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Question: in math225 we learned that systems of linear ordinary homogeneous...

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(6 points) In MATH225, we learned that systems of linear, ordinary, homogeneous, constant-coefficient differential equations can be solved by finding the eigenvalues and eigenvectors of the systems coefficient matrix. As a reminder, consider the system: 2(t)+4 3 -2 The coefficient matrix for this system is A - The eigenvalues for this 0.8944 matrix are 2 and X2-5. The corresponding eigenvectors are vi- an 0.4472 0.7071 The general solution ot the system is -0.7071 0.8944 0.70715t 0.7071 y(t) ci e + C2 -0.4472 What we learned in MATH225 about small linear systems generalizes to larger linear sys- tems. Using MATLAB to find the eigenvalues and eigenvectors, find the general solution of the following system of three differential equations

In MATH225, we learned that systems of linear, ordinary, homogeneous, constant-coefficient differential equations can be solved by finding the eigenvalues and eigenvectors of the system’s coefficient matrix. As a reminder, consider the system: y 0 1 (t) = 3y1 − 2y2 y 0 2 (t) = −y1 + 4y2 The coefficient matrix for this system is A =   3 −2 −1 4  . The eigenvalues for this matrix are λ1 = 2 and λ2 = 5. The corresponding eigenvectors are v1 =   −0.8944 −0.4472   and v2 =   0.7071 −0.7071  . The general solution ot the system is y(t) = c1   −0.8944 −0.4472   e 2t + c2   0.7071 −0.7071   e 5t What we learned in MATH225 about small linear systems generalizes to larger linear systems. Using Matlab to find the eigenvalues and eigenvectors, find the general solution of the following system of three differential equations: y 0 1 (t) = −2y1 − 4y2 + 2y y 0 2 (t) = −2y1 + y2 + 2y3 y 0 3 (t) = 4y1 + 2y2 + 5y3

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