A&M Texas University
Department of Mathematics
Volodymyr Nekrashevych
Fall 2011
MATH 308 Homework 9
9.1. Find the fundamental matrix (t) = eAt satisfying (0) = I for the system
x=
3 2
2 2
x.
4 2
8 4
x.
9.2. Find the general solution of the system
x=
9.3. Find the general solution of the system
100
x = 4 1 0 x.
362
9.4. Find the general solution of the system
x=
2 5
1 2
x+
et
t
.
9.5. Solve the previous system in MatLab using the initial condition x(0) =
plot the solution (x1 (t), x2 (t)) on the interval [0, 4 ].
1
1
and
9.1 Let us solve the system at rst. The characteristic polynomial is 2
2, hence the eigenvalues are = 1 and = 2.
For = 1, the eigenvectors are solutions of the system
4 2
2 1
1
2
so we can take v =
,
. The corresponding solution is x(t) = et
et
.
2et
For = 2:
1
2
2
1
so we can take v =
0
0
v=
2
4
0
0
v=
e t
2et
1
2
(0) =
and
(0)1 =
1
3
2e2t
e2t
2
1
1
2
=
,
. The corresponding solution is x(t) =
A fundamental matrix is (t) =
1
2
2e2t
e2t
.
. Then
,
2
1
.
It follows that
(t) =
et
2e2t
2e2t
et
1/3
2/3
2/3
1/3
=
2et 2e2t
3
4e2t et
3
4e2t et
3
2et 2e2t
3
.
9.2 The characteristic polynomial is 2 . 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
is, for example, w =
1/4
0
. Then, the corresponding solution of the system
1
2
A general solution of the system is
of dierential equation is x(t) = t
x(0) = c1
1
2
+
+ c2 t
1
2
1/4
0
+ c2
.
1/4
0
.
9.3 The eigenvalues of the matrix are the entries on the diagonal (since it is
triangular).
1
The eigenvalue = 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 = 2 we solve the
a
4a b
3a + 6 b
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 of polynomial the homogeneous system is 2 + 1,
hence the eigenvalues are = i. Let us nd the eigenvector associated with
= i.
(2 i)a 5b = 0
a + (2 i)b = 0
A solution is
2+i
1
. Hence a complex solution of the homogeneous system
2
is
x1
x2
2+i
1
= eit
= (cos t + i sin t)
2+i
1
=
2 cos t sin t + i(2 sin t + cos t)
cos t + i sin t
2 cos t sin t
cos t
=
2 sin t + cos t
sin t
+i
.
It follows that the general solution of the homogeneous system is
x = c1
2 cos t sin t
cos t
+ c2
2 sin t + cos t
sin t
,
and the fundamental matrix is
(t) =
2 cos t sin 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,
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 indenite integral
(one can also take a denite integral and then add the general solution of the
homogeneous system to it):
(t)
et
t
1 (s)
(t)
dt =
et sin t + 2t sin t + t cos t
et cos t 2t cos t + t sin t
(t)
The integrals
t cos t dt and
t cos t dt =
dt =
et sin t + 2t sin t + t cos t dt
et cos t 2t cos t + t sin t dt
t sin t dt are computed by parts:
t d sin t = t sin t
sin t dt = t sin t + cos t + c,
and
t sin t dt =
t d cos t = t cos t +
3
cos t dt = t cos t + sin t + c.
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
1 (1+i)t
e
+c=
1+i
et (cos t + sin t)
et (sin t cos t)
1i t
(e cos t + iet sin t) + c =
+i
+ c.
2
2
2
(cos t + i sin t)et dt =
et+it dt =
e(1+i)t dt =
Taking the real and imaginary parts, we get
et cos t dt = et (cos t + sin t)/2 + c
and
et sin t dt = et (sin t cos t)/2 + c.
Thus, we have that the general solution of the system is
(t)
et (cos t sin t)/2 2t cos t + 2 sin t + t sin t + cos t + c1
et (cos t + sin t)/2 2t sin t 2 cos t t cos t + sin t + c2
2 cos t sin t 2 sin t + cos t
cos t
sin t
3t
2 e 5t
1t
2 e 2t +
1
+ c1
=
et (cos t sin t)/2 + (1 2t) cos t + (2 + t) sin t + c1
et (cos t + sin t)/2 + (1 2t) sin t (2 + t) cos t + c2
2 cos t sin t
cos t
+ c2
2 sin t + cos t
sin t
.
A dierent approach is to use 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
=