Document 6535933

Sample Solutions of Assignment 5 for MATH3270A
a
Note: Any problems about the sample solutions, please email Mr.Xiao Yao (yxiao
math.cuhk.edu.hk) directly.
October,2013
Section 3.6
In each of problems use the methods of variation of parameters to find a particular solution of
the given differential equation. Then check your answer by using the method of undetermined
coefficients.
1. y 00 − 5y 0 + 6y = 4et
3. y 00 + 2y 0 + y = 6e−t
Answer: 1. It is easy to know that the solution of homogeneous equation is
y¯ = c1 e2t + c2 e3t
According the methods of variation of parameters, we can assume a particular solution is y(t) =
u1 (t)e2t + u2 (t)e3t ,then
(
u01 e2t + u02 e3t = 0
2u01 e2t + 3u02 e3t = 4et
Hence
(
u01 = −4e−t
u02 = 4e−2t
Solve this equation,
u1 =
4e−t
u2 = −2e−2t
Then y(t) = 2et
3. It is easy to know that the solution of homogeneous equation is
y¯ = (c1 + c2 t)e−t
1
2
According the methods of variation of parameters, we can assume a particular solution is y(t) =
u1 (t)e−t ,then
u001 = 6
Solve this equation,
u1 = 3t2
Then y(t) = 3t2 e−t
In each of Problems find the general solution of the given differential equation.
00
5. y + y = 2 tan t, 0 < t <
π
2
00
7. y + 4y 0 + 4y = 2t−2 e−2t , t > 0
9. 4y 00 + y = 8 sec(t/2)
11. y 00 − 7y 0 + 10y = g(t)
Answer: 5.The characteristic equation is
r2 + 1 = 0
Thus the possible values of r are r1 = i and r2 = −i, and the general solution of the equation is
y(t) = c1 cos t + c2 sin t.
W (y1 , y2 ) = 1 then a particular solution of the original equation is
Y (t) = − cos t
Z
2 sin t tan tdt + sin t
Z
2 cos t tan tdt
= −2 cos t ln tan t + sec t, 0 < t <
π
2
Hence,
y(t) = c1 cos t + c2 sin t − 2 cos t ln tan t + sec t, 0 < t <
are the general solutions of the original equation.
π
2
3
Answer: 7. The characteristic equation is
r2 + 4r + 4 = 0
Thus the possible values of r are r1 = −2 and r2 = −2, and the general solution of the equation
is
y(t) = c1 e−2t + c2 te−2t .
W (y1 , y2 ) = e−4t then a particular solution of the original equation is
Y (t) = −y1 (t)
Z
Z
y2 (t)g(t)
y1 (t)g(t)
dt + y2 (t)
dt
W (y1 , y2 )
W (y1 , y2 )
= −2e−2t ln t − e−2t
Hence,
y(t) = c1 e−2t + c2 te−2t − 2e−2t ln t
are the general solutions of the original equation.
9. It is easy to know that the solution of homogeneous equation is
1
1
y¯ = c1 cos t + c2 sin t
2
2
According the methods of variation of parameters, we can assume a particular solution is u(t) =
u1 (t) cos 12 t + u2 (t) sin 12 t,then
(
u01 cos 12 t + u02 sin 21 t = 0
sin 12 t + 21 u02 cos 12 t = 8 sec(t/2)
− 12 u01
Hence
(
u01 = −4 sin 21 t sec 12 t
u02 = 4 cos 12 t sec 12 t
Solve this equation,
u1 = 8 ln cos 12 t
u2 =
4t
Then u(t) = 8(ln cos 12 t) cos 21 t + 4t sin 12 t. The general solution is
1
1
1
1
1
y(t) = y¯ + u = c1 cos t + c2 sin t + 8(ln cos t) cos t + 4t sin t.
2
2
2
2
2
4
Answer: 11. The characteristic equation is
r2 − 7r + 10 = 0
Thus the possible values of r are r1 = 2 and r2 = 5, and the general solution of the equation is
y(t) = c1 e2t + c2 e5t .
