Homework6 Solutions
§3.7
In Problem 1 through 4 use the method of variation of parameters to find a particular solution
of the given differential equation. Then check your answer by using the method of undetermined
coeffients.
1. y 00 − 5y 0 + 6y = 2et
Solution: Characteristic equation is r2 − 5r + 6 = 0, therefore r1 = 2, r2 = 3, and y1 (t) = e2t ,
y2 (t) = e3t . Wronskian
¯
¯ 2t
¯ e 2e2t ¯
¯ = e5t
W (y1 , y2 )(t) = ¯¯ 3t
e 3e3t ¯
Using the formula given in the text, we have
Z 3t
Z 2t
e · 2et
e · 2et
2t
3t
Y (t) = −e
dt
+
e
dt
e5t
e5t
Z
Z
2t
−t
3t
e dt + 2e
e−2t dt = et .
= −2e
2. y 00 − y − 2y = 2e−t
Solution: Characteristic equation is r2 −r−2 = 0, therefore r1 = −1, r2 = 2, and y1 (t) = e−t ,
y2 (t) = e2t . Wronskian
¯ −t
¯
¯ e
−e−t ¯¯
¯
W (y1 , y2 )(t) = ¯ 2t
= 3et
e
2e2t ¯
Using the formula given in the text, we have
Z −t
Z 2t
e · 2e−t
e · 2e−t
2t
−t
dt
+
e
dt
Y (t) = −e
2et
3et
2
2
= − te−t − e−t
3
9
We can choose Y (t) = − 23 te−t .
In problem 5–12 find the general solution of the given equation.
5. y 00 + y = tan t, 0 < t < π/2.
Solution: Characteristic equation is r2 + 1 = 0, therefore
y2 (t) = sin t. Wronskian
¯
¯ cos t − sin t
W (y1 , y2 )(t) = ¯¯
sin t cos t
1
r1 = i, r2 = −i, and y1 (t) = cos t,
¯
¯
¯ = 1.
¯
Using the formula given in the text,
Z
Y (t) = − cos t
Z
= − cos t
Z
= − cos t
Z
= − cos t
we have
Z
sin t tan t dt + sin t cos t tan t dt
Z
sin2 t
dt + sin t sin t dt
cos t
sin2 t
d(sin t) − sin t cos t
1 − sin2 t
1
(
− 1) d(sin t) − sin t cos t
1 − sin2 t
1 1 + sin t
) − sin t cos t
= − cos t(− sin t + ln
2 1 − sin t
1
1 + sin t
1 + sin t
= − cos t ln
= − cos t ln
,
2
1 − sin t
cos t
or Y (t) = − cos t ln(sec t + tan t). Note the range for t guarantees the function inside ln is
positive.
10. y 00 − 2y 0 + y = et /(1 + t2 )
Solution: Characteristic equation is r2 − 2r + 1 = 0, therefore r1 = r2 = 1, and y1 (t) = et ,
y2 (t) = tet . Wronskian
¯ t
¯
t
¯ e
¯
e
¯ = e2t
W (y1 , y2 )(t) = ¯¯ t t
te e + tet ¯
Using the formula given in the text, we have
Z
Z
1
t
t
t
dt
+
te
dt
Y (t) = −e
1 + t2
1 + t2
1
= − et ln(1 + t2 ) + tet arctan t .
2
In problems 13–20 verify that the given functions y1 and y2 are solutions to the corresponding
homogeneous equation; then find a particular solution of the given nonhomogeneous equation.
13. t2 y 00 − 2y = 3t2 − 1, t > 0, y1 = t2 , y2 = t−1
Solution: It’s easy to verify that the given functions are solutions to the homogeneous
equation. We will concentrate on finding a particular solution. First rewrite the equation
so that the leading coefficient is 1.
y 00 −
The Wronskian is
2
1
y =3− 2 .
2
t
t
¯ 2
¯ t
2t
W (y1 , y2 )(t) = ¯¯ −1
t
−1/t2
2
¯
¯
¯ = −3
¯
and
Z
t−1 (3 − 1/t2 )
Y (t) = −t
dt + t−1
−3
2
t
1
1
= (3 ln t + 2 ) − (t3 − t)
3
2t
3t
2
t
1
= t2 ln t − + .
