9.1 Let us solve the system at first. The characteristic polynomial is λ2 −
λ − 2, hence the eigenvalues are λ = −1 and λ = 2.
For λ = −1, the eigenvectors are solutions of the system
4 −2
0
v=
,
2 −1
0
1
1
so we can take v =
. The corresponding solution is x(t) = e−t
=
2
2
−t e
.
2e−t
For λ = 2:
1 −2
0
v=
,
2 −4
0
2t 2
2e
so we can take v =
. The corresponding solution is x(t) =
.
1
e2t
−t
e
2e2t
A fundamental matrix is Ψ(t) =
. Then
−t
2e
e2t
1 2
,
Ψ(0) =
2 1
and
−1
Ψ(0)
1
=−
3
It follows that
−t
e
2e2t
−1/3
Φ(t) =
2e2t e−t
2/3
1
−2
2/3
−1/3
−2
1
=
.
4e2t −e−t
3
2e−t −2e2t
3
2e−t −2e2t
3
4e2t −e−t
3
!
.
2
9.2 The characteristic
polynomial is λ . So, the only eigenvalue is 0. An
1
eigenvector is v =
. The corresponding solution is constant x(t) = v.
2
A solution of
4 −2
1
w=
8 −4
2
1/4
is, for example, w =
. Then, the corresponding solution of the system
0
1
1/4
of differential equation is x(t) = t
+
.
2
0
A general solution of the system is
1
1
1/4
x(0) = c1
+ c2 t
+ c2
.
2
2
0
9.3 The eigenvalues of the matrix are the entries on the diagonal (since it is
triangular).
1
The eigenvalue λ = 1 is repeated. Its eigenvector is found from the system

= 0
 0
−4a
= 0

3a + 6b + c = 0
We see that a =
 0, and
 b, c are any numbers such that 6b + c = 0. We get
0
an eigenvector  1 . All the other eigenvectors are proportional to it.
−6
Therefore, we have to solve the system

= 0
 0
−4a
= 1

3a + 6b + c = −6
We can take a = −1/4, b = 0, and c = −6 + 3/4 = −21/4. We get in this way
two solutions of the system:






−1/4
0
0
0 .
x1 (t) = et  1  , x2 (t) = tet  1  + et 
−21/4
−6
−6
For the eigenvalue λ = 2 we solve the

 −a
−4a − b

3a + 6b
system
= 0
= 0
= 0

0
An eigenvector is  0 , and the corresponding solution is
1


0
x3 (t) =  0  .
e2t

The general solution is then x(t) = c1 x1 (t) + c2 x2 (t) + c3 x3 (t).
9.4. The characteristic polynomial of the homogeneous system is λ2 + 1,
hence the eigenvalues are λ = ±i. Let us find the eigenvector associated with
λ = i.
(2 − i)a − 5b = 0
a + (−2 − i)b = 0
2+i
A solution is
. Hence a complex solution of the homogeneous system
1
2
is
x1
x2
= eit
2+i
2+i
= (cos t + i sin t)
=
1
1
2 cos t − sin t + i(2 sin t + cos t)
=
cos t + i sin t
2 cos t − sin t
2 sin t + cos t
+i
.
cos t
sin t
It follows that the general solution of the homogeneous system is
2 cos t − sin t
2 sin t + cos t
x = c1
+ c2
,
cos t
sin t
and the fundamental matrix is
2 cos t − sin t
Ψ(t) =
cos t
2 sin t + cos t
sin t
.
Its determinant is
(2 cos t − sin t) sin t − (2 sin t + cos t) cos t = − sin2 t − cos2 t = −1.
Hence,
Ψ−1 (t) =
− sin t 2 sin t + cos t
cos t −2 cos t + sin t
.
Then the general solution of the equation is given by the indefinite integral
(one can also take a definite integral and then add the general solution of the
homogeneous system to it):
Z
Ψ(t)
−1
Ψ
(s)
et
t
dt =
Z −et sin t + 2t sin t + t cos t
Ψ(t)
dt =
et cos t − 2t cos t + t sin t
R
−et sin t + 2t sin t + t cos t dt
R
Ψ(t)
et cos t − 2t cos t + t sin t dt
R
R
The integrals t cos t dt and t sin t dt are computed by parts:
Z
Z
Z
t cos t dt = t d sin t = t sin t − sin t dt = t sin t + cos t + c,
and
Z
Z
t sin t dt = −
Z
t d cos t = −t cos t +
3
cos t dt = −t cos t + sin t + c.
R
R
The integrals cos tet dt and sin tet dt can be also computed using integration by parts (twice, and then solving a linear equation in terms of the integral),
or by computing
Z
Z
Z
1 (1+i)t
e
+c=
1+i
et (cos t + sin t)
et (sin t − cos t)
1−i t
(e cos t + iet sin t) + c =
+i
+ c.
2
2
2
t
(cos t + i sin t)e dt =
e
t+it
dt =
e(1+i)t dt =
Taking the real and imaginary parts, we get
Z
et cos t dt = et (cos t + sin t)/2 + c
and
Z
et sin t dt = et (sin t − cos t)/2 + c.
Thus, we have that the general solution of the system is
et (cos t − sin t)/2 − 2t cos t + 2 sin t + t sin t + cos t + c1
Ψ(t)
=
et (cos t + sin t)/2 − 2t sin t − 2 cos t − t cos t + sin t + c2
t
2 cos t − sin t 2 sin t + cos t
e (cos t − sin t)/2 + (1 − 2t) cos t + (2 + t) sin t + c1
=
cos t
sin t
et (cos t + sin t)/2 + (1 − 2t) sin t − (2 + t) cos t + c2
3 t
2 cos t − sin t
2 sin t + cos t
2 e − 5t
+ c1
+ c2
.
1 t
cos
t
sin t
e
−
2t
+
1
2
A different
to use
approach is the Laplace transform with initial conditions
1
0
x(0) =
and x(0) =
, and then take linear combination of the two
0
1
solutions.
4