University:
California State University, Northridge
Course:
MATH 351 | Differential Equations
Academic year:

2014
Views:

202

Pages:

3
Author:

Hadassah Chase

The method of variation of parameters for the nonhomogeneous nth order linear differential equation For the nonhomogeneous nth order linear differential equation L[y ] = y (n) + p1 (t)y (n−1) + p2 (t)y (n−2) + · · · + pn−1 (t)y 0 + pn (t)y = g (t) (1) Suppose that we know a fundamental set of solutions y1 (t), y2 (t), . . . yn (t) (2) y (n) + p1 (t)y (n−1) + p2 (t)y (n−2) + · · · + pn−1 (t)y 0 + pn (t)y = 0. (3) of the homogeneous equation Then the general solution of the homogeneous equation is yc (t) = c1 y1 (t) + c2 y2 (t) + · · · + cn yn (t) (4) Then a particular solution is yp (t) = u1 (t)y1 (t) + u2 (t)y2 (t) + · · · + un (t)yn (t), 0 where the uk , k = 1, 2, . . . , n are determined by the n equations (5) y1 u10 + y2 u20 + · · · + yn un0 y10 u10 + y20 u20 + · · · + yn0 un0 .. . (n−1) 0 (n−1) 0 (n−1) 0 y1 u1 + y2 u2 + · · · + yn un = = = 0, 0, .. . g (t), (6) 0 0 The system (6) is a linear algebraic system for the unknown quantities u1 , . . . , un . Using Cramer’s rule, we can write the solution of the system of equations (6) in the form 0 uk = Wk (t) , k = 1, 2, . . . , n, W (t) (7) where W (t) is the Wronskian of y1 , y2 , . . . , yn , i.e., W (t) = W (y1 , y2 , . . . , yn )(t). (8) Wk is the determinant obtained by replacing the kth column of the Wronskian by the column consisting of the right-hand side of (6), that is, the column consisting of (0, 0, . . . , g (t)). With this notation a particular solution of Eq. (1) is given by Z t n X Wk (s) yp (t) = yk (t) ds (9) t0 W (s) k=1 where t0 is arbitrary. When n = 2, see Section 3.6; When n = 3, the particular solution is yp (t) = u1 (t)y1 (t) + u2 (t)y2 (t) + u3 (t)y3 (t), (10) where y1 , y2 and y3 constitute a fundamental set of solutions of the associated homogeneous DE and u1 , u2 , u3 are determined from y1 u10 + y2 u20 + y3 u30 = 0, y10 u10 + y20 u20 + y30 u30 = 0, y100 u10 + y200 u20 + y300 u30 = g (t), (11) that is, u10 = W2 W3 W1 , u20 = , u30 = , W W W (12) where W1 = 0 y2 y3 y1 0 y3 y1 y2 0 y1 y2 y3 0 y20 y30 , W2 = y10 0 y30 , W3 = y10 y20 0 , and W = y10 y20 y30 g (t) y200 y300 y100 y200 g (t) y100 y200 y300 y100 g (t) y300