MATH 215/255
Fall 2014
Assignment 4
§2.2, §2.4, §2.5
Solutions to selected exercises can be found in [Lebl], starting from page 303.
• 2.2.9: Solve y 00 + 9y 0 = 0 for y(0) = 1, y 0 (0) = 1.
Answer.
The characteristic equation of y 00 + 9y 0 = 0 is:
r2 + 9r = 0.
Solve r2 + 9r = 0, we have
r1 = 0,
and r2 = −9.
So the general solution to y 00 + 9y 0 = 0 is:
y(x) = C1 + C2 e−9x .
Then
y 0 (x) = −9C2 e−9x .
Since y(0) = 1 and y 0 (0) = 1, then
1 = C1 + C2 ,
So
C1 =
10
,
9
and
− 9C2 = 1.
1
and C2 = − .
9
So the solution to y 00 + 9y 0 = 0 for y(0) = 1, y 0 (0) = 1 is:
y(x) =
10 e−9x
−
.
9
9
• 2.2.102: Find the general solution to y 00 − 6y 0 + 9y = 0.
Answer.
The characteristic equation of y 00 − 6y 0 + 9y = 0 is:
r2 − 6r + 9 = 0.
Solve r2 − 6r + 9 = 0, we have
r1 = r2 = 3.
So the general solution to y 00 − 6y 0 + 9y = 0 is:
y(x) = C1 e3x + C2 xe3x .
• 2.2.105: Find the solution to z 00 (t) = −2z 0 (t) − 2z(t), z(0) = 2, z 0 (0) = −2.
Answer.
Rewrite the equation:
z 00 + 2z 0 + 2z = 0.
The characteristic equation of z 00 + 2z 0 + 2z = 0 is:
r2 + 2r + 2 = 0.
Solve r2 + 2r + 2 = 0, we have
r1 = −1 + i,
and r2 = −1 = i.
So the general solution to z 00 + 2z 0 + 2z = 0 is:
z(t) = C1 e−t cos(t) + C2 e−t sin(t).
Then
z 0 (t) = −C1 e−t cos(t) − C1 e−t sin(t) − C2 e−t sin(t) + C2 e−t cos(t).
Since z(0) = 2 and z 0 (0) = −2, then
2 = C1 + 0,
and
− 2 = −C1 + C2 .
So
C1 = 2,
and C2 = 0.
Therefore, the solution to z 00 (t) = −2z 0 (t) − 2z(t), z(0) = 2, z 0 (0) = −2 is:
z(t) = 2e−t cos(t) .
• 2.4.2: Consider a mass and spring system with a mass m = 2, spring constant k = 3,
and damping constant c = 1. a) Set up and find the general solution of the system.
b) Is the system underdamped, overdamped or critically damped? c) If the system is
not critically damped, find a c that makes the system critically damped.
Answer.
a). Let x(t) be the displacement of the mass at time t. By the assumption,
our differential equation is:
mx00 + cx0 + kx = 2x00 + x0 + 3x = 0.
The characteristic equation of 2x00 + x0 + 3x = 0 is:
2r2 + r + 3 = 0.
Solving 2r2 + r + 3 = 0, we have
√
√
−1 + 1 − 4 · 2 · 3
−1 + i 23
r1 =
=
,
4
4
2
√
−1 − i 23
and r2 =
.
4
So the general solution to 2x00 + x0 + 3x = 0 is:
x(t) = C1 e
− 4t
cos
√ !
t
23
t + C2 e− 4 sin
4
√
23
t
4
!
.
b). Since 1 − 4 · 2 · 3 = −23 < 0, the system is underdamped .
c). In order to make the system to be critically damped, that is,
c2 − 4 · 2 · 3 = 0.
Then
c=
√
√
24 = 2 6 .
• 2.4.6: Suppose you wish to measure the friction a mass of 0.1 kg experiences as it
slides along a floor (you wish to find c). You have a spring with spring constant k = 5
N/m. You take the spring, you attach it to the mass and fix it to a wall. Then you pull
on the spring and let the mass go. You find that the mass oscillates with frequency 1
Hz. What is the friction?
