# Math 315 - Homework #4 Solutions

#### Document technical information

Format pdf
Size 371.5 kB
First found May 22, 2018

#### Document content analysis

Category Also themed
Language
English
Type
not defined
Concepts
no text concepts found

#### Transcript

```Math 315 - Homework #4 Solutions
Note: B&D refers to our course textbook written by Boyce and DiPrima.
Section 3.6
1. (B&D Problem 4) Use the method of variation of parameters to find a particular solution of the following differential equation. Then check your answer by using the method of undetermined coefficients.
4y 00 − 4y 0 + y = 16et/2
Solution: We first find the solution of the associated homogeneous equation. We find the roots of the
characteristic equation:
4r2 − 4r + 1 = 0
=⇒
(2r − 1)2 = 0
=⇒
r=
1
2
(repeated root)
Therefore, the complementary solution is
yc (t) = c1 et/2 + c2 tet/2 .
Before we proceed to find a particular solution, let’s put the differential equation in standard form:
1
y 00 − y + y = 4et/2 .
4
Therefore, the nonhomogeneous term that we will be working with is g(t) = 4et/2 . According to the
complementary solution, our particular solution will have the form
yp (t) = u1 (t)et/2 + u2 (t)tet/2 .
The Wronskian of y1 = et/2 and y2 = tet/2 is
et/2
W (y1 , y2 ) = 1
et/2
2
Furthermore, the determinants W1 and
0
tet/2
W1 = 4et/2 et/2 + 1 tet/2
2
W2 are
= −4tet
tet/2
et/2 + 12 tet/2
and
= et .
et/2
W2 = 1
et/2
2
4et/2
Z
Z
0
= 4et
Therefore,
Z
u1 (t) =
W1
dt =
W
Z
−4t dt = −2t
2
and
u2 (t) =
It follows that
yp (t) = (−2t2 )et/2 + (4t)tet/2
After simplifying, we obtain
yp (t) = 2t2 et/2
1
W2
dt =
W
4 dt = 4t.
Let’s verify our result by applying the method of undetermined coefficients. According to the complementary solution, the nonhomogeneous term, 16et/2 , is also a solution of the associated homogeneous
equation. Since the characteristic equation has a repeated real root, we assume a particular solution
of the form yp (t) = At2 et/2 . The first and second derivatives are
1
yp0 (t) = 2Atet/2 + At2 et/2
2
and
1
yp00 (t) = 2Aet/2 + 2Atet/2 + At2 et/2 .
4
Therefore,
1
1
4yp00 − 4yp0 + yp = 4 2Aet/2 + 2Atet/2 + At2 et/2 − 4 2Atet/2 + At2 et/2 + At2 et/2
4
2
= 8Aet/2 + (8A − 8A)tet/2 + (A − 2A + A)t2 et/2
t/2
= 8Aet/2 = |16e
{z }
g(t)
Therefore, we must have A = 2, and so yp (t) = 2t2 et/2 .
2. (B&D Problems 5 and 10) In each of Problems (a) and (b), find the general solution of the given
differential equation.
(a) y 00 + y = tan t,
0 < t < π/2
Solution: Note that the complementary solution is yc (t) = c1 cos t + c2 sin t. Therefore, a particular solution has the form yp (t) = u1 (t) cos t + u2 (t) sin t. The Wronskian of cos t and sin t is
W (cos t, sin t) = 1. Furthermore,
0
sin2 t
cos2 t − 1
sin t =
−
W1 = =
= cos t − sec t
tan t cos t cos t
cos t
and
cos t
0
W2 = − sin t tan t
= sin t.
Therefore,
Z
u1 (t) =
W1
dt =
W
Z
(cos t − sec t) dt = sin t − ln | sec t + tan t|
and
Z
u2 (t) =
W2
dt =
W
Z
sin t dt = − cos t.
Hence, our particular solution is
yp (t) = (sin t − ln | sec t + tan t|) cos t − cos t sin t = −(cos t) ln | sec t + tan t|.
It follows that the general solution is
y(t) = c1 cos t + c2 sin t − (cos t) ln | sec t + tan t| .
2
(b) y 00 − 2y 0 + y = et /(1 + t2 )
Solution: The characteristic equation of the corresponding homogeneous equation has r = 1 as
a repeated root. Therefore, yc (t) = c1 et + c2 tet . The Wronskian of et and tet is
t
e
tet = e2t .
