# E find the general solution to the matrix dierential

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Unformatted text preview: x2 . ∂x2 It now follows from Newton’s second law of motion that Force = Mass × Acceleration, and hence d2 x1 d2 x2 , F2 = m 2 . dt2 dt Thus we obtain a second-order system of diﬀerential equations, F1 = m m d2 x1 = −k1 x1 + k2 (x2 − x1 ) = −(k1 + k2 )x1 + k2 x2 , dt2 d2 x2 = k2 (x1 − x2 ) − k3 x2 = k2 x1 − (k2 + k3 )x2 . dt2 We can write this system in matrix form as m m d2 x = Ax, dt2 where −(k1 + k2 ) k2 k2 −(k2 + k3 ) A= 50 . (2.14) Note that A is indeed a symmetric matrix. The potential energy is given by the expression 1 x V (x1 , x2 ) = − x1 x2 A 1 . x2 2 Example. Let us consider the special case of the mass-spring system in which m = k1 = k2 = k3 = 1, so that the system (2.14) becomes d2 x = dt2 −2 1 1 x. −2 (2.15) To ﬁnd the eigenvalues, we must solve the characteristic equation det −2 − λ 1 1 −2 − λ = (λ + 2)2 − 1 = 0, which yields λ = −2 ± 1. The eigenvalues in this case are λ1 = −1, and λ2 = −3. The eigenspace corresponding to the eigenvalue −1 is 1 1 W−1 = {b ∈ R2 : (A + I )b = 0} = . . . = span . It follows from the argument i...
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