If I have this equation:
$\frac{\partial^2u}{\partial x^2}=\frac{\partial^2u}{\partial t^2}$
And this general solution:
$u(x,t)=\sum^\infty_{n=-\infty}\cos k_nx(C_n\cos k_nt+D_n\sin k_nt)$
Would it then be wrong to write the above solution with only positive values of $n$? In my text book they often write the result from a superposition with only positive values of $n$, becasue the negative values of n already are included in the terms obtained for positive values of $n$.
The boundary condition:
$\frac{\partial u(x,t)}{\partial x}|_{x=0} = \frac{\partial u(x,t)}{\partial x}|_{x=L} = 0$
To get from $\frac{\partial^2u}{\partial x^2}=\frac{\partial^2u}{\partial t^2}$ to the solution I have used separation of variables and superposition.
Where I found:
U(x,t) = X(x)T(t)
And
X^{''}(x) = -k^{2}X(x)
T^{''}(t) = -k^{2}T(t)
Which gives me:
$X(x) = Acos(kx) + Bsin(kx)$
$T(t) = Ccos(kt) + Dsin(kt)$
Using the boundary condition I get: B = 0 and $k = k_{n} = \dfrac{n\pi}{L}$
Then I use superposition to get the solution:
$u(x,t)=\sum^\infty_{n=-\infty}\cos k_nx(C_n\cos k_nt+D_n\sin k_nt)$