W (et , tet ) = t
e tet + et Furthermore,
W1 = 0
et
1 + t2
tet
2t
= − te
1 + t2
tet + et t
e
W2 = et
and
0
et
1 + t2
2t
= e
1 + t2 .
Therefore,
Z
u1 (t) =
W1
dt = −
W
Z
1
t
dt = − ln(1 + t2 )
2
1+t
2
and
Z
W2
dt =
W
u2 (t) =
Z
1
dt = tan−1 t
1 + t2
Hence, our particular solution is
1
yp (t) = − et ln(1 + t2 ) + tet tan−1 t.
2
It follows that the general solution is
1
y(t) = c1 et + c2 tet − et ln(1 + t2 ) + tet tan−1 t .
2
3. (B&D Problems 13 and 16) In each of Problems (a) and (b), verify that the given functions y1
and y2 satisfy the corresponding homogeneous equation; then find a particular solution of the given
nonhomogeneous equation.
(a) t2 y 00 − 2y = 3t2 − 1,
y1 (t) = t2 , y2 (t) = t−1
t > 0;
Solution: Plugging y1 into the corresponding homogeneous equation yields
t2 (2) − 2t2 = 0.
Doing the same for y2 yields
t2 (2t−3 ) − 2t−1 = 2t−1 − 2t−1 = 0.
Therefore, both y1 and y2 are solutions of the corresponding homogeneous equation. Next, we
seek a particular solution of the nonhomogeneous equation of the form yp (t) = u1 (t)t2 + u2 (t)t−1 .
Let’s write the differential equation in standard form:
y 00 −
2
1
y = 3 − 2.
2
t
t
Note that
2
−1
W (t , t
Furthermore,
0
W1 = 3 − 1/t2
1/t
−1/t2
2
t
) = 2t
= −1 + 1
3 t3
3
1/t
−1/t2
and
= −3.
2
t
W2 = 2t
0
3 − 1/t2
= 3t2 − 1.
Therefore,
W1
dt =
W
Z u2 (t) =
W2
dt =
W
Z t3
1
1
2
dt = − + t.
−t +
3
3
3
Z
u1 (t) =
1
1
− 3
t
3t
dt = ln t +
1
6t2
and
Z
Hence,
yp (t) =
1
ln t + 2
6t
3
t
t 1
1
t2
t + − +
= + t2 ln t − .
3
3 t
2
3
2
Since −t2 /3 is a solution of the associated homogeneous equation, it may be removed from yp (t).
Therefore, we can take our particular solution to be
1
+ t2 ln t .
2
yp (t) =
(b) (1 − t)y 00 + ty 0 − y = 2(t − 1)2 e−t ,
0 < t < 1;
y1 (t) = et
y2 (t) = t
Solution: Plugging y1 into the corresponding homogeneous equation yields
(1 − t)et + tet − et = et − tet + tet − et = 0.
Doing the same for y2 yields
(1 − t)(0) + t(1) − t = 0 + t − t = 0.
Therefore, both y1 and y2 are solutions of the corresponding homogeneous equation. Next, we
seek a particular solution of the nonhomogeneous equation of the form yp (t) = u1 (t)et + u2 (t)t.
Let’s write the differential equation in standard form:
y 00 +
1
t
y0 −
y = 2(1 − t)e−t .
1−t
1−t
Note that
t
e
W (e , t) = t
e
t
1
t
Furthermore,
0
W1 = 2(1 − t)e−t
t
1
= −2t(1 − t)e−t
= et − tet = (1 − t)et .
t
e
W2 = t
e
and
0
2(1 − t)e−t
= 2(1 − t).
Therefore,
Z
u1 (t) =
W2
dt =
W
Z
−2t(1 − t)e−t
dt = −2
et (1 − t)
Z
1
te−2t dt = te−2t + e−2t
2
Z
2e−t dt = −2e−t .
and
Z
u2 (t) =
W2
dt =
W
Z
2(1 − t)
dt =
et (1 − t)
It follows that
1 −2t t
1
−2t
yp (t) = te
+ e
e − (2e−t )t = −te−t + e−t
2
2
or
1
yp (t) = − (2t − 1)e−t .
2
4
Note: Acceleration due to gravity: g ≈ 9.8 m/s2 = 980 cm/s2 (or about 32 ft/s2 )
Section 3.7
4. (B&D Problem 6) A mass of 100 g stretches a spring 5 cm. If the mass is set in motion from
its equilibrium position with a downward velocity of 10 cm/s, and if there is no damping, determine
the position u(t) of the mass at any time t. When does the mass first return to its equilibrium position?
