hw6-solutions

# In other words the general solution of the dierential

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Unformatted text preview: fundamental system of solutions. In other words, the general solution of the diﬀerential equation y ′ (t) = Ay (t) is given by y (t) = a1 2 cos(t) − sin(t) cos(t) + 2 sin(t) + a2 cos(t) sin(t) for suitable constants a1 , a2 . The initial condition y (0) = 3 2 implies a1 = 2 and a2 = −1. Thus, y (t) = 3 cos(t) − 4 sin(t) . 2 cos(t) − sin(t) Problem 9: We consider the diﬀerential equation y ′ (t) = 1 −5 y (t) 1 −3 with initial condition y (0) = 1 . 1 We begin by computing the eigenvalues of the coeﬃcient matrix A: 0 = det λ−1 5 = (λ − 1)(λ + 3) + 5 = λ2 + 2λ + 2. −1 λ + 3 Therefore, the eigenvalues are −1 + i and −1 − i. The vector v= 2+i 1 is an eigenvector of A with eigenvalue i. Consequently, the function x(t) = e(−1+i)t 2+i 1 is a solution of the diﬀerential equation x′ (t) = Ax(t). We now split x(t) into its real and imaginary pa...
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## This note was uploaded on 02/09/2014 for the course MATH 53 taught by Professor Staff during the Winter '08 term at Stanford.

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