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MATH 215/255
Fall 2014
Assignment 5
§2.5, §2.6
Solutions to selected exercises can be found in [Lebl], starting from page 303.
• 2.5.7: a) Find a particular solution of y 00 − 2y 0 + y = ex using the method of variation
of parameters.
b) Find a particular solution using the method of undetermined coefficients.
Answer.
In both cases, we first solve the complementary equation
yc00 − 2yc0 + yc = 0.
The characteristic equation (r − 1)2 = 0 has a unique root r = 1, hence
yc = C1 ex + C2 xex .
(1)
a) Following the recommended method, we try
yp = u1 ex + u2 xex
= (u1 + xu2 )ex ,
where u1 , u2 are arbitrary functions (for now). We have
yp0 = (u01 + u2 + xu02 + u1 + xu2 )ex .
We assume that u01 + xu02 = 0 so that
yp0 = (u1 + (x + 1)u2 )ex .
Using u01 = −xu02 , we may also compute
yp00 = (u01 + u2 + (x + 1)u02 + u1 + (x + 1)u2 )ex ,
= (u1 + (x + 2)u2 + u02 )ex .
Therefore
yp00 − 2yp0 + yp = (u1 + (x + 2)u2 + u02 ) − 2(u1 + (x + 1)u2 ) + (u1 + xu2 )
= u02 ex .
We want to make the righ-hand side equal ex , so that at this point we have to solve
u01 + xu02 = 0,
u02 = 1.
The second equation is satisfied for u2 = x, and the first then becomes u01 = −x,
which is satisfied for u1 = −x2 /2. Our particular solution is now
yp = − 12 x2 ex + x2 ex = 21 x2 ex .
Recalling the complementary solution (1), we have found the general form of the
solution:
y = C1 ex + C2 xe−x + 21 x2 ex .
b) The right-hand side is an exponential ex , so we try first yp = Aex . This fails quite
dramatically however, since we always obtain yp00 − 2yp0 + yp = 0. A similar experiment
with y = Axex would yield the same result. We try finally yp = Ax2 ex , for which
yp0 = (Ax2 + 2Ax)ex ,
yp00 = (Ax2 + 4Ax + 2A)ex .
Then
h
i
yp00 − 2yp0 + yp = (Ax2 + 4Ax + 2A) − 2(Ax2 + 2Ax) + Ax2 ex = 2Aex .
We take therefore A = 1/2, so that after adding the complementary part (1), we find
the same general solution as before:
y = C1 ex + C2 xe−x + 12 x2 ex .
• 2.5.9 For an arbitrary constant c find a particular solution to y 00 − y = ecx . Make
sure to handle every possible real c.
Answer.
The complementary equation yc00 − yc = 0 has an associated characteristic
equation r2 = 1, with roots ±1. Its general solution is therefore
yc = C1 ex + C2 e−x .
The right-hand side of y 00 − y = ecx is an exponential, so that we first try to find a
special solution yp = Aecx . When c 6= ±1, we succeed since
yp00 − yp = ecx
⇔
(c2 − 1)Aecx = ecx
⇔
A=
1
.
c2 − 1
When c = ±1, we have yp00 − yp = 0, and we apply the usual trick of trying yp = Axecx
instead. Computing the derivatives
yp0 = (cAx + A)ecx ,
yp00 = (c2 Ax + 2cA)ecx ,
2
we find that
yp00 − yp = (c2 Ax + 2cA − Ax)ecx = 2cAecx .
1
In that case we naturally take A = 2c
. Adding the complementary part, we eventually
obtain the general solution:
(
C1 ex + C2 e−x + (c2 − 1)−1 ecx if c 6= ±1,
y=
C1 ex + C2 e−x + (2c)−1 ecx
if c = ±1.
• 2.5.101: Find a particular solution to y 00 − y 0 + y = 2 sin(3x).
Answer.
The right-hand side is a sine function and the left-hand side features odd
and even derivates, so we try
yp = A cos(3x) + B sin(3x).
The derivatives are
yp0 = −3A sin(3x) + 3B cos(3x)
yp00 = −9A cos(3x) − 9B sin(3x).