W (y1 , y2 ) = 3e7t then a particular solution of the original equation is
Y (t) = −y1 (t)
=
Z
Z
Z
y2 (t)g(t)
y1 (t)g(t)
dt + y2 (t)
dt
W (y1 , y2 )
W (y1 , y2 )
[e5(t−s) − e2(t−s) ]g(s)ds
Hence,
y(t) = c1 e2t + c2 e5t +
Z
[e5(t−s) − e2(t−s) ]g(s)ds
are the general solutions of the original equation.
In each of the problems verify that the given functions y1 and y2 satisfy the corresponding
homogeneous equation; then find a particular solution of the given nonhomogeneous equation.
00
13. t2 y − 2y = 4t2 − 3, t > 0; y1 = t2 , y2 = t−1
00
0
15. ty − (1 + t)y + y = t2 e2t , y1 = 1 + t, y2 = et
00
0
00
0
3
1
1
18. x2 y + xy + (x2 − 0.25)y = 3x 2 sin x, x > 0; y1 = x− 2 sin x, y2 = x− 2 cos x
1
1
20. x2 y + xy + (x2 − 0.25)y = g(x), x > 0; y1 = x− 2 sin x, y2 = x− 2 cos x
Answer: 13. It is easy to check that y1 and y2 satisfy
00
t2 y − 2y = 0, t > 0.
W (y1 , y2 ) = −3 and let
2
Y (t) = −t
Z
Z 2
t−1 (4t2 − 3)
t (4t2 − 3)
−1
+t
−3
−3
2
1
= t4 − t2 ln t + t2 , t > 0.
5
3
5
then a particular solution of the original equation is
2
1
y(t) = t4 − t2 ln t + t2 , t > 0.
5
3
15. It is easy to check that y1 and y2 satisfy
00
0
ty (1 + t)y + 4y = 0
W (y1 , y2 ) = tet and let
Z
et (t2 e2t )
(1 + t)(t2 e2t )
t
+
e
tet
tet
5
5
1
= ( t2 − t + )e2t .
2
4
4
then a particular solution of the original equation is
1
5
5
y(t) = ( t2 − t + )e2t .
2
4
4
18. It is easy to check that y1 and y2 satisfy
Y (t) = −(1 + t)
Z
00
0
x2 y + xy + (x2 − 0.25)y = 0
W (y1 , y2 ) = − x1 and let
Y (t) = −y1 (t)
Z
Z
y2 (t)g(t)
y1 (t)g(t)
dt + y2 (t)
dt
W (y1 , y2 )
W (y1 , y2 )
3
= − x1/2 cos x.
2
then a particular solution of the original equation is
3
y(t) = − x1/2 cos x.
2
20. From 18, we know that y1 and y2 satisfy
00
0
x2 y + xy + (x2 − 0.25)y = 0
W (y1 , y2 ) = − x1 and let
Y (t) = −y1 (t)
Z
=x
Z
y2 (t)g(t)
y1 (t)g(t)
dt + y2 (t)
dt
W (y1 , y2 )
W (y1 , y2 )
− 21
Z
1
t 2 sin(x − t)g(t)dt.
6
then a particular solution of the original equation is
1
y(t) = x− 2
Z
1
t 2 sin(x − t)g(t)dt.
21. Show that the solution of the initial value problem
L[y] = y 00 + p(t)y 0 + q(t)y = g(t), y(t0 ) = y0 , y 0 (t0 ) = y00
(i)
can be written as y = u(t) + v(t), where u and v are solutions of two initial value problems
L[u] = 0, u(t0 ) = y0 , u0 (t0 ) = y00
(ii)
L[v] = g(t), v(t0 ) = 0, v 0 (t0 ) = 0
(iii)
respectively. In other words, the nonhomogeneities in the differential equation and in the initial
conditions can be dealt with respectively. Observe that u is easy to find if a fundamental set of
solution of L[u] = 0 is known.
Answer: Suppose y is a solution of equation (i), u is a solution of equation (ii), Let v = y − u,
then
L[y] = y 00 + p(t)y 0 + q(t)y = g(t), y(t0 ) = y0 , y 0 (t0 ) = y00
L[u] = u00 + p(t)u0 + q(t)u = 0, u(t0 ) = y0 , u0 (t0 ) = y00
Hence
L[v] = L[y − u] = g(t) and v(t0 ) = 0, v 0 (t0 ) = 0.
