Higher Order Homogeneous Differential Equations With Constant Coefficients   We now investigate how to solve higher order homogeneous linear differential equations with constant coefficients.  The good news is that we use the same technique that we used for second order differential equations.  We demonstrate this by an example. Example Solve         y''' - 4y'' + y' + 6y  =  0   Solution We make the assumption that the solution has the form        y  =  ert Then         y' =  rert         y''  =  r2ert         y'''  =  r3ert Substituting back into the differential equation, we get         r3ert - 4(r2ert) + rert + 6ert   =  0 Dividing by ert, we get the characteristic equation         r3 - 4r2 + r + 6  =  0 We can factor to find the roots of the equation.  A calculator can efficiently do this, or you can use the rational root theorem to get         (r + 1)(r - 2)(r - 3)  =  0        r  =  -1        r  =  2        or       r  =  3 The general solution is        y  =  c1e-t + c2e2t + c3e3t   Example Find the solution to         y(v) - 4y'''  =  0        y(0)  =  2    y'(0)  =  3    y''(0)  =  1    y'''(0)  =  -1,    y(iv)(0)  =  1   Solution We have the characteristic equation         r5 + 4r3  =  0    Which has a root of order 3 at         r  =  0 and complex roots        r  =  2i     and    r  =  -2i We use what we have learned about repeated roots and complex roots to get the general solution.  Since the order of the repeated root is 3, we have         y1  =  1,        y2  =  t,        y3  =  t2    For order 2, multiply by t and t2 The complex roots give the other two solutions         y3  =  cos(2t)       and       y5  =  sin(2t) The general solution is         y  =  c1 + c2t + c3t2 + c4 cos(2t) + c5 sin(2t) Now Find the first four derivatives         y'  =  c2 + 2c3t - 2c4 sin(2t) + 2c5 cos(2t)         y''  =  2c3 - 4c4 cos(2t) - 4c5 sin(2t)         y'''  =  8c4 sin(2t) - 8c5 cos(2t)         y(iv)  =  16c4 cos(2t) + 16c5 sin(2t) Now plug in the initial conditions to get         2  =  c1 + c4         3  =  c2  + 2c5         1  =  2c3 - 4c4         -1  =  8c5         1  =  16c4   We can solve this either by hand or by using a calculator to get                 c1  =   31/16        c2  =  23/4          c3  =  5/8          c4   =  1/16       c5  = 1/8 The solution is         y   =   31/16  +  23/4 t  +  5/8 t2  +  1/16 cos(2t)  +  1/8 sin(2t)   Back to the Differential Equations Home Page