17.2 Nonhomogeneous Linear Equations. 27 September 2007

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1 17.2 Nonhomogeneous Linear Equations 27 September 2007

2 Nonhomogeneous Linear Equations The differential equation to be studied is of the form ay (x) + by (x) + cy(x) = G(x) (1) where a 0, b, c are given constants and G(x) 0 is a continuous function on some open interval I R. Form of the General Solution Let y p (x) be any particular solution of the equation (1). If y 1 (x) and y 2 (x) are independent solutions of the complementary homogeneous equation and c 1, c 2 are constants, then y(x) = y p (x) + c 1 y 1 (x) + c 2 y 2 (x) (2) is the general solution of the nonhomogeneous equation.

3 Method of Undetermined Coefficients This method for finding a particular solution only applies when G(x) is a finite sum of terms of of the form p 1 (x)e rx, p 2 (x)e rx cos βx, p 3 (x)e rx sin βx, (3) where p 1 (x), p 2 (x), p 3 (x) are polynomials.

4 Recipe from Table 17.1 on page The first column lists the type of term present in G(x). The term auxiliary equation refers to az 2 + bz + c = 0. The last column lists the type of term to be included as a summand in the attempted solution y p (x). The constants A, B, C, D, E, F are to be determined by substitution into the equation (1). Ae rx, if r is not a root; e rs Axe rx, if r is a simple root; Ax 2 e rx, if r is a double root. sin kx, cos kx B cos kx + C sin kx, if k is not a root; Dx 2 + Ex + F, if 0 is not a root; px 2 + qx + m Dx 3 + Ex 2 + F x, if 0 is a simple root; Dx 4 + Ex 3 + F x 2, if 0 is a double root.

5 Variation of Parameters This general method presupposes that two independent solutions of the associated homogenous problem y 1 (x), y 2 (x) are known. The method is based on trying to find a solution of the form y p (x) = v 1 (x)y 1 (x) + v 2 (x)y 2 (x), (4) where the unknown functions are to be determined by using equation (1).

6 The method leads to the following two equations for the derivatives of these unknown functions v 1 y 1 + v 2 y 2 = 0 (5) v 1 y 1 + v 2 y 2 = a 1 G(x). (6) The assumption of independence made above regarding y 1 (x), y 2 (x) is the condition y 1 y 2 y 2y 1 0, (7) which guarantees that these equations uniquely determine the functions v 1 and v 2.

7 Once these functions are determined the functions v 1 and v 2 are determined by two antiderivative operations. The integration constants supplied by these antiderivatives guarantee that the general solution to equation (1) is obtained when these constants are varied.

8 Summary of Examples y 2y 3y = 1 x 2, r ± { 1, 3} (1) y(x) = c e x + c + e 3x x x y y = 2 sin x, r ± {0, 1} (2) y(x) = c + c + e x + cos x sin x y 3y + 2y = 5e x, r ± {1, 2} (3) y(x) = c e x + c + e 2x 5xe x

9 y 6y + 9y = e 3x, r ± {3} (4) y(x) = (c 0 + c 1 x) e 3x x2 e 3x y y = 5e x sin 2x, r ± {0, 1} (5) y(x) = c + c + e x + 5xe x 1 10 cos 2x + 1 sin 2x 5 y + y = tan x, r ± {± 1} (6) y(x) = c 1 cos x + c 2 sin x cos x ln sec x + tan x y + y 2y = xe x, r ± { 2, 1} (7) ( y(x) = c e 2x + c + e x ) 6 x2 e x

10 Exercise y + y + y = 4e x (cos x sin x) (1) y(0) = 0, y (0) = 1 r ± { 1 ± 1} y 1 (x) = e x cos x, y 2 (x) = e x sin x 0 = v 1 y 1 + v 2 y 2 8e x (cos x sin x) = v 1 y 1 + v 2 y 2 v 1 (x) = 8e2x sin x(cos x sin x) v 2 (x) = 8e2x cos x(cos x sin x) v 1 (x) = 2e 2x (1 sin 2x) + C 1 v 2 (x) = 4e 2x (cos x) 2 + C 2 y p (x) = 2e x cos x (2)

11 y(x) = e x (C 1 cos x + C 2 sin x) + 2e x cos x (3) y(0) = C = 0 y (0) = C 1 + C = 1 C 1 = 2 C 2 = 3 y(x) = e x ( 2 cos x 3 sin x) + 2e x cos x (4)

12 Variation of Parameters (Details) Assume that y 1 and y 2 are independent solutions of the homogeneous problem: ay 1 + by 1 + cy 1 = 0, ay 2 + by 2 + cy 2 = 0. (1) If one seeks a particular solution of the form y(x) = v 1 (x)y 1 (x) + v 2 (x)y 2 (x) (2) then y = ( v 1 y 1 + v 2 y ( 2) + v1 y 1 + v 2y 2 (3) 0 =. v 1 y 1 + v 2 y 2 (Imposed condition!) (4) y = v 1 y 1 + v 2y 2 (Simplified expression!) (5) y = ( v 1 y 1 + ) ( v 2 y 2 + v1 y 1 + v 2y 2 ) (6) )

13 and ay + by + cy = G(x) (7) ) = a ( v 1 y 1 + v 2 y 2 +a ( v 1 y 1 + v 2y 2 ) ) +b ( v 1 y 1 + v 2y 2 +c (v 1 y 1 + v 2 y 2 )

14 ay + by + cy = G(x) (8) Using equations (1) : = a ( v 1 y 1 + v 2 y 2 ( +v 1 ay 1 + by 1 + cy 1 +v 2 ( ay 2 + by 2 + cy 2 ) ) ) (9) (10) ay + by + cy = a ( v 1 y 1 + v 2 y 2) = G(x) (11) SUMMARY: v 1 (x)y 1(x) + v 2 (x)y 2(x) = 0 (12) v 1 (x)y 1 (x) + G(x) v 2 (x)y 2 (x) = (13) a

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