<s>[INST] Finding particular solution to 1D - wave equation given general solution. Given the general solution of the wave equation: 
$\displaystyle u(x,t) = \sum_{n=0}^\infty b_n \sin\left[ct\left(n+\frac{1}{2}\right) \pi\right]\sin\left[\pi\left(n+\frac{1}{2}\right) x\right]$
Find the particular solution given that:
$\displaystyle \frac{\partial}{\partial t}u(x,0) = x$   for $0\leq x \leq 1$
First Attempt:
$\displaystyle u_t(x,0) = \sum_{n=0}^\infty b_nc\left(n+\frac{1}{2}\right)\pi\cdot \sin\left[\pi\left(n+\frac{1}{2}\right) x\right] = x $
Then we find $b_n$ as shown below: 
$b_n = \frac{2}{\pi} \int_0^\pi \pi x c(n+\frac{1}{2})) \cdot sin\big(\pi(n+\frac{1}{2}))\big)dx $
which simplifies to:
$2c(n+\frac{1}{2})\Big(\Big[-x\pi (n+\frac{1}{2})   \cdot   cos\big(\pi x(n+\frac{1}{2})\big)\Big]_0^\pi +  \pi (n+\frac{1}{2}) \int_0^\pi cos\big( x\pi (n+\frac{1}{2})\big)\Big)$
But is this the right way to go?
Second Attempt:
$\displaystyle u(x,t) = \sum_{n=0}^\infty b_n \sin\left[ct\left(n+\frac{1}{2}\right) \pi\right]\sin\left[\pi\left(n+\frac{1}{2}\right) x\right]$
let $\alpha = (n+\frac{1}{2})$
$\displaystyle u(x,t)  = \sum_{n=0}^\infty b_n \cdot sin\big(ct\alpha \pi\big) \cdot sin\big(\pi\alpha x\big)$
$\displaystyle u_t(x,t) = \sum_{n=0}^\infty b_n \cdot c\pi \alpha \cdot cos\big(ct\alpha \pi\big) \cdot sin\big(\pi\alpha x\big)$
How do I calculate $b_n$ from here?
Me attempting to find $b_n$:
$\displaystyle b_n = \frac{2}{\pi} \int_0^1 \pi x c\alpha \cdot sin\big(\pi\alpha\big)dx $
goes to: 
$b_{n} = \frac{2}{(c \pi^{2} \alpha)} \int_{0}^{1} x \sin (\alpha \pi x) dx$
 [/INST] $$\displaystyle u(x,t) = \sum_{n=0}^\infty b_n \sin\left[ct\left(n+\frac{1}{2}\right) \pi\right]\sin\left[\pi\left(n+\frac{1}{2}\right) x\right]$$
$$\displaystyle u(x,t) = \sum_{n=0}^\infty \frac{b_n}{2}\left[ \cos\left[\pi\left(n+\frac{1}{2}\right)(x-ct) \right]-\cos\left[\pi\left(n+\frac{1}{2}\right) (x+ct)\right] \right]$$
 $$u(x,t) =f(x-ct)-f(x+ct)$$ 
Where $$f(y)=\sum_{n=0}^\infty \frac{b_n}{2} \cos\left[\pi\left(n+\frac{1}{2}\right)(y) \right]$$
Now the boundary condition($\displaystyle \frac{\partial}{\partial t}u(x,0) = x$) gives $f(x)= -\frac{x^2}{4c}+d$ which gives $$u(x,t) =\frac{(x+ct)^2}{4c} -\frac{(x-ct)^2}{4c}$$
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