Solution: Recall the model for a spring-mass system with no applied force to the mass:
mu00 (t) + γu0 (t) + ku(t) = 0
where m, γ, and k represent the mass, damping constant, and spring constant, respectively. The spring
constant is given by
k=
(100 g)(980 cm/s2 )
mg
2
=
= 19600 g/s .
L
5 cm
Since there is no damping, the differential equation that describes the motion of the mass is
100u00 + 19600u = 0
or
u00 + 196u = 0
where u(t) represents the position of the mass in cm at time t in seconds. The characteristic equation
is 100r2 + 196 = 0, which has a pair of imaginary roots, r = ±14i. Hence,
u(t) = c1 cos 14t + c2 sin 14t.
Since the motion of the mass begins at its equilibrium position, we have u(0) = 0. Applying this initial
condition yields u(0) = c1 = 0. Therefore,
u(t) = c2 sin 14t.
The velocity of the mass at time t in seconds is
u0 (t) = 14c2 cos 14t cm/s.
The initial velocity of the mass is u0 (0) = 10 cm/s. Therefore, u0 (0) = 14c2 = 10 which implies that
c2 = 10/14 = 5/7. It follows that
u(t) =
5
sin 14t .
7
The mass first returns to its equilibrium position when 14t = π, i.e. t = π/14.
5
5. (B&D Problem 11) A spring is stretched 10 cm by a force of 3 N. A mass of 2 kg is hung from a spring
and is also attached to a viscous damper that exerts a force of 3 N when the velocity of the mass is 5
m/s. If the mass is pulled down 5 cm below its equilibrium position and given an initial velocity of 10
cm/s, determine its position u(t) at any time t. Find the quasi frequency µ and the ratio of µ to the
natural frequency of the corresponding undamped motion.
Solution: The spring constant is given by
k=
3N
force that stretches the spring
=
= 30 N/m.
length of the stretch caused by the force
0.1 m
Moreover, we are given that the damping force, Fd , is 3 N when the velocity, u0 (t) is 5 m/s. Since
Fd = γu0 (t), the damping constant is
γ=
3
3N
= N-s/m.
5 m/s
5
Therefore, the differential equation that describes the motion of the mass is
2u00 + 3/5u0 + 30u = 0
or
10u00 + 3u0 + 150u = 0.
The characteristic equation is
2
10r + 3r + 150 = 0
=⇒
3
r=− ±
20
√
−5991
20
≈ −0.15 ± 3.87i
Therefore,
u(t) = e−0.15t c1 cos(3.87t) + c2 sin(3.87t) .
The velocity of the mass is
u0 (t) = −0.15e−0.15t c1 cos(3.87t) + c2 sin(3.87t) +e−0.15t [−3.87c1 sin(3.87t) + 3.87c2 cos(3.87t)].
The initial conditions are u(0) = 0.05 m and u0 (0) = 0.1 m/s. Hence,
u(0) = c1 = 0.05
and
u0 (0) = −0.15c1 + 3.87c2 = 0.1
=⇒
c2 =
0.1 + 0.15(0.05)
≈ 0.027778
3.87
Therefore,
u(t) = e−.15t 0.05 cos(3.87t) + 0.027778 sin(3.87t)
.
Note that we can also express u(t) in the form u(t) = Re−γt/(2m) cos(µt − δ), where R =
δ = tan−1 (c2 /c1 ). The quasi frequency is µ = 3.87 rad/s. It follows that
u(t) = 0.057198e−0.15t cos(3.87t − 0.5071) .
6
p
c21 + c22 and
Note that the general solution of the undamped system
2u00 + 30u = 0
is
√
√
u(t) = c1 cos( 15 t) + c2 sin( 15 t).
Therefore, the natural frequency is ω0 =
the natural frequency is
√
15 rad/s. It follows that the ratio of the quasi frequency to
µ
3.87
= √ ≈ 0.99923 .
ω0
15
6. (B&D Problem 12) A series circuit has a capacitor of 10−5 F, a resistor of 3 × 102 Ω, and an inductor
of 0.2 H. The initial charge on the capacitor is 10−6 C and there is no initial current. Find the charge
Q on the capacitor at any time t.