We have then
yp00 − yp0 + yp = (−8A − 3B) cos(3x) + (−8B + 3A) sin(3x).
We are reduced to solving
−8A − 3B = 0
−8B + 3A = 2.
Quick linear algebra gives A = 6/73, B = −16/73. Therefore we have found a
particular solution
yp =
6
73
cos(3x) −
16
73
sin(3x).
• 2.5.103: Solve y 00 + 2y 0 + y = x2 , y(0) = 1, y 0 (0) = 2.
Answer.
The associated homogeneous equation has characteristic equation (r +
1)2 = 0 with a unique real root r = −1; therefore its general solution (complementary
solution to the original non-homogeneous equation) is
yc = C1 e−x + C2 xe−x .
3
The right-hand side of the complete equation is a polynomial of order 2, so we look
for a particular solution of the form yp = Ax2 + Bx + C. Then
yp00 + 2yp0 + yp = 2A + 2(2Ax + B) + (Ax2 + Bx + C)
= Ax2 + (4A + B)x + (2A + 2B + C).
We want the right-hand side to be x2 , so we solve
A = 1,
4A + B = 0,
2A + 2B + C = 0
and obtain A = 1, B = −4, C = 6 so that yp = x2 − 4x + 6. The general solution and
its derivative are therefore
y = C1 e−x + C2 xe−x + x2 − 4x + 6
y 0 = −C1 e−x + C2 (1 − x)e−x + 2x − 4.
Inserting the initial values y(0) = 1 and y 0 (0) = 2, we have 1 = C1 + 6 and 2 =
−C1 + C2 − 4, which gives C1 = −5 and C2 = 1. The final solution is therefore:
y = (x − 5)e−x + x2 − 4x + 6.
• 2.6.1: Derive a formula for xsp when the equation is mx00 + cx0 + kx = F0 sin(ωt).
Answer.
We carry out a computation very similar to that in Lebl’s book. We first
renormalize the equation to
x00 + 2px0 + ω02 x =
with ω0 =
F0
sin(ωt)
m
p
k/m, p = c/2m.
The particular solution we try, and its derivatives, are
x = A cos(ωt) + B sin(ωt),
x0 = −ωA sin(ωt) + ωB cos(ωt),
x00 = −ω 2 A cos(ωt) − ω 2 B sin(ωt).
After some rearranging, the left-hand side of the original equation becomes
x00 + 2px0 + ω02 x = (ω02 − ω 2 )A + 2pωB cos(ωt) + (ω02 − ω 2 )B − 2pωA sin(ωt)
4
We now solve
(ω02 − ω 2 )A + 2pωB = 0,
(ω02 − ω 2 )B − 2pωA =
F0
,
m
and obtain, after somewhat unpleasant linear algebra,
A=
F0
−2pω
·
,
2
m (2pω) + (ω02 − ω 2 )2
B=
ω02 − ω 2
F0
·
.
m (2pω)2 + (ω02 − ω 2 )2
The particular solution we obtain is thus
xsp =
F0 −2pω cos(ωt) + (ω02 − ω 2 ) sin(ωt)
.
·
m
(2pω)2 + (ω02 − ω 2 )2
Alternative solution. Observe, that the differential equation can be rewritten as mx00 +
cx0 + kx = F0 cos(ωt − π/2). Further, the introduction of τ = t − π/(2w) leads to
mx00 + cx0 + kx = F0 cos(ωτ ), where the derivatives are now taken with respect to τ .
The solution for this equation was covered in class and can be written as
F0 /m
xsp = p 2
cos(wτ − γ),
(w0 − w2 )2 + (2pw)2
where tan γ = 2pw/(w02 − w2 ). The last step is to replace τ by t − π/(2w), which gives
F0 /m
xsp = p 2
sin(wt − γ).
(w0 − w2 )2 + (2pw)2
It is easy to check using trigonometric identities that both answers are identical.
• 2.6.2: Derive a formula for xsp when the equation is mx00 + cx0 + kx = F0 cos(ωt) +
F1 cos(3ωt).