3
2
Z
2
t2 (3 − 1/t2 )
dt
−3
15. ty 00 − (1 + t)y 0 + y = t2 e2t , t > 0, y1 = 1 + t, y2 = et
Solution: Rewrite the equation so that the leading coefficient is 1.
y 00 −
The Wronskian is
1+t 0 1
y + y = te2t .
t
t
¯
¯
¯ 1+t 1 ¯
¯ = tet
W (y1 , y2 )(t) = ¯¯
et
et ¯
and
Z
Z
et (te2t )
(1 + t)te2t
t
Y (t) = −(1 + t)
dt
+
e
dt
tet
tet
Z
Z
= −(1 + t) e2t dt + et (1 + t)et dt
1 + t 2t
e + et ((1 + t)et − et )
2
1
1
1
= − e2t + te2t = e2t (t − 1) .
2
2
2
=−
§3.8
In problems 1–4 determine ω0 , R, and δ so as to write the given expression in the form u =
R cos(ω0 t − δ).
1. u = 3 cos 2t + 4 sin 2t
Solution:
3
4
3
u = 5( cos 2t + sin 2t) = 5 cos(2t − arccos )
5
5
5
Therefore R = 5, ω0 = 2, δ = arccos(3/5).
√
2. u = − cos t + 3 sin t
Solution:
√
1
2π
3
u = 2(− cos t +
sin t) = 2 cos(t −
).
2
2
3
Therefore R = 2, ω0 = 1, δ = 2π/3.
3
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/sec, and if there is no damping, determine the
position u of the mass at any time t. When does the mass first return to its equilibrium
position?
Solution: m = 100 = 0.1 kg, L = 5 = 0.05 m, u(0) = 0 m, u0 (0) = 10 = 0.1 m/sec.
Therefore k = 0.1 × 9.8/0.05 = 98/5 N/m. The equation for the system is
1 00 98
u + u=0
10
5
Solving the characteristic equation we get r = ±14i. Therefore the general solution is
u(t) = C1 cos 14t + C2 sin 14t .
Using the initail condition we get C1 = 0, C2 = 1/140. Thus
u(t) =
1
sin 14t .
140
Setting u(t) = 0 we get t = π/14 sec as the time when the mass first return to its equilibrium
position.
13. A certain vibrating system satisfies the equation u00 + γu0 + u = 0. Find the value of the
damping coefficient γ for which the quasi period of the damped motion is 50% geater than
the period of the corresponding undamped motion.
Solution: We can solve the equation to find the quasi period directly. But since there is a
formula in the text for the ratio of the quasi period to the period of the undamped system
we simply use that one. By this formula (28) in the text, and notice that m = 1, k = 1, we
have
µ
¶−1/2 µ
¶−1/2
Td
γ2
3
γ2
= 1−
= 150% = ,
= 1−
T
4km
4
2
√
solving for γ we get γ = 32 5.
17. A mass weighing 8 lb stretches a sping 1.5 in. 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: m = 8/32 = 1/4 lb·sec2 /ft, L = 1.5/12 = 1/8 ft. Thus k = 8/(1/8) = 64 lb/ft.
Thus the equation is
1 00
u + γu0 + 64u = 0 .
4
So in order for the system to be critically damped we need
γ 2 = 4mk = 64
or γ = 8 lb·sec/ft.
4
19. Assume the system decribed by the equation mu00 + γu0 + ku = 0 is critically damped or
overdamped. Show that the mass can pass through the equilibrium position at most once,
regardless of the initial condition.
Solution: First assume the system is critically damped, then the general solution to the
equation is u = C1 ert + C2 tert where r is the double root to the characteristic equation, and
r is negative. Set u = 0 and we get
ert (C1 + C2 t) = 0 .
Clearly this equation has no solution if C2 = 0 and exactly one solution if C2 6= 0.
For the overdamped case the general solution is u = C1 er1 t + C2 er2 t and the proof is similar.
§3.9
6. A mass of 6 kg stretches a spring 10 cm. The mass is acted on by an external force of
10 sin(t/2) N and moves in a medium that imparts a viscous force of 2 N when the speed
of the mass is 4 cm/sec. If the mass is set in mostion from its equilibrium position with an
initial velocity of 3 cm/sec, formulate the initial value problem describing the motion of the
mass.
