Suppose that "real parts of the characteristic roots are non-positive and the roots with zero real part have multiplicity 1". Then look at the solutions corresponding to each characteristic root. If a+ ib is a root, then solutions are of the form and for some constant C. Obviously cos(bx) and sin(bx) are bounded. What about with a negative and x positive?
If a+ bi is a multiple root, you can have solutions of the form and . You will need to show that , again with a negative and x positive, is bounded for all n.
If the real part is 0, then the root is simply bi and the solution is C(cos(bx)+ i sin(bx)) which is clearly bounded.
If bi is a multiple root, we could have solutions of the form and which are not bounded. That's why you need "roots with zero real part have multiplicity 1".