Sample Solutions of Assignment 5 for MAT3270A
Note: Any problems to the sample solutions, email Ms. Zhang Rong
a
(rzhangmath.cuhk.edu.hk)
directly.
October 9,2014
Part 1: Pages 144-145
Problem 2,4,7 on page 144
Find the general solution of the given differential equation
(2). y 00 + 5y 0 + 6y = 0
(4). 3y 00 − 4y 0 + y = 0
(7). y 00 − 8y 0 + 9y = 0
Answer: (2)The characteristic equation is :
r2 + 5r + 6 = 0
Thus r1 = −2,r2 = −3.
The general solution is y = c1 e−2t + c2 e−3t .(Revised)
(4)The characteristic equation is :
3r2 − 4r + 1 = 0
1
Thus r1 = ,r2 = 1.
3
1
The general solution is y = c1 e 3 t + c2 et .
(7)The characteristic equation is :
r2 − 8r + 9 = 0
√
√
Thus r1 = 4 + 7,r2 = 4 − 7.
√
The general solution is y = c1 e(4+
7)t
+ c2 e(4−
Problem 9,12,14,16 on page 144
Answer: (9). The characteristic equation is
r2 + 2r − 3 = 0
1
√
7)t
.
2
Thus the possible values of r are r = 1 and r = −3, and the general
solution of the equation is
y(t) = c1 et + c2 e−3t .
From the initial value, we have
c1 + c2 = 1
(1)
c1 − 3c2 = 1,
hence c1 = 1 and c2 = 0. The solution of the equation is
y(t) = et .
y(t) −→ ∞, as t −→ ∞.
(12).The characteristic equation is :
r2 + 3r = 0
The roots are r = 0, −3.Hence the general solution is
y = c1 + c2 e−3t
According to the initial condition,the specific solution is
y = 1 − e−3t
y(t) −→ ∞, as t −→ ∞.
(14).The characteristic equation is :
2r2 + r − 4 = 0
The roots are r =
√
√
−1+ 33 −1− 33
,
.Hence
4
4
y = c1 e
√
−1+ 33
t
4
the general solution is
+ c2 e
√
−1− 33
t
4
According to the initial condition,the specific solution is
4 −1+√33
4 −1−√33
y= √ e 4 t−√ e 4 t
33
33
3
y(t) −→ ∞, as t −→ ∞.
(16). The characteristic equation is
9r2 − 1 = 0
1
The roots are r = ± . Hence the general solution is
3
1
1
y = c1 e 3 t + c2 e − 3 t
Accounting for the initial conditions, the specific solution is
y(t) = −e
t+2
3
+ 2e−
t+2
3
y(t) −→ −∞, as t −→ ∞.
Problem 20 on page 144
Find the solution of the initial value problem
1
2
Then determine the maximum value of the solution and also find the
2y 00 − 3y 0 + y = 0, y(0) = 2, y 0 (0) =
point where the solution is zero.
Answer: The characteristic equation is
2r2 − 3r + 1 = (2r − 1)(r − 1) = 0
1
Thus the possible values of r are r1 = and r2 = 1, and the general
2
solution of the equation is
1
t
y(t) = c1 e 2 + c2 et .
Using the first initial condition, we obtain
c1 + c2 = 2.
Using the second initial condition, we obtain
1
1
c1 + c2 = .
2
2
4
Solving above equations we find that c1 = 3 and c2 = −1. Hence,
1
t
y(t) = 3e 2 − et .
Since
1
1
t
t
9
3
9
t
y(t) = 3e 2 − e = − (e 2 − )2 ≤ ,
4
2
4
9
3
when t = 2 ln , y(t) reach the maximum value . Solving y(t) =
2
4
1
t
3e 2 − et = 0, we know that the zero point of solution is t = 2 ln 3.
Problem 22 on page 144
00
0
Solve the initial value problem 4y − y = 0, y(0) = 2, y (0) = β. Then
find β so that the solution approaches zero as t → ∞.
Answer: The the characteristic equation is
4r2 − 1 = (2r + 1)(2r − 1) = 0
Thus the possible values of r are r1 =
1
2
and r2 =
−1
,
2
and the general
solution of the equation is
1
1
y(t) = c1 e 2 t + c2 e− 2 t .
Using the first initial condition, we obtain
c1 + c2 = 2.
