Explanation Given: The differential equation is, y ‴ − 3 y ″ + 3 y ′ − y = 2 x − 2 e x , x > 0 . Approach: A differential equation of the form y ‴ + a 1 y ″ + a 2 y ′ + a 3 y = F Here, a 1 , a 2 , a 3 , and F are assumed to be continuous on the interval I . Let y 1 , y 2 and y 3 be linearly independent solutions to the associated homogenous equation y ‴ + a 1 y ″ + a 2 y ′ + a 3 y = 0 The auxiliary equation of the above differential equation can be written as, r 3 + a 1 r 2 + a 2 r + a 3 = 0 The linearly independent solution can be written as, y 1 ( x ) = e r x , y 2 ( x ) = x e r x and y 3 ( x ) = x 2 e r x ... ( 1 ) The particular solution of the non-homogenous differential equation is given as, y p ( x ) = u 1 y 1 + u 2 y 2 + u 3 y 3 Here, r is a real and repeated root. The formula to calculate u 1 , u 2 and u 3 is given as, y 1 u ′ 1 + y 2 u ′ 2 + y 3 u ′ 3 = 0 y ′ 1 u ′ 1 + y ′ 2 u ′ 2 + y ′ 3 u ′ 3 = 0 y ″ 1 u ′ 1 + y ″ 2 u ′ 2 + y ″ 3 u ′ 3 = F The general solution of the differential equation is given as, y ( x ) = c 1 e r x + c 2 x e r x + c 3 x 2 e r x + y p ( x ) Calculation: The associated homogenous differential equation is, y ‴ − 3 y ″ + 3 y ′ − y = 0 Compare the standard differential equation with the above differential equation. Then, a 1 = − 3 a 2 = 3 a 3 = − 1 The auxiliary equation of the above differential equation can be written as, r 3 − 3 r 2 + 3 r − 1 = 0 Calculate the roots of the above equation. ( r − 1 ) 3 = 0 r = 1 , 1 , 1 Since the roots are repeated then, The linearly independent solution can be written as, Substitute 1 for r in equation ( 1 ) . y 1 ( x ) = e x , y 2 ( x ) = x e x and y 3 ( x ) = x 2 e x The particular solution of the non-homogenous differential equation can be written as, y p ( x ) = e x u 1 + x e x u 2 + x 2 e x u 3 Calculate u 1 , u 2 and u 3

Differential Equations and Linear Algebra (4th Edition)

4th Edition

ISBN: 9780321964670

Author: Stephen W. Goode, Scott A. Annin

Publisher: PEARSON

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