22. By choosing the lower limit of the integration in Equ.(28) in the text as the initial point t0 ,
show that Y (t) becomes
Y (t) =
Z t
t0
y1 (s)y2 (t) − y1 (t)y2 (s)
g(s)ds.
0
0
y1 (s)y2 (s) − y1 (s)y2 (s)
Show that Y (t) is a solution of the initial value problem
0
L[y] = g(t), y(t0 ) = 0, y (t0 ) = 0.
Thus Y can be identified with v in problem 21.
7
Answer:
Y (t) = −y1 (t)
=
Z t
t0
Z
y2 (s)g(s)
y1 (s)g(s)
ds + y2 (t)
ds
W (y1 , y2 )(s)
W (y1 , y2 )(s)
Z
Z t
−y1 (t)y2 (s)g(s)
y2 (t)y1 (s)g(s)
ds +
ds
0
0
0
0
y1 (s)y2 (s) − y2 (s)y1 (s)
t0 y1 (s)y2 (s) − y2 (s)y1 (s)
Z t
=
t0
y1 (s)y2 (t) − y2 (s)y1 (t)
g(s)ds
0
0
y1 (s)y2 (s) − y2 (s)y1 (s)
Hence,
Z t0
Y (0) =
t0
y1 (s)y2 (t) − y2 (s)y1 (t)
g(s)ds = 0
0
0
y1 (s)y2 (s) − y2 (s)y1 (s)
From
Y (t) = −y1 (t)
Z t
t0
Z t
y1 (s)g(s)
y2 (s)g(s)
ds + y2 (t)
ds,
W (y1 , y2 )(s)
t0 W (y1 , y2 )(s)
we get
0
0
Y (t) = −y1 (t)
Z t
t0
Z t
y1 (s)g(s)
y2 (s)g(s)
0
ds + y2 (t)
ds
W (y1 , y2 )(s)
t0 W (y1 , y2 )(s)
then
0
0
Y (t0 ) = −y1 (t0 )
00
00
Y (t) = −y1 (t)
00
Z t0
t0
Z t
t0
Z t0
y2 (s)g(s)
y1 (s)g(s)
0
ds + y2 (t0 )
ds = 0
W (y1 , y2 )(s)
t0 W (y1 , y2 )(s)
Z t
y2 (s)g(s)
y1 (s)g(s)
00
ds + y2 (t)
ds + g(t)
W (y1 , y2 )(s)
t0 W (y1 , y2 )(s)
0
Hence, Y (t) + p(t)Y (t) + q(t)Y (t)
00
0
= [y2 (t) + p(t)y2 (t) + q(t)y2 (t)]
Z t
t0
y1 (s)g(s)
ds
W (y1 , y2 )(s)
Z t
y2 (s)g(s)
ds + g(t) ≡ g(t).
t0 W (y1 , y2 )(s)
Therefore, Y (t) is a solution of the initial value problem
00
0
−[y1 (t) + p(t)y1 (t) + q(t)y1 (t)]
0
L[y] = g(t), y(t0 ) = 0, y (t0 ) = 0.
8
28. The method of reduction of order (Section 3.4) can also be used for the nonhomogeneous
equation
00
0
y (t) + p(t)y (t) + q(t)y(t) = g(t)
(i)
provided one solution y1 of the corresponding homogeneous equation is known. Let y = v(t)y1 (t)
and show that y satisfies Equ.(i) if and only if
00
0
0
y1 v + [2y1 (t) + p(t)y1 (t)]v = g(t)
(ii)
0
Equation (ii) is a first order linear equation for v . Solving this equation, integrating the result,
and then multiplying by y1 (t) lead to the general solution of Eq.(i).
