial velocity.
Let $w(x,t)=y(x,t)+\frac{g}{2v^2}(\pi x-x^2)$ (the second term is the "steady-state" or "long-time" solution which appears after the Fourier series in the desired answer). Then $$v^2w_{xx}=v^2y_{xx}-g=y_{tt}=w_{tt},$$ and so $w$ satisfies the homogeneous wave equation $w_{tt}=v^2w_{xx}$ with homogeneous Dirichlet BCs (that is, $w(0,t)=w(\pi,t)=0$ for $t>0$), and initial conditions $$w(x,0)=\frac{g}{2v^2}(\pi x-x^2), \quad w_t(x,0)=0 \quad \text{for } x \in [0,\pi].$$ This is now a standard separation of variables problem.
