Undetermined Coefficients For Higher Order Differential Equations   We have already seen how to solve a second order linear nonhomogeneous differential with constant coefficients where the "g" function generates a UC-set.  If you need a review of this click here.  For higher order nonhomogeneous differential equation, the exact same method will work.     Example Solve         y''' + y''  =  cos(2t)        y(0)  =  1,      y'(0)  =  2,      y''(0)  =  3 Solution We first solve the homogeneous differential equation         y''' + y''  =  0 The characteristic equation is          r3 + r2  =  0  Factoring gives          r2(r + 1)  =  0         r  =  0 (repeated twice)        or        r  =  -1 The homogeneous solution is          yh  =  c1 + c2t + c3e-t  The UC-set generated by g(t)  =  cos(2t) is         {cos(2t), sin(2t)} Notice that the UC-set does not intersect the homogeneous solution set.  We can write         yp  =  Acos(2t) + Bsin(2t)         yp'  =  -2Acos(2t) + 2Bsin(2t)         yp''  =  -4Asin(2t) - 4Bcos(2t)         yp'''  =  -8Acos(2t) + 8Bsin(2t) Plugging back into the original differential equation gives           [-8Acos(2t) + 8Bsin(2t)] + [-4Asin(2t) - 4Bcos(2t)]  =  cos(2t)     Combining like terms gives          (-8A - 4B)cos(2t) + (8B - 4A)sin(2t)  =  cos(2t) Equating coefficients gives          -8A - 4B  =  1         -4A + 8B  =  0 We can use matrices to solve and get         A  =  -.1      and        B  =  -0.05 The general solution is thus         y  =  c1 + c2t + c3e-t - .1cos(2t) - 0.05sin(2t) Now take derivatives to get         y'  =  c2 - c3e-t + .2sin(2t) - 0.1cos(2t)         y''  =  c3e-t + .4cos(2t) + 0.2sin(2t) Plug in the initial values to get         1  =  c1 + c3 - .1                c1 + c3  =  1.1         2  =  c2 - c3 - 0.1                c2 - c3  =  2.1         3  =  c3 + .4                        c3  =  2.6 This is a system of 3 equations and 3 unknowns which can be solve by matrices or easily by hand.  We get         c1  =  -3.6        c2  =  4.7        c3  =  2.6 The final solution is          y  =  -3.6 + 4.7t + 2.6e-t - .1cos(2t) - 0.05sin(2t)   Example Solve y(iv) +4y''  =  16 + 15et   Solution We first find the solution to the homogeneous differential equation.  The characteristic equations is          r4 + 4r2  =  0        r2(r2 + 4)  =  0 The roots are          r  =  0  (repeated twice)              r  =  2i        r  =  -2i The homogeneous solution is          yh  =  c1 + c2t + c3sin(2t) + c4cos(2t) Since g(t) is a sum of two terms, we can work each term separately.  The UC-Set for the function g(t)  =  16 is          {1} Since this is a solution to the homogeneous equation, we multiply by t to get          {t} This is also a solution to the homogeneous equation, so multiply by t again to get         {t2} which is not a solution of the homogeneous equation.  We write         yp1  =  At2         yp1'  =  2At      yp1''  =  2A         yp1'''  =  2A        yp1(iv)  =  0           Substituting back in, we get         0 +4(2A)  =  16         A  =  2 Hence         yp1  =  2t2 Now we work on the second piece.  The UC-set for g(t)  =  15et is          {et} Since this in not a solution to the homogeneous equation, we get         yp2  =  Aet        yp2'  =  Aet        yp2''  =  Aet        yp2'''  =  Aet        yp2(iv)  =  Aet     Plugging back into the original equation gives         Aet  + 4Aet  =  15et          5A  =  15         A  =  3 Hence         yp2  =  3et  The general solution to the nonhomogeneous differential equation is          y  =  c1 + c2t + c3sin(2t) + c4cos(2t) + 2t2 + 3et    Back to the Differential Equations Home Page