Answer: Let y = v(t)y1 (t), then
0
0
0
y (t) = v (t)y1 (t) + v(t)y1 (t)
00
00
0
00
0
y (t) = v (t)y1 (t) + 2v (t)y1 (t) + v(t)y1 (t)
So,
00
0
y (t) + p(t)y (t) + q(t)y(t)
00
0
0
00
0
= y1 v + [2y1 (t) + p(t)y1 (t)]v + v(y1 (t) + p(t)y1 (t) + q(t)y1 (t))
= g(t)
if y1 is the solution of the corresponding homogeneous equation and
00
0
0
y1 v + [2y1 (t) + p(t)y1 (t)]v = g(t).
In each of the problems use the method outlined in Problem 28 to solve the given differential
equation.
00
0
30. t2 y + 7ty + 5y = 3t, t > 0; y1 (t) = t
00
0
32. (1 − t)y + ty − y = 2(t − 1)2 e−t , 0 < t < 1; y1 (t) = et
Answer: 30. Use the method in Problem 28, the original equation can be written as
7 0
5
3
00
y (t) + y (t) + 2 y(t) = .
t
t
t
9
Let v satisfies
00
0
0
y1 v + [2y1 (t) + p(t)y1 (t)]v = g(t)
00
0
then v + 5t v = 3 and v = 14 t2 + c1 t−4 .
00
0
So y2 = y1 v = 14 t + c1 t−5 is the solution of t2 y + 7ty + 5y = 3t, t > 0.
Hence, the general solution are
1
y2 = c1 t−5 + c2 t−1 + t.
4
32.Use the method in Problem 28, the original equation can be written as
t
1
00
y (t) +
y0 −
y = 2(1 − t)e−t .
(1 − t)
(1 − t)
Let v satisfies
00
0
0
y1 v + [2y1 (t) + p(t)y1 (t)]v = g(t)
00
0
t
)v = 2(1 − t)e−2t and v = ( 21
1−t
= ( 12 − t)e−t + c1 t is the solution
then v + (2 +
− t)e−2t + c1 te−t .
So y2 = y1 v
of (1 − t)y + ty − y = 2(t − 1)2 e−t , 0 < t <
00
0
1; y1 (t) = et .
Hence, the general solution are
1
y2 = c1 t + c2 et + ( − t)e−t .
2
Section 3.7
In the following problem, determine ω0 , R and δ so as to write the given expression in the form
u = R cos(ω0 t − δ).
√
2. u = − cos t + 3 sin t
3. u = 2 cos 3t − 4 sin 3t
Answer: 2. R =
√
Hence u = 2 cos(t −
√
1 + 3 = 2 and ω0 = 1, δ = arctan(− 3) = − 2π
3
2π
).
3
10
√
√
42 + 22 = 2 5 and δ = arctan( −4
)∼
= −1.1071
2
√
Hence u = 2 5 cos(3t − δ) with δ = arctan( −4
)∼
= −1.1071.
2
3. R =
7.A mass weighing 3 lb stretches a spring 3 in. If the mass is pushed upward, contracting the
spring a distance of 1 in, and then set in motion with a downward velocity of 4 ft/sec, and if
there is no damping, find the position u of the mass at any time t. Determined the frequency,
period, amplitude, and phase of the motion.
Answer: The spring constant is k = 3/(1/4) = 12lb/f t. Mass m = 3/32lb − s2 /f t. Since there
is no damping, the equation of motion is
3 00
u + 12u = 0
32
u00 + 128u = 0
√
√
The general solution isu(t) = A cos 8 2t + B sin 8 2t. Invoking the initial conditions, we have
√
√
1
1
cos 8 2t + √ sin 8 2t
12
2 2
√
√
√
√
R = 1219 f t, δ = π − α tan(6/ 2)rad, w0 = 8 2rad/s, andT = π/(4 2)sec.
u(t) = −
8. A series circuit has a capacitor of 0.25 × 10−6 farad and an inductor of 1 henry. If the initial
charge on the capacitor is 2 × 10−6 coulomb and there is no initial current, find the charge Q on
the capacitor at any time t.
Answer: By Kirchhoff’s Law,
L
dI Q
+ =0
dt C
So we have
LQ00 +
1
Q = 0.
