Explanation Approach: To solve the given homogeneous differential solution, do the following. Solve the auxiliary equation for the homogeneous differential equations From Newton’s method numerical technique is applied as follows: r n + 1 = r n − f ( r n ) f ′ ( r n ) r 1 = r 0 − f ( r 0 ) f ′ ( r 0 ) Use various iterations to calculate roots for the auxiliary equation. r n + 1 = r n − f ( r n ) f ′ ( r n ) r 1 = r 0 − f ( r 0 ) f ′ ( r 0 ) Change the values r 0 and find the values or r n to find the root of the equation. General solution is obtained with the help of roots. Calculation: Consider D = d d x in the given equation. y ( 4 ) + 2 y ‴ − 3 y ″ − y ′ + 1 2 y = 0 ( D 4 + 2 D 3 − 3 D 2 − D + 1 2 ) y = 0 Auxiliary equation of the above homogeneous equation is as follows. f ( r ) = 0 r 4 + 2 r 3 − 3 r 2 − r + 1 2 = 0 Differentiate above equation with respect to r f ′ ( r ) = 4 r 3 + 6 r 2 − 6 r − 1 (From Newton’s method) r n + 1 = r n − f ( r n ) f ′ ( r n ) r 1 = r 0 − f ( r 0 ) f ′ ( r 0 ) For, r 0 = 0 r 0 = 1 r 0 = − 1 r 0 = − 3 Take r 0 = 0 f ( r 0 ) = r 0 4 + 2 r 0 3 − 3 r 0 2 − r 0 + 1 2 f ′ ( r 0 ) = 4 r 0 3 + 6 r 0 2 − 6 r 0 − 1 r 1 = r 0 − r 0 4 + 2 r 0 3 − 3 r 0 2 − r 0 + 1 2 4 r 0 3 + 6 r 0 2 − 6 r 0 − 1 = 0 + 1 2 = 0.5 r 2 = r 1 − f ( r 1 ) f ′ ( r 1 ) Take r 1 = 0.5 r 2 = r 1 − f ( r 1 ) f ′ ( r 1 ) = r 1 − r 1 4 + 2 r 1 3 − 3 r 1 2 − r 1 + 1 2 4 r 1 3 + 6 r 1 2 − 6 r 1 − 1 = 0.5 − 0.21875 = 0.28125 Similarly, calculate value of r 3 , r 4 , r 5 , r 6 , r 7 , r 8 , and r 9 . r [ 1 ] 0.498505481066269 r [ 2 ] 0.281573887091018 r [ 3 ] 0.296399737129932 r [ 4 ] 0.296324805937571 r [ 5 ] 0.296324780002779 r [ 6 ] 0.296324779994422 r [ 7 ] 0.296324779994419 r [ 8 ] 0.296324779994419 r [ 9 ] 0.296324779994419 r 4 , r 5 , r 6 , r 7 , r 8 , and r 9 values are approximately equal. Therefore, root is 0.296324 Similarly, For r 0 = 1 r 1 = r 0 − f ( r 0 ) f ′ ( r 0 ) f ( r 0 ) = r 0 4 + 2 r 0 3 − 3 r 0 2 − r 0 + 1 2 f ′ ( r 0 ) = 4 r 0 3 + 6 r 0 2 − 6 r 0 − 1 This gives, r 1 = r 0 − r 0 4 + 2 r 0 3 − 3 r 0 2 − r 0 + 1 2 4 r 0 3 + 6 r 0 2 − 6 r 0 − 1 Calculate the values r 1 , r 2 , r 3 , r 4 , r 5 , r 6 , r 7 , r 8 , and r 9 r [ 1 ] 1.16616783078326 r [ 2 ]

Fundamentals of Differential Equations and Boundary Value Problems

7th Edition

ISBN: 9780321977106

Author: Nagle, R. Kent

Publisher: Pearson Education, Limited

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Chapter 6.2, Problem 27E

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The general solution for the differential equation y(4)+2y‴−3y″−y′+12y=0 using numerical technique.

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The general solution for the differential equation y ( 4 ) + 2 y ‴ − 3 y ″ − y ′ + 1 2 y = 0 using numerical technique.

Solution Summary: The author explains how Newton's method solves the homogeneous differential equation using numerical technique.

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The general solution for the differential equation y(4)+2y‴−3y″−y′+12y=0 using numerical technique.

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The general solution for the differential equation y ( 4 ) + 2 y ‴ − 3 y ″ − y ′ + 1 2 y = 0 using numerical technique.

Solution Summary: The author explains how Newton's method solves the homogeneous differential equation using numerical technique.

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To find: The general solution for the differential equation y(4)+2y‴−3y″−y′+12y=0 using numerical technique.

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Chapter 6 Solutions

To find:

The general solution for the differential equation y(4)+2y‴−3y″−y′+12y=0 using numerical technique.