Explanation Given: The given non-homogeneous system is, X ′ = ( 1 − 1 1 3 ) X + ( t t + 1 ) and X ( 0 ) = ( 3 2 ) Calculation: For the given system, the coefficient matrix is A = ( 1 − 1 1 3 ) and F ( t ) = ( t t + 1 ) . Since, the characteristic equation for the matrix A is | ( A − λ I ) | = 0 where I is the Identity matrix. Substitute the values of matrix A and I in above expression. | 1 − λ − 1 1 3 − λ | = 0 ( 1 − λ ) ( 3 − λ ) + 1 = 0 λ 2 − 4 λ + 4 = 0 ( λ − 2 ) 2 = 0 Therefore, the Eigen value of the above system is λ 1 = 2 with multiplicity 2. Now, to obtain the Eigen vector K 1 corresponding to the Eigen value λ 1 = 2 . Solve the equation ( A − λ 1 I ) K 1 = 0 where, K 1 = ( x 1 y 1 ) . Substituting the values of A , λ ` 1 and K 1 in the above equation gives, [ ( 1 − 1 1 3 ) − ( 2 0 0 2 ) ] ⋅ ( x 1 y 1 ) = ( 0 0 ) ( − 1 − 1 1 1 ) ⋅ ( x 1 y 1 ) = ( 0 0 ) Applying the row transformation, R 1 → R 1 + R 2 in above system, ( 0 0 1 1 ) ⋅ ( x 1 y 1 ) = ( 0 0 ) ( 0 x 1 + y 1 ) = ( 0 0 ) Solving above equations gives, x 1 = − y 1 Assume that y 1 = n for some scalar n which implies x 1 = − n . Therefore, the value of ( x 1 y 1 ) is, ( x 1 y 1 ) = ( − n n ) = n ( − 1 1 ) Therefore the eigenvector K 1 corresponding to the Eigen value λ 1 = 2 is ( − 1 1 ) . Now, to obtain the second Eigen vector K 2 . Solve the equation ( A − λ 2 I ) K 2 = K 1 where K 2 = ( x 2 y 2 ) . Substituting the values of A , λ ` 2 , K 2 and K 1 in the above equation gives, [ ( 1 − 1 1 3 ) − ( 2 0 0 2 ) ] ⋅ ( x 2 y 2 ) = ( − 1 1 ) ( − 1 − 1 1 1 ) ⋅ ( x 2 y 2 ) = ( − 1 1 ) Applying the row transformation, R 1 → R 1 + R 2 in above system, ( 0 0 1 1 ) ⋅ ( x 2 y 2 ) = ( − 1 1 ) ( 0 x 2 + y 2 ) = ( 0 1 ) Solving above equations gives, x 2 = 1 − y 2 Assume that y 2 = m for some scalar m which implies x 2 = 1 − m . Therefore, the value of ( x 2 y 2 ) is, ( x 2 y 2 ) = ( 1 − m m ) = m ( − 1 1 ) + ( 1 0 ) Since, from the above calculation, the eigenvector K 1 obtained is ( − 1 1 ) . Therefore the second eigenvector K 2 is ( 1 0 ) . The complementary function for the given non homogeneous system is, X c = c 1 K 1 e λ 1 t + c 2 [ K 1 t + K 2 ] e λ 1 t . Substitute the value of λ 1 , K 1 and K 2 in above equation, X c = c 1 ( − 1 1 ) e 2 t + c 2 [ ( − 1 1 ) t + ( 1 0 ) ] e 2 t Since the function F ( t ) is given as F ( t ) = ( t t + 1 ) = ( 1 1 ) t + ( 0 1 ) Therefore, to obtain the particular solution X p for the given system, assume that X p is given of the form, X p = ( a 2 b 2 ) t + ( a 1 b 1 ) , where a 1 , b 1 are constants

A First Course in Differential Equations with Modeling Applications (MindTap Course List)

11th Edition

ISBN: 9781305965720

Author: Dennis G. Zill

Publisher: Cengage Learning

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Differential Equations

Have you ever observed the movement of the satellites and rockets or how the planets go around the sun in their orbits? How will you determine their motion? Definitely they follow some scientific laws and these kinds of problems are framed as differentia…

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Chapter 8.3, Problem 10E

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To solve: The initial value problem.

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Solution Summary: The author calculates the coefficient matrix for the given non-homogeneous system. Substitute the values of matrix A and I in above expression.

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To solve: The initial value problem.

Chapter 8 Solutions