Explanation Procedure used: Suppose n × n constant matrix A has n linearly independent eigenvectors u 1 , u 2 , ⋯ , u n . The eigenvalues can be finding by using characteristics equation | A − r I | = 0 . Here, r denotes the eigenvalues. The corresponding the vector of the eigenvalues are called eigenvectors. It can be found by substitute the eigenvalues in the formula ( A − r I ) u = 0 . The general solution is, x ( t ) = c 1 e r 1 t u 1 + c 2 e r 2 t u 2 . Calculation: Consider the given system, t x ′ ( t ) = [ 1 3 − 1 5 ] x ( t ) , t > 0 . Here, A = [ 1 3 − 1 5 ] Let t = e u and t x ′ ( t ) = x ′ ( u ) x ′ ( u ) = [ 1 3 − 1 5 ] x ( u ) , t > 0 Substitute the value of matrix A in the characteristic equation and solve for r : | [ 1 3 − 1 5 ] − r [ 1 0 0 1 ] | = 0 | [ 1 3 − 1 5 ] − [ r 0 0 r ] | = 0 | [ 1 − r 3 − 1 5 − r ] | = 0 ( 1 − r ) ( 5 − r ) + 3 = 0 Further simplify as: 5 − r − 5 r + r 2 + 3 = 0 r 2 − 6 r + 8 = 0 r 2 − 4 r − 2 r + 8 = 0 ( r − 4 ) ( r − 2 ) = 0 Equate first factor to 0. ( r − 4 ) = 0 r = 4 Equate second factor to 0. ( r − 2 ) = 0 r = 2 The eigenvalues are r 1 = 2 , r 2 = 4 . Find the eigenvector corresponding to eigenvalue r 1 = 2 by using the formula ( A − r I ) u = 0 : ( A − r I ) u = ( [ 1 3 − 1 5 ] − 2 [ 1 0 0 1 ] ) [ u 1 u 2 ] = ( [ 1 3 − 1 5 ] − [ 2 0 0 2 ] ) [ u 1 u 2 ] = [ − 1 3 − 1 3 ] [ u 1 u 2 ] The matrix equation is: [ − 1 3 − 1 3 ] [ u 1 u 2 ] = [ 0 0 ] It can be write in scalar equation as, − u 1 + 3 u 2 = 0

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 9.5, Problem 43E

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The general solution for the system tx′(t)=[13−15]x(t), t>0.

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Chapter 9.5, Problem 42E

Chapter 9.5, Problem 44E

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Chapter 9 Solutions

Fundamentals of Differential Equations and Boundary Value Problems

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The general solution for the system t x ′ ( t ) = [ 1 3 − 1 5 ] x ( t ) , t > 0 .

Solution Summary: The author explains how to find the general solution for the given system by using the characteristics equation left|A-rIright|=0.

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To find:

The general solution for the system tx′(t)=[13−15]x(t), t>0.

Want to see the full answer?

Chapter 9 Solutions

The general solution for the system t x ′ ( t ) = [ 1 3 − 1 5 ] x ( t ) , t &gt; 0 .

Solution Summary: The author explains how to find the general solution for the given system by using the characteristics equation left|A-rIright|=0.

Videos

To find: The general solution for the system tx′(t)=[13−15]x(t), t>0.

Want to see the full answer?

Chapter 9 Solutions

The general solution for the system t x ′ ( t ) = [ 1 3 − 1 5 ] x ( t ) , t > 0 .

To find:

The general solution for the system tx′(t)=[13−15]x(t), t>0.