Explanation Procedure used: 1. Substitute y ″ = r 2 , y ′ = r and y = 1 in the general equation, a y ″ + b y ′ + c y = 0 and obtain the characteristic equation as a r 2 + b r + c = 0 . 2. Find the roots of the characteristic equation. 3. If roots are real. Case-1 The general solution if r = r 1 = r 2 , is y ( t ) = c 1 e r t + c 2 t e r t . Case-2 The general solution if r 1 ≠ r 2 is y ( t ) = c 1 e r 1 t + c 2 e r 2 t . 4. The general solution if roots are imaginary in form of r 1 , 2 = α ± β i is y ( t ) = e a t ( c 1 cos β t + c 2 sin β t ) . For each function in f i ( t ) in f ( t ) , determine the associate set of functions F i ; 1 ≤ i ≤ m can be obtained as, 5. The associate function if f i ( t ) = t m is F = { 1 , t , t 2 , ... , t m } . 6. The associate function if f i ( t ) = t m e α t is F = { e α t , e α t t , e α t t 2 , ... , e α t t m } . 7. The associate function if f i ( t ) = t m e α t cos β t or f i ( t ) = t m e α t sin β t is, F = { e α t cos β t , e α t t cos β t , e α t t 2 cos β t , ... , e α t t m cos β t , e α t sin β t , e α t t sin β t , e α t t 2 sin β t , ... , e α t t m sin β t } 8. The general solution if roots are imaginary in form of r 1 , 2 = α ± β i is y ( t ) = e a t ( c 1 cos β t + c 2 sin β t ) . The particular solution can be obtained by taking the linear combination of all functions in the associated sets. Calculation: Consider the differential equation y ″ − 6 y ′ + 8 y = − 256 t 3 . The corresponding homogenous equation is y ″ − 6 y ′ + 8 y = 0 . The corresponding characteristic equation of the homogeneous equation of the given differential equation is as follows. r 2 − 6 r + 8 = 0 Solve the characteristic equation and obtain the roots as shown below. r 2 − 6 r + 8 = 0 ( r − 2 ) ( r − 4 ) = 0 r = 2 , 4 Here, the two roots are distinct and real. Observe that r 1 ≠ r 2 , so apply Case-2. Hence, the solution of y ″ − 6 y ′ + 8 y = 0 is obtained as follows. y h = c 1 e ( 2 ) t + c 2 e ( 4 ) t = c 1 e 2 t + c 2 e 4 t where c 1 and c 2 are arbitrary constant. Here, the forcing function is − 256 t 3 . So, the associated set of functions is F = { t 3 , t 2 , t , 1 }

5th Edition

ISBN: 9780128149485

Author: Abell, Martha L. L.

Publisher: Elsevier Science

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Chapter 4.3, Problem 32E

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The general solution of the equation y″−6y′+8y=−256t3.

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A factorization A = PDP 1 is not unique. For A= 7 2 -4 1 1 1 5 0 2 1 one factorization is P = D= and P-1 30 = Use this information with D₁ = to find a matrix P₁ such that - -1 -2 0 3 1 - - 1 05 A-P,D,P P1 (Type an integer or simplified fraction for each matrix element.)

Matrix A is factored in the form PDP 1. Use the Diagonalization Theorem to find the eigenvalues of A and a basis for each eigenspace. 30 -1 - 1 0 -1 400 0 0 1 A= 3 4 3 0 1 3 040 3 1 3 0 0 4 1 0 0 003 -1 0 -1 Select the correct choice below and fill in the answer boxes to complete your choice. (Use a comma to separate vectors as needed.) A basis for the corresponding eigenspace is { A. There is one distinct eigenvalue, λ = B. In ascending order, the two distinct eigenvalues are λ₁ ... = and 2 = Bases for the corresponding eigenspaces are { and ( ), respectively. C. In ascending order, the three distinct eigenvalues are λ₁ = = 12/2 = and 3 = Bases for the corresponding eigenspaces are {}, }, and { respectively.

Chapter 4 Solutions

Introductory Differential Equations

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The general solution of the equation y ″ − 6 y ′ + 8 y = − 256 t 3 .

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Chapter 4 Solutions