C
for the charge Q. The initial conditions are
Q(0) = Q0 , Q0 (0) = I(0) = 0
11
It is easy to see that the solution is
Q(t) = Q0 cos √
t
= 2 × 10−6 cos 2000t(coulomb)
LC
12. A series circuit has a capacitor of 10−5 farad, a resistor of 3 × 102 ohms, and an inductor of
0.2 henry. The initial charge on the capacitor is 3 × 10−6 coulomb and there is no initial current.
Find the charge Q on the capacitor at any time t.
Answer: By Kirchhoff’s Law,
L
dI
Q
+ RI + = 0
dt
C
So we have
LQ00 + RQ0 +
1
Q = 0.
C
for the charge Q. The initial conditions are
Q(0) = Q0 , Q0 (0) = I(0) = 0
It is easy to see that the solution is
Q(t) = −3 × 10−6 e−1000t + 6 × 10−6 e−500t
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 25% greater than the
period of the corresponding undamped motion.
Answer: √
u00 + γu0 + u = 0, we know the characteristic equation is r2 + γr + 1 = 0, the solution
−γ± γ 2 −4
is r =
. For γ 2 − 4 < 0, the general solution is
2
√
√
4 − γ2
4 − γ2
− γ2
− γ2
u(t) = c1 e cos
t + c2 e sin
t.
2
2
√
4−γ 2
The quasi-period is 2π/( 2 ).
For the undamped motion
u00 + u = 0,
12
√
the period is 2π/(
4
)
2
= 2π.
Since the quasi-period of the damped motion is 25% greater than the period of the corresponding
undamped motion, so
√
2π/(
5
4 − γ2
) = × 2π,
2
4
hence γ = 65 .
15. Show that the solution of the initial value problem
mu00 + γu0 + ku = 0.
u0 (t0 ) = u00
u(t0 ) = u0 ,
can be expressed as the sum u = v + w, where v satisfies the initial conditions v(t0 ) = u0 , v 0 (t0 ) =
0, w satisfies the initial conditions w(t0 ) = 0, w0 (t0 ) = u00 , and both v and w satisfy the same
differential equation as u. This is another instance of superposing soluti6ons of simpler problems
to obtain the solution of a more general problem.
Answer: The general solution of the system is u(t) = A cos γ(t − t0 ) + B sin γ(t − t0 ). Invoking
the initial conditions, we have u(t) = u0 cos γ(t − t0 ) + (u00 /γ) sin γ(t − t0 ). Clearly, the functions
v = u0 cos γ(t − t0 )andw = (u00 /γ) sin γ(t − t0 ) satisfy the given criteria.
16.Show that A cos ω0 t + B cos ω0 t can be written in the form r sin(ω0 t − θ). Determine r and θ
in term of A and B If R cos(ω0 t − δ) = r sin(ω0 t − θ),determined the relationship among R, r, δ,
and θ.
Answer: Let r =
√
A2 + B 2 and choose θ s.t.
cos θ =
A
,
r
sin θ =
B
r
Then
A cos ω0 t + B cos ω0 t = r cos θ · cos ω0 t + r sin θ · sin ω0 t = r sin(ω0 t − θ).
√
Hence r = A2 + B 2 and θ = arctan B
or θ = arctan B
+ π.
A
A
Let f = R cos(ω0 t − δ), g = r sin(ω0 t − θ) and assume that f = g It is easy to know that
the maximum values of f, g are |R|, |r| respectively, hence R = ±r. And then, cos(ω0 t − δ) =
± sin(ω0 t − θ) = ± cos(ω0 t − θ − π/2). hence δ = kπ + π/2 + θ, k ∈ Z.
13
20. Assume that the system described by the equation mu00 + γu0 + ku = 0 is critically damped
and the initial conditions are u(0) = u0 , u0 (0) = v0 . If v0 = 0, show that u → 0 as t → ∞, but
that u is never zero. If u0 is positive, determine a condition on v0 that will assure that the mass
passes through it equilibrium position after it released.
√
Answer: Since the equation is critically damped, γ = 2 km. Hence the general solution is
√k
u(t) = (c1 + c2 t)e− m t , and we always have u(t) → 0 as t → ∞.
