MATH 215/255
Fall 2015
Assignment 5
§2.5.3, §2.6, §6.1
1 00
y + y 0 + y = e−2t sec(2t), − π4 < t < π4 .
4
Hint: To use the method of variation of parameters, rewrite the DE so that the
coefficient of y 00 is 1 before solving the boxed system on page 74 of [Lebl].
1. Find a particular solution of
Answer.
Rewrite the DE as
y 00 + 4y 0 + 4y = 4e−2t sec(2t).
The characteristic equation is r2 + 4r + 4 = 0 with r = −2, −2. The solutions of the
homogeneous equation is yc (t) = c1 e−2t + c2 te−2t . A particular solution is given by
yp (t) = u1 (t)y1 (t) + u2 (t)y2 (t), with u01 and u02 satisfying
e−2t u01 + te−2t u02 = 0
−2e−2t u01 + (1 − 2t)e−2t u02 = 4e−2t sec(2t).
They can be simplified as
(1) · 2 + (2) gives u02 =
Z
u1 = −
u01 + tu02 = 0,
(1)
−2u01 + (1 − 2t)u02 = 4 sec(2t).
(2)
4
cos(2t) .
4t
By (1), u01 = −tu2 = − cos(2t)
. Thus
4t
dt,
cos(2t)
Hence one particular solution is
Z t
yp (t) = −e−2t
Z
u2 =
4
dt = 2 ln | sec 2t + tan 2t|.
cos(2t)
4s
ds + 2te−2t ln | sec 2t + tan 2t|.
cos(2s)
1’ (Intended problem 1 – a factor of 2 was missing in the DE)
1
Find a particular solution of y 00 + y 0 + 2y = e−2t sec(2t), − π4 < t < π4 .
4
Hint: To use the method of variation of parameters, rewrite the DE so that the
coefficient of y 00 is 1 before solving the boxed system on page 74 of [Lebl].
Answer.
Rewrite the DE as
y 00 + 4y 0 + 8y = 4e−2t sec(2t).
The characteristic equation is r2 + 4r + 8 = 0 with r = −2 ± 2i. The solutions of the
homogeneous equation is yc (t) = c1 e−2t cos 2t + c2 e−2t sin 2t. A particular solution is
given by yp (t) = u1 (t)y1 (t) + u2 (t)y2 (t), with u01 and u02 satisfying
e−2t cos(2t)u01 + e−2t sin(2t)u02 = 0
e−2t (−2 cos 2t − 2 sin 2t)u01 + e−2t (2 cos 2t − 2 sin 2t)u02 = 4e−2t sec(2t).
They can be simplified as
cos(2t)u01 + sin(2t)u02 = 0,
(3)
(− cos 2t − sin 2t)u01 + (cos 2t − sin 2t)u02 = 2 sec(2t).
(4)
(3) ∗ (cos 2t − sin 2t) − (4) ∗ sin(2t) gives u01 = −2 tan 2t. Thus
Z
u1 = −2 tan 2t dt = ln | cos 2t| + c1 .
(3) ∗ (cos 2t + sin 2t) + (4) ∗ cos(2t) gives u02 = 2. Thus
Z
u2 = 2 dt = 2t + c2 .
Hence one particular solution is yp (t) = e−2t cos(2t) ln | cos 2t| + 2te−2t sin 2t.
2. Find the general solution of y 00 + 4y = 3 csc 2t,
0 < t < π/2.
Answer.
The solutions of the homogeneous equation is yc (t) = c1 cos 2t + c2 sin 2t.
A particular solution is given by yp (t) = u1 (t)y1 (t) + u2 (t)y2 (t), with u01 and u02
satisfying
cos(2t)u01 + sin(2t)u02 = 0,
(5)
(−2 sin 2t)u01 + (2 cos 2t)u02 = 3 csc(2t).
(6)
(4) ∗ (2 cos 2t) − (4) ∗ sin(2t) gives 2u01 = −3. Thus
Z
3
3
u1 = − dt = − t + c1 .
