Math 215
Spring 2010
Assignment 5
You are highly recommended to own a textbook, Boyce-DiPrima or another, to see a comprehensive treatment of the subjects and more examples.
§3.8: 6, 8, 17(a,b,d),
§6.1: 10, 13, 21,
§6.2: 4, 10, 12, 23
• (§3.8: 6) A mass of 5 kg stretches a spring 10 cm. The mass is acted on by an external
force of 10 sin(t/2) N (newtons) and moves in a medium that imparts a viscous force
of 2 N when the speed of the mass is 4 cm/s. If the mass is set in motion from
its equilibrium position with an initial velocity of 3 cm/s, formulate the initial value
problem describing the motion of the mass.
Answer.
Using MKS units, the spring constant is k = 5(9.8)/0.1 = 490 N/m, and
the damping coefficient is γ = 2/0.04 = 50 N-s/m. The equation of motion is
5u00 + 50u0 + 490u = 10 sin(t/2).
The initial conditions are u(0) = 0 and u0 (0) = 0.03 m/s.
• (§3.8: 8)
(a) Find the solution of the initial value problem in Problem 6 (the previous problem).
(b) Identity 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 of 2 cos ωt of frequency ω, find
the value of ω for which the amplitude of the forced response is maximum.
√
√
Answer. (a) The homogeneous solution is uc (t) = c1 e−5t cos 73t + c2 e−5t sin 73t.
By the method of undetermined coefficients, a particular solution is
U (t) =
1
[−160 cos(t/2) + 3128 sin(t/2)].
153281
Hence the general solution is u(t) = uc (t) + U (t).
Invoking the initial conditions, we find that c1 =
(b) The transient part is uc (t) =
The steady state part is U (t).
160
153281
1
−5t cos
153281 [160e
√
383443 73
153281·7300 .
√
√
383443 73 −5t
e sin 73t].
7300
and c2 =
√
73t +
(c) skipped; it is periodic and looks like a sine curve.
(d) Using formulas (11) and (12) in the textbook on page 208, R =
√2
D
where
D = m2 (k/m − ω 2 )2 + γ 2 ω 2 = 25(98 − ω 2 )2 + 2500ω 2 .
(See next problem for direct computation, which is how this formula is derived.) We
may write D(ω) = 25g(ω 2 ) with g(t) = (98 − t)2 + 100t. max R occurs as min D and
min g occur. The graph of g(t) is an upward parabola whose minimum occurs at a
critical point . Set√g 0 (t) = −2(98 − t) + 100 = 0, we get t = 48 = ω 2 . Thus max R
occurs when ω = 4 3.
• (§3.8: 17(a,b,d)) Consider a vibrating system described by the initial value problem
1
u00 + u0 + 2u = 2 cos ωt,
4
u(0) = 0,
u0 (0) = 2.
(a) Determine the steady state part of the solution of this problem.
(b) Find the amplitude A the steady state solution in terms of ω.
(d) Find the maximum value of A and the frequency ω for which it occurs.
Answer. (a) The steady state is the particular solution of the form U (t) = c1 cos ωt+
c2 sin ωt. Substitution into the equation yields
ω
ω
(2 − ω 2 )c1 + c2 = 2, − c1 + (2 − ω 2 )c2 = 0.
4
4
One can solve c1 =
32(2−ω 2 )
,
D
c2 =
8ω
D
with D = 16(2 − ω 2 )2 + ω 2 .
(b) The amplitude is
q
8p
8√
8
c21 + c22 =
[4(2 − ω 2 )]2 + (ω)2 =
D=√ .
D
D
D
(d) 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 whose minimum occurs at
2
a critical point . Set g 0 (t) = −32(2 − t) + 1 = 0, we get t = 63
32 = ω . Thus maximal
q
√64
amplitude occurs when ω = 63
32 , and the maximal amplitude is 127 .
• (§6.1: 10) 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
b
1
−
)=
Lf = (Le(a+b)t − Le(a−b)t ) = (
.
2
2 s−a−b s−a+b
(s − a)2 − b2
• (§6.1: 13) 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
2
Z
∞
• (§6.1: 21) Determine whether the integral
Answer.
Since (t2 + 1)−1 < t−2 and
integral also converges.
RA
1
(t2 + 1)−1 dt converges or diverges.
0
t−2 dt = 1 − A−1 converges as A → ∞, our
• (§6.2: 4) Find the inverse Laplace transform of the function F (s) =
Answer.
3s
.
s2 − s − 6
Using partial fractions,
F (s) =
s2
3s
9/5
6/5
=
+
.
−s−6
s−3 s+2
Hence L−1 [F (s)] = 59 e3t + 65 e−2t .
• (§6.2: 10) Find the inverse Laplace transform of the function F (s) =
s2
2s − 3
.
+ 2s + 10
Answer.
Note the denominator s2 +2s+10 is irreducible over the reals. Completing
the square, s2 + 2s + 10 = (s + 1)2 + 9. Now convert the function to a rational function
of the variable ξ = s + 1, that is
2(s + 1) − 5
2s − 3
=
= G(s + 1),
s2 + 2s + 10
(s + 1)2 + 9
with G(s) = 2s−5
. We find out L−1 G = 2 cos 3t −
s2 +9
L[eat f (t)] = L[f (t)]s→s−a ,
L−1 F = e−t (2 cos 3t −
5
3
sin 3t. Using the formula
5
sin 3t).
3
• (§6.2: 12) Use Laplace transform to solve the initial value problem
y 00 + 3y 0 + 2y = 0;
y(0) = 1,
y 0 (0) = 0.
Answer.
Let Ly = Y . We have Ly 0 = sY − 1, Ly 00 = s(sY − 1) − 0. Thus the
Laplace transform of the ODE is
s2 Y − s + 3(sY − 1) + 2Y = 0,
i.e.,
(s2 + 3s + 2)Y = s + 3.
Solving for Y , we get
Y (s) =
s2
s+3
2
1
=
−
.
+ 3s + 2
s+1 s+2
Hence y(t) = L−1 Y = 2e−t − e−2t .
3
• (§6.2: 23) Use Laplace transform to solve the initial value problem
y 00 + 2y 0 + y = 4e−t ;
y(0) = 2,
y 0 (0) = −1.
Answer.
Let Ly = Y . We have Ly 0 = sY − 2, Ly 00 = s(sY − 2) − (−1). Thus the
Laplace transform of the ODE is
s2 Y − 2s + 1 + 2(sY − 2) + Y =
that is, (s2 + 2s + 1)Y = 2s + 3 +
Y (s) =
4
s+1 .
4
,
s+1
Solving for Y , we get
2s + 3
4
2
4
1
+
=
+
.
+
2
3
2
(s + 1)
(s + 1)
s + 1 (s + 1)
(s + 1)3
Note that
µ
−1
L
2
1
4
+ 2+ 3
s s
s
¶
= 2 + t + 2t2 .
By the translation formula for Laplace transform, we get
y(t) = L−1 Y = e−t (2 + t + 2t2 ).
4