Math 308 Week 5 Solutions
There is a solution manual to Chapter 4 online: www.pearsoncustom.com/tamu math/.
This online solutions manual contains solutions to some of the suggested problems. Here
are solutions to suggested problems that cannot be found in the online solutions manual.
NSS 4.1
1. Use the form A cos Ωt + B sin Ωt to find synchronous solutions to the following forced
oscillator equation:
y ′′ + 2y ′ + 4y = 3 sin(5t)
Answer:
First, we take derivatives of y = A cos Ωt + B sin Ωt:
y ′ = −AΩ sin Ωt + BΩ cos Ωt
y ′′ = −AΩ2 cos Ωt − BΩ2 sin Ωt
In order for y to be a solution, we will need Ω = 5 (because of the sin(5t) in the
differential equation). Thus, setting Ω = 5, and plugging y, y ′ , and y ′′ into the
differential equation, we get
−25A cos 5t − 25B sin 5t + 2 (−5A sin 5t + 5B cos 5t) + 4 (A cos 5t + B sin 5t) = 3 sin 5t
Combining sin 5t terms and cos 5t, we get:
(−25B − 10A + 4B) sin 5t + (−25A + 10B + 4A) cos 5t = 3 sin 5t
The coefficients of sin 5t and cos 5t on the two sides of the equation need to be equal.
Thus, the coefficient of cos 5t on the left side needs to equal 0, and the coefficient of
sin 5t on the left side needs to equal 3. This gives the following equations:
−25B − 10A + 4B = 3
−25A + 10B + 4A = 0
These equations simplify to:
−10A − 21B = 3
−21A + 10B = 0
We could solve these by hand, but instead we’ll ask Matlab to solve them. The
command is:
>> sol = solve('-10*A - 21*B = 3', '-21*A + 10*B = 0', 'A', 'B')
>> sol.A
>> sol.B
We get A = −30/541 and B = −63/541. Thus, the synchronous solution is
1
y(t) = −
63
30
cos(5t) −
sin 3t
541
541
2. Use the form A cos Ωt + B sin Ωt to find synchronous solutions to the following forced
oscillator equation:
y ′′ + 2y ′ + 4y = 5 sin(3t)
Answer: If we use the same method as in (1), we get that Ω = 3 and A and B
satisfy the equations:
−9B − 6A + 4B = 5
−9A + 6B + 4A = 0
These equations simplify to
−6A − 5B = 5
−5A + 6B = 0
Using Matlab to solve these equations, we get A = −30/60 and B = −25/61. Thus,
the synchronous solution is
y(t) = −
30
25
cos(3t) −
sin 3t
60
61
3. Use the form A cos Ωt + B sin Ωt to find synchronous solutions to the following forced
oscillator equation:
y ′′ + 2y ′ + 4y = 3 cos(2t) + 4 sin(2t)
Answer: If we use the same method as in (1), we get that Ω = 2 and A and B
satisfy the equations:
−4B − 4A + 4B = 4
−4A + 4B + 4A = 3
These equations simplify to
−4A = 4
4B = 3
Thus, A = −1 and B = 3/4. Thus, the synchronous solution is
2
y(t) = − cos(2t) +
3
sin 2t
4
4. Undamped oscillators that are driven at resonance have unusual (and nonphysical)
solutions.
(a) To investigate this, find the synchronous solution A cos Ωt + B sin Ωt to the
generic forced oscillator equation
my ′′ + by ′ + ky = cos Ωt
(b) Sketch graphs of the coefficients A and B, as functions of Ω, for m = 1, b = 0.1,
k = 25.
(c) Now set b = 0 in your formulas for A and B, and resketch the graphs in part
(b) with m = 1, k = 25. What happens at Ω = 5? Notice that the amplitudes
of the synchronous solutions grow without bound as Ω approaches 5.
(d) Show directly, by substituting the form A cos Ωt + B sin Ωt
pinto equation (5),
that when b = 0 there are no synchronous solutions if Ω = k/m.
p
(e) Verify that (2mΩ)−1 t sin Ωt solves equation (5) when b = 0 and Ω = k/m.
Notice that this nonsynchronous solution grows in time, without bound.
