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MATH 215/255
Fall 2014
Assignment 4
В§2.2, В§2.4, В§2.5
Solutions to selected exercises can be found in [Lebl], starting from page 303.
• 2.2.9: Solve y + 9y = 0 for y(0) = 1, y (0) = 1.
Answer.
The characteristic equation of y + 9y = 0 is:
r2 + 9r = 0.
Solve r2 + 9r = 0, we have
r1 = 0,
and r2 = в€’9.
So the general solution to y + 9y = 0 is:
y(x) = C1 + C2 eв€’9x .
Then
y (x) = в€’9C2 eв€’9x .
Since y(0) = 1 and y (0) = 1, then
1 = C1 + C2 ,
So
C1 =
10
,
9
and
в€’ 9C2 = 1.
1
and C2 = в€’ .
9
So the solution to y + 9y = 0 for y(0) = 1, y (0) = 1 is:
y(x) =
10 eв€’9x
в€’
.
9
9
• 2.2.102: Find the general solution to y − 6y + 9y = 0.
Answer.
The characteristic equation of y в€’ 6y + 9y = 0 is:
r2 в€’ 6r + 9 = 0.
Solve r2 в€’ 6r + 9 = 0, we have
r1 = r2 = 3.
So the general solution to y в€’ 6y + 9y = 0 is:
y(x) = C1 e3x + C2 xe3x .
• 2.2.105: Find the solution to z (t) = −2z (t) − 2z(t), z(0) = 2, z (0) = −2.
Answer.
Rewrite the equation:
z + 2z + 2z = 0.
The characteristic equation of z + 2z + 2z = 0 is:
r2 + 2r + 2 = 0.
Solve r2 + 2r + 2 = 0, we have
r1 = в€’1 + i,
and r2 = в€’1 = i.
So the general solution to z + 2z + 2z = 0 is:
z(t) = C1 eв€’t cos(t) + C2 eв€’t sin(t).
Then
z (t) = в€’C1 eв€’t cos(t) в€’ C1 eв€’t sin(t) в€’ C2 eв€’t sin(t) + C2 eв€’t cos(t).
Since z(0) = 2 and z (0) = в€’2, then
2 = C1 + 0,
and
в€’ 2 = в€’C1 + C2 .
So
C1 = 2,
and C2 = 0.
Therefore, the solution to z (t) = в€’2z (t) в€’ 2z(t), z(0) = 2, z (0) = в€’2 is:
z(t) = 2eв€’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:
mx + cx + kx = 2x + x + 3x = 0.
The characteristic equation of 2x + x + 3x = 0 is:
2r2 + r + 3 = 0.
Solving 2r2 + r + 3 = 0, we have
в€љ
в€љ
в€’1 + 1 в€’ 4 В· 2 В· 3
в€’1 + i 23
r1 =
=
,
4
4
2
в€љ
в€’1 в€’ i 23
and r2 =
.
4
So the general solution to 2x + x + 3x = 0 is:
x(t) = C1 e
в€’ 4t
cos
в€љ
23
t
4
в€љ
+ C2 e
в€’ 4t
sin
23
t
4
.
b). Since 1 в€’ 4 В· 2 В· 3 = в€’23 < 0, the system is underdamped .
c). In order to make the system to be critically damped, that is,
c2 в€’ 4 В· 2 В· 3 = 0.
Then
c=
в€љ
в€љ
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:
0.1x + cx + 5x = 0.
Rewrite the equation as
x + 10cx + 50x = 0.
By the assumption, we know that our motion is underdamped with frequency 1 Hz,
which implies that (10c)2 в€’ 4 В· 50 < 0 and
в€љ
4 В· 50 в€’ (10c)2
200 в€’ 100c2
1 В· 2ПЂ =
=
= 50 в€’ 25C 2 .
2
2
So we get 25c2 = 50 в€’ 4ПЂ 2 , that is, c =
2в€’
4ПЂ 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 mx = в€’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 x (0) = 2. Therefore the initial value problem we have to solve is
x (t) +
k
x(t) = 0,
m
x(0) = 0,
3
x (0) = 2.
The general solution is
x(t) = A cos
k
x + B sin
m
k
x .
m
The initial conditions give
0 = x(0) = A,
2 = x (0) =
k
B,
m
so A = 0 and B = 2 m/k. Therefore the solution is
m
sin
k
x(t) = 2
k
x .
m
A maximum displacement of 3 m means that
4m
8
m
≤ 3 =⇒ k ≥
= .
k
9
9
|x(t)| ≤ 3 ⇐⇒ 2
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 x = в€’kx
or equivalently x +(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 x (0) = в€’1. Therefore
we are looking at the initial value problem
x (t) +
k
x(t) = 0,
M
x(0) = 0, x (0) = в€’1.