Using the second initial condition, we obtain
2c1 − c2 = 2β.
By solving above equations we find that c1 = β + 1 and c2 = 1 − β.
Hence,
1
1
y(t) = (β + 1)e 2 t + (1 − β)e− 2 t .
From y(t) → 0 as t → ∞, we find β = −1.
5
(24). Determine the values of α, if any, for which all solution tend
to zero as t → ∞; also determine the values of α, if any, for which
all(nonzero) solutions become unbounded as t → ∞.
Answer: The characteristic equation is
r2 + (3 − α)r − 2(α − 1) = (r + 2)(r + 1 − α) = 0
Thus the possible values of r are r1 = −2 and r2 = α − 1, and the
general solution of the equation is
y(t) = c1 e−2t + c2 e(α−1)t .
So when α < 1, all solutions tend to zero as t → ∞; when α > 1, all
solutions become unbounded as t → ∞.
Problem 28 on page 145
Consider the equation ay 00 + by 0 + c = 0, where a, b, and c are constants
with a > 0. Find conditions on a, b, and c such that the roots of the
characteristic equation are:
(a) real, different, and negative.
(b) real with opposite signs.
(c) real, different, and positive.
Answer: The characteristic equation is
ar2 + br + c = 0
(a) The conditions are b2 > 4ac, b > 0 and c > 0.
(b) The condition is c < 0.
(c) The conditions are b2 > 4ac, b < 0 and c > 0.
6
Part 2:164-165
Problems1,3,5 on page 164
Answer:
1.e(cos 3 + i sin 3)
3.cos 3π + i sin 3π = −1
5. 21−i = exp[(1 − i)ln2] = exp[ln 2] exp[− ln 2i] = 2(cos ln 2 − i sin ln 2)
Problem 8,10,13,16 on page 164
Answer:
8.The characateristic equation is :
r2 − 2r + 8 = 0
The roots are r = 1 ±
√
7i,hence the general solution is:
√
y = c1 e(1+
7i)t
√
+ c2 e(1−
7i)t
10.The characateristic equation is :
r2 + 4r + 5 = 0
The roots are r = 2 ± i,hence the general solution is:
y = c1 e(2+i)t + c2 e(2−i)t
13.The characateristic equation is :
r2 + 6r + 9.25 = 0
1
The roots are r = −3 ± i,hence the general solution is:
2
1
1
y = c1 e(−3+ 2 i)t + c2 e(−3− 2 i)t
7
16.The characateristic equation is :
r2 + 6r + 11.25 = 0
3
The roots are r = −3 ± i,hence the general solution is:
2
3
3
y = c1 e(−3+ 2 i)t + c2 e(−3− 2 i)t
Problem 18,20 on page 164
Answer: 18. The characteristic equation is
r2 + 4r + 5 = 0
Thus the possible values of r are r1 = −2 + i and r2 = −2 − i, and the
general solution of the equation is
y(t) = e−2t (c1 cos t + c2 sin t).
Substitute the initial data, we have
c1 = 1, c2 − 2c1 = 1
Then c1 = 1, c2 = 3. The solution is
y(t) = e−2t (cos t + 3 sin t)
20. 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.
Substitute the initial data, we have
√
√
c1 = 1 + 3, c2 = 3 − 1
8
Then the solution is
y(t) = (1 +
√
√
3) cos t + ( 3 − 1) sin t
27. Show that W (eλt cos µt, eλt sin µt) = µe2λt
Answer: Let
y1 = eλt cos µt
y2 = eλt sin µt
0
0
W (eλt cos µt, eλt sin µt) = y2 y1 − y1 y2
= e2λt cos µt(λ sin µt + µ sin µt) − e2λt sin µt(λ cos µt − µ sin µt)
= µe2λt
33. If the functions y1 and y2 are linearly independent solutions of
00
0
y + p(t)y + q(t) = 0, show that between consecutive zeros of y1 there
is one and only one zero of y2 . Note that this result is illustrated by
00
the solutions y1 = cos t and y2 = sin t of the equation y + y = 0.
Answer: Assume the two consecutive zeros of y1 are t1 and t2 , and
t1 < t2 . Then y1 (t) < 0 or y1 (t) > 0 for all t1 < t < t2 .
Case A: We consider the case y1 (t) > 0 for all t1 < t < t2 .
