Explanation Formula used: The Newton’s method to find the solution of the equation is, r n + 1 = r n − f ( r n ) f ′ ( r n ) …… ( 2 ) Here, the root of the equation is r and the auxiliary function is f ( r ) . The general solution of the third order differential equation is given by, y = c 1 e r 1 t + c 2 e r 2 t + c 3 e r 3 t …… ( 3 ) Here, the roots are r 1 , r 2 and r 3 . The constants are c 1 , c 2 and c 3 . Calculation: The differential equation is given by, 3 y ‴ + 18 y ″ + 13 y ′ − 19 y = 0 …… ( 1 ) The auxiliary equation for the equation ( 1 ) . f ( r ) = 3 r 3 + 18 r 2 + 13 r − 19 …… ( 4 ) Differentiate the above equation. f ′ ( r ) = 9 r 2 + 36 r + 13 Substitute 9 r 2 + 36 r + 13 for f ′ ( r ) and 3 r 3 + 18 r 2 + 13 r − 19 for f ( r ) in the equation ( 2 ) for obtaining the root. r n + 1 = r n − 3 ( r n ) 3 + 18 ( r n ) 2 + 13 ( r n ) − 19 9 ( r n ) 2 + 36 ( r n ) + 13 …… ( 5 ) Here, the number of iteration is shown as n . Let us assume r 0 = 0.5 . Substitute 0.5 for r 0 and r 0 for r n in equation ( 5 ) . r 1 = r 0 − 3 ( r 0 ) 3 + 18 ( r 0 ) 2 + 13 ( r 0 ) − 19 9 ( r 0 ) 2 + 36 ( r 0 ) + 13 = 0.5 − 3 ( 0.5 ) 3 + 18 ( 0.5 ) 2 + 13 ( 0.5 ) − 19 9 ( 0.5 ) 2 + 36 ( 0.5 ) + 13 = 0.5 − ( − 7.625 ) 33.25 = 0.7291 Substitute 0.7291 for r 1 and 1 for n in the equation ( 5 ) . r 2 = r 1 − 3 ( r 1 ) 3 + 18 ( r 1 ) 2 + 13 ( r 1 ) − 19 9 ( r 1 ) 2 + 36 ( r 1 ) + 13 = 0.7291 − 3 ( 0.7291 ) 3 + 18 ( 0.7291 ) 2 + 13 ( 0.7291 ) − 19 9 ( 0.7291 ) 2 + 36 ( 0.7291 ) + 13 = 0.7291 − 1.2096 44.03188 = 0.70164 Substitute 0.70164 for r 2 and 2 for n in the equation ( 5 ) . r 3 = r 2 − 3 ( r 2 ) 3 + 18 ( r 2 ) 2 + 13 ( r 2 ) − 19 9 ( r 2 ) 2 + 36 ( r 2 ) + 13 = 0.70164 − 3 ( 0.70164 ) 3 + 18 ( 0.70164 ) 2 + 13 ( 0.70164 ) − 19 9 ( 0.70164 ) 2 + 36 ( 0.70164 ) + 13 = 0.70119 Substitute 0.70119 for r 3 and 3 for n in the equation ( 5 ) . r 4 = r 3 − 3 ( r 3 ) 3 + 18 ( r 3 ) 2 + 13 ( r 3 ) − 19 9 ( r 3 ) 2 + 36 ( r 3 ) + 13 = 0.70119 − 3 ( 0.70119 ) 3 + 18 ( 0.70119 ) 2 + 13 ( 0.70119 ) − 19 9 ( 0.70119 ) 2 + 36 ( 0.70119 ) + 13 = 0.701196 The value of r 3 and r 4 are approximately same. Therefore, the first root is r 1 as 0.70119 . Let us assume r 0 = − 1 . Substitute − 1 for r 0 and r 0 for r n in equation ( 5 ) . r 1 = r 0 − 3 ( r 0 ) 3 + 18 ( r 0 ) 2 + 13 ( r 0 ) − 19 9 ( r 0 ) 2 + 36 ( r 0 ) + 13 = − 1 − 3 ( − 1 ) 3 + 18 ( − 1 ) 2 + 13 ( − 1 ) − 19 9 ( − 1 ) 2 + 36 ( − 1 ) + 13 = − 2.21506 Substitute − 2.21506 for r 1 and 1 for n in the equation ( 5 ) . r 2 = r 1 − 3 ( r 1 ) 3 + 18 ( r 1 ) 2 + 13 ( r 1 ) − 19 9 ( r 1 ) 2 + 36 ( r 1 ) + 13 = − 2.21506 − 3 ( − 2.21506 ) 3 + 18 ( − 2.21506 ) 2 + 13 ( − 2.21506 ) − 19 9 ( − 2.21506 ) 2 + 36 ( − 2 To determine (b) To find: The general solution of the differential equation y i v − 5 y ″ + 5 y = 0 . To determine (c) To find: The general solution of the differential equation y v − 3 i v − 5 y ‴ + 15 y ″ + 4 y ′ − 12 y = 0 .

Fundamentals of Differential Equations and Boundary Value Problems

7th Edition

ISBN: 9780321977106

Author: Nagle, R. Kent

Publisher: Pearson Education, Limited

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Question

Chapter 4.2, Problem 45E

To determine

(a)

To find:

The general solution of the differential equation 3y‴+18y″+13y′−19y=0.

To determine

(b)

To find:

The general solution of the differential equation yiv−5y″+5y=0.

To determine

(c)

To find:

The general solution of the differential equation yv−3iv−5y‴+15y″+4y′−12y=0.

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Chapter 4.2, Problem 44E

Chapter 4.2, Problem 46E

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Chapter 4 Solutions

Fundamentals of Differential Equations and Boundary Value Problems

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(a) To find: The general solution of the differential equation 3 y ‴ + 18 y ″ + 13 y ′ − 19 y = 0 .

Solution Summary: The author explains the Newton's method to find the solution of the differential equation.

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Videos

(a)

To find:

The general solution of the differential equation 3y‴+18y″+13y′−19y=0.

(b)

To find:

The general solution of the differential equation yiv−5y″+5y=0.

(c)

To find:

The general solution of the differential equation yv−3iv−5y‴+15y″+4y′−12y=0.

Want to see the full answer?

Chapter 4 Solutions

(a) To find: The general solution of the differential equation 3 y ‴ + 18 y ″ + 13 y ′ − 19 y = 0 .

Solution Summary: The author explains the Newton's method to find the solution of the differential equation.

Concept explainers

Videos

(a) To find: The general solution of the differential equation 3y‴+18y″+13y′−19y=0.

(b) To find: The general solution of the differential equation yiv−5y″+5y=0.

(c) To find: The general solution of the differential equation yv−3iv−5y‴+15y″+4y′−12y=0.

Want to see the full answer?

Chapter 4 Solutions

(a)

To find:

The general solution of the differential equation 3y‴+18y″+13y′−19y=0.

(b)

To find:

The general solution of the differential equation yiv−5y″+5y=0.

(c)

To find:

The general solution of the differential equation yv−3iv−5y‴+15y″+4y′−12y=0.