Answer:
(a) First, we take derivatives of y = A cos Ωt + B sin Ωt:
y ′ = −AΩ sin Ωt + BΩ cos Ωt
y ′′ = −AΩ2 cos Ωt − BΩ2 sin Ωt
Plugging y, y ′, and y ′′ into the differential equation, we get
m −AΩ2 cos Ωt − BΩ2 sin Ωt +b (−AΩ sin Ωt + BΩ cos Ωt)+k (A cos Ωt + B sin Ωt) = cos Ωt
Combining sin Ωt terms and cos Ωt terms, we get:
−mΩ2 B − bΩA + kB sin Ωt + −mΩ2 A + bΩB + kA cos Ωt = cos Ωt
The coefficients of sin Ωt and cos Ωt on the two sides of the equation need to be
equal. Thus, the coefficient of sin Ωt on the left side needs to equal 0, and the
coefficient of cos Ωt on the left side needs to equal 1. This gives the following
equations:
−mΩ2 B − bΩA + kB = 0
−mΩ2 A + bΩB + kA = 1
These equations are equivalent to:
bΩA + (k − mΩ2 )B = 0
(k − mΩ2 )A + bΩB = 1
We’ll ask Matlab to solve these equations for A and B. We’ll use w instead of
Ω. The command is:
3
>> sol = solve('b*w*A+(k-m*w^2)*B=0', '(k-m*w^2)*A+b*w*B=1', 'A', 'B')
>> sol.A
>> sol.B
We get
k − mΩ2
−b2 Ω2 + k 2 − 2kmΩ2 + m2 Ω4
−bΩ
B =
2
2
2
−b Ω + k − 2kmΩ2 + m2 Ω4
A =
Thus, the synchronous solution is
y=
k − mΩ2
−bΩ
cos(Ωt)+ 2 2
sin(Ωt)
2
2
2
2
2
4
2
−b Ω + k − 2kmΩ + m Ω
−b Ω + k − 2kmΩ2 + m2 Ω4
(b) We can have Matlab plot these graphs. Here are the commands to substitute
the appropriate values in for m, b, and k (using the fact we already have A and
B stored in Matlab as sol.A and sol.B):
>>
>>
>>
>>
A
A
A
A
=
=
=
=
sol.A
subs(A, 'm', 1)
subs(A, 'b', 0.1)
subs(A, 'k', 25)
Here are the (mostly identical) commands to substitute values for B:
>>
>>
>>
>>
B
B
B
B
=
=
=
=
sol.B
subs(B, 'm', 1)
subs(B, 'b', 0.1)
subs(B, 'k', 25)
We get that
25 − Ω2
−5001Ω2 /100 + 625 + Ω4
−Ω/10
B =
−5001Ω2 /100 + 625 + Ω4
A =
We graph these with ezplot:
>> ezplot(A)
>> ezplot(B)
4
For A, we get the graph:
(25−w2)/(−5001/100 w2+625+w4)
0.3
0.2
0.1
0
−0.1
−0.2
−0.3
−6
−4
−2
0
w
2
4
6
For B, we get the graph:
−1/10 w/(−5001/100 w2+625+w4)
0.025
0.02
0.015
0.01
0.005
0
−0.005
−0.01
−0.015
−0.02
−0.025
−6
−4
−2
0
w
2
4
From these graphs, we can’t really tell what is happening near Ω = −5 and
5
Ω = 5. If you instead use the plot command, and play around with zoom, you
will discover that the graph does not have any asymptotes and is continuous.
Here are some graphs for A.
1500
1000
500
0
−500
−1000
−1500
−6
−4
−2
0
2
4
6
1500
1000
500
0
−500
−1000
−1500
−5.1
−5.05
−5
6
−4.95
−4.9
−4.85
Here are the commands used to get these graphs. The first graph is on the
interval [−6, 6], and the commands are:
>> w = -6:.0001:6;
>> y = (25-w.^2)./(-5001*w.^2/100 + 625 + w.^4);
>> plot(w,y)
The second graph is on the interval from [−5.1, −4.8] and the commands are:
>> w = -5.1:.0001:-4.8;
>> y = (25-w.^2)./(-5001*w.^2/100 + 625 + w.^4);
>> plot(w,y)
(c) We can modify the above Matlab commands to give us the formulas for A and
B when m = 1, b = 0, and k = 25. We get that B = 0 and
A=
25 − Ω2
625 − 50Ω2 + Ω4
The graph for A is
(25−w2)/(625−50 w2+w4)
0.25
0.2
0.15
0.1
0.05
0
−0.05
−0.1
−0.15
−0.2
−10
−8
−6
−4
−2
0
w
2
4
6
8
10
This time there really are vertical aymptotes at Ω = 5 and Ω = −5 (we know
this, because if we plug Ω = 5 or Ω = −5 into the equation for A, we get 0 in
the denominator).
p
(d) We are considering the differential equation my ′′ +ky = cos Ωt with Ω = k/m.
