Program Explanation Given information: The value of Ais [ 4 − 1 5 − 2 ] , f ( t ) = [ 18 e 2 t 30 e 2 t ] , x ( 0 ) = [ 0 0 ] and e A t = 1 4 [ e − t + 5 e 3 t e − t − e 3 t − 5 e − t + 5 e 3 t 5 e − t − e 3 t ] . Explanation: The initial value problem can be solving by the variation of parameters method as shown below. x ( t ) = e A t x 0 + e A t ∫ 0 t e − A s f ( s ) d s The determinant of e A t can be obtain as, | e A t | = 1 4 [ e − t + 5 e 3 t e − t − e 3 t − 5 e − t + 5 e 3 t 5 e − t − e 3 t ] = 1 4 | ( e − t + 5 e 3 t ) ( − 5 e − t + 5 e 3 t ) − ( e − t − e 3 t ) ( 5 e − t − e 3 t ) | = e 2 t The determinant of e A t is not 0. Therefore the inverse of the matrix is shown below. e − A t = 1 | e − A t | adj ( e − A t ) = 1 4 e 2 t [ e − t + 5 e 3 t e − t − e 3 t − 5 e − t + 5 e 3 t 5 e − t − e 3 t ] = 1 4 [ 5 e − 3 t − e t e t − e − 3 t 5 e − 3 t − 5 e t 5 e t − e − 3 t ] The value of ∫ 0 t e − A s f ( s ) d s can be obtained as, ∫ 0 t e − A s f ( s ) d s = ∫ 0 t 1 4 [ 5

Pearson eText for Differential Equations: Computing and Modeling Tech Update -- Instant Access (Pearson+)

5th Edition

ISBN: 9780137442782

Author: C. Edwards, David Penney

Publisher: PEARSON+

See similar textbooks

Concept explainers

Linear Programming Concepts

Linear programming is a famous mathematical modeling tool for determining the best distribution of scarce resources among competing demands. Machinery, time, personnel, raw materials, money, and space are all resources. It is used to find the most optima…

Question

Chapter 5.7, Problem 21P

Program Plan Intro

Program Description:Purpose of problem is to solve the initial value problem x′=Ax+f(t) and x(a)=xa by using method of variation of parameters.

Expert Solution & Answer

Want to see the full answer?

Check out a sample textbook solution

See solution

Chapter 5.7, Problem 20P

Chapter 5.7, Problem 22P

Students have asked these similar questions

Let C be the intersection of the cylinder x² + y² = 2.95 with the plane z = 1.13x, with the clockwise orientation, as viewed from above. Then the value of cos (₤23 COS 2 y dx xdy+3 z dzis 3 z dz) is 0.131 -0.108 -0.891 -0.663 -0.428 0.561 -0.332 -0.387

2 x² + 47 The partial fraction decomposition of f(x) g(x) can be written in the form of + x3 + 4x2 2 C I where f(x) = g(x) h(x) = h(x) + x +4

The partial fraction decomposition of f(x) 4x 7 g(x) + where 3x4 f(x) = g(x) = - 52 –10 12x237x+28 can be written in the form of

Chapter 5 Solutions

Pearson eText for Differential Equations: Computing and Modeling Tech Update -- Instant Access (Pearson+)

Ch. 5.1 - Let A=[2347] and B=[3451]. Find (a) 2A+3B; (b)...Ch. 5.1 - Prob. 2P Ch. 5.1 - Find AB and BA given A=[203415] and B=[137032].Ch. 5.1 - Prob. 4P Ch. 5.1 - Prob. 5P Ch. 5.1 - Prob. 6P Ch. 5.1 - Prob. 7P Ch. 5.1 - Prob. 8P Ch. 5.1 - Prob. 9P Ch. 5.1 - Prob. 10P

