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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
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