2
2
(4) ∗ (2 sin 2t) + (4) ∗ cos(2t) gives 2u02 = 3 cot 2t. Thus
Z
3
3
cot(2t) dt = ln | sin 2t| + c2 .
u2 =
2
4
Hence a particular solution is yp (t) = − 23 t cos 2t + 43 (sin 2t) ln | sin 2t|. The general
solution is
3
3
y(t) = c1 cos 2t + c2 sin 2t − t cos 2t + (sin 2t) ln | sin 2t|.
2
4
3. Find the general solution of y 00 − 5y 0 + 6y = g(t). Here g(t) is an arbitrary continuous
function.
Answer.
Two linearly independent solutions of the homogeneous DE are y1 (t) = e3t
and y2 (t) = e2t . A particular solution is given by yp (t) = u1 (t)y1 (t) + u2 (t)y2 (t), with
u01 and u02 satisfying
e3t u01 + e2t u02 = 0,
3e3t u01 + 2e2t u02 = g(t).
2
Thus u01 (t) = e−3t g(t) and u02 (t) = −e−2t g(t), and hence
Z t
Z t
−3s
2t
3t
e−2s g(s) ds.
e g(s) ds − e
yp (t) = e
The general solution is y(t) = c1 e3t + c2 e2t + yp (t).
4. Verify that the given functions y1 and y2 satisfy the corresponding homogeneous
equation; then find a particular solution of the given nonhomogeneous equation.
ty 00 − (1 + t)y 0 + y = t2 e2t ,
t > 0;
y1 (t) = 1 + t,
y2 (t) = et .
Hint: To use the method of variation of parameters, first rewrite the equation as
1
0
2t
y 00 − 1+t
t y + t y = te ; then solve the same boxed system on page 74 of [Lebl].
Answer.
ty100 − (1 + t)y10 + y1 = 0 − (1 + t) + (1 + t) = 0 and ty200 − (1 + t)y20 + y2 =
t
te − (1 + t)et + et = 0.
The equation can be rewritten as
y 00 −
1+t 0 1
y + y = te2t .
t
t
A particular solution is given by yp (t) = u1 (t)y1 (t) + u2 (t)y2 (t), in which
(1 + t)u01 + et u02 = 0,
u01 + et u02 = te2t .
Thus u01 = −e2t , u02 = (1 + t)et , and
Z
u1 (t) = −e2t dt = −e2t /2,
Z
u2 (t) =
(1 + t)et dt = tet .
Hence a particular solution is yp (t) = −(1 + t)e2t /2 + et tet = 21 e2t (t − 1).
5. A mass of 5 kg stretches a spring 9.8 cm. The mass is acted on by an external force
of 85 sin(2t) N (newtons) and moves in a medium that imparts a viscous force of 3
N when the speed of the mass is 5 cm/s. Suppose the mass is set in motion from its
equilibrium position with an initial velocity of 25 cm/s.
(a) Formulate the initial value problem describing the motion of the mass in MKS
units (meter/kilogram/second). Assume the gravity constant is 9.8 m/s2 .
(b) Find the solution of the initial value problem in (a).
(c) Identity the transient and steady periodic parts of the solution.
(d) If the given external force is replaced by a force of 85 cos ωt of frequency ω, find
the value of ω for which the amplitude of the forced response is maximum.
F0
Hint: By [Lebl, p.80], the amplitude is C = p
.
2
m (2pω) + (ω02 − ω 2 )2
Answer. (a) Using MKS units, the spring constant is k = 5 · 9.8/0.098 = 500 N/m,
and the damping coefficient is c = 3/0.05 = 60 N-s/m. The equation of motion is
5x00 + 60x0 + 500x = 85 sin(2t),
3
with initial conditions
x(0) = 0,
x0 (0) = 0.25.
(b) The DE has characteristic equation 5r2 + 60r + 500 = 0, thus r = −6 ± 8i. The
complementary solution is
xc (t) = c1 e−6t cos 8t + c2 e−6t sin 8t.
By the method of undetermined coefficients, a particular solution is
xp (t) = −
1
1
cos(2t) + sin(2t).
