Review Problems for Exam 2, Math 308-503, Fall 2014
Notes: The exam will take place in class on Friday, November 21. No calculators, notes,
textbooks or other aids will be allowed. You will have the entire class period to work on the
exam, but exams will be collected promptly at the end of the period, so please be sure to arrive
on time to ensure that you have sufficient time to complete the exam. There will be no makeup
exam unless it is for a pre-excused absence or a legitimate emergency. The exam will officially
cover Sections 6.1-6.6, 7.1-7.6, and 7.8 (that is, homework assignments 7-12).
As on the first exam, you will be expected to know integrals, derivatives, and basic properties of
standard functions: polynomials and powers, exponential (ex ), logarithmic (ln x), and basic trig
(sin x, cos x, tan x). In addition, you should be able to use basic differentiation rules (product,
quotient, chain rule) and “undo” these rules in order to calculate integrals and be able to use
partial fractions in order to solve problems involving Laplace Transforms. A copy of Table 6.2.1
(Elementary Laplace Transforms) will be included in the exam (possibly with a line or two
deleted if I decide to ask you to calculate a Laplace transform from the definition), so you need
not memorize the chart.
1. Find the Laplace transform of t2 e2t . Remember to specify the values of s for which your
answer is valid.
RA
R∞
Solution: L{t2 e2t } = 0 e−st t2 e2t dt = limA→∞ 0 t2 e(2−s)t dt. To integrate by parts, let
R
1 (2−s)t
1 2 (2−s)t
u = t2 , du = 2tdt, dv = e(2−s)t dt, and v = 2−s
e
. Then t2 e(2−s)t dt = 2−s
t e
−
R
2
1 2 (2−s)t
2t
2
(2−s)t
(2−s)t
(2−s)t
te
dt = 2−s t e
− (2−s)2 e
+ (2−s)3 e
. (We have omitted the details
2−s
1
of the last integration by parts.) Thus L{t2 e2t } = limA→∞ [ (2−s
)A2 e(2−s)A −
2
e(2−s)A
(2−s)3
−
2
(2−s)3
=
2
.
(s−2)3
2A
e(2−s)t
(2−s)2
+
The answer is valid for 2 − s < 0, or s > 2.
2. Let h(t) = 1 if π ≤ t < 2π, and h(t) = cos(2t) otherwise. Find an expression for h(t) (t ≥ 0)
that involves step functions.
Solution: h(t) = (1 − uπ (t)) cos(2t) + (uπ (t) − u2π (t)) + u2π (t) cos(2t). The logic here is to split
the real line into three parts (t < π, π ≤ t < 2π, and t ≥ 2π), and multiply the definition of h
on each interval by a step function which switches “on” at the left endpoint (if it is finite) and
“off” at the right endpoint.
3. Let h(t) = t if 0 ≤ t < π, and h(t) = t + sin(t − π) for t ≥ π. Find the Laplace transform of
h.
Solution: First we express h(t) in terms of step functions. h(t) = t + uπ (t) sin(t − π). Thus
L{h(t)} = L{t} + L{uπ (t) sin(t − π)} = s12 + e−πs L{sin t} = s12 + e−πs s21+1 .
s
.
4. Find the inverse Laplace transform of s2 −6s+10
Solution: We first complete the square to find that s2 − 6s + 10 = (s − 3)2 + 1. Because
we have a factor of s − 3 in the denominator, we also want one in the numerator, and so we
s−3
3
s
= (s−3)
L{cos t} = s2s+1 , and L{ect f (t)} = F (t − c) where
write: s2 −6s+10
2 +1 + (s−3)2 +1 .
s−3
3
3t cos t, and similarly L−1 {
L{f (t)} = F (s), so L−1 { (s−3)
} = 3e3t sin t. Thus
2 +1 } = e
(s−3)2 +1
s
L−1 { s2 −6s+10
} = e3t (cos t + 3 sin t).
5. Use Laplace transforms to solve y 00 + 3y 0 + 2y = e2t , y(0) = 0, y 0 (0) = 1.
1
Solution: L{y 00 + 3y 0 + 2y} = [s2 L{y} − sy(0) − y 0 (0)] + 3[sL{y} − y(0)] + 2L{y} = (s2 +
1+(s−2)
1
1
1
3s + 2)L{y} − 1, and L{e2t } = s−2
. Thus L{y} = (s2 +3s+2)(s−2)
+ s2 +3s+2
= (s+2)(s+1)(s−2)
=
s−1
(s+2)(s+1)(s−2) . Partial fractions gives s−1 = a(s+1)(s−2)+b(s+2)(s−2)+c(s+2)(s+1), where
s−1
a
b
c
2
2
2
(s+2)(s+1)(s−1) = s+2 + s+1 + s−2 . Expanding yields s−1 = a(s −s−2)+b(s −4)+c(s +3s+2),
so a + b + c = 0, −a + 3c = 1, and −2a − 4b + 2c = −1. Adding four times the first to the
third equation gives 2a + 6c = −1, which when added with the twice the second gives 12c = 1
1
or c = 12
. Thus a = 3c − 1 = − 34 , and b = −a − c = 23 .
