Explanation Given: The differential equation is, y ″ + 4 y ′ + 4 y = 4 e − 2 x 1 + x 2 + 2 x 2 − 1 . Approach: A differential equation of the form y ″ + a 1 y ′ + a 2 y = F Here, a 1 , a 2 , and F are assumed to be continuous on the interval I . Let y 1 and y 2 be linearly independent solutions to the associated homogenous equation y ″ + a 1 y ′ + a 2 y = 0 The auxiliary equation of the above differential equation can be written as, r 2 + a 1 r + a 2 = 0 The linearly independent solution can be written as, y 1 ( x ) = e r x and y 2 ( x ) = x 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 Here, r is a real and repeated root. The formula to calculate u 1 and u 2 is given as, y 1 u ′ 1 + y 2 u ′ 2 = 0 y ′ 1 u ′ 1 + y ′ 2 u ′ 2 = F The general solution of the differential equation is given as, y ( x ) = c 1 e r x + c 2 x e r x + y p ( x ) Calculation: The associated homogenous differential equation is, y ″ + 4 y ′ + 4 y = 0 Compare the standard differential equation with the above differential equation. Then, a 1 = 4 a 2 = 4 The auxiliary equation of the above differential equation can be written as, r 2 + 4 r + 4 = 0 Calculate the roots of the above equation. ( r + 2 ) 2 = 0 r = − 2 , − 2 Since the roots are repeated then, The linearly independent solution can be written as, Substitute − 2 for r in equation ( 1 ) . y 1 ( x ) = e − 2 x and y 2 ( x ) = x e − 2 x The particular solution of the non-homogenous differential equation can be written as, y p ( x ) = e − 2 x u 1 + x e − 2 x u 2 Calculate u 1 and u 2 . e − 2 x u 1 ′ + x e − 2 x u 2 ′ = 0 ... ( 2 ) − 2 e − 2 x u 1 ′ + e − 2 x ( 1 − 2 x ) u 2 ′ = 4 e − 2 x 1 + x 2 + 2 x 2 − 1 ... ( 3 ) Solve equation ( 2 ) and equation ( 3 ) then, u 1 ′ = − x ( 4 1 + x 2 + 2 x 2 e 2 x − e 2 x ) and u 2 ′ = 4 1 + x 2 + 2 x 2 e 2 x − e 2 x Integrate both side of the above equation

Differential Equations and Linear Algebra (4th Edition)

4th Edition

ISBN: 9780321964670

Author: Stephen W. Goode, Scott A. Annin

Publisher: PEARSON

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Differential Equations and Linear Algebra (4th Edition)

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Solution Summary: The author explains that the general solution of the differential equation is y(x)=c_1e-2x,2, and F are assumed to be continuous

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