I am trying to solve the following boundary value problem, where $0<x<\pi$:
$$\frac{\partial^2 u}{\partial t^2}=9\frac{\partial^2u}{\partial t^2}$$ $$\frac{\partial u}{\partial x}(0,t)=\frac{\partial u}{\partial x}(\pi,t)=0$$ $$u(x,0)=0, \frac{\partial u}{\partial t}(x,0)=-2\sin x$$
By letting $u(x,t)=\phi(x)h(t)$, substituting into the PDE, and separating variables, I obtained the following ODEs:
$$Spatial \; ODE:\frac{d^2 \phi}{dt^2}=-\lambda \phi$$ $$Time \; ODE: \frac{d^2 h}{dt^2}=-9\lambda h$$
Using the Spatial ODE, the given boundary conditions imply that for $\lambda>0$, the corresponding eigenfunctions are $\phi_n(x)=\cos nx$, and for $\lambda=0, \phi_0(x)=1$. The general solution to the Time ODE is $h(t)=c_1\cos 3nt+c_2\sin 3nt$. So, the product solution can be expressed as:
$$u(x,t)=\sum_{n=0}^\infty A_n\cos nx\cos 3nt+\sum_{n=0}^\infty B_n\cos nx\sin 3nt$$
Using the first initial condition, $A_n=0, \forall \; n$. The second initial condition yields: $$-2\sin x=\sum_{n=0}^\infty 3n B_n\cos nx=\sum_{n=1}^\infty 3n B_n\cos nx$$
Using orthogonality of cosines yields that $B_n=\frac{8}{3n\pi(n^2-1)}$ when $n\geq 1$ is even and $0$ when $n \geq 1$ is odd.
The product solution follows, but this is where I'm confused. From the Spatial ODE and the boundary conditions, we determined that $\lambda=0$ is indeed an eigenvalue, but it does not appear in the product solution. Is this a discrepancy, or is this just because the actual value of $B_0$ doesn't matter since it will be multiplied by an expression equivalent to $0$ in both $u(x,t)$ and $\frac{\partial u}{\partial t}(x,t)$ for $n=0$? Furthermore, does the infinite series I derived from the second initial condition still equate to $-2\sin x$, as usually Fourier cosine series are from $n=0$ to infinity?