Example 9.14.  Solve     with  .  

Solution 9.14.

    The characteristic equation    has complex roots and hence the general solution is  .   Use the initial conditions and form the linear system  

        

          

then solve for the constants and get and .  Hence the solution is

        .

We leave it for the reader to verify that this can be written as  

        .

(c) 2006 John H. Mathews, Russell W. Howell