Explanation Given: The third order differential equation y ‴ − y ″ + y ′ − y = 0 and y ( 0 ) = 2 , y ′ ( 0 ) = − 1 , y ″ ( 1 ) = − 2 . Procedure used: Consider a linear homogenous n th order equation with constant real coefficient. a y n + b y n − 1 + … + c y ′ + d y = 0 1. The corresponding characteristic equation is, a r n + b r n − 1 + … + c r + d = 0 Let the roots of the equation be r 1 , r 2 , … , r n . The, the different conditions are, a. Roots are real and distinct: Then, n linearly independent solutions are, y 1 = e r 1 t , y 2 = e r 2 t ,…, y n = e r n t . b. Roots are repeated: If the root r = r 1 of the characteristics equation has multiplicity k , that is, ( r − r 1 ) k is a factor of the characteristic equation. Then k linearly independent solutions are, y 1 = e r 1 t , y 2 = t e r 1 t ,…, y k = t k − 1 e r 1 t c. Roots are complex conjugate: If the roots are r 1 = α + β i and r 2 = α − β i with multiplicity k , then 2 k linearly independent solutions associated with r 1 , 2 = α ± β i are, y 1 = e a t cos β t , y 2 = e a t sin β t , y 3 = t e a t cos β t ,…, y 2 k − 1 = t k − 1 e a t cos β t , y 2 k = t k − 1 e a t sin β t . Calculation: Consider the differential equation. y ‴ − y ″ + y ′ − y = 0 The corresponding characteristic equation is, r 3 − r 2 + r − 1 = 0 Find the roots of the above equation. r 3 − r 2 + r − 1 = 0 r 2 ( r − 1 ) + 1 ( r − 1 ) = 0 ( r − 1 ) ( r 2 + 1 ) = 0 r − 1 = 0 or r 2 = − 1 r = 1 or r = ± i Since the degree of the auxiliary equation is 3, there are 3 roots of the equation. Therefore, the solutions to the auxiliary equation are ( 1 , i , − i ) . The roots of the auxiliary equation are real and conjugated complex numbers. Therefore, the general solution to the differential equation is, y = c 1 e r t + e a t ( c 2 cos b t + c 3 sin b t ) y = c 1 e ( 1 ) t + e ( 0 ) t ( c 2 cos ( 1 ) t + c 3 sin ( 1 ) t ) y = c 1 e t + c 2 cos t + c 3 sin t Differentiate the above solution and obtain the value of y ′ as shown below: y = c 1 e t + c 2 cos t + c 3 sin t y ′ =

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ISBN: 9780470458327

Author: William E. Boyce, Richard C. DiPrima

Publisher: Wiley, John & Sons, Incorporated

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Chapter 4.2, Problem 32P

To determine

The solution of the initial value problem y‴−y″+y′−y=0 with initial conditions y(0)=2,y′(0)=−1,y″(1)=−2.

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Chapter 4.2, Problem 31P

Chapter 4.2, Problem 33P

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Chapter 4 Solutions

Elementary Differential Equations

Ch. 4.1 - In each of Problems 1 through 6, determine...Ch. 4.1 - In each of Problems 1 through 6, determine...Ch. 4.1 - In each of Problems 1 through 6, determine...Ch. 4.1 - In each of Problems 1 through 6, determine...Ch. 4.1 - In each of Problems 1 through 6, determine...Ch. 4.1 - Prob. 6P Ch. 4.1 - In each of Problems 7 through 10, determine...Ch. 4.1 - Prob. 8P Ch. 4.1 - In each of Problems 7 through 10, determine...Ch. 4.1 - Prob. 10P

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The solution of the initial value problem y ‴ − y ″ + y ′ − y = 0 with initial conditions y ( 0 ) = 2 , y ′ ( 0 ) = − 1 , y ″ ( 1 ) = − 2 .

The solution of the initial value problem y‴−y″+y′−y=0 with initial conditions y(0)=2,y′(0)=−1,y″(1)=−2.

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Chapter 4 Solutions