Solution: m = 5 kg, L = 0.1m, γ = 2/4 N·sec/cm = 50 N·sec/m, k = 5 × 9.8/0.1 = 490
N/m. Therefore the equation is
5u00 + 50u0 + 490u = 10 sin t/2 .
Initial condition is u(0) = 0, u0 (0) = 3/100 m/sec.
8. (a) Find the solution of the initial value problem in Problem 6.
(b) Identify the transient and steady-state parts of the solution.
(c) Plot the graph of the steady-state solution.
(d) If the given external force is replaced by a force 2 cos ωt of frequency ω, find the value
of ω for which the amplitude of the forced response is maximum.
Solution
√
(a) Solving the characteristic equation we get r = −5 ± 73i, so the general solution to
the homogeneous equation is
√
√
e−5t (C1 cos 73t + C2 sin 73t) .
One way to find a particular solution is using undetermined coefficients, but that’s a
bit tedious. What we will do here is to use equation (8) and (9) in the text. Let’s√first
rewrite the equation as u00 + 10u0 + 98u = 2 sin t/2. Thus m = 1, γ = 10, ω0 = 73,
ω = 1/2, F0 = 2. Then according to equation (9)
q
∆ = m2 (ω02 − ω 2 )2 + γ 2 ω 2 = 72.92 ,
5
m(ω 2 −ω 2 )
0
and R = F0 /∆ = 0.02743, δ = cos−1
= cos−1 72.75
= 0.06830. However we
∆
72.92
cannot say u(t) = R cos(ωt − δ) is a particular solution to our equation since that’s a
solution to equation (1) in the text in which the right hand side is F0 cos ωt, while in
our equation the right hand side is F0 sin ωt. To remedy the situation we consider the
function u(t) = R sin(ωt − δ). It’s easy to check that this is a particular solution to
our equation. Therefore
√
√
u(t) = e−5t (C1 cos 73t + C2 sin 73t) + 0.02743 sin(0.5t − 0.06830)
is the general solution. Using the initial condition we get C1 = 0.00187 m = 0.187 cm,
C2 = 0.0062 m = 0.62 cm.
√
√
(b) The transient part is u(t) = e−5t (0.187 cos 73t + 0.62 sin 73t) cm, steady-state part
is u(t) = 2.743 sin(0.5t − 0.06830) cm.
(c) Use your calculator (or computer :).
(d) The amplitude of the forced response R is given as F0 /∆,p
therefore in order for it to
reach maximum ∆ should reach the minimum. Note ∆ = m2 (ω02 − ω 2 )2 + γ 2 ω 2 . To
find the minimum we take the derivative of ∆2 with respect to ω and set it to 0. Thus
2m2 (ω02 − ω 2 )(−2ω) + 2γ 2 ω = 0
r
√
γ2
ω = ω02 −
=
23 .
2m2
9. If an undamped spring-mass system with a mass that weighs 6 lb and a spring constant 1
lb/in. is suddenly set in motion at t = 0 by an external force of 4 cos 7t lb, determine the
position of the mass at any time and draw a graph of the displacement versus t.
3 00
u +12u = 4 cos 7t
Solution: m = 6/32 = 3/16 lb·sec2 /ft, k = 12 lb/ft, thus the equation is 16
with initial condition u(0) = 0, u0 (0) = 0. Rewrite the equation as u00 + 64u = 64
cos 7t we
3
√
see that ω0 = 64 = 8. Using equation (5) in the text we have the solution
u=
=
F0
(cos ωt − cos ω0 t)
− ω2)
m(ω02
64
(cos 7t − cos 8t) .
45
18. 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.
(b) 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:
(a) Again using equation (5) we get the solution
u(t) =
3
(cos ωt − cos t) .
1 − ω2
6
(b) We will not show the graph here, but we can describe what happens as ω gets closer
to 1. Rewrite the solution as
u(t) =
3
1−ω
1+ω
sin
t sin
t
2
2(1 − ω )
2
2
3
1−ω
t. As ω gets
We may view this as a vibration with a varying amplitude 2(1−ω
2 ) sin
2
3
closer to 1 the term 2(1−ω2 ) becomes very large, which means the maximal amplitude
is very big and also the period according to which the amplitude changes gets very big
too since it’s equal to 4π/(1 − ω).
7