MA 2326 Differential Equations
Instructor: Petronela Radu
Wednesday, March 27, 2013
Solutions to assignment 4
1. (5 points)
t2
2t
t
1
(a) Show that Φ(t) =
is a fundamental matrix for the system Y0 = A(t)Y where A(t) =
0
1
on any interval I not including the origin.
−2/t2 2/t
(b) Does the fact that detΦ(0) = 0 contradict Abel’s Theorem?
Solution: To show that Φ is a fundamental solution we need to check that every column is a solution
of the system and that the columns are linearly independent, i.e. detΦ(t) 6= 0 on I.
2 0 2 0 t
0
1
t
t
0
1
t
Indeed, it is easy to check that
=
and that
=
.
2t
−2/t2 2/t
2t
1
−2/t2 2/t
1
Also detΦ(t) = −t2 6= 0 for all t nonzero. Note that this does not contradict Abel’s theorem, as the
point zero does not belong to the interval I (the solution is not well defined at t = 0).
2 1
0
2. (10 points) Consider the system Y = AY + G(t) where A =
,
0 2
2t
y1 (t)
sin t
e
te2t
Y(t) =
, G(t) =
. Verify that Φ(t) =
is a fundamental matrix of
y2 (t)
cos t
0
e2t
1
Y0 = AY. Find that solution Y of the non homogeneous system for which Y(0) =
.
−1
2t e
Solution: As is in the previous problem it is a simple computation to show that both vectors
0
2t te
and
satisfy the homogeneous system. In addition, detΦ(t) = e4t 6= 0 for all t, hence Φ is a
e2t
fundamental matrix.
The variation of parameters formula gives us that the non homogenous solution that satisfies the given
initial condition can be written as
Y (t) = Yh (t) + Yp (t)
(1)
where Yh (t) is a solution to the homogeneous system that satisfies the IC; hence, we can take
Yh (t) = Φ(t)Y(0)
and
Z
Yp (t) = Φ(t)
(2)
t
Φ−1 (s)G(s)ds.
(3)
0
We compute Φ−1 (t) = e−4t
e2t
0
−te2t
e2t
= e−2t
Z t
Yp (t) = Φ(t)
0
1
0
−t
1
, hence
e−2s sin s − e−2s s cos s
e−2s cos s
ds.
To compute the integrals we use the formulas (5), (6) and (7) from the Appendix with a = 2, so after
simplifications we obtain
−2t
1
1
e (2t cos t − (2 + t) sin t)
−2 sin t + 2te2t
=
.
Yp (t) = Φ(t)
e−2t (sin t − 2 cos t) + 2
5
5 sin t − 2 cos t + 2e2t
Hence,
2t
Y (t) = e
1−t
−1
1
+
5
−2 sin t + 2te2t
sin t − 2 cos t + 2e2t
.
3. (10 points) Use the variation of constants formula to find the solution of the system
(
y10 = y1 + y2 + sin t
y20 = 2y1 + cos t
with y1 (0) = 1, y2 (0) = 1.
Solution: The system can be written in vectorial form as Y 0 = AY + B(t) where
y1
1 1
sin t
Y =
,A=
, B(t) =
y2
2 0
cos t
A fundamental matrix corresponding to the homogeneous system can be obtained by computing etA .
First, the eigenvalues are solutions of
λ(λ − 1) − 2 = 0
1
so λ1 = 2 and λ2 = −1. An eigenvector corresponding to λ1 is
and an eigenvector for λ2 can be
1
1
chosen as
.
−2
2
0
Therefore, A is diagonalizable and it can be written as A = P
P −1 , where
0 −1
1
1
1
3
P =
1
−2
and we compute
P −1 =
2
1
1
−1
We have that a fundamental matrix of the system is
2t
1
e
0
2e2t + e−t
−1
etA = P etD P −1 = P
P
=
−t
0 e
3 2e2t − 2e−t
e2t − e−t
2t
e + 2e−t
.
The solution of the system is given by formulas (1), (2), (3). Fist compute
Z Z t
etA t
(2e−2s + es ) sin s + (e−2s − es ) cos s
sin s
ds
Yp = etA
e−sA
ds =
cos s
(2e−2s − 2es ) sin s + (e−2s + 2es ) cos s
3 0
0
Using formulas (5) and (6) from the Appendix we obtain
−2t
etA
e (sin t − 45 cos t) + 54 − et cos t + 1
Yp =
e−2t (sin t − 45 cos t) + 45 + 2et cos t − 2
3
The homogeneous solution is given by
Yh = etA
so Y = Yp + Yh , the sum of the above solutions.
