I was hoping someone could help me understand the following proof:
For the initial boundary problem:
$$ u_{tt} = a^2u_{xx} \hspace{10 mm} 0 < x < L \hspace{10 mm} t>0 $$ $$ u_x(0,t) = 0 \hspace{10 mm} u(L,t) = 0\hspace{10 mm}t>0 $$ $$ u(x,0) = f(x)\hspace{10 mm} u_t(x,0)=g(x)\hspace{10 mm}0\leq x \leq L $$
The formal solution is: $$ u(x,t) = \sum\limits_{i=1}^\infty (\alpha_n \cos(\frac{(2n-1)\pi at}{2L})+\frac{2L \beta_n}{(2n-1)\pi a}\sin(\frac{(2n-1)\pi at}{2L}))\cos(\frac{(2n-1)\pi x}{2L}) $$
Our textbook just states this, but I was hoping someone could walk me through the steps needed to piece this together so I can understand?
Thank You!