0
0
In this case, obviously, y1 (t1 ) ≥ 0 and y1 (t2 ) ≤ 0.
0
0
W (y1 , y2 ) = y2 y1 − y1 y2 6= 0 for all t, because y1 and y2 are linearly
00
0
independent solutions of y + p(t)y + q(t) = 0 , then W (y1 , y2 )(t) > 0
or W (y1 , y2 )(t) < 0 for all t.
9
If W (y1 , y2 )(t) > 0 for all t.
We get
0
W (y1 , y2 )(t1 ) = −y1 (t1 )y2 (t1 ) > 0
0
W (y1 , y2 )(t2 ) = −y1 (t2 )y2 (t2 ) > 0
Hence y2 (t1 ) > 0 and y2 (t2 ) < 0, so there is one zero of y2 .
If W (y1 , y2 )(t) < 0 for all t.
We get
0
W (y1 , y2 )(t1 ) = −y1 (t1 )y2 (t1 ) < 0
0
W (y1 , y2 )(t2 ) = −y1 (t2 )y2 (t2 ) < 0
Hence y2 (t1 ) < 0 and y2 (t2 ) > 0, so there is one zero of y2 .
Case B: We consider the case y1 (t) < 0 for all t1 < t < t2 .
0
0
In this case, obviously, y1 (t1 ) < 0 and y1 (t2 ) > 0.
0
0
W (y1 , y2 ) = y2 y1 − y1 y2 6= 0 for all t, because y1 and y2 are linearly
00
0
independent solutions of y + p(t)y + q(t) = 0 , then W (y1 , y2 )(t) > 0
or W (y1 , y2 )(t) < 0 for all t.
If W (y1 , y2 )(t) > 0 for all t.
We get
0
W (y1 , y2 )(t1 ) = −y1 (t1 )y2 (t1 ) > 0
0
W (y1 , y2 )(t2 ) = −y1 (t2 )y2 (t2 ) > 0
Hence y2 (t1 ) < 0 and y2 (t2 ) > 0, so there is one zero of y2 .
If W (y1 , y2 )(t) < 0 for all t.
We get
0
W (y1 , y2 )(t1 ) = −y1 (t1 )y2 (t1 ) < 0
0
W (y1 , y2 )(t2 ) = −y1 (t2 )y2 (t2 ) < 0
Hence y2 (t1 ) > 0 and y2 (t2 ) < 0, so there is one zero of y2 .
We need to show there is only one zero of y2 between t1 and t2 .
If otherwise, then there is at least two zero of y2 between t1 and
t2 , we can select two consecutive zeros of y2 ,say t3 and t4 , such that
10
t1 < t3 < t4 < t2 . Then from above proof we can show there is another
zero of y1 , say t5 such that t1 < t3 < t5 < t4 < t2 . This contradict that
t1 and t2 are two consecutive zeros of y1 .
Part 3:Pages 172-175
In each of problems find the general solution of the given differential
equation
2. 9y 00 + 12y 0 + 4y = 0
4. 4y 00 − 12y 0 + 9y = 0
6. y 00 − 10y 0 + 25y = 0
10. 2y 00 + 2y 0 + y = 0
Answer:
2. The characteristic equation is
9r2 + 12r + 4 = 0
Thus the possible values of r are r1 = r2 = − 23 , and the general solution
of the equation is
2t
y(t) = (c1 + c2 t)e− 3 .
4. The characteristic equation is
4r2 − 12r + 9 = 0
Thus the possible values of r are r1 = r2 = 32 , and the general solution
of the equation is
3t
y(t) = (c1 + c2 t)e 2 .
6. The characteristic equation is
r2 − 10r + 25 = 0
11
Thus the possible values of r are r1 = r2 = 5, and the general solution
of the equation is
y(t) = (c1 + c2 t)e5t .
10. The characteristic equation is
2r2 + 2r + 1 = 0
Thus the possible values of r are r1 =
−1+i
2
and r2 =
−1−i
,
2
and the
general solution of the equation is
t
y(t) = e− 2 (c1 cos
t
t
+ c2 sin ).
2
2
Problem 12 on page 173
Answer: The characteristic equation is
r2 − 6r + 9 = 0
Thus the possible values of r are r1 = r2 = 3, and the general solution
of the equation is
y(t) = (c1 + c2 t)e3t
Invoking the initial conditions, we have
y(t) = 2te3t
From the figure, we know that as t increases to ∞, the solution tends
to ∞,.
