MATH 215/255
Fall 2014
Assignment 7
§6.3, §6.4, §3.1, §3.3
Solutions to selected exercises can be found in [Lebl], starting from page 303.
• 6.3.3: Find the solution to
mx00 + cx0 + kx = f (t), x(0) = 0, x0 (0) = 0
for an arbitrary function f (t), where m > 0, c > 0, k > 0, and c2 − 4km > 0 (system
is overdamped). Write the solution as a definite integral.
Answer.
Applying the Laplace transform to the equation and denoting L{x(t)} =
X(s) and L{f (t)} = F (s), we obtain
m[s2 X(s) − sx(0) − x0 (0)] + c[sX(s) − x(0)] + kX(s) = F (s).
Under the imposed initial condition, it becomes
ms2 X(s) + csX(s) + kX(s) = F (s).
Thus,
X(s) =
1
F (s).
ms2 + cs + k
Therefore,
−1
x(t) = L
1
2
ms + cs + k
∗f
(t).
Let us write
ms2 + cs + k = m[(s + p)2 − ω12 ]
where p =
c
2m ,
ω12 = p2 −
k
m
=
c2 −4km
4m2
> 0. Note that
1
sinh(ω1 t)
−1
L
=
,
2
2
ω1
s − ω1
and hence,
−1
L
1
(s + p)2 − ω12
= e−pt
sinh(ω1 t)
.
ω1
We thus obtain the solution
1
x(t) =
m
Z
0
t
e−pτ
sinh(ω1 τ )
f (t − τ )dτ.
ω1
• 6.3.4: Find the solution to
mx00 + cx0 + kx = f (t), x(0) = 0, x0 (0) = 0
for an arbitrary function f (t), where m > 0, c > 0, k > 0, and c2 − 4km < 0 (system
is underdamped). Write the solution as a definite integral.
Answer.
Following the process for Exercise 6.3.3, we have
X(s) =
1
F (s).
ms2 + cs + k
Therefore,
−1
x(t) = L
1
2
ms + cs + k
∗f
(t).
Let us write
ms2 + cs + k = m[(s + p)2 + ω12 ]
where p =
c
2m ,
ω12 =
k
m
− p2 =
4km−c2
4m2
L−1
> 0. Note that
1
s2 + ω12
=
sin(ω1 t)
,
ω1
and hence,
−1
L
1
(s + p)2 + ω12
= e−pt
sin(ω1 t)
.
ω1
We thus obtain the solution
x(t) =
1
m
Z
t
e−pτ
0
sin(ω1 τ )
f (t − τ )dτ.
ω1
• 6.3.104: Solve x000 + x0 = f (t), x(0) = 0, x0 (0) = 0, x00 (0) = 0 using convolution.
Write the solution as a definite integral.
Answer.
Applying the Laplace transform to the equation, denoting L{x(t)} = X(s)
and L{f (t)} = F (s), and using the imposed initial condition, we obtain
s3 X(s) + sX(s) = F (s).
Thus,
1
F (s).
s(s2 + 1)
1
−1
x(t) = L
∗ f (t).
s(s2 + 1)
X(s) =
Write
1
1
s
= − 2
,
+ 1)
s s +1
s(s2
and recall
s
1
L−1 { } = 1, L−1 { 2
} = cos(t).
s
s +1
We thus obtain the solution
Z t
x(t) =
[1 − cos(τ )]f (t − τ )dτ.
0
2
• 6.4.1: Solve (find the impulse response) x00 + x0 + x = δ(t), x(0) = 0, x0 (0) = 0.
Answer.
Denote L{x(t)} = X(s). Applying the Laplace transform to the equation
and use the initial condition, we obtain
s2 X(s) + sX(s) + X(s) = 1.
Thus,
X(s) =
=
=
1
+s+1
1
1 2
(s + 2 ) +
1
s2
3
4
√
3 2
2 )
3
2
√
.
