Explanation Given: The given condition is: Two tanks, each holding 1 L of liquid, are connected by a pipe through which liquid flows from tank A into tank B at a rate of 3 − α L / min ( 0 < α < 3 ) . The liquid inside each tank is kept well stirred. Pure water flows into tank A at a rate of 3 L / min . Solution flows out of tank A at α L / min and out of tank B at 3 − α L / min . Figure ( 1 ) Approach: Suppose n × n constant matrix A has n linearly independent eigenvectors u 1 , u 2 , ⋯ , u n . Let r i be the eigenvalues corresponding to u i . Then, { e r 1 t u 1 , e r 2 t u 2 , ⋯ , e r n t u n } is a fundamental solution set and X ( t ) = [ e r 1 t u 1 , e r 2 t u 2 , ⋯ , e r n t u n ] is called fundamental matrix. Consequently, a general solution of X ′ = A X is: x ( t ) = X ( t ) X − 1 ( t ) X ( 0 ) Here, X ( 0 ) is the initial value. Calculation: Consider the given condition: Two tanks each holding 1 L of liquid, then the salt concentration in tank A is x 1 ( t ) 1 kg / L and the salt flow in pipes at the rate of ( 3 − α ) x 1 ( t ) 1 kg / min . The lower pipe brings salt into tank A is α x 1 ( t ) 1 kg / L . The rate of change of the mass of salt in tank A is: d x 1 ( t ) d t = input rate − output rate = 0 − ( 3 − α ) x 1 ( t ) 1 − α x 1 ( t ) 1 = − [ 3 x 1 ( t ) − α x 1 ( t ) + α x 1 ( t ) ] = − 3 x 1 ( t ) The rate of change of the mass of salt in tank B is: d x 2 ( t ) d t = input rate − output rate = ( 3 − α ) x 1 ( t ) 1 − ( 3 − α ) x 2 ( t ) 1 = ( 3 − α ) x 1 ( t ) − ( 3 − α ) x 2 ( t ) A system of differential can be written as: x ′ 1 ( t ) = − 3 x 1 ( t ) and x ′ 2 ( t ) = ( 3 − α ) x 1 ( t ) − ( 3 − α ) x 2 ( t ) The system in matrix form: [ x 1 ( t ) x 2 ( t ) ] ′ = [ − 3 0 ( 3 − α ) − ( 3 − α ) ] [ x 1 ( t ) x 2 ( t ) ] Here, the matrix A = [ − 3 0 ( 3 − α ) − ( 3 − α ) ] . Substitute the value of matrix A in the characteristic equation and solve for r : | [ − 3 0 ( 3 − α ) − ( 3 − α ) ] − r [ 1 0 0 1 ] | = 0 | [ − 3 0 ( 3 − α ) − ( 3 − α ) ] − [ r 0 0 r ] | = 0 | [ − 3 − r 0 ( 3 − α ) − ( 3 − α ) − r ] | = 0 ( − 3 − r ) ( − ( 3 − α ) − r ) = 0 The first factor is: ( − 3 − r ) = 0 r = − 3 r = 0 The second factor is: ( − ( 3 − α ) − r ) = 0 r = α − 3 The eigenvalues are r 1 = − 3 , r 2 = α − 3 . Find the eigenvector corresponding to eigenvalue r 1 = − 3 by using the formula ( A − r I ) u = 0 : ( A − r I ) u = ( [ − 3 0 ( 3 − α ) − ( 3 − α ) ] − ( − 3 ) [ 1 0 0 1 ] ) [ u 1 u 2 ] = ( [ − 3 0 ( 3 − α ) − ( 3 − α ) ] + [ 3 0 0 3 ] ) [ u 1 u 2 ] = [ 0 0 ( 3 − α ) α ] [ u 1 u 2 ] The matrix equation is: [ 0 0 ( 3 − α ) α ] [ u 1 u 2 ] = [ 0 0 ] It can be write in scalar equation as, ( 3 − α ) u 1 + α u 2 = 0 . Choose the arbitrary value of u 2 = 1 , then u 1 = α α − 3 . The eigenvector can be expressed as: u 1 = [ α α − 3 1 ] Find the eigenvector corresponding to eigenvalue r 2 = α − 3 by using the formula ( A − r I ) u = 0 : ( A − r I ) u = ( [ − 3 0 ( 3 − α ) − ( 3 − α ) ] − ( α − 3 ) [ 1 0 0 1 ] ) [ u 1 u 2 ] = ( [ − 3 0 ( 3 − α ) − ( 3 − α ) ] − [ α − 3 0 0 α − 3 ] ) [ u 1 u 2 ] = [ − α 0 ( 3 − α ) 0 ] [ u 1 u 2 ] The matrix equation is: [ − α 0 ( 3 − α ) 0 ] [ u 1 u 2 ] = [ 0 0 ] It can be write in scalar equation as, u 1 = 0 and u 2 is free. Choose the arbitrary value of u 2 = 1

Fundamentals Of Differential Equations And Boundary Value Problems, Books A La Carte Edition (7th Edition)

7th Edition

ISBN: 9780321977182

Author: Nagle, R. Kent, Saff, Edward B., Snider, Arthur David

Publisher: PEARSON

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Chapter 9.5, Problem 46E

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The mass of the salt in each tank at time t≥0, the manner in which the mass of salt in tank A depends on the choice of α and the maximum mass of salt in tank B.

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Chapter 9.5, Problem 45E

Chapter 9.5, Problem 48E

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Fundamentals Of Differential Equations And Boundary Value Problems, Books A La Carte Edition (7th Edition)

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The mass of the salt in each tank at time t ≥ 0 , the manner in which the mass of salt in tank A depends on the choice of α and the maximum mass of salt in tank B.

Solution Summary: The author explains how the mass of salt in each tank at time tge 0 depends on the choice of alpha and the maximum mass in tank B.

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The mass of the salt in each tank at time t≥0, the manner in which the mass of salt in tank A depends on the choice of α and the maximum mass of salt in tank B.

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Chapter 9 Solutions