From u(0) = u0 , u0 (0) = v0 , we can easily get
s
u(t) = (u0 + (v0 + u0
If v0 = 0, we get u(t) = u0 (1 +
q
√k
k
)t)e−
m
m
t
√k
k
)t)e− m t
m
hence u → 0 as t → ∞, but that u is never zero.
Assume that there exist t0 s.t. u(t0 ) = 0, t0 ≥ 0, we have u0 + (v0 + u0
necessary and sufficient condition for such t0 exists is v0 + u0
s
v0 < −u0
q
k
m
q
k
)t
m 0
= 0. then the
< 0, i.e.
k
.
m
24. The position of a certain spring-mass system satisfied the initial value problem
3 00
0
u + ku = 0, u(0) = 2, u (0) = v.
2
If the period and amplitude of the resulting motion are observed to be π and 3, respectively,
determine the value of k and v.
Answer: The period of the motion is
T = 2π(
m 1
3/2 1
) 2 = 2π(
) 2 = π.
k
k
00
So we get k = 6 and the equation can written as u + 4u = 0.
Obviously, the general solution of this equation is
u(t) = A cos 2t + B sin 2t.
From u(0) = 2, then A = 2.
From the amplitude of the resulting motion is 3, R =
√
√
A2 + B 2 = 3 and then B = ± 5.
14
Hence,
√
u(t) = 2 cos 2t + ± 5 sin 2t
and
√
0
v = u (0) = ±2 5
26. Consider the initial value problem
mu00 + γu0 + ku = 0,
u(0) = u0 ,
u0 (0) = v0
Assume that γ 2 < 4km.
(a) Solve the initial value problem.
(b) Write the solution in the form u(t) = R exp(−γt/2m) cos(µt − δ). Determine R in terms of
m,γ,k,µ0 , and v0 .
(c) Investigate the dependence of R on the damping coefficient γ for fixed values of the other
parameters.
Answer:
(a) The general solution is
√
√
4km − γ 2
4km − γ 2
u=e
t + B sin
t]
[A cos
2m
2m
By u(0) = u0 and u0 (0) = v0 we obtain that
2m
γu0
A = u0 , B = √
(v
+
)
0
2m
4km − γ 2
(b)
γt
− 2m
u(t) = R exp(−γt/2m) cos(µt − δ),
where
R=
√
A2 + B 2 =
v
u
u 4m2 v02
t
+ 4mv0 γu0 + 4kmu20
4km − γ 2
√
4km − γ 2
µ=
2m
B
2m
v0
γ
δ = arctan = arctan √
( +
)
2
A
4km − γ u0 2m
(c) R increases when γ increases.
15
30.In the absence of damping the motion of a spring-mass system satisfies the initial value
problem
00
0
mu + ku = 0, u(0) = a, u (0) = b
(a) Show that the kinetic energy initially imparted to the mass is
energy initially stored in the spring is
(ka2 +mb2 )
2
ka2
2
mb2
,
2
and that the potential
, so that initially the total energy in the system is
.
(b) Solve the given initial value problem.
(c) Using the solution in part (b), determine the total energy in the system at any time t. Your
result should confirm the principle of conservation of energy for this system.
0
0
Answer: (a)u (0) = b ⇒ the kinetic energy initially imparted to the mass is
u(0) = a ⇒ the potential energy initially stored in the spring is
total energy in the system is
k(u(0))2
2
=
ka2
,
2
m(u (0))2
2
u(t) = Acosω0 t + Bsinω0 t
k
m
⇒ ω0 =
q
k
,
m
0
u(0) = a ⇒ A = a and u (0) = b ⇒ B =
q
q
b
ω0
= √b k . Hence,
m
q
u(t) = acos( k/mt) + b m/ksin( k/mt)
(c) The total energy in the system at any time t is
1
1
0
E(t) = m(u (t))2 + k(u(t))2
2
2
when t = 0, E(0) =
(mb2 +ka2 )
2
31.
Answer: Based on Newton’s second law, with the Positive direction to the right,
ΣF = mu00
where
ΣF = −ku − γu0 .
mb2
;
2
then the initially
(ka2 +mb2 )
.
