Explanation 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: Consider the given differential equation, y ‴ − 4 y ″ + 7 y ′ − 6 y = 0 The initial conditions are, y ( 0 ) = 1 , y ′ ( 0 ) = 0 , y ″ ( 0 ) = 0 The auxiliary equation corresponding to the given differential equation is, r 3 − 4 r 2 + 7 r − 6 = 0 By inspection r = 2 is a root of auxiliary equation ( r − 2 ) ( r 2 − 2 r + 3 ) = 0 ( r − 2 ) ( ( r − 1 ) 2 + 2 ) = 0 The roots of the auxiliary equation are, r 1 = 2 r 2 = 1 + 2 i , r 3 = 1 − 2 i Thus, the general solution of the given equation is, y ( x ) = C 1 e 2 x + C 2 e x cos ( 2 x ) + C 3 e x sin ( 2 x ) … ( 1 ) Differentiate equation ( 1 ) with respect to x y ′ ( x ) = 2 C 1 e 2 x + C 2 e x cos ( 2 x ) − 2 C 2 e x sin ( 2 x ) + C 3 e x sin ( 2 x ) + 2 C 3 e x cos ( 2 x ) … ( 2 ) Differentiate equation ( 2 ) with respect to x y ″ ( x ) = 4 C 1 e 2 x + C 2 e x cos ( 2 x ) − 2 C 2 e x sin ( 2 x ) − 2 C 2 e x sin ( 2 x ) − 2 C 2 e x cos ( 2 x ) + C 3 e x sin ( 2 x ) + 2 C 3 e x cos ( 2 x ) + 2 C 3 e x cos ( 2 x ) − 2 C 3 e x sin ( 2 x ) … ( 3 ) Put 1 for y ( 0 ) in equation ( 1 )

7th Edition

ISBN: 9780135902738

Author: Nagle

Publisher: PEARSON

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Chapter 6.2, Problem 21E

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To solve:

The initial value problem y‴−4y″+7y′−6y=0

y(0)=1,y′(0)=0,y″(0)=0

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Chapter 6.2, Problem 20E

Chapter 6.2, Problem 22E

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MYLAB MATH-W/ETEXT F/FUND.DIFF.EQUAT.

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The initial value problem y ‴ − 4 y ″ + 7 y ′ − 6 y = 0 y ( 0 ) = 1 , y ′ ( 0 ) = 0 , y ″ ( 0 ) = 0

Solution Summary: The author explains how to solve the nth order linear homogeneous differential equation.

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To solve:

The initial value problem y‴−4y″+7y′−6y=0

y(0)=1,y′(0)=0,y″(0)=0

Want to see the full answer?

Chapter 6 Solutions

The initial value problem y ‴ − 4 y ″ + 7 y ′ − 6 y = 0 y ( 0 ) = 1 , y ′ ( 0 ) = 0 , y ″ ( 0 ) = 0

Solution Summary: The author explains how to solve the nth order linear homogeneous differential equation.

Concept explainers

Videos

To solve: The initial value problem y‴−4y″+7y′−6y=0 y(0)=1,y′(0)=0,y″(0)=0

Want to see the full answer?

Chapter 6 Solutions

To solve:

The initial value problem y‴−4y″+7y′−6y=0

y(0)=1,y′(0)=0,y″(0)=0