3 2
1 2
)
+
(
2
2 )
(s + 21 )2 + (
√
=
Therefore, the solution is
• 6.4.6: Compute L−1 { s
2
√
3 (s +
√
3t
2 −t
).
x(t) = √ e 2 sin(
2
3
2 +s+1
s2
}.
Answer.
L−1 {
s2 + s + 1
1
1
} = L−1 {1 + + 2 }
2
s
s s
1
1
−1
= L {1} + L−1 { } + L−1 { 2 }
s
s
= δ(t) + 1 + t.
• 6.4.103: Suppose that L(x) = δ(t), x(0) = 0, x0 (0) = 0, has the solution x(t) = cos(t)
for t > 0. Find (in closed form) the solution to L(x) = sin(t), x(0) = 0, x0 (0) = 0 for
t > 0.
Answer.
We know that if x(t) satisfies L(x) = δ(t), then
(L(x) ∗ f )(t) = (δ ∗ f )(t)
⇒ L(x ∗ f )(t) = (δ ∗ f )(t) = f (t).
Therefore, x ∗ f is the solution to L(x) = f (t) = sin(t), and
Z t
Z t
(x ∗ f ) =
x(τ )f (t − τ )dτ =
cos(τ ) sin(t − τ )dτ
0
0
Z t
=
cos(τ )[sin(t) cos(τ ) − sin(τ ) cos(t)]dτ
0
Z t
Z t
2
= sin(t)
cos (τ )dτ − cos(t) [cos(τ ) sin(τ )]dτ
0
=
0
t sin(t) sin(t) sin(2t) cos(t) sin2 (t)
+
−
.
2
4
2
3
• 3.1.2: Find the general solution of x01 = x2 − x1 + t, x02 = x2 .
Answer.
Method 1: The second equation involves only x2 and can be solved first. We obtain
x2 (t) = c1 et .
Substitute this x2 into the first equation, we obtain x01 = c1 et − x1 + t or
x01 + x1 = c1 et + t.
This first-order equation has an integrating factor r(t) = et . We multiply r(t) to the
equation to get
et x01 + et x1 = c1 et et + tet ,
i.e.
d(et x1 (t))
= c1 e2t + tet .
dt
We integrate to obtain
et x1 (t) =
Thus,
x1 (t) =
c1 2t
e + tet − et + c2 .
2
c1 t
e + t − 1 + c2 e−t .
2
Therefore the general solution is
x1 (t) =
c1 t
e + t − 1 + c2 e−t , x2 (t) = c1 et .
2
Method 2: Taking the derivative of the first equation yields
x001 = x02 − x01 + 1.
By substitution, we have
x001 = x2 − x01 + 1
= x1 − t + 1
Thus, x1 (t) is the solution to the second-order equation
x001 − x1 = 1 − t.
(1)
The general solution to the associated homogeneous equation
x001 − x1 = 0
is x1 (t) = c1 et + c2 e−t . A particular solution to (1) is x1 (t) = t − 1. Therefore, the
general solution to (1) is
x1 (t) = c1 et + c2 e−t + t − 1.
4
From x2 = x01 + x1 − t, we compute
x2 (t) = c1 et − c2 e−t + 1 + (c1 et + c2 e−t + t − 1) − t
= 2c1 et .
The general solution is
x1 (t) = c1 et + c2 e−t + t − 1, x2 (t) = 2c1 et .
• 3.1.4: Write ay 00 + by 0 + cy = f (x) as a first order system of ODEs.
Answer.
Assume a 6= 0. Let z = y 0 . Then
z 0 = y 00
f (x) b 0 c
=
− y − y
a
a
a
f (x) b
c
=
− z − y.
a
a
a
Thus, we have the first-order system
0
y =z
b
c
z 0 = f (x)
a − a z − a y.
• 3.1.102: Solve y 0 = 2x, x0 = x + y, x(0) = 1, y(0) = 3.
