To determine The general solution of the equation, x ′ = ( 2 − 5 1 − 2 ) x + ( 0 cos t ) , 0 < t < π . Explanation The given system of equations is, x ′ = ( 2 − 5 1 − 2 ) x + ( 0 cos t ) , 0 < t < π It is noted that, the given system is a non-homogeneous system. Consider the homogeneous part. That is, x ′ = ( 2 − 5 1 − 2 ) x This is of the form, x ′ = A x , where A = ( 2 − 5 1 − 2 ) . The eigenvalues r and the eigenvectors ξ satisfy the equation, | A − r I | ξ = 0 . That is, | A − r I | ξ = 0 | 2 − r − 5 1 − 2 − r | ( ξ 1 ξ 2 ) = ( 0 0 ) ( 2 − r ) ( − 2 − r ) + 5 = 0 r 2 + 1 = 0 The roots of the above characteristic equation are obtained as, r 1 = i and r 2 = − i . Thus, the solutions are, r 1 = i and r 2 = − i . Set r 1 = i in the coefficient matrix. That is, ( 2 − i − 5 1 − 2 − i ) On solving, ( 2 − i − 5 1 − 2 − i ) ( ξ 1 ξ 2 ) = ( 0 0 ) , the following equation is obtained. ( 2 − i ) ξ 1 − 5 ξ 2 = 0 Let ξ 1 = 5 , then the above equation becomes, ξ 2 = 2 − i . Then, the eigen vector formed is, ξ ( 1 ) = ( 5 2 − i ) . Set r 2 = − i in the coefficient matrix. That is, ( 2 + i − 5 1 − 2 + i ) On solving, ( 2 + i − 5 1 − 2 + i ) ( ξ 1 ξ 2 ) = ( 0 0 ) , the following equation is obtained. ( 2 + i ) ξ 1 − 5 ξ 2 = 0 Let ξ 1 = 5 , then the above equation becomes, ξ 2 = 2 + i . Then, the eigen vector formed is, ξ ( 2 ) = ( 5 2 + i ) . Therefore, the eigenvector matrix is, T = ( 5 5 2 − i 2 + i ) . Therefore, the Jordan form of the coefficient matrix is, J = T − 1 A T = ( i 0 0 − i ) Obtain the general solution to the homogeneous system to obtain the fundamental matrix. Consider the first eigenvector and it’s corresponding solution. x ( 1 ) ( t ) = e i t ( 5 2 − i ) = ( cos t + i sin t ) ( 5 2 − i ) = ( 5 ( cos t + i sin t ) ( 2 − i ) ( cos t + i sin t ) ) = ( 5 cos t + 5 i sin t 2 cos t + 2 i sin t − i cos t + sin t ) Split the above matrix into real and imaginary part. x ( 1 ) ( t ) = ( 5 cos t + i 5 sin t 2 cos t + sin t + i ( 2 sin t − cos t ) ) = ( 5 cos t 2 cos t + sin t ) + i ( 5 sin t 2 sin t − cos t ) Therefore, the fundamental matrix is, Ψ ( t ) = ( 5 cos t 5 sin t 2 cos t + sin t 2 sin t − cos t ) The inverse of the fundamental matrix is obtained as, Ψ − 1 ( t ) = ( 1 5 ( cos t − 2 sin t ) sin t 1 5 ( 2 cos t + sin t ) − cos t ) Use the method of variation of parameters to obtain the solution of the IVP

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ISBN: 9780470458327

Author: William E. Boyce, Richard C. DiPrima

Publisher: Wiley, John & Sons, Incorporated

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Chapter 7.9, Problem 11P

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The general solution of the equation, x′=(2−51−2)x+(0cost), 0<t<π.

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Chapter 7.9, Problem 10P

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The general solution of the equation, x ′ = ( 2 − 5 1 − 2 ) x + ( 0 cos t ) , 0 &lt; t &lt; π .

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The general solution of the equation, x′=(2−51−2)x+(0cost), 0<t<π.

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The general solution of the equation, x ′ = ( 2 − 5 1 − 2 ) x + ( 0 cos t ) , 0 < t < π .