1 1
. Its eigenvalues are 1 and 2.
0 2
For λ = 1, the eigenvectors are solutions of the system
b = 0
b = 0
0
so we can take
.
1
For λ = 2, the corresponding system is
−a + b = 0
0
= 0
1
and we can take
.
1
0
1
It follows that the general solution is c1 et
+ c2 e2t
.
1
1
8.2 The characteristic polynomial is λ2 − 1 = 0, the eigenvalues are λ = ±1.
For λ = 1:
a−b
= 0
3a − 3b = 0
1
so we can take
.
1
For λ = −1:
3a − b = 0
3a − b = 0
1
so we can take
.
3
The general solution is
1
1
t
−t
x(t) = c1 e
+ c2 e
.
1
3
8.1 The matrix of the system is
8.3 The characteristic polynomial is λ2 + 2λ + 2. The roots are λ = −1 ± i.
Let us find an eigenvector for λ = −1 + i.
(1 − i)a + b
= 0
−2a + (−1 − i)b = 0
We can take a = 1, b = −1 + i. (Note that (−1 − i)(−1 + i) = 2.) The
corresponding solution of the system is then
1
1
e(−1+i)t
= e−t (cos t + i sin t)
=
−1 + i
−1 + i
e−t cos t + ie−t sin t
=
−e−t cos t − ie−t sin t + ie−t cos t − e−t sin t
e−t cos t
e−t sin t
+
i
.
−e−t cos t − e−t sin t
−e−t sin t + e−t cos t
1
It follows that the general solution is
e−t cos t
e−t sin t
x(t) = c1
+ c2
.
−e−t cos t − e−t sin t
−e−t sin t + e−t cos t
8.4 The characteristic polynomial is λ2 + 3λ + 2. The eigenvalues are −1
and −2.
Let us find the eigenvectors.
For λ = −1:
2a − 2b = 0
3a − 3b = 0
1
we can take
.
1
For λ = −2:
3a − 2b = 0
3a − 2b = 0
2
we can take
.
3
The general solution is:
1
2
x = c1 e−t
+ c2 e−2t
.
1
3
The initial condition gives:
3
1
2
c1 + 2c2
= c1
+ c2
=
.
1
1
3
c1 + 3c2
Solving the corresponding system, we get c1 = 7, c2 = −2. Consequently, the
solution of the initial value problem is
−t
7e − 4e−2t
x(t) =
.
7e−t − 6e−2t
Solutions converge to the zero vector as t → ∞. (The corresponding equilibrium point is a stable node.)
8.5 The characteristic polynomial is λ2 + 2λ + 5. The eigenvalues are λ =
−1 ± 2i. Eigenvectors of λ = −1 + 2i are found from the system
−2i −4
0
v=
.
1
−2i
0
2i
We can take v =
. The corresponding solution is
1
2i
x(t) = e(−1+2i)t
=
1
2ie−t cos 2t − 2e−t sin 2t
=
e−t cos 2t + ie−t sin 2t
−t
−2e−t sin 2t
2e cos 2t
+
i
.
e−t cos 2t
e−t sin 2t
2
The general solution is then
−t
−2e−t sin 2t
2e cos 2t
x(t) = c1
+
c
.
2
e−t cos 2t
e−t sin 2t
Substitution of the initial condition gives
0
2
4
c1
+ c2
=
.
1
0
−3
It follows that c1 = −3 and c2 = 2, so that the solution of the initial value
problem is
6e−t sin 2t + 4e−t cos 2t
6 sin 2t + 4 cos 2t
−t
x(t) =
=
e
.
−3e−t cos 2t + 2e−t sin 2t
2 sin 2t − 3 cos 2t
The solution also converges to zero as t → ∞ (the trajectories spiral towards
zero).
3