as functions cosh ⁡ ⁡ ( s ) is even we get two solutions: s 1 − − s 2 = s 3 − − s 4 s 1 − − s 2 = − − s 3 + s 4 from the last equation, leading to a non-trivial solution, we solve s 2 and substitute into the dependent equation, we obtain e s 1 and then e s 2 , we have: e s 1 = e s 4 m 1 e s 3 + m 2 e s 4 m 1 e s 4 + m 2 e s 3 e s 2 = e s 3 m 1 e s 3 + m 2 e s 4 m 1 e s 4 + m 2 e s 3