Answer.
It is enough to find particular solutions u, v to mu00 +cu0 +ku = F0 cos(ωt)
00
and mv + cv 0 + kv = F0 cos(3ωt) and add them up. Particular solutions to these two
equations are found in the course. We define
F0
ω02 − ω 2
·
,
m (2pω)2 + (ω02 − ω 2 )2
F1
ω02 − 9ω 2
A1 =
·
,
m (6pω)2 + (ω02 − 9ω 2 )2
A0 =
2pω
F0
·
,
2
m (2pω) + (ω02 − ω 2 )2
F1
6pω
B1 =
·
.
m (6pω)2 + (ω02 − 9ω 2 )2
B0 =
A particular solution to the equation under study is then
xsp = A0 cos(ωt) + B0 sin(ωt) + A1 cos(3ωt) + B1 sin(3ωt).
5
• 2.6.3: Fix parameters F0 , k, m > 0. Consider the equation mx00 + cx0 + kx =
F0 cos(ωt). For what values of c (in terms of F0 , k, m) will there be no practical
resonance? In other words, for what values of c is there no maximum of C(ω) for
ω > 0?
Answer.
We recall from the course that at fixed c, k, m, F0 , the amplitude C(ω)
has derivative
F0
−2ω(ω 2 + 2p2 − ω02 )
,
·
m ((2pω)2 + (ω02 − ω 2 )2 )3/2
p
where as usual p = c/2m and ω0 = k/m. Note that F0 , m, ω and the denominator
are always positive. When 2p2 − ω02 > 0, C 0 is negative on ]0, ∞[ and C is strictly
non-increasing there, so that it does not possess a maximum. When 2p2 − ω02 < 0, the
sign of C 0 (ω) is equal to that of
q
q
2
2
2
2
2
ω + ω02 − 2p2 ,
−(ω + 2p − ω0 ) = − ω − ω0 − 2p
C 0 (ω) =
and therefore C possesses a maximum at ω02 − 2p2 (it increases before this point and
decreases after).
In conclusion, C does not possess a maximum if and only if
2p2 > ω02
⇔
c2
k
>
2
2m
m
⇔
√
c>
2km.
• 2.6.4: Fix parameters F0 , c, k > 0. Consider the equation mx00 +cx0 +kx = F0 cos(ωt).
For what values of m (in terms of F0 , c, k) will there be no practical resonance?
Answer.
We just need to reproduce the condition obtain in Exercise 2.6.3 above,
and rewrite it as an inequality for m:
√
c>
2km
⇔
m6
c2
.
2k
• 2.6.101: A mass m = 4 is attached to a spring with k = 4 and a damping constant
c = 1. Suppose that F0 = 2. Using the forcing function F0 cos(ωt), find the ω that
causes practical resonance and find the amplitude.
Weqare considering the usual equation mx00 + cx0 + kx = F0 cos(ωt). We
k
c
compute ω0 = m
= 1 and p = 2m
= 18 . The practical resonance frequency is given
by
r
r
q
2
31
ω = ω02 − 2p2 = 1 −
=
.
64
32
Answer.
6
The amplitude at this frequency is given by
F0
1
·p
2
m
(2pω) + (ω02 − ω 2 )2
1
1
= ·q
2
31
+ ( 1 )2
C(ω) =
16·32
32
16
=√ .
63
• 2.6.102: Derive a formula for xsp when the equation is mx00 +cx0 +kx = F0 cos(ωt)+K,
where K is some constant.
Answer.
We know a solution to mu00 + cu0 + ku = F0 cos(ωt) from the course, and
we only need to add to it a solution v to mv 00 + cv 0 + kv = K. The right-hand side
of this last equation is a constant, so we try v = A, and find that A = K/k is a good
choice. The particular solution we obtain is
F0
ω02 − ω 2
2pω
K
xsp =
cos(ωt) +
sin(ωt) + .
2
2
2
2
m (2pω)2 + (ω02 − ω 2 )2
k
(2pω) + (ω0 − ω )
7