Answer.
It’s easy to see that our differential equation is:
0.1x00 + cx0 + 5x = 0.
Rewrite the equation as
x00 + 10cx0 + 50x = 0.
By the assumption, we know that our motion is underdamped with frequency 1 Hz,
which implies that (10c)2 − 4 · 50 < 0 and
p
√
4 · 50 − (10c)2
200 − 100c2 p
1 · 2π =
=
= 50 − 25C 2 .
2
2
r
So we get
25c2
= 50 −
4π 2 ,
that is, c =
2−
4π 2
≈ 0.648. Hence the friction is
25
0.648 .
• 2.4.101: A mass of 2 kilograms is on a spring with spring constant k newtons per
meter with no damping. Suppose the system is at rest and at time t = 0 the mass is
kicked and starts traveling at 2 meters per second. How large does k have to be to so
that the mass does not go further than 3 meters from the rest position?
Answer.
The only force acting on the spring is the spring compression force so the
differential equation is mx00 = −kx where m = 2 kg is the mass and k is the spring
constant. The spring is initially uncompressed so x(0) = 0, and it was kicked at a
velocity of 2 m/s so x0 (0) = 2. Therefore the initial value problem we have to solve is
x00 (t) +
k
x(t) = 0,
m
x(0) = 0,
3
x0 (0) = 2.
The general solution is
r
x(t) = A cos
r
k
k
x + B sin
x .
m
m
The initial conditions give
r
0
0 = x(0) = A,
2 = x (0) =
k
B,
m
p
so A = 0 and B = 2 m/k. Therefore the solution is
r
r
m
k
x(t) = 2
sin
x .
k
m
A maximum displacement of 3 m means that
r
4m
8
m
≤ 3 =⇒ k ≥
= .
|x(t)| ≤ 3 ⇐⇒ 2
k
9
9
Therefore to ensure the mass doesn’t go farther than 3 m from the rest position we
must have
8
k ≥ N/m .
9
• 2.4.103: A 5,000 kg railcar hits a bumper (a spring) at 1 m/s, and the spring compresses by 0.1 m. Assume no damping. a) Find k. b) Find out how far does the
spring compress when a 10,000 kg railcar hits the spring at the same speed. c) If the
spring would break if it compresses further than 0.3 m, what is the maximum mass
of a railcar that can hit it at 1 m/s? d) What is the maximum mass of a railcar that
can hit the spring without breaking at 2 m/s?
Answer.
a) Let M = 5000 kg be the mass of the railcar. Since the only force acting on the railcar
is the spring compression force, this means that the differential equation is M x00 = −kx
or equivalently x00 +(k/M )x = 0. Now the spring is initially uncompressed so x(0) = 0
and since the railcar is hitting the spring at 1 m/s this means x0 (0) = −1. Therefore
we are looking at the initial value problem
x00 (t) +
k
x(t) = 0,
M
x(0) = 0, x0 (0) = −1.
The general solution is
r
r
k
k
x(t) = A cos
t + B sin
t .
M
M
p
The initial conditions give A = 0 and B = −1 M/k so the solution is
r
r
M
k
x(t) = −
sin
t .
k
M
4
Since the spring compresses to 0.1 m this means that the maximum of |x(t)| is 0.1.
But note that the maximum of the sine term is 1 so this means that
r
M
= 0.1 =⇒ k = 5000/(0.1)2 = 500, 000.
k
Therefore the spring constant is
k = 500, 000 N/m .
b) If we instead have M = 10, 000 kg, then the maximum compression is
r
r
M
10000
1
=
= √ .
k
500000
5 2
Therefore the maximum compression is
1
√ = 0.14142 m .