Solution: Recall the differential equation that describes the flow of electric current in the circuit with
no impressed voltage:
LQ00 + RQ +
1
Q=0
C
where L, R, and C represent the inductance, resistance, and capacitance, respectively. Since L = 0.2,
R = 300, and C = 10−5 , we have
0.2Q00 + 300Q0 + 100000Q = 0
or
Q00 + 1500Q0 + 500000Q = 0.
The characteristic equation is
2
r + 1500r + 500000 = 0
=⇒
r=
=
−1500 ±
p
15002 − 4(500000)
2
−1500 ± 500
.
2
Therefore, the roots of the characteristic equation are r1 = −1000 and r2 = −500. Hence,
Q(t) = c1 e−1000t + c2 e−500t .
We differentiate Q to obtain the current at time t:
I(t) = Q0 (t) = −1000c1 e−1000t − 500c2 e−500t .
The initial conditions are
Q(0) = 10−6 C (coulombs)
and
7
I(0) = Q0 (0) = 0 A (amperes).
Applying these initial conditions yields
Q(0) =
c1 +
c2 = 10−6
I(0) = −1000c1 − 500c2 = 0
Solving the above system of equations yields c1 = −10−6 and c2 = 2 × 10−6 . It follows that
Q(t) = −10−6 e−1000t + (2 × 10−6 )e−500t
or
Q(t) = 10−6 2e−500t − e−1000t .
7. (B&D Problem 17) A mass weighing 8 lb stretches a spring 1.5 inches. The mass is also attached to
a damper with coefficient γ. Determine the value of γ for which the system is critically damped; be
sure to give the units for γ.
Solution: Since the weight of the mass is given by w = mg, we have
m=
w
8 lb
1
1
=
= lb · s2 /ft = slug.
2
g
32 ft/s
4
4
Moreover, the spring constant is
k=
8 lb
8 lb
mg
=
=
= 64 lb/ft.
L
1.5 in
1/8 ft
It follows that
u(t) =
1 00
u + γu0 + 64u = 0.
4
The system is critically damped when γ 2 − 4(1/4)(64) = 0. Let’s solve for γ:
2
γ −4
1 lb · s2
4 ft
lb
64
=0
ft
=⇒
=⇒
8
γ 2 = 64
lb2 · s2
ft2
γ = 8 lb · s/ft .
Section 3.8
8. A mass weighing 4 lb stretches a spring 1.5 inches. The mass is displaced 2 inches in the positive
direction from its equilibrium position and released with no initial velocity. Assume that there is no
damping and that the mass is acted on by an external force of 2 cos(3t) lb.
(a) Formulate and solve the initial value problem describing the motion of the mass.
Solution: Since the mass weighs 4 lb, we have
mg = 4 lb
=⇒
m=
1
1
4 lb
= lb · s2 /ft = slug.
2
32 ft/s
8
8
Moreover, since the mass stretches the spring 1.5 inches, the spring constant is
k=
mg
4 lb
4 lb
=
=
= 32 lb/ft.
L
1.5 in
1/8 ft
The initial value problem that describes the motion of the mass is
1 00
u + 32u = 2 cos(3t),
8
u(0) = 1/6 ft, u0 (0) = 0 ft/s.
The differential equation can be written as
u00 + 256u = 16 cos(3t).
Note that the complementary solution is
uc (t) = c1 cos(16t) + c2 sin(16t).
We apply the method of undetermined coefficients to find a particular solution of the nonhomogeneous equation. Assume a particular solution of the form up (t) = A cos(3t) + B sin(3t). Plugging
up into the differential equation and collecting like terms yields
u00p + 256up = 247A cos(3t) + 247B sin(3t) = 16 cos(3t) .
| {z }
g(t)
It follows that A = 16/247 and B = 0, and so
up (t) =
16
cos(3t).
247
Therefore, the general solution is
u(t) = c1 cos(16t) + c2 sin(16t) +
16
cos(3t).
247
Differentiating yields
u0 (t) = −16c1 sin(16t) + 16c2 cos(16t) −
48
sin(3t).
247
After applying the initial conditions, we obtain
u(0) = c1 +
16
1
=
247
6
=⇒
c1 =
9
151
,
1482
u0 (0) = 16c2 = 0
=⇒
c2 = 0.
It follows that
u(t) =
151
16
cos(16t) +
cos(3t) .
1482
247
(b) Use MAPLE or other graphing software to plot the graph of the solution.
Solution:
Figure 1: Plot of the solution.