Ch. 5.1 - Prob. 11P Ch. 5.1 - Prob. 12P Ch. 5.1 - Prob. 13P Ch. 5.1 - Prob. 14P Ch. 5.1 - Prob. 15P Ch. 5.1 - Prob. 16P Ch. 5.1 - Prob. 17P Ch. 5.1 - Prob. 18P Ch. 5.1 - Prob. 19P Ch. 5.1 - Prob. 20P Ch. 5.1 - Prob. 21P Ch. 5.1 - Prob. 22P Ch. 5.1 - Prob. 23P Ch. 5.1 - Prob. 24P Ch. 5.1 - Prob. 25P Ch. 5.1 - Prob. 26P Ch. 5.1 - Prob. 27P Ch. 5.1 - Prob. 28P Ch. 5.1 - Prob. 29P Ch. 5.1 - Prob. 30P Ch. 5.1 - Prob. 31P Ch. 5.1 - Prob. 32P Ch. 5.1 - Prob. 33P Ch. 5.1 - Prob. 34P Ch. 5.1 - Prob. 35P Ch. 5.1 - Prob. 36P Ch. 5.1 - Prob. 37P Ch. 5.1 - Prob. 38P Ch. 5.1 - Prob. 39P Ch. 5.1 - Prob. 40P Ch. 5.1 - Prob. 41P Ch. 5.1 - Prob. 42P Ch. 5.1 - Prob. 43P Ch. 5.1 - Prob. 44P Ch. 5.1 - Prob. 45P Ch. 5.2 - Prob. 1P Ch. 5.2 - Prob. 2P Ch. 5.2 - Prob. 3P Ch. 5.2 - Prob. 4P Ch. 5.2 - Prob. 5P Ch. 5.2 - Prob. 6P Ch. 5.2 - Prob. 7P Ch. 5.2 - Prob. 8P Ch. 5.2 - Prob. 9P Ch. 5.2 - Prob. 10P Ch. 5.2 - Prob. 11P Ch. 5.2 - Prob. 12P Ch. 5.2 - Prob. 13P Ch. 5.2 - Prob. 14P Ch. 5.2 - Prob. 15P Ch. 5.2 - Prob. 16P Ch. 5.2 - Prob. 17P Ch. 5.2 - Prob. 18P Ch. 5.2 - Prob. 19P Ch. 5.2 - Prob. 20P Ch. 5.2 - Prob. 21P Ch. 5.2 - Prob. 22P Ch. 5.2 - Prob. 23P Ch. 5.2 - Prob. 24P Ch. 5.2 - Prob. 25P Ch. 5.2 - Prob. 26P Ch. 5.2 - Prob. 27P Ch. 5.2 - Prob. 28P Ch. 5.2 - Prob. 29P Ch. 5.2 - Prob. 30P Ch. 5.2 - Prob. 31P Ch. 5.2 - Prob. 32P Ch. 5.2 - Prob. 33P Ch. 5.2 - Prob. 34P Ch. 5.2 - Prob. 35P Ch. 5.2 - Prob. 36P Ch. 5.2 - Prob. 37P Ch. 5.2 - Prob. 38P Ch. 5.2 - Prob. 39P Ch. 5.2 - Prob. 40P Ch. 5.2 - Prob. 41P Ch. 5.2 - Prob. 42P Ch. 5.2 - Prob. 43P Ch. 5.2 - Prob. 44P Ch. 5.2 - Prob. 45P Ch. 5.2 - Prob. 46P Ch. 5.2 - Prob. 47P Ch. 5.2 - Prob. 48P Ch. 5.2 - Prob. 49P Ch. 5.2 - Prob. 50P Ch. 5.3 - Prob. 1P Ch. 5.3 - Prob. 2P Ch. 5.3 - Prob. 3P Ch. 5.3 - Prob. 4P Ch. 5.3 - Prob. 5P Ch. 5.3 - Prob. 6P Ch. 5.3 - Prob. 7P Ch. 5.3 - Prob. 8P Ch. 5.3 - Prob. 9P Ch. 5.3 - Prob. 10P Ch. 5.3 - Prob. 11P Ch. 5.3 - Prob. 12P Ch. 5.3 - Prob. 13P Ch. 5.3 - Prob. 14P Ch. 5.3 - Prob. 15P Ch. 5.3 - Prob. 16P Ch. 5.3 - Prob. 17P Ch. 5.3 - Prob. 18P Ch. 5.3 - Prob. 19P Ch. 5.3 - Prob. 20P Ch. 5.3 - Prob. 21P Ch. 5.3 - Prob. 22P Ch. 5.3 - Prob. 23P Ch. 5.3 - Prob. 24P Ch. 5.3 - Prob. 25P Ch. 5.3 - Prob. 26P Ch. 5.3 - Prob. 27P Ch. 5.3 - Prob. 28P Ch. 5.3 - Prob. 29P Ch. 5.3 - Prob. 30P Ch. 5.3 - Prob. 31P Ch. 5.3 - Prob. 32P Ch. 5.3 - Prob. 33P Ch. 5.3 - Verify Eq. (53) by substituting the expressions...Ch. 5.3 - Prob. 35P Ch. 5.3 - Prob. 36P Ch. 5.3 - Prob. 37P Ch. 5.3 - Prob. 38P Ch. 5.3 - Prob. 39P Ch. 5.3 - Prob. 40P Ch. 5.4 - Prob. 1P Ch. 5.4 - Prob. 2P Ch. 5.4 - Prob. 3P Ch. 5.4 - Prob. 4P Ch. 5.4 - Prob. 5P Ch. 5.4 - Prob. 6P Ch. 5.4 - Prob. 7P Ch. 5.4 - Prob. 8P Ch. 5.4 - Prob. 9P Ch. 5.4 - Prob. 10P Ch. 5.4 - Prob. 11P Ch. 5.4 - Prob. 12P Ch. 5.4 - Prob. 13P Ch. 5.4 - Prob. 14P Ch. 5.4 - Prob. 15P Ch. 5.4 - Prob. 16P Ch. 5.4 - Prob. 17P Ch. 5.4 - Prob. 18P Ch. 5.4 - Prob. 19P Ch. 5.4 - Prob. 20P Ch. 5.4 - Prob. 21P Ch. 5.4 - Prob. 22P Ch. 5.4 - Prob. 23P Ch. 5.4 - Prob. 24P Ch. 5.4 - Prob. 25P Ch. 5.4 - Prob. 26P Ch. 5.4 - Prob. 27P Ch. 5.4 - Prob. 28P Ch. 5.4 - Prob. 29P Ch. 5.5 - Prob. 1P Ch. 5.5 - Prob. 2P Ch. 5.5 - Prob. 3P Ch. 5.5 - Prob. 4P Ch. 5.5 - Prob. 5P Ch. 5.5 - Prob. 6P Ch. 5.5 - Prob. 7P Ch. 5.5 - Prob. 8P Ch. 5.5 - Prob. 9P Ch. 5.5 - Prob. 10P Ch. 5.5 - Prob. 11P Ch. 5.5 - Prob. 12P Ch. 5.5 - Prob. 13P Ch. 5.5 - Prob. 14P Ch. 5.5 - Prob. 15P Ch. 5.5 - Prob. 16P Ch. 5.5 - Prob. 17P Ch. 5.5 - Prob. 18P Ch. 5.5 - Prob. 19P Ch. 5.5 - Prob. 20P Ch. 5.5 - Prob. 21P Ch. 5.5 - Prob. 22P Ch. 5.5 - Prob. 23P Ch. 5.5 - Prob. 24P Ch. 5.5 - Prob. 25P Ch. 5.5 - Prob. 26P Ch. 5.5 - Prob. 27P Ch. 5.5 - Prob. 28P Ch. 5.5 - Prob. 29P Ch. 5.5 - Prob. 30P Ch. 5.5 - Prob. 31P Ch. 5.5 - Prob. 32P Ch. 5.5 - Prob. 33P Ch. 5.5 - Prob. 34P Ch. 5.5 - Prob. 35P Ch. 5.5 - Prob. 36P Ch. 5.6 - Prob. 1P Ch. 5.6 - Prob. 2P Ch. 5.6 - Prob. 3P Ch. 5.6 - Prob. 4P Ch. 5.6 - Prob. 5P Ch. 5.6 - Prob. 6P Ch. 5.6 - Prob. 7P Ch. 5.6 - Prob. 8P Ch. 5.6 - Prob. 9P Ch. 