Explanation Given: The given differential equations are, x 1 ' = 2 x 1 + 4 x 2 , x 2 ' = − 4 x 1 − 6 x 2 Approach: If the roots of auxiliary equation are real and same, then, The general solution of the given differential equation is given by, x ( t ) = ( c 1 + c 2 t ) e r t Calculation: Consider the equations. x 1 ' = 2 x 1 + 4 x 2 , x 2 ' = − 4 x 1 − 6 x 2 Rewrite the given system in operator form to solve the equations as follows, ( D − 2 ) x 1 − 4 x 2 = 0 … ( 1 ) And, 4 x 1 + ( D + 6 ) x 2 = 0 … ( 2 ) Multiply equation ( 1 ) with ( D + 6 ) , and equation ( 2 ) by 4 . ( D + 6 ) ( D − 2 ) x 1 − 4 ( D + 6 ) x 2 = 0 … ( 3 ) And, 16 x 1 + 4 ( D + 6 ) x 2 = 0 … ( 4 ) Add equation ( 3 ) , ( 4 ) , and simplify as follows. ( ( D + 6 ) ( D − 2 ) x 1 − 4 ( D + 6 ) x 2 ) + 16 x 1 + 4 ( D + 6 ) x 2 = 0 ( D + 6 ) ( D − 2 ) x 1 + 16 x 1 = 0 ( D 2 + 4 D − 12 ) x 1 + 16 x 1 = 0 ( D 2 + 4 D + 4 ) x 1 = 0 The auxiliary equation can be written as follows, r 2 + 4 r + 4 = 0 r 2 + 2 r + 2 r + 4 = 0 r ( r + 2 ) + 2 ( r + 2 ) = 0 ( r + 2 ) ( r + 2 ) = 0 Hence, r = − 2 , − 2 The roots of auxiliary equations are real and same

Differential Equations and Linear Algebra (4th Edition)

4th Edition

ISBN: 9780321964670

Author: Stephen W. Goode, Scott A. Annin

Publisher: PEARSON

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Solve the linear system of equations attached using Gaussian elimination (not Gauss-Jordan) and back subsitution. Remember that: A matrix is in row echelon form if Any row that consists only of zeros is at the bottom of the matrix. The first non-zero entry in each other row is 1. This entry is called aleading 1. The leading 1 of each row, after the first row, lies to the right of the leading 1 of the previous row.

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The given system of differential equations.