Yes, 0 is a perfectly good eigenvalue. I presume you know that if \(\displaystyle \lambda\) (real) then the solution will involve \(\displaystyle Ce^{\lambda t}\). If \(\displaystyle \lambda= 0\), that solution will be \(\displaystyle Ce^{0t}= Ce^0= C\). In other words, if 0 is an eigenvalue, then one of the solutions will be "x(t)= constant".