Let a string with linear density $\rho$ and tension $k$ Its left and right hand ends $ [-\pi,\pi]$ are held fixed at height zero (Maybe not at $t=0$, but it are for $t>0$). Initial velocity $v_0(x)\equiv 0$ and $u_0(x)= x$. Find $u(x,t)$
What I did
I find by the boundary conditions
\begin{equation} u(x,t)=\sum_{j=1}^\infty\left[\alpha_j\cos jct +\beta_j\sin jct\right]\sin jx \end{equation} And by the Initial conditions \begin{equation} u_0(x)=\sum_{j=1}^\infty \alpha_j\sin jx\hspace{1.5cm}\text{if}\hspace{0.15cm}x\in(-\pi,\pi) \end{equation} \begin{equation} v_0(x)=\sum_{j=1}^\infty \beta_jcj\sin jx \hspace{1cm}\text{if}\hspace{0.15cm}x\in(-\pi,\pi) \end{equation}
First I tried to find the Fourier series for $u_0(x)=x$
$$a_n=\frac{1}{\pi}\int_{-\pi}^{\pi}x\cos n\; dx=0\hspace{1cm}a_0=\frac{1}{\pi}\int_{-\pi}^{\pi}x \; dx=0$$
$$b_n=\frac{1}{\pi}\int_{-\pi}^{\pi}x \sin nx \; dx=\frac{2(-1)^{n+1}}{n}$$ So $$u_0(x)=\sum_{n=1}^\infty\frac{2(-1)^{n+1}}{n}\sin(nx)$$
And $$u_0(x)=\sum_{j=1}^\infty \alpha_j\sin(jx)$$
So $$\alpha_j=\frac{2(-1)^{j+1}}{j}$$
And because $v_0(x)\equiv 0 \Rightarrow\beta_j=0$
$$u(x,t)=\sum_{j=1}^\infty\frac{2(-1)^{j+1}}{j}\cos(jct)\sin(jx)$$
Is this correct?