We want to show that there are no solutions of the form y = A cos Ωt + B sin Ωt.
7
First, we take derivatives of y = A cos Ωt + B sin Ωt:
y ′ = −AΩ sin Ωt + BΩ cos Ωt
y ′′ = −AΩ2 cos Ωt − BΩ2 sin Ωt
Plugging y and y ′′ into the differential equation, we get
m −AΩ2 cos Ωt − BΩ2 sin Ωt + k (A cos Ωt + B sin Ωt) = cos Ωt
Combining sin Ωt terms and cos Ωt terms, we get:
−mΩ2 B + kB sin Ωt + −mΩ2 A + kA cos Ωt = cos Ωt
The coefficients of sin Ωt and cos Ωt on the two sides of the equation need to be
equal. Thus, the coefficient of sin Ωt on the left side needs to equal 0, and the
coefficient of cos Ωt on the left side needs to equal 1. This gives the following
equations:
−mΩ2 B + kB = 0
−mΩ2 A + kA = 1
These equations are equivalent to:
(k − mΩ2 )B = 0
(k − mΩ2 )A = 1
p
If Ω = k/m, then these equations are equivalent to 0 = 0 and 0 · A = 1. Thus,
there are no solutions of the form y = A cos Ωt + B sin Ωt.
t sin Ωt
(e) We want to show that y =
is a solution to the differential equation
2mΩ
p
my ′′ + ky = cos Ωt when Ω = k/m.
First, we take derivatives of y =
t sin Ωt
2mΩ
1
(sin Ωt + Ωt cos Ωt)
2mΩ
1
1
=
Ω cos Ωt + Ω cos Ωt − Ω2 t sin Ωt =
(2 cos Ωt − Ωt sin Ωt)
2mΩ
2m
y′ =
y ′′
We substitute y ′′ and y into the differential equation:
kt sin Ωt
1
(2 cos Ωt − Ωt sin Ωt) +
= cos Ωt
2
2mΩ
This is equivalent to
cos Ωt +
Ω
k
t sin Ωt = cos Ωt
−
2mΩ
2
8
Since Ω =
p
k/m, we can simplify
Ω
k
− . It becomes
2mΩ
2
Ω
k − mΩ2
k − m(k/m)
k
− =
=
=0
2mΩ
2
2mΩ
2mΩ
t sin Ωt
Thus, we have cos Ωt = cos Ωt, so y =
is a solution to the differential
2mΩ
equation.
This problem was about resonance. In an undamped forced mass-spring system,
resonance occurs when the forcing function is at the same period as the natural
period
pof the undamped unforced mass-spring system (in this problem, this was when
Ω = k/m). As we saw in part (e) of the problem, solutions grow without bound.
When there is a small bit of damping (as in part (b) with b = 0.1) and the forcing
function is close to the same period as the natural period, solutions initially look
similar to the undamped solutions. The amplitude of the oscillations increase before
approaching the synchronous solution. As we saw in part (b), the synchronous solutions near resonance have rather large amplitudes (the values A and B determine
the amplitudes).
NSS 4.2
1. linear, homogeneous, variable coefficients
2. linear, nonhomogeneous, variable coefficients
4. linear, homogeneous, variable coefficients
5. linear, nonhomogeneous, variable coefficients
6. nonlinear
7. linear, nonhomogeneous, constant coefficients
8. nonlinear
10. Let L[y](x) = y ′′(x) − 4y ′(x) + 3y(x). Compute
(a) L[x2 ]
(b) L[e3 x]
(c) L[erx ]
Answer:
(a) Here we want y = x2 . We compute y ′ (x) = 2x and y ′′ (x) = 2. Plugging these
into the operator L, we get
L[x2 ] = 2 − 8x + 3x2
9
(b) Here we want y = e3x . We compute y ′(x) = 3e3x and y ′′(x) = 9e3x . Plugging
these into the operator L, we get
L[x2 ] = 9e3x − 12e3x + 9e3x = 6e3x
(c) Here we want y = erx . We compute y ′(x) = re3x and y ′′(x) = r 2 e3x . Plugging
these into the operator L, we get
L[x2 ] = r 2 erx − 4rerx + 3erx = (r 2 − 4r + 3)erx
16. Discuss the existence and uniqueness of a solution to the following differential equation that satisfies the initial conditions y(1) = y0 , y ′(1) = y1 , where y0 and y1 are real
constants.