1
1
Thus y = − 34 L−1 { s+2
} + 23 L−1 { s+1
}+
1 −1
1
12 L { s−2 }
= − 34 e−2t + 32 e−t +
1 2t
12 e .
6. Use Laplace transforms to solve y 00 + 4y = u2 (t)sin(t − 2), y(0) = 0, y 0 (0) = 0.
1
1
−2s
. In order to do a partial
Solution: We have (s2 + 4)L{u} = e−2s 1+s
2 , or L{u} = e
(s2 +1)(s2 +4)
1
A
B
2
2
fractions decomposition we write (s2 +1)(s
2 +4) = s2 +1 + s2 +4 , or A(s + 4) + B(s + 1) = 1. Thus
A + B = 0 and 4A + B = 1, or B = −A and 3A = 1. Thus A = 1/3 and B = −1/3, and
L{u} = e−2s
1 1
1 1
1 2
1 1
− e−2s 2
= e−2s 2
− e−2s 2
.
2
3s +1
3s +4
3s +1
6s +4
Table 6.2.1 then yields u(t) = 13 u2 (t) sin(t − 2) − 61 u2 (t) sin 2(t − 2).
7. Use Laplace transforms to solve y 00 + 5y 0 + 6y = δ(t − π), y(0) = 1, y(0) = 0.
Solution: We have s2 L{y} − sy(0) − y 0 (0) + 5[sL{y} − y(0)] + 6L{y} = e−πs , or
1
s+5
1
(s + 3) + 2
+
= e−πs
+
s2 + 5s + 6 s2 + 5s + 6
(s + 3)(s + 2) (s + 3)(s + 2)
1
1
2
= e−πs
+
+
.
(s + 3)(s + 2) (s + 2) (s + 3)(s + 2)
L{y} = e−πs
Partial fractions gives A(s + 3) + B(s + 2) = 1, or A + B = 0 and 3A + 2B = 1. Thus A = 1
1
1
1
and B = −1, or (s+3)(s+2)
= s+2
− s+3
. Thus
L{y} = e−πs
1
1
3
2
− e−πs
+
−
.
s+2
s+3 s+2 s+3
Taking the inverse transform yields y(t) = uπ (t)e−2(t−π) − uπ (t)e−3(t−π) + 3e−2t − 2e−3t .
8. Use Laplace transforms to solve y 00 + 7y 0 + 12y = g(t), y(0) = 2, y 0 (0) = −1. Express your
solution as a convolution integral.
Solution: We have s2 L{y} − sy(0) − y 0 (0) + 7[sL{y} − y(0)] + 12L{y} = L{g(t)} := G(s). Thus
L{y} = G(s)
s2
2s + 13
1
2(s + 4) + 5
1
+ 2
= G(s)
+
.
+ 7s + 12 s + 7s + 12
(s + 3)(s + 4) (s + 3)(s + 4)
Partial fractions yields A(s + 4) + B(s + 3) = 1, or A + B = 0, 4A + 3B = 1, or A = 1, B = −1.
1
1
1
Thus (s+4)(s+3)
= s+3
− s+4
, and
L{y} = G(s)
Finally, y(t) =
Rt
0
e−3(t−τ ) g(τ )dτ −
1
1
7
5
− G(s)
+
−
.
s+3
s+4 s+3 s+4
Rt
0
e−4(t−τ ) g(τ )dτ + 7e−3t − 5e−4t .
2
9. Transform the following second-order initial value problem into an initial value problem for
a first-order system:
u00 + sin(u)u0 + t2 u3 = sin t, u(0) = 1, u0 (0) = 3.
Solution: Let x1 (t) = u(t) and x2 (t) = u0 (t) = x01 (t). Then we have x01 = x2 , x02 = − sin t −
sin x1 + t2 x31 with initial conditions x1 (0) = 1, x2 (0) = 3.
−1 5
−1
10. Find A for A =
. Solution:
2 3
1
1
3 −5
−3 5
−1
A =
=
.
2 1
−3 − 10 −2 −1
13
11. a) Find the eigenvalues λ1 and λ2 and the corresponding eigenvectors ~x(1) and ~x(2) of the
matrix
−3 43
.
