Explanation Given: The differential equation is given as, y ‴ + 2 y ″ − 19 y ′ − 20 y = 0 . Approach: Consider the linear homogeneous n th order differential equation, a n y ( n ) ( x ) + a n − 1 y ( n − 1 ) ( x ) + a n − 2 y ( n − 2 ) ( x ) + … … + a 0 y ( x ) = 0 . The general solution for this differential equation is given by the steps as follows: Step 1 Assume y = e r x to be the solution of the n th order linear homogeneous differential equation. Step 2 Calculate the subsequent derivatives of y and substitute in the differential equation. Step 3 An Auxiliary equation in terms of r is formed. Here r is the root of the equation. Step 4 If the roots r 1 , … … , r n of the auxiliary equation are real and distinct, then the general solution to the differential equation is given as, y ( x ) = C 1 e r 1 x + C 2 e r 2 x + … … + C n e r n x . Here C 1 , C 2 , … … , C n are arbitrary constants. Step-5 If the roots of the auxiliary equation are complex numbers such as r = ( α ± i β ) , then the general solution to the differential equation is given as, y ( x ) = C 1 e α x ( C 2 cos β x ± C 3 sin β x ) . Here C 1 , C 2 , … … , C n are arbitrary constants. Step-6 If the roots of the auxiliary equation are repeated such as r 1 is a repeated root of multiplicity m , then the general solution to the differential equation is given as, y ( x ) = C 1 e r 1 x + C 2 x e r 1 x + C 3 x 2 e r 1 x + … … + C m x m − 1 e r 1 x . Here C 1 , C 2 , … … , C n are arbitrary constants. Calculation: The differential equation is, y ‴ + 2 y ″ − 19 y ′ − 20 y = 0 ... ( 1 ) The auxiliary equation for the equation ( 1 ) is given as, m 3 + 2 m 2 − 19 m − 20 = 0 ... ( 2 ) Consider ( m + 1 ) as a factor of the equation ( 2 ) as ( − 1 ) 3 + 2 × ( − 1 ) 2 − 19 ( − 1 ) − 20 = 0 . Therefore, m = − 1 is a root of the equation ( 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 4E

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The general solution for the given differential equation y‴+2y″−19y′−20y=0.

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Chapter 6.2, Problem 3E

Chapter 6.2, Problem 5E

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Fundamentals of Differential Equations and Boundary Value Problems

Ch. 6.1 - In Problems 1-6, determine the largest interval...Ch. 6.1 - In Problems 1-6, determine the largest interval...Ch. 6.1 - In Problems 1-6, determine the largest interval...Ch. 6.1 - In Problems 1-6, determine the largest interval...Ch. 6.1 - In Problems 1-6, determine the largest interval...Ch. 6.1 - In Problems 1-6, determine the largest interval...Ch. 6.1 - Prob. 7E Ch. 6.1 - In Problems7-14, determine whether the given...Ch. 6.1 - In Problems7-14, determine whether the given...Ch. 6.1 - In Problems7-14, determine whether the given...

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The general solution for the given differential equation y ‴ + 2 y ″ − 19 y ′ − 20 y = 0 .

Solution Summary: The author explains how the general solution to the differential equation is given by the steps as follows.

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To find:

The general solution for the given differential equation y‴+2y″−19y′−20y=0.

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Chapter 6 Solutions

The general solution for the given differential equation y ‴ + 2 y ″ − 19 y ′ − 20 y = 0 .

Solution Summary: The author explains how the general solution to the differential equation is given by the steps as follows.

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To find: The general solution for the given differential equation y‴+2y″−19y′−20y=0.

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Chapter 6 Solutions

To find:

The general solution for the given differential equation y‴+2y″−19y′−20y=0.