1
1
= e2t
1
1
4. (a) (5 points) Show that the equation
y 00 + py 0 + qy = 0
where p and q are constants is equivalent to the system Y0 = AY with A =
0
−q
1
−p
and
compute the eigenvalues λ1 , λ2 of A.
y1
. Then the given system is
y2
equivalent to the system Y0 = AY. The eigenvalues are solutions to the equation
Solution: Let y1 := y and y2 := y 0 , then denote by Y =
λ(λ + p) + q = 0
√
−p+
p2 −4q
−p−
√
p2 −4q
so, λ1 =
and λ2 =
.
2
2
(b) (7 points) Compute a fundamental matrix for the system in part (a) if λ1 6= λ2 and construct the
general solution in this case.
Solution: if the eigenvalues
are different, the matrix is diagonalizable. The first eigenvector
can
1
1
be chosen to be
, while a choice for the second eigenvector is taken to be
. Then
λ1 λ2
λ2 −1
1
1
λ1
0
1
. We obtain
A = P DP −1 with P =
,D=
, P −1 = λ2 −λ
1
−λ1 1
λ1 λ2
0 λ2
that
λt
λt
e 1
0
e 1
0
tA
tD −1
e = Pe P = P
=P
0
eλ2 t
0
eλ2 t
so a fundamental matrix is given by
1
λ2 eλ1 t − λ1 eλ2 t
eλ2 t − eλ1 t
tA
e =
.
λ2 − λ1 λ1 λ2 (eλ1 t − eλ2 t ) λ2 eλ2 t − λ1 eλ1 t
(c) (8 points) Compute a fundamental matrix for the system in part (a) if λ1 = λ2 and construct the
general solution in this case.
Solution:
If λ1 = λ2 = λ = −p/2 the eigenspace is generated in this case by the eigenvector
1
1
with a generalized eigenvector W satisfying (A − λI)W = V so given by
. The
λ
λ+1
matrix A can be written in Jordan form as
A = P JP −1
1
1
λ 1
λ + 1 −1
with P =
,J =
, P −1 =
. In order to compute etA note
λ λ+1
0
λ
−λ
λ
1 t
tJ
λt
first that we have that e = e
. Hence a fundamental matrix is given by
0 1
1 t
λ + 1 − λ(t + 1) −1 + λt + λ
etA = eλt P
P −1 = eλt
0 1
−λ2 t
λ2 (t + 1)
5. By converting to an equivalent system, find the general solution of the scalar equations
(a) (10 points) y 00 − y = f (t)
Solution: Let u := y and v := y 0 , then the equation is equivalent to the system:
0 u
0 1
u
0
=
+
.
v
1 0
v
f (t)
1
1
)
Note that the eigenvalues of the matrix A are λ1 = 1 (with a choice for the eigenvector
1
and λ2 = −1 with an eigenvector
. In fact, we fall into the case 4b) above so we get that
−1
a fundamental matrix is given by
1 et + e−t et − e−t
.
Φ(t) = etA =
2 et − e−t et + e−t
To obtain the non homogeneous solution we apply formulas (1), (2), (3), so we get that
Z
Y (t) = Φ(t)Y (0) + Φ(t)
t
Φ−1 (s)G(s)ds
(4)
0
where Φ−1 (t) = e−tA =
1
2
et + e−t
−(et − e−t )
t
−t
−(e − e )
et + e−t
and G(s) =
0
f (s)
.
(b) (10 points) y 00 + y = f (t).
Solution: The system is now
u
v
0
=
0
−1
1
0
u
v
+
0
f (t)
.
The eigenvalues are λ1 = i and λ2 = −i and by applying again the results of 4b) we obtain that
a fundamental matrix is given by
−1 −ieit − ie−it
e−it − eit
cos t
sin t
tA
=
.
Φ(t) = e =
eit − e−it
− sin t cos t
−ie−it − ieit
2i
The final solution is given by formula (4) with
Φ(t) = e−tA =
1
cos t − sin t
sin t
cos t
.
Appendix
Let
Z
I(t) :=
eat sin tdt
Then after a double integration by parts (write the trigonometric function as a derivative) we get
I(t) =
eat
(− cos t + a sin t) + C.
1 + a2
(5)
Similarly, we obtain
Z
J(t) :=
eat cos tdt =
eat
(sin t + a cos t) + C.
1 + a2
Using I and J we compute
Z
Z
Z
teat cos tdt = tJ 0 (t)dt = tJ(t) − J(t)dt = tJ(t) −
1
a
I(t) −
J(t) + C.
1 + a2
1 + a2
(6)
(7)