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**Unformatted text preview: **− λt) y (0) + t Ay (0)
= e−t (1 + t)
= e−t 3
−9
+t
−1
−5 3 − 6t
.
−1 − 6t Problem 9: Let us consider the diﬀerential equation
y ′ (t) = 2
3
−2 with initial condition
y (0) = 3
2 −1 y (t) 3
.
−2 We ﬁrst compute the eigenvalues of A:
0 = det 1
9
−3
2
= (λ − 2)(λ + 1) + = λ2 − λ + .
λ+1
4
4 λ−2
3
2 1
Therefore, A has a single eigenvalue of multiplicity two (λ = 2 ). Applying
the solution formula discussed in lecture gives y (t) = eλt (1 − λt) y (0) + t Ay (0)
t = e− 2
t = e− 2 t
3
3
(1 − )
+t
−5
2 −2
2
3 + 3t
2
.
−2 − 3t
2 Problem 13: The eigenvalues of A are given by
λ= a+d
±
2 (a + d)2
− (ad − bc).
4 (The calculation is the same as for Problem 23 in Section 3.4.) The statement that all solutions of the ODE approach the origin as
t → ∞ is equivalent to the statement that both λ1 and λ2 have negative
real part.
It is useful to break the discussion into two cases: 2
Case 1: Let us ﬁrst consider the case that (a+d) ≥ ad − bc. In this
4
case, the eigenvalues of A are real numbers. I...

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