5.6 - Prob. 10P Ch. 5.6 - Prob. 11P Ch. 5.6 - Prob. 12P Ch. 5.6 - Prob. 13P Ch. 5.6 - Prob. 14P Ch. 5.6 - Prob. 15P Ch. 5.6 - Prob. 16P Ch. 5.6 - Prob. 17P Ch. 5.6 - Prob. 18P Ch. 5.6 - Prob. 19P Ch. 5.6 - Prob. 20P Ch. 5.6 - Prob. 21P Ch. 5.6 - Prob. 22P Ch. 5.6 - Prob. 23P Ch. 5.6 - Prob. 24P Ch. 5.6 - Prob. 25P Ch. 5.6 - Prob. 26P Ch. 5.6 - Prob. 27P Ch. 5.6 - Prob. 28P Ch. 5.6 - Prob. 29P Ch. 5.6 - Prob. 30P Ch. 5.6 - Prob. 31P Ch. 5.6 - Prob. 32P Ch. 5.6 - Prob. 33P Ch. 5.6 - Prob. 34P Ch. 5.6 - Prob. 35P Ch. 5.6 - Prob. 36P Ch. 5.6 - Prob. 37P Ch. 5.6 - Prob. 38P Ch. 5.6 - Prob. 39P Ch. 5.6 - Prob. 40P Ch. 5.7 - Prob. 1P Ch. 5.7 - Prob. 2P Ch. 5.7 - Prob. 3P Ch. 5.7 - Prob. 4P Ch. 5.7 - Prob. 5P Ch. 5.7 - Prob. 6P Ch. 5.7 - Prob. 7P Ch. 5.7 - Prob. 8P Ch. 5.7 - Prob. 9P Ch. 5.7 - Prob. 10P Ch. 5.7 - Prob. 11P Ch. 5.7 - Prob. 12P Ch. 5.7 - Prob. 13P Ch. 5.7 - Prob. 14P Ch. 5.7 - Prob. 15P Ch. 5.7 - Prob. 16P Ch. 5.7 - Prob. 17P Ch. 5.7 - Prob. 18P Ch. 5.7 - Prob. 19P Ch. 5.7 - Prob. 20P Ch. 5.7 - Prob. 21P Ch. 5.7 - Prob. 22P Ch. 5.7 - Prob. 23P Ch. 5.7 - Prob. 24P Ch. 5.7 - Prob. 25P Ch. 5.7 - Prob. 26P Ch. 5.7 - Prob. 27P Ch. 5.7 - Prob. 28P Ch. 5.7 - Prob. 29P Ch. 5.7 - Prob. 30P Ch. 5.7 - Prob. 31P Ch. 5.7 - Prob. 32P Ch. 5.7 - Prob. 33P Ch. 5.7 - Prob. 34P

Knowledge Booster

Learn more about

Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, computer-science and related others by exploring similar questions and additional content below.

Program Description: Purpose of problem is to solve the initial value problem x ′ = A x + f ( t ) and x ( a ) = x a by using method of variation of parameters.

Solution Summary: The author explains that the initial value problem can be solved by the variation of parameters method as shown below.

Concept explainers

Program Description:Purpose of problem is to solve the initial value problem x′=Ax+f(t) and x(a)=xa by using method of variation of parameters.

Want to see the full answer?

Chapter 5 Solutions