A=
−5 1
= λ2 + 2λ + 43 = 0, which has roots
Solution: The characteristic equation is (λ + 3)(λ − 1) + 15
4
3
1
(2)
(1)
, respectively.
and ~x =
λ1 = −3/2, λ2 = −1/2. The eigenvectors are ~x =
10
2
(2)
b) Show that the eigenvectors ~x(1)
and ~x from part a) are linearly independent. Solution: We
1 3
= 4 6= 0. Thus ~x(1) and ~x(2) are linearly independent.
compute det ~x(1) ~x(2) = det
2 10
1
t3
(2)
.
and ~x (t) =
12. Let
=
2
t
t
a) Compute the Wronskian of ~x(1) and ~x(2) . Solution: W (~x(1) , ~x(2) = t4 − t2 .
~x(1) (t)
b) In what intervals are ~x(1) and ~x(2) linearly independent? Solution: ~x(1) and ~x(2) are linearly
independent when t 6= 0, 1, −1.
p11 (t) p12 (t)
c) Let P (t) =
, and assume that ~x(1) and ~x(2) are both solutions of the system
p21 (t) p22 (t)
~x0 (t) = P (t)~x(t). What conclusions can you draw about the matrix P (t)? Solution: P (t) must
be discontinuous at t = 0, 1, −1 because the Wronskian of two solutions to a system ~x0 = P ~x
with continuous coefficients P must be either always 0 or never 0.
1 −2
13. The eigenvalues of the matrix A =
are λ1 = −2 and λ2 = −1, and the corre3 −4
2
1
sponding eigenvectors are ~x(1) =
and ~x(2) =
.
3
1
2
0
−2t
a) Find the general solution to the system ~x (t) = A~x. Solution: ~x(t) = c1 e
+
3
1
c2 e−t
.
1
3
b) Draw a phase portrait for the above system and characterize the equilibrium point
0
0
in
terms of stability and type (node, spiral, etc.)
Figure 1: Phase portrait for Problem 8b.
~x0 (t)
2
3
+
. Solution: We have ~x(0) = c1
3
4
c) Solve the initial value problem
= A~x, ~x(0) =
3
1
. Thus 2c1 + c2 = 3 and 3c1 + c2 = 4, or c1 = c2 = 1. Thus the solution to the
=
c2
4
1
1
2
−t
−2t
.
+e
initial value problem is ~x(t) = e
1
3
5 −3
has a single eigenvalue λ = 2 with algebraic multiplicity 1.
3 −1
a) Find all linearly independent eigenvectors of A.
Solution: We write Aξ = 2ξ, or 5ξ1 −3ξ2 = 2ξ1 , 3ξ1 −ξ2 = 2ξ2 . Both equations
give 3ξ1 −3ξ2 = 0,
1
.
or ξ1 = ξ2 . Thus there is only one linearly independent eigenvector
1
14. The matrix A =
b) Find the general solution to the system ~x0 (t) = A~x.
3 −3
,
3 −3
so we
have3η1 − 3η2 = 1 (from both lines rows of the equation). Thus setting η2 = 0 gives
1/3
η=
. The general solutions is then
0
1
1
1/3
2t
2t
2t
~x(t) = c1 e
+ c2 te
+e
.
1
1
0
Solution: We first find the generalized eigenvector by solving (A−λI)η = ξ. A−λI =
c) Characterize the equilibrium point
0
0
in terms of stability and type (node, spiral, etc.)
Solution: Unstable, improper node.
−3 −7
15. Let A =
.
1 −1
4
a) Find the eigenvalues and associated eigenvectors of A.
Solution: The eigenvalue
equation is (−3
− λ)(−1 − λ) + 7 = 0, or λ2 + 4λ + 10 = 0. Thus
√
√
√
the eigenvalues are −4± 216−40 = −2 ± i 224 = −2 ± i 6. Solving the eigenvector equation with
√
eigenvalue −1 + ı 6 (details omitted) gives
"
# "
#
1
0
1
√
√
ξ = −1−i 6 =
+i
.
− 17
− 76
7
b) Find the general solution to the system ~x0 (t) = A~x.
Solution: A solution to the equation is
#
#
"
"
√
√
√
0
0
1
1
√
√
~x(t) = e(−1+i 6)t
= e−t (cos 6t + i sin 6t)
+i
+i
− 17
− 17
− 76
− 76
"
"
√ #
√ #
cos
6t
6t
sin
−t
−t
√
√
√
√
√
√
=e
+ ie
.
6
6
1
1
− 7 cos 6t + 7 sin 6t
− 7 sin 6t − 7 cos 6t
Taking the real and complex parts as a fundamental solution set as we did in class, we find that
the general solution is
"
"
√ #
√ #
sin
6t
6t
cos
−t
−t
√
√
√
√
√
√
+ c2 e
.
~x(t) = c1 e
6
6
1
1
− 7 cos 6t + 7 sin 6t
− 7 sin 6t − 7 cos 6t
0
in terms of stability and type (node, spiral, etc.)
0
Answer: The origin is an asymptotically stable spiral point.
c) Characterize the equilibrium point
5