The general solution is
x(t) = A cos
k
t + B sin
M
k
t .
M
The initial conditions give A = 0 and B = в€’1 M/k so the solution is
x(t) = в€’
M
sin
k
4
k
t .
M
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
M
= 0.1 =в‡’ 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
M
=
k
10000
1
= в€љ .
500000
5 2
Therefore the maximum compression is
1
в€љ = 0.14142 m .
5 2
c) In order to have a maximum compression of 0.3 m we need to constrain M according
to
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 x (0) = в€’2 so the solution becomes
M
sin
k
x(t) = в€’2
k
t .
M
Therefore to keep the spring from breaking we have to constrain M according to
2
M
≤ 0.3 =⇒ M ≤ (0.3)2 (500000)/4 = 11250.
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 − y − 6y = e2x .
Answer.
The characteristic equation of y в€’ y в€’ 6y = 0 is:
r2 в€’ r в€’ 6 = 0.
5
Solve r2 в€’ r в€’ 6 = 0, then
r1 = 3,
and r2 = в€’2.
Let y(x) = Ae2x be a particular solution to y в€’ y в€’ 6y = e2x , then
y (x) = 2Ae2x ,
and y (x) = 4Ae2x .
So we have
4Ae2x в€’ 2Ae2x в€’ 6Ae2x = e2x .
That is,
в€’4Ae2x = e2x .
So
1
A=в€’ .
4
Therefore, a particular solution of y в€’ y в€’ 6y = e2x is:
y(x) = в€’
e2x
.
4
• 2.5.3: Find a particular solution of y − 4y + 4y = e2x .
Answer.
The characteristic equation of y в€’ 4y + 4y = 0 is:
r2 в€’ 4r + 4 = 0.
Solve r2 в€’ 4r + 4 = 0, then
r1 = r2 = 2.
Let y(x) = Ax2 e2x be a particular solution to y в€’ 4y + 4y = e2x , then
y (x) = 2Axe2x + 2Ax2 e2x ,
and y (x) = 2Ae2x + 4Axe2x + 4Axe2x + 4Ax2 e2x .
So we have
2Ae2x + 4Axe2x + 4Axe2x + 4Ax2 e2x в€’ 4 2Axe2x + 2Ax2 e2x + 4Ax2 e2x = e2x .
That is,
2Ae2x = e2x .
So
1
A= .
2
Therefore, a particular solution of y в€’ 4y + 4 = e2x is:
y(x) =
6
x2 e2x
.
2
• 2.5.102: a) Find a particular solution to y + 2y = ex + x3 . b) Find the general
solution.
Answer.
a). The characteristic equation to y + 2y = 0 is:
r2 + 2 = 0.
Solve r2 + 2 = 0, then
r1 =
в€љ
2i,
в€љ
and r2 = в€’ 2i.
Let y(x) = Aex + Bx3 + Cx2 + Dx + E be particular solution to y + 2y = ex + x3 ,
then
y (x) = Aex + 3Bx2 + 2Cx + D, and y (x) = Aex + 6Bx + 2C.
Then
Aex + 6Bx + 2C + 2 Aex + Bx3 + Cx2 + Dx + E = ex + x3 .
That is, we have
(3A в€’ 1)ex + (2B в€’ 1)x3 + 2Cx2 + (6B + 2D)x + 2C + 2E = 0.
So we get
3A в€’ 1 = 0,
Then
2B в€’ 1 = 0,
1
A= ,
3
2C = 0,
1
B= ,
2
6B + 2D = 0,
C = 0,
3
D=в€’ ,
2
2C + 2E = 0.
E = 0.
That is, a particular solution to y + 2y = ex + x3 is:
y(x) =
ex x3 3
+
в€’ x.
3
2
2
b). By the computations in part a), we know that the general solution to y + 2y = 0
is:
в€љ
в€љ
y(x) = C1 cos( 2x) + C2 sin( 2x).
So the general solution to y + 2y = ex + x3 is:
в€љ
в€љ
ex x3 3
y(x) = C1 cos( 2x) + C2 sin( 2x) +
+
в€’ x.
3
2
2
7
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