24
6
Hence the general solution is x(t) = xc (t) + xp (t).
1
1
24 and c2 = 48 .
1 −6t
sin 8t. The steady
48 e
Invoking the initial conditions, we find that c1 =
(c) The transient part is xc (t) =
is xp (t).
1 −6t
cos 8t +
24 e
(d) Note 2p = 12 and ω02 = 100. By the amplitude formula, C =
17
√
D
periodic part
where
D = (2pω)2 +(ω02 −ω 2 )2 = (12ω)2 +(100−ω 2 )2 = ω 4 −56ω 2 +104 = (ω 2 −28)2 +9216.
The maximum
of C occurs when the minimum of D occurs, which is when ω 2 = 28,
√
i.e., ω = 28 ≈ 5.29.
6. Consider a vibrating system described by the initial value problem
1
x00 + x0 + 2x = 2 cos ωt,
4
x(0) = 0,
x0 (0) = 2.
(a) Determine the steady periodic part of the solution of this problem.
(b) Find the amplitude C of the steady periodic part of the solution in terms of ω.
(c) Find the maximum value of C and the frequency ω for which it occurs.
Answer. (a) The steady periodic part is the particular solution of the form xp (t) =
c1 cos ωt + c2 sin ωt. Substitution into the equation yields
(2 − ω 2 )c1 +
One can solve c1 =
32(2−ω 2 )
,
D
ω
c2 = 2,
4
c2 =
8ω
D
ω
− c1 + (2 − ω 2 )c2 = 0.
4
with D = 16(2 − ω 2 )2 + ω 2 .
(b) The amplitude is
q
8p
8√
8
C = c21 + c22 =
[4(2 − ω 2 )]2 + (ω)2 =
D=√ .
D
D
D
(c) Maximal amplitude occurs when min D occurs. We may write D(ω) = g(ω 2 ) with
g(t) = 16(2−t)2 +t, whose graph is an upward parabola. Set g 0 (t)
q= −32(2−t)+1 = 0,
we get t =
63
32
= ω 2 . Thus maximal amplitude occurs when ω =
amplitude is C =
√8
D
=
q 8
g( 63
)
32
=
√64 .
127
4
63
32 ,
and the maximal
7. Recall that cosh bt = (ebt + e−bt )/2 and sinh bt = (ebt − e−bt )/2. Find the Laplace
transform of f (t) = eat sinh bt, where a, b are real constants.
Answer.
Since f (t) = (e(a+b)t − e(a−b)t )/2, we have
1
1
1
1
b
Lf = (Le(a+b)t − Le(a−b)t ) = (
−
)=
.
2
2 s−a−b s−a+b
(s − a)2 − b2
8. Recall that cos bt = (eibt + e−ibt )/2 and sin bt = (eibt − e−ibt )/2i. Assuming that
the necessary elementary integration formulas extend to this case, find the Laplace
transform of f (t) = eat sin bt, where a, b are real constants.
Answer.
Lf =
Since f (t) = (e(a+ib)t − e(a−ib)t )/2i, we have
1
1
1
1
b
(Le(a+ib)t − Le(a−ib)t ) = (
−
)=
.
2i
2i s − a − ib s − a + ib
(s − a)2 + b2
Z
9. Determine whether the integral
∞
(t2 + 1)−1 dt converges or diverges.
0
Answer.
Since (t2 + 1)−1 < t−2 and
integral also converges.
RA
1
t−2 dt = 1 − A−1 converges as A → ∞, our
10. Determine the range of real s so that the integral converges:
Z ∞ 2t
Z ∞ 2t
e
e
(a)
e−st dt
(b)
e−st dt
2
1
+
t
1
+
t
0
0
Answer.
Both integrals converge
R ∞ 1 for s > 2 and diverge for s < 2. When s = 2,
The integral in (a) becomes 0 1+t
dt which diverges, and the integral in (b) becomes
R∞ 1
0 1+t2 dt which converges.
To summarize, (a) converges for 2 < s < ∞ and (b) converges for 2 ≤ s < ∞.
5