2
(b) Obviously, the general solution of the equation is
ω02 =
=
16
Hence the equation of motion is mu00 + γu0 + ku = 0. The only difference in this problem is that
the equilibrium position is located at the unstretched configuration of the spring.
Section 3.8
In each of problem written the given expression as a product of two trigonometric function of
different frequencies.
2. sin 9t − sin 6t
4. sin 4t + sin 5t
Answer:
2.
2 sin
3t
15t
cos
.
2
2
4.
2 sin(9t/2) cos(t/2)
6. A mass of 5kg stretches a spring 10cm. The mass is acted on by an external force of
10 sin( 2t )N , and moves in a medium that imparts a viscous force of 4N when the speed of the
mass is 8cm/sec, if the mass is set in motion from its equilibrium position with an initial velocity
of 3cm/sec, formulate the initial value problem describing the motion of the mass.
Answer:
u00 + 10u0 + 98u = 2 sin(t/2),
where u in m, t in sec.
u(0) = 0,
u0 (0) = 0.03,
17
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 8 cos 7t lb, determined the position
of the mass at any time and draw a graph of the displacement versus t.
Answer: The spring constant is k=12 lb/ft and hence the equation of motion is
6 00
u + 12u = 8 cos 7t,
32
that is, u00 + 64u =
128
3
cos 7t. The initial conditions are u(0) = 0, u0 (0) = 0. The general
solution is u(t) = A cos 8t + B sin 8t +
− 128
cos 8t +
45
64
45
cos 7t =
256
45
128
45
cos 7t. Invoking the initial conditions, we have u(t) =
sin(t/2) sin(15t/2).
11. A spring is stretched 6 in, by a mass that weighs 8lb, the mass is attached to a dashpot
mechanism that has a damping constat of 0.25 lb-sec/ft and is affected on by an external force
of 4 cos 2tlb.
(a) Determine the steady-state response of this system.
(b) If the given mass is replaced by a mass m, determine the value of m for which the amplitude
of the steady-state response is maximum.
Answer: (a) m =
8lb
32f t/sec2
=
1 lb−sec2
,
4
ft
k=
8lb
1
ft
2
2
.
= 16 flbt , γ = 0.25 lb−sec
ft
From the equation
mu00 + γu0 + ku = 4 cos 2t,
we have
1 00 1 0
u + u + 16u = 4 cos 2t
4
4
Let U (t) = A cos 2t + B sin 2t , we deduced that
U (t) =
8
(30 cos 2t + sin 2t)f t,
901
tinsec.
(b) If the given mass is replaced by a mass m, the equation becomes
1
mu00 + u0 + 2mu = 4 cos 2t.
4
Let U (t) = A cos 2t + B sin 2t, we have A = 4mB, B =
8
,
1−16m2
hence we have R =
we want the steady-state response is maximum, by computation, we need m = 14 lb.
√
A2 + B 2 ,
18
12. Answer: The equation of motion is
u00 + u0 + 3u = 3 cos 3t − 2 sin 3t.
Since the system is damped, the steady state response is equal to the particular solution. Using
the method of undetermined coefficients, we obtain
7
4
uss (t) =
sin 3t −
cos 3t.
15
15
Further, we find that R =
q
13
45
q
13/45 and δ = arctan(−7/4). Hence we can write uss (t) =
cos(3t − π + arctan(7/4)).
16. A series circuit has a capacitor of 0.25×10−6 farad, a resistor of 5×103 ohms, and an inductor
of 1henry. The initial charge on the capacitor is zero. If a 12volt buttery is connected to the
circuit and the circuit is closed at t = 0, determine the charge on the capacitor at t = 0.001,
t = 0.01 sec, and at any time t. Also determine the limiting charge as t −→ ∞.
Answer: First we have
1
Q = E(t),
C
we deduce that Q00 + 5 × 103 Q0 + 4 × 106 Q = 12, Q(0) = 0, Q0 (0) = 0.
LQ00 + RQ0 +
Then Q(t) = 10−6 (e−4000t − 4e−1000t + 3), Q(0.001) ' 1.5468 × 10−6 , Q(0.01) ' 2.9998 × 10−6 ,
and Q(t) −→ 3 × 10−6 as t −→ ∞.