5 2
c) In order to have a maximum compression of 0.3 m we need to constrain M according
to
r
M
≤ 0.3 =⇒ M ≤ (0.3)2 (500000) = 45000.
k
Therefore to keep the bumper from breaking the mass of the railcar can be at most
45, 000 kg .
d) If instead of going at 1 m/s the railcar is going at 2 m/s, the initial condition
changes to x0 (0) = −2 so the solution becomes
r
r
M
k
x(t) = −2
sin
t .
k
M
Therefore to keep the spring from breaking we have to constrain M according to
r
M
≤ 0.3 =⇒ M ≤ (0.3)2 (500000)/4 = 11250.
2
k
So to keep the bumper from breaking the mass of the railcar travelling at 2 m/s can
be at most
11, 250 kg .
• 2.5.2: Find a particular solution of y 00 − y 0 − 6y = e2x .
Answer.
The characteristic equation of y 00 − y 0 − 6y = 0 is:
r2 − r − 6 = 0.
5
Solve r2 − r − 6 = 0, then
r1 = 3,
and r2 = −2.
Let y(x) = Ae2x be a particular solution to y 00 − y 0 − 6y = e2x , then
y 0 (x) = 2Ae2x ,
and y 00 (x) = 4Ae2x .
So we have
4Ae2x − 2Ae2x − 6Ae2x = e2x .
That is,
−4Ae2x = e2x .
So
1
A=− .
4
Therefore, a particular solution of y 00 − y 0 − 6y = e2x is:
y(x) = −
e2x
.
4
• 2.5.3: Find a particular solution of y 00 − 4y 0 + 4y = e2x .
Answer.
The characteristic equation of y 00 − 4y 0 + 4y = 0 is:
r2 − 4r + 4 = 0.
Solve r2 − 4r + 4 = 0, then
r1 = r2 = 2.
Let y(x) = Ax2 e2x be a particular solution to y 00 − 4y 0 + 4y = e2x , then
y 0 (x) = 2Axe2x + 2Ax2 e2x ,
and y 00 (x) = 2Ae2x + 4Axe2x + 4Axe2x + 4Ax2 e2x .
So we have
2Ae2x + 4Axe2x + 4Axe2x + 4Ax2 e2x − 4 2Axe2x + 2Ax2 e2x + 4Ax2 e2x = e2x .
That is,
2Ae2x = e2x .
So
1
A= .
2
Therefore, a particular solution of y 00 − 4y 0 + 4 = e2x is:
y(x) =
6
x2 e2x
.
2
• 2.5.102: a) Find a particular solution to y 00 + 2y = ex + x3 . b) Find the general
solution.
Answer.
a). The characteristic equation to y 00 + 2y = 0 is:
r2 + 2 = 0.
Solve r2 + 2 = 0, then
r1 =
√
2i,
√
and r2 = − 2i.
Let y(x) = Aex + Bx3 + Cx2 + Dx + E be particular solution to y 00 + 2y = ex + x3 ,
then
y 0 (x) = Aex + 3Bx2 + 2Cx + D, and y 00 (x) = Aex + 6Bx + 2C.
Then
Aex + 6Bx + 2C + 2 Aex + Bx3 + Cx2 + Dx + E = ex + x3 .
That is, we have
(3A − 1)ex + (2B − 1)x3 + 2Cx2 + (6B + 2D)x + 2C + 2E = 0.
So we get
3A − 1 = 0,
Then
2B − 1 = 0,
1
A= ,
3
2C = 0,
1
B= ,
2
6B + 2D = 0,
C = 0,
3
D=− ,
2
2C + 2E = 0.
E = 0.
That is, a particular solution to y 00 + 2y = ex + x3 is:
y(x) =
ex x3 3
+
− x.
3
2
2
b). By the computations in part a), we know that the general solution to y 00 + 2y = 0
is:
√
√
y(x) = C1 cos( 2x) + C2 sin( 2x).
So the general solution to y 00 + 2y = ex + x3 is:
√
√
ex x3 3
y(x) = C1 cos( 2x) + C2 sin( 2x) +
+
− x.
3
2
2
7