**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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