Variation of Parameters for Higher Order Differential Equations For second order nonhomogeneous differential equations, we saw that if the function g(x) does not generate a UC-Set, then we must use the method of variation of parameters.  To review this method click here.  It turns out that the method of variation of parameters naturally generalizes to higher order differential equations. Let          L(y)  =  y(n) + p1y(n-1) + ... + pn-1y' + pny  =  g(t) and suppose that the general solution to L(y)  =  0  is given by          yh  =  c1y1 + c2y2 + ... + cnyn  As with variation of parameters for second order equations, we assume that         yp  =  u1y1 + u2y2 + ... + unyn  where the ui are functions of t.  Following the prior discussion, we see that we need only one yp but have labeled n functions of t.  We therefore may impose n - 1 conditions on the u's.  The conditions that we impose are         u1y1' + u2y2' + ... + unyn'          u1'y1' + u2'y2' + ... + un'yn'          u1''y1' + u2''y2' + ... + un''yn'               |             |                  |         u1(n-1) y1' + u2(n-1) y2' + ... + un(n-1) yn'  Now take derivatives to get         yp'  =  u1y1' + u1'y1 + u2y2' +u2'y2 + ... + unyn' + un'yn =  u1'y1 + u2'y2 + ... + un'yn          yp''  =  u1'y1' + u1''y1 + u2'y2' +u2''y2 + ... + un'yn' + un''yn =  u1''y1 + u2''y2 + ... + un''yn              |            |            |         yp(n-1)  =   u1(n-1)y1 + u2(n-1)y2 + ... + un(n-1)yn  Plugging in to the original differential equation and noting that each of the y's are solution to the differential equation, we get         y1(n-1)u1' + y2(n-1)u2' + ... + yn(n-1)un'  =  g(t) Notice that this gives a system of n equations and n unknowns.  We can write this as Since the Wronskian is never zero, we can take its inverse.  Taking the inverse of matrix where the entries are functions is not easy.  Fortunately, we only need the last column of the inverse, which can be found with much less pain using the adjoint formula.   Example Solve          y''' + y'  =  sec t   Solution We first find the homogeneous solution.  The characteristic equation is          r3 + r  =  0         r(r2 + 1)  =  0         r  =  0,        r  =  i,         r  =  -i We conclude          yh  =  c1 + c2 cos t + c3 sin t We have         yp  =  v1 + v2 cos t + v3 sin t and the Wronskian matrix is The Wronskian is          |W|  =  1(sin2 t + cos2 t) - cos t (0 - 0) + sin t (0 - 0)  =  1 The matrix equation takes the following form So that          u1'  =  sec t        u2'  =  -1        u3'  =  -tan t Integrating, we get                  u1'  =  ln|sec t + tan t|       u2'  =  -t        u3'  =  ln|cos t| The final solution is         y  =  c1 + c2 cos t + c3 sin t + ln|sec t + tan t|  -t cos t + (sin t) ln|cos t|   Back to the Differential Equations Home Page