Higher Order Differential Equations   Recall that the order of a differential equation is the highest derivative that appears in the equation.  So far we have studied first and second order differential equations.  Now we will embark on the analysis of higher order differential equations.  We will restrict our attention to linear differential equations.  The general linear differential equation can be written as                         dny                   dn-1y                             dy             L(y)  =               +  p1(t)              + . . .  + pn-1(t)             +  pn(t)y  =  g(t)                           dt                      dt                                 dt The good news is that all the results from second order linear differential equation can be extended to higher order linear differential equations.  We list without proof the results If p1, ... pn are continuous on an interval [a,b] then there is a unique solution to the initial value problem, where instead of the initial conditions y(0)  =  y0 and y'(0)  =  y'0 , we need the initial conditions         y(0)   =  y0 ,   y'(0)  = y'0,   y''(0)  =  y''0,    ...   ,   y(n-1)(0)  =  y(n-1)0  L(y) is a linear transformation, that is          L(c1y1+ x2y2 + ... + cnyn)  =  c1L(y1) + c2L(y2) + ... + cnL(yn) If y1, y2 ... yn are solutions to L(y)  =  0, then c1y1+ x2y2 + ... + cnyn is also a solution. For differentiable functions y1, y2 ... yn, we define the Wronskian by         Differentiable functions y1, y2 ... yn are linearly independent if the Wronskian is nonzero for some t in [a,b]. L(y)  =  0 has n linearly independent solutions. If yh is the general solution to L(y)  =  0 and if yp is a particular solution to L(y)  =  g(t), then yh + yp is the general solution to L(y)  =  g(t). Abel's theorem still holds.  That is, if y1, y2 ... yn are linearly independent solutions of L(y)  =  0, then         The method of reduction of order still works.  If y1 is a solution to L(y1)  =  0, then L(v1y1) can be reduced to a differential equation of degree n - 1.   Example Determine if the following functions are linearly independent         y1  =  e2x         y2  =  sin(3x)        y3  =  cos x   Solution First take derivatives         y'1  =  2e2x         y'2  =  3cos(3x)        y'3  =  -sin x         y''1  =  4e2x         y''2  =  9sin(3x)        y''3  =  -cos x The Wronskian is         Now plug in x   =  0 (or any other value for x) to get         (1)(-3 - 0) - (0)(-2 + 0) + (1)(0 - 12)  =  -15 In particular, since this is a nonzero number, we can conclude that the three functions are linearly independent.   Example Use Abel's theorem to find the Wronskian of the differential equation         ty(iv) + 2 y''' - tety'' + (t3 - 4t)y' + t2sin ty  =  0   Solution We first divide by t to get         y(iv) + 2/t y''' - ety'' + (t2 - 4)y' +  t sin t y  =  0 Now take the integral of 2/t to get         2ln t The Wronskian is thus        c e2ln t  =  c t2   Back to the Differential Equations Home Page