x(x − 3)y ′′ + 2xy ′ − y = x2
Answer: First, we put the differential equation in standard form:
y ′′ +
1
x
2x
y′ −
y=
x(x − 3)
x(x − 3)
x−3
2
1
is continuous for all x 6= 0, 3. The function
is conx(x − 3)
x(x − 3)
x
tinuous for all x 6= 0, 3, and the function
is continuous for all x 6= 3.
x−3
Thus, unique solutions exist on the intervals (−∞, 0), (0, 3), and (3, ∞). Since our
inital condition is at x = 1, there is a unique solution to the given initial value
problem on the interval (0, 3).
The function
18. Discuss the existence and uniqueness of a solution to the following differential equation that satisfies the initial conditions y(1) = y0 , y ′(1) = y1 , where y0 and y1 are real
constants.
x2 y ′′ + y = cos x
Answer: First, we put the differential equation in standard form:
y ′′ +
y
cos x
= 2
2
x
x
1
cos x
The function 2 is continuous for all x 6= 0, and the function
is continuous for
x
x2
all x 6= 3.
Thus, unique solutions exist on the intervals (−∞, 0) and (0, ∞). Since our inital
condition is at x = 1, there is a unique solution to the given initial value problem on
the interval (0, ∞).
23. Express the given operator L using the differential operator D.
L[y] = y ′′ + 7y ′ + 12y
Answer: L = D 2 + 7D + 12
10
24. Express the given operator L using the differential operator D.
L[z] = 2z ′′ + z ′ − z
Answer: L = 2D 2 + D − 1
26. Express the given operator L using the differential operator D.
L[y] = x2 y ′′ − xy ′ + y
Answer: L = x2 D 2 − x + 1
30. Boundary Value Problems. When the values of a solution to a differential equation are specified at two different points, these conditions are called boundary conditions. (In contrast, initial conditions specify the values of a function and its
derivative at the same point.) Given that every solution to
(14)
y ′′ + y = 0
is of the form
y(x) = c1 cos x + c2 sin x,
where c1 and c2 are constants, show that
(a) there is a unique solution to (14) that satisfies the boundary conditions y(0) = 2
and y(π/2) = 0;
(b) there is no solution to (14) that satisfies y(0) = 2 and y(π) = 0;
(c) there are infinitely many solutions to (14) that satisfy y(0) = 2 and y(π) = −2.
Answer:
(a) We want to find all functions of the form y(x) = c1 cos x + c2 sin x that satisfy
the conditions y(0) = 2 and y(π/2) = 0. This means we want to find c1 and c2
that satisfy:
c1 cos(0) + c2 sin(0) = 2
and
c1 cos(π/2) + c2 sin(π/2) = 0
These first equation becomes c1 = 2 and the second equation becomes c2 = 0.
Thus, we see that there is exactly one solution to the differential equation that
satisfies the given boundary condition: y(x) = 2 cos x.
(b) We want to find all functions of the form y(x) = c1 cos x + c2 sin x that satisfy
the conditions y(0) = 2 and y(π) = 0. This means we want to find c1 and c2
that satisfy:
c1 cos(0) + c2 sin(0) = 2
and
c1 cos(π) + c2 sin(π) = 0
These first equation becomes c1 = 2 and the second equation becomes −c1 = 0.
Thus, we see that there are no solutions to the differential equation satsifying
the given boundary conditions.
11
(c) We want to find all functions of the form y(x) = c1 cos x + c2 sin x that satisfy
the conditions y(0) = 2 and y(π) = −2. This means we want to find c1 and c2
that satisfy:
c1 cos(0) + c2 sin(0) = 2
and
c1 cos(π) + c2 sin(π) = −2
These first equation becomes c1 = 2 and the second equation becomes c1 = −2.
Thus, we see that any function of the form y(x) = 2 cos x + c2 sin x is a solution
to the differential equation satisfying the given boundary conditions, so there
are infinitely many solutions.
12