Explanation Given: The equation one is d x 1 d t = 3 x 1 + 2 x 2 . The equation two is d x 2 d t = 3 x 1 + 8 x 2 . The value of x 1 ( 0 ) = 3 . The value of x 2 ( 0 ) = 23 . Approach: First put the differential equation in the matrix with the coefficient of x 1 in the first column and the coefficient of x 2 in the second column. Determine the eigenvalues, eigenvectors and also the x and y components of the eigenvector. Substitute the values in the expression [ x 1 x 2 ] = [ v 1 v 2 ] [ C 1 e λ 1 t C 2 e λ 2 t ] to calculate the equations for x 1 and x 2 . Substitute the values of x 1 and x 2 in the expressions of the general solutions to obtain the values of constant C 1 and C 2 . Substitute the values of constant terms in the general expressions to obtain the particular solution. Calculation: The matrix form of the differential equations is, M = [ 3 2 3 8 ] Calculate the eigenvalues by subtracting λ from the diagonals. M − λ I = [ 3 − λ 2 3 8 − λ ] Take the determinate of the matrix. det ( M − λ I ) = [ 3 − λ 2 3 8 − λ ] = λ 2 − 11 λ + 24 − 6 0 = λ 2 − 11 λ + 18 0 = ( λ − 9 ) ( λ − 2 ) λ 1 = 9 λ 2 = 2 Calculate the eigenvector, v by using the equations, ( M − λ I ) v = [ 3 − 9 2 3 8 − 9 ] [ x y ] = [ 0 0 ] Thus, the eigenvector is v = [ x y ] Substitute the first eigenvalue to get the solution, [ 3 − 9 2 3 8 − 9 ] [ x y ] = [ 0 0 ] [ − 6 2 3 − 1 ] [ x y ] = [ 0 0 ] Therefore the first solution is, v 1 = [ x y ] = [ 1 2 ] Substitute the second eigenvalue to get the solution, [ 3 − 2 2 3 8 − 2 ] [ x y ] = [ 0 0 ] [ 1 2 3 6 ] [ x y ] = [ 0 0 ] Therefore the second solution is, v 2 = [ x y ] = [ 3 − 1 ] The general expression is, [ x 1 x 2 ] = [ v 1 v 2 ] [ C 1 e λ 1 t C 2 e λ 2 t ] Substitute the values in the above general expression

Pearson eText for Calculus for the Life Sciences -- Instant Access (Pearson+)

2nd Edition

ISBN: 9780137553457

Author: Raymond Greenwell, Nathan Ritchey

Publisher: PEARSON+

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Chapter 11.4, Problem 11E

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The solutions for the differential equations dx1dt=3x1+2x2 and dx2dt=3x1+8x2.

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Chapter 11.4, Problem 10E

Chapter 11.4, Problem 12E

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Chapter 11 Solutions

Pearson eText for Calculus for the Life Sciences -- Instant Access (Pearson+)

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The solutions for the differential equations d x 1 d t = 3 x 1 + 2 x 2 and d x 2 d t = 3 x 1 + 8 x 2 .

Solution Summary: The author explains how to find the solutions for the differential equations dx_1&, and the equation 2.

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To find:

The solutions for the differential equations dx1dt=3x1+2x2 and dx2dt=3x1+8x2.

Want to see the full answer?

Chapter 11 Solutions

The solutions for the differential equations d x 1 d t = 3 x 1 + 2 x 2 and d x 2 d t = 3 x 1 + 8 x 2 .

Solution Summary: The author explains how to find the solutions for the differential equations dx_1&, and the equation 2.

Concept explainers

Videos

To find: The solutions for the differential equations dx1dt=3x1+2x2 and dx2dt=3x1+8x2.

Want to see the full answer?

Chapter 11 Solutions

To find:

The solutions for the differential equations dx1dt=3x1+2x2 and dx2dt=3x1+8x2.