Explanation Given: The values D y = 4 , y = − 2 and x = 0 . Formula used: The auxiliary equation of a linear homogeneous differential equation of the form a 0 D 2 y + a 1 D y + a 2 y = 0 with constant coefficients is given by a 0 m 2 + a 1 m + a 2 = 0 . The general solution of the auxiliary equation with complex roots is given by y = e α x ( c 1 sin β x + c 2 cos β x ) . The general solution of the auxiliary equation with double roots m is given by y = e m x ( c 1 + c 2 x ) . The general solution of the auxiliary equation with two distinct roots m 1 and m 2 , is given by y = c 1 e m 1 x + c 2 e m 2 x . The solution of a linear non-homogeneous differential equation of the form a 0 D 2 y + a 1 D y + a 2 y = b where the a ’s are constants and b is a function of x or is a constant, is given by y = y c + y p . The y c is called the complementary solution obtained by solving the corresponding homogeneous equation and where y p is the particular solution. The different forms of b and their corresponding particular solution to be used is shown below: b y p ( x ) a e β x A e β x a cos ( β x ) A cos ( β x ) + B sin ( β x ) b sin ( β x ) A cos ( β x ) + B sin ( β x ) a cos ( β x ) + b sin ( β x ) A cos ( β x ) + B sin ( β x ) n th degree polynomial A n x n + A n − 1 x n − 1 + A 1 x + A 0 Calculation: The given differential equation is D 2 y − 2 D y + y = x e 2 x − e 2 x . Compare the given equation with a 0 D 2 y + a 1 D y + a 2 y = b and obtain the values of a 0 , a 1 , a 2 , b as shown below: a 0 = 1 , a 1 = − 2 , a 2 = 1 , b = x e 2 x − e − 2 x . Write the auxiliary equation of the homogeneous equation D 2 y − 2 D y + y = 0 as, m 2 − 2 m + 1 = 0 Solve the auxiliary equation and obtain the roots as follows: m 2 − 2 m + 1 = 0 m = 1 Thus, the auxiliary equation has double roots m = 1 . Using the above formula, obtain the complementary solution of the homogeneous equation D 2 y − 2 D y + y = 0 as, y c = e x ( c 1 + c 2 x ) Let y p = A e 2 x + B x e 2 x ⋯ ⋯ ( 1 ) Differentiate y p and obtain the value of D y p , and D 2 y p as follows: y p = A e 2 x + B x e 2 x D y ′ p = 2 A e 2 x +2 B x e 2 x + B e 2 x D 2 y ″ p = 4 A e 2 x +4 B x e 2 x + 4 B e 2 x Substitute y p = A e 2 x + B x e 2 x , D y ′ p = 2 A e 2 x +2 B x e 2 x + B e 2 x , and D 2 y ″ p = 4 A e 2 x +4 B x e 2 x + <

Pearson eText for Basic Technical Mathematics with Calculus -- Instant Access (Pearson+)

11th Edition

ISBN: 9780137554843

Author: Allyn Washington, Richard Evans

Publisher: PEARSON+

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Chapter 31.9, Problem 36E

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The particular solution the differential equation D2y−2Dy+y=xe2x−e2x.

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Chapter 31.9, Problem 35E

Chapter 31.9, Problem 37E

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Chapter 31 Solutions

Pearson eText for Basic Technical Mathematics with Calculus -- Instant Access (Pearson+)

Ch. 31.1 - Show that is a solution of . Is it the general...Ch. 31.1 - Prob. 1E Ch. 31.1 - In Exercises 1 and 2, show that the indicated...Ch. 31.1 - In Exercises 3–6, determine whether the given...Ch. 31.1 - Prob. 4E Ch. 31.1 - In Exercises 3–6, determine whether the given...Ch. 31.1 - Prob. 6E Ch. 31.1 - In Exercises 7–10, show that each function y =...Ch. 31.1 - Prob. 8E Ch. 31.1 - In Exercises 7–10, show that each function y =...

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The particular solution the differential equation D 2 y − 2 D y + y = x e 2 x − e 2 x .

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The particular solution the differential equation D2y−2Dy+y=xe2x−e2x.

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