Explanation Given: The given differential equations are, u ' ( t ) = − 2 u ( t ) + 2 v ( t ) − 3 w ( t ) v ' ( t ) = 2 u ( t ) + v ( t ) − 6 w ( t ) w ' ( t ) = − u ( t ) − 2 v ( t ) The initial values are, u ( 0 ) = 3 v ( 0 ) = − 1 w ( 0 ) = 3 Approach: Consider a system of differential equation u ' ( t ) = a u ( t ) + b v ( t ) + c w ( t ) v ' ( t ) = d u ( t ) + e v ( t ) + f w ( t ) w ' ( t ) = g u ( t ) + h v ( t ) + i w ( t ) The above equation can be written in matrix form. x ( t ) = [ u ( t ) v ( t ) w ( t ) ] x ' ( t ) = A x ( t ) x ' ( t ) = [ u ' ( t ) v ' ( t ) w ' ( t ) ] A = [ a b c d e f g h i ] Calculation: Here, A = [ − 2 2 − 3 2 1 6 − 1 − 2 0 ] and x 0 = [ 3 − 1 3 ] . Calculate the characteristic polynomial for A . p ( t ) = | A − t I | Substitute [ − 2 2 − 3 2 1 6 − 1 − 2 0 ] for A and [ 1 0 0 0 1 0 0 0 1 ] for I in above equation. p ( t ) = [ − 2 2 − 3 2 1 − 6 − 1 − 2 0 ] − t [ 1 0 0 0 1 0 0 0 1 ] = [ − 2 2 − 3 2 1 − 6 − 1 − 2 0 ] − [ t 0 0 0 t 0 0 0 t ] = [ − 2 − t 2 − 3 2 1 − t − 6 − 1 − 2 − t ] On solving the equation further, p ( t ) = ( − 2 − t ) ( − t + t 2 − 12 ) − 2 ( − 2 t − 6 ) − 3 ( − 4 + 1 − t ) = t 3 + t 2 − 21 t − 45 = ( t + 3 ) ( t + 3 ) ( t − 5 ) The eigen values of A are the roots of p ( t ) = 0 . Therefore, the eigenvalues of A are λ 1 = − 3 , λ 2 = − 3 and λ 3 = 5 . Corresponding eigenvectors can be calculated using, A x = λ 1 x , A x = λ 2 x and A x = λ 3 x . Where x is an eigenvector. Consider A x = λ 1 x . Here λ 1 = − 3 . ( A + 3 I ) x = 0 ( [ − 2 2 − 3 2 1 − 6 − 1 − 2 0 ] + [ 3 0 0 0 3 0 0 0 3 ] ) x = 0 [ 1 2 − 3 2 4 − 6 − 1 − 2 3 ] [ x 1 x 2 x 3 ] = [ 0 0 0 ] Reduce the augmented matrix as follows. [ 1 2 − 3 0 2 4 − 6 0 − 1 − 2 3 0 ] → R 1 + R 2 R 2 − 2 R 1 [ 1 2 − 3 0 0 0 0 0 0 0 0 0 ] The equation becomes, x 1 + 2 x 2 − 3 x 3 = 0 x 1 = − 2 x 2 + 3 x 3 Hence, x = [ x 1 x 2 x 3 ] = [ − 2 x 2 + 3 x 3 x 2 x 3 ] = [ − 2 1 0 ] x 2 + [ 3 0 1 ] x 3 Hence an eigenvector corresponding to λ 1 = λ 2 = − 3 are, u 1 = [ − 2 1 0 ] u 2 = [ 3 0 1 ] Consider ( A − λ 3 I ) x = 0 . Here λ 3 = 5 . ( A − 5 I ) x = 0 ( [ − 2 2 − 3 2 1 − 6 − 1 − 2 0 ] − [ 5 0 0 0 5 0 0 0 5 ] ) x = 0 [ − 7 2 − 3 2 − 4 − 6 − 1 − 2 − 5 ] [ x 1 x 2 x 3 ] = [ 0 0 0 ] Reduce the augmented matrix as follows

Introduction to Linear Algebra (Classic Version) (5th Edition) (Pearson Modern Classics for Advanced Mathematics Series)

5th Edition

ISBN: 9780134689531

Author: Lee Johnson, Dean Riess, Jimmy Arnold

Publisher: PEARSON

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Differential Equations

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Chapter 4.8, Problem 18E

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The initial value problem u'(t)=−2u(t)+2v(t)−3w(t), u(0)=3v'(t)=2u(t)+v(t)−6w(t), v(0)=−1w'(t)=−u(t)−2v(t), w(0)=3

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Chapter 4.8, Problem 17E

Chapter 4.8, Problem 19E

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MODELING REAL LIFE Your checking account has a constant balance of $500. Let the function $m$ represent the balance of your savings account after $t$ years. The table shows the total balance of the accounts over time. Year, $t$ Total balance 0 1 2 3 4 5 $2500 $2540 $2580.80 $2622.42 $2664.86 $2708.16 a. Write a function $B$ that represents the total balance after $t$ years. Round values to the nearest hundredth, if necessary. $B\left(t\right)=$ Question 2 b. Find $B\left(8\right)$ . About $ a Question 3 Interpret $B\left(8\right)$ . b represents the total balance checking and saving accounts after 8 years the balance would be 16 / 10000 Word Limit16 words written of 10000 allowed Question 4 c. Compare the savings account to the account, You deposit $9000 in a savings account that earns 3.6% annual interest compounded monthly. A = 11998.70 SINCE 9000 is the principal ( 1+0.036/12)12 times 8 gives me aproxtimately 1997 14 / 10000 Word Limit14 words written of 10000 allowed Skip to…

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Introduction to Linear Algebra (Classic Version) (5th Edition) (Pearson Modern Classics for Advanced Mathematics Series)

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The initial value problem u ' ( t ) = − 2 u ( t ) + 2 v ( t ) − 3 w ( t ) , u ( 0 ) = 3 v ' ( t ) = 2 u ( t ) + v ( t ) − 6 w ( t ) , v ( 0 ) = − 1 w ' ( t ) = − u ( t ) − 2 v ( t ) , w ( 0 ) = 3

Solution Summary: The author explains that the initial value problem is lu'(t)+2e-3t

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The initial value problem u'(t)=−2u(t)+2v(t)−3w(t), u(0)=3v'(t)=2u(t)+v(t)−6w(t), v(0)=−1w'(t)=−u(t)−2v(t), w(0)=3

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The initial value problem u ' ( t ) = − 2 u ( t ) + 2 v ( t ) − 3 w ( t ) , u ( 0 ) = 3 v ' ( t ) = 2 u ( t ) + v ( t ) − 6 w ( t ) , v ( 0 ) = − 1 w ' ( t ) = − u ( t ) − 2 v ( t ) , w ( 0 ) = 3

Solution Summary: The author explains that the initial value problem is lu'(t)+2e-3t

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To solve: The initial value problem u'(t)=−2u(t)+2v(t)−3w(t), u(0)=3v'(t)=2u(t)+v(t)−6w(t), v(0)=−1w'(t)=−u(t)−2v(t), w(0)=3

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Chapter 4 Solutions

To solve:

The initial value problem u'(t)=−2u(t)+2v(t)−3w(t), u(0)=3v'(t)=2u(t)+v(t)−6w(t), v(0)=−1w'(t)=−u(t)−2v(t), w(0)=3