(c) If the given external force is replaced by a force of 4 sin(ωt) of frequency ω, find the value of ω
for which resonance occurs.
Solution: Resonance occurs whenever ω = ω0 . In this case, resonance occurs when ω = 16 rad/s
9. A spring-mass system has a spring constant of 3 N/m. A mass of 2 kg is attached to the spring, and
the motion takes place in a viscous fluid that offers a resistance numerically equal to the magnitude
of the instantaneous velocity. If the system is driven by an external force of (3 cos 3t − 2 sin 3t) N,
Solution: We are given that the damping force is Fd = |u0 (t)|. Since Fd = γu0 (t) (in this case, Fd ≥ 0),
the damping constant is γ = 1. Since m = 2, k = 3, and F (t) = 3 cos 3t − 2 sin 3t, the differential
equation that describes the motion of the mass is
2u00 + u0 + 3u = 3 cos 3t − 2 sin 3t.
The steady state response is a particular solution of the nonhomogeneous equation. We will find this
by the method of undetermined coefficients. Assume a solution of the form up (t) = A cos 3t + B sin 3t.
Then
yp0 = −3A sin 3t + 3B cos 3t
yp00 = −9A cos 3t − 9B sin 3t
After plugging these functions into the differential equation and simplifying, we obtain
2u00p + u0p + 3up = (−15A + 3B ) cos 3t + (−15B − 3A) sin 3t = 3 cos 3t − 2 sin 3t
|
{z
}
|
{z
}
|
{z
}
−2
3
10
g(t)
The solution of the system
−15A + 3B = 3
−3A − 15B = −2
is (A, B) = (−1/6, 1/6). Therefore,
1
1
up (t) = − cos 3t + sin 3t.
6
6
Note that
√
p
p
R = A2 + B 2 = (−1/6)2 + (1/6)2 =
2
6
Furthermore, since the point (A, B) = (−1/6, 1/6) is located in the second quadrant of the Cartesian
plane, we have
3π
1/6
= tan−1 (−1) =
.
δ = tan−1 (B/A) = tan−1
−1/6
4
Hence,
√
up (t) =
3π
2
cos 3t −
.
6
4
10. Consider the forced but undamped system described by the initial value problem
u00 + u = 3 cos(ωt),
u(0) = 0, u0 (0) = 0.
(a) Find the solution u(t) for ω 6= 1.
Solution: The complementary solution is
uc (t) = c1 cos t + c2 sin t.
Assume a particular solution of the form up (t) = A cos(ωt) + B sin(ωt). Plugging up into the
differential equation yields
u00p + up = −ω 2 A cos(ωt) − ω 2 B sin(ωt) + A cos(ωt) + B sin(ωt)
= (1 − ω 2 )A cos(ωt) + (1 − ω 2 )B sin(ωt) = 3 cos(ωt).
Therefore, we must have A = 3/(1 − ω 2 ) and B = 0. It follows that
up (t) =
3
cos(ωt)
1 − ω2
and so
u(t) = c1 cos t + c2 sin t +
3
cos(ωt).
1 − ω2
Hence,
u0 (t) = −c1 sin t + c2 cos t −
11
3ω
sin(ωt).
1 − ω2
Applying the initial conditions yields
u(0) = c1 +
3
=0
1 − ω2
=⇒
c1 = −
3
1 − ω2
and
u0 (0) = c2 = 0.
Therefore,
u(t) =
3 cos(ωt) − cos t .
2
1−ω
(b) Use MAPLE or other graphing software to plot the solution u(t) versus t for ω = 0.7, ω = 0.8, and
ω = 0.9. Describe how the response u(t) changes as ω varies in this interval. What happens as ω
takes on values closer and closer to 1? Note that the natural frequency of the unforced system is
ω0 = 1.
Solution:
Figure 2: Plot of the solution for ω = 0.7.
Figure 3: Plot of the solution for ω = 0.8.
12
Figure 4: Plot of the solution for ω = 0.9.
We observe that a beat begins to form when 0.7 < ω < 0.8. We obtain a beat for each of the
values ω = 0.8 and ω = 0.9. The amplitude and period of the beat corresponding to ω = 0.9
are greater than the amplitude and period of the beat corresponding to ω = 0.8. In fact, as ω
approaches ω0 = 1, the amplitude and the period of a beat increases. The system resonates when
ω = ω0 = 1.
Figure 5: Plot of the solution for ω = 1.
13
```