Program Explanation Given information: The differential equation is y ( 4 ) − 4 y ″ = x 2 and the initial condition is y ( 0 ) = y ′ ( 0 ) = 1 , y ″ ( 0 ) = y ( 3 ) = − 1 . Calculation: The given differential equation is y ( 4 ) − 4 y ″ = x 2 . Now, obtain the partial solution of the differential equation y ( 4 ) − 4 y ″ = x 2 using the method of undetermined coefficient. Consider y as e r x be the solution of homogeneous differential equation. Therefore, the characteristic equation corresponding to homogeneous differential equation y ( 4 ) − 4 y ″ = 0 is r 4 − 4 r 2 = 0 . Now, the roots of the characteristic equation is calculated as, r 4 − 4 r 2 = 0 r 2 ( r 2 − 4 ) = 0 r 2 ( r − 2 ) ( r + 2 ) = 0 r = 0 , r = 0 , r = 2 , r = − 2 Therefore, the roots of the equations are r 1 = 0 , r 2 = 0 , r 3 = 2 and r 4 = − 2 . The complementary function of the differential equation can be written as, y c ( x ) = c 1 + c 2 x + c 3 e 2 x + c 4 e − 2 x From the given differential equation y ( 4 ) − 4 y ″ = x 2 , the value of f ( x ) = x 2 . Therefore, the derivative of x 2 involves the term constant. The right hand side of the given differential equation is x 2 . Consider trial particular solution as y p ( x ) = A x 2 + B x + C . It can be seen that the complementary solution and particular solution has same terms. Therefore, to eliminate the duplication multiply the assumed particular solution by x 2 . Now differentiate y p ( x ) = A x 4 + B x 2 + C x 2 with respect to x . y ′ p ( x ) = 4 A x 3 + 3 B x 2 + 2 C x Now differentiate y ′ p ( x ) = 4 A x 3 + 3 B x 2 + 2 C x with respect to x . y ″ p ( x ) = 12 A x 2 + 6 B x + 2 C Now differentiate y ″ p ( x ) = 12 A x 2 + 6 B x + 2 C with respect to x . y p ( 3 ) ( x ) = 24 A x + 6 B Now differentiate y p ( 3 ) ( x ) = 24 A x + 6 B with respect to x . y p ( 4 ) ( x ) = 24 A Substitute the values of y p ( 4 ) ( x ) and y ″ p ( x ) in the differential equation y ( 4 ) − 4 y ″ = x 2 to obtain the value of A . 24 A − 4 ( 12 A x 2 + 6 B x + 2 C ) = x 2 24 A − 48 A x 2 − 24 B x − 8 C = x 2 Compare coefficient of x 2 , x and constant to obtain the value of A . − 48 A = 1 , − 24 B = 0 , 24 A − 8 C = 0 Now, solve the system of linear equation to obtain the value of A . A = − 1 48 , B = 0 , C = − 1 16 Substitute − 1 48 for A , 0 for B and − 1 16 for C in y p ( x ) = A x 4 + B x 2 + C x 2 to obtain the particular solution of the differential equation y ( 4 ) − 4 y ″ = x 2 . y p ( x ) = − 1 48 x 4 + ( 0 ) x 2 − 1 16 x 2 y p ( x ) = − 1 48 x 4 − 1 16 x 2 Therefore, the particular solution is y p ( x ) = − 1 48 x 4 − 1 16 x 2 . The general solution of the given differential equation y ( 4 ) − 4 y ″ = x 2 can be written as, y ( x ) = c 1 + c 2 x + c 3 e 2 x + c 4 e − 2 x − 1 48 x 4 − 1 16 x 2 ........ (1) Differentiate y ( x ) = c 1 + c 2 x + c 3 e 2 x + c 4 e − 2 x − 1 48 x 4 − 1 16 x 2 with respect to x and obtain the first derivative as, y ′ ( x ) = c 2 + 2 c 3 e 2 x − 2 c 4 e − 2 x − 1 12 x 3 + 1 8 x

5th Edition

ISBN: 9780135904206

Author: Edwards

Publisher: PEARSON

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A naïve solution for the above-stated problem is to evaluate all of the possible combinations of different pieces of varying size to find the maximum revenue-generating combination. Although this method seems the easiest and quick to practice, it has exp…

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Chapter 3.5, Problem 36P

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Program Description: Purpose of problem is to obtain the solution y(x) of the differential equation y(4)−4y″=x2 .

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Chapter 3.5, Problem 35P

Chapter 3.5, Problem 37P

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Chapter 3 Solutions

DIFFERENTIAL EQUATIONS+..-MYLAB ACCESS

Ch. 3.1 - In Problems 1 through 16, a homogeneous...Ch. 3.1 - Prob. 2P Ch. 3.1 - Prob. 3P Ch. 3.1 - Prob. 4P Ch. 3.1 - Prob. 5P Ch. 3.1 - Prob. 6P Ch. 3.1 - Prob. 7P Ch. 3.1 - Prob. 8P Ch. 3.1 - Prob. 9P Ch. 3.1 - Prob. 10P

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Program Description: Purpose of problem is to obtain the solution y ( x ) of the differential equation y ( 4 ) − 4 y ″ = x 2 .

Solution Summary: The author explains that the problem is to obtain the solution of the differential equation y(4)-4y′′=x2.

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Program Description: Purpose of problem is to obtain the solution y(x) of the differential equation y(4)−4y″=x2 .

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