Answer.
Differentiating the first equation and using the second equation lead to
y 00 = 2x0 = 2x + 2y = y 0 + 2y.
That is,
y 00 − y 0 − 2y = 0.
The characteristic equation for this homogeneous second-order ODE is
r2 − r − 2 = 0,
and the characteristic roots are r1 = 2, r2 = −1. Therefore, the general solution is
y(t) = c1 e2t + c2 e−t .
From y 0 = 2x, we obtain
x(t) = c1 e2t −
c2 −t
e .
2
As x(0) = 1, y(0) = 3, we have
c1 −
c2
= 1, c1 + c2 = 3.
2
Thus, c1 = 5/3, c2 = 4/3. The solution is therefore
5
2
5
4
x(t) = e2t − e−t , y(t) = e2t + e−t .
3
3
3
3
5
• 3.3.1: Write the system x01 = 2x1 − 3tx2 + sin t, x02 = et x1 + 3x2 + cos t in the vector
form ~x0 = P (t)~x + f~(t), where P (t) is a 2 × 2 matrix.
Answer.
x01
x02
=
2 −3t
et
3
x1
x2
+
sin t
cos t
.
• 3.3.2: Consider linear system
0
~x =
1 3
3 1
~x.
a) Verify that the system has the following two solutions
1
1
4t
−2t
.
, ~x(t) = e
~x(t) = e
1
−1
b) Show that they are linearly independent and write down the general solution.
c) Write down the general solution in the form x1 (t) =?, x2 (t) = ? (i.e. write down a
formula for each element of the solution).
Answer.
a) Let
~x(t) = e
−2t
Then
0
−2t
1
−1
.
1
−1
~x (t) = −2e
.
On the other hand,
−2
1
1 3
1 3
1
1 3
−2t
−2t
−2t
.
=e
~x =
e
=e
2
−1
3 1
3 1
3 1
−1
Therefore,
~x0 =
1 3
3 1
1
1
~x.
Next, let
~x(t) = e
4t
Then
~x0 (t) = 4e4t
1
1
.
.
On the other hand,
1 3
1 3
1
1 3
1
4
4t
4t
4t
~x =
e
=e
=e
.
1
3 1
1
4
3 1
3 1
Therefore,
0
~x =
6
1 3
3 1
~x.
b) Consider
c1 e−2t
1
−1
+ c2 e4t
1
1
= 0,
i.e.,
c1 e−2t + c2 e4t
−c1 e−2t + c2 e4t
= 0.
In other words, c1 e−2t + c2 e4t = 0, −c1 e−2t + c2 e4t = 0 for all t. If we let t = 0, then
c1 + c2 = 0, −c1 + c2 = 0. Thus c2 = c1 = 0. Therefore they are linearly independent.
The general solution of this equation is
1
1
4t
−2t
.
+ C2 e
~x(t) = C1 e
1
−1
c)
x1 (t) = C1 e−2t + C2 e4t , x2 (t) = −C1 e−2t + C2 e4t .
• 3.3.104: a) Write x01 = 2tx2 , x02 = 2tx2 in matrix notation. b) Solve and write the
solution in matrix notation.
Answer.
a)
x01
x02
=
0 2t
0 2t
x1
x2
.
b) The 2nd equation involves only x2 , and can be solved first. First, x2 (t) ≡ 0 is a
solution. Next, we consider
x02
= 2t
x2
⇒ ln |x2 (t)| = t2 + c1
2
⇒ x2 (t) = cet ,
where c = ec1 or −ec1 . We may now express the general solution to x02 = 2tx2 by
2
x2 (t) = C2 et and allow C2 to be any real number. Substituting this x2 (t) to the first
equation, we have
2
x01 = 2C2 tet .
We thus find
2
x1 (t) = C2 et + C3 .
Let us write the solution in matrix form
#
"
2
C2 e t + C3
~x(t) =
.
2
C2 e t
7