Problem 13 on page 173
Answer: The characteristic equation is
9r2 + 6r + 82 = 0
Thus the possible values of r are r1 = − 13 + 12 i, r2 = − 13 − 12 i, and the
general solution of the equation is
1
1
1
y(t) = (c1 cos t + c2 sin t)e− 3 t
2
2
12
20 000
15 000
10 000
5000
0.5
1.0
1.5
2.0
2.5
3.0
Figure 1. Figs for problem (12): y(t)
Invoking the initial conditions, we have
1
1
1
y(t) = (−3 cos t + 2 sin t)e− 3 t
2
2
From the figure, we know that as t increases to ∞, the solution tends
to 0.
Problem 16 on page 173
Answer: The characteristic equation is
r2 − r + 0.25 = 0
Thus the possible values of r are r1 = r2 = 12 , and the general solution
of the equation is
1
y(t) = (c1 + c2 t)e 2 t
Invoking the initial conditions, we have
1
y(t) = [2 + (b − 1)t]e 2 t
The critical value b = 1 , as t increases to ∞, the solution tends to
+∞,when b > 1, the solution tends to −∞,when b < 1.
Problem 18 on page 173
Consider the initial value problem
9y 00 + 12y 0 + 4y = 0, y(0) = a > 0, y 0 (0) = −1.
13
(a) Solve the initial value problem.
(b)Find the critical value of a that separates solutions that become
negative from those that are always positive.
Answer: (a)The characteristic equation is
9r2 + 12r + 4 = 0
Thus the possible values of r are r1 = r2 = − 23 , and the general solution
of the equation is
2
y(t) = (c1 + c2 t)e− 3 t .
From y(0) = a, we get c1 = a.
0
From y (0) = −1, we get c2 = 23 a − 1.
Therefore,
2
2
y(t) = (a + ( a − 1)t)e− 3 t .
3
(b)Since y(t) > 0 ⇔ a + ( 23 a − 1)t > 0. That is, a >
t
=
1 + 23 t
3
9
−
, for all t ≥ 0. So a > 32 , and so the critical value of a is
2 2(2t + 3)
3
.
2
Problem 20 on page 173
Answer: (a) The characteristic equation is
r2 + 2ar + a2 = 0
Thus the possible values of r are r1 = r2 = −a, and the general solution
of the equation is
y(t) = (c1 + c2 t)e−at
(b)Using Abel’s formula ,we have
Z
W (y1 , y2 )(t) = Cexp[−
p(t)dt] = Ce−2at
(c) y1 (t) = e−at ,use the result of part (b),we have
e−at y20 + ae−at y2 = Ce−2at
14
Solve the above ODE ,thus y2 = te−at
Problem 23,25,27 on page 174
Answer: 23. Let y2 (t) = t3 v(t), then
0
0
y = v t3 + 3t2 v
00
00
0
y = v t3 + 6t2 v + 6tv
So we get
00
0
tv + 2v = 0.
Then
v(t) = c1 t−1 + c2
and so
y(t) = c1 t2 + c2 t3 .
From y1 (t) = t3 , we find the second solution is y2 (t) = t2
25. Let y2 (t) = t−1 v(t), then
0
0
y = v t−1 − t−2 v
00
00
0
y = v t−1 − 2t−2 v + 2t−3 v
So we get
00
0
tv + v = 0.
Then
v(t) = c1 ln t + c2
and so
y(t) = c1 t−1 ln t + c2 t−1 .
From y1 (t) = t−1 , we find the second solution is y2 (t) = t−1 ln t
27.Let y(x) = v(x) sin x2 , then
0
0
y = v sin x2 + 2vx cos x2 ,
00
00
0
y = v sin x2 + 4v x cos x2 + 2v(cos x2 − 2x2 sin x2 ).
15
So we get
00
0
v x sin x2 + (4x2 cos x2 − sin x2 )v = 0.
Then
v(x) = c1 cot x2 + c2
and so
y(t) = c1 cos x2 + c2 sin x2 .
From y1 (t) = sin x2 , we find the second solution is y2 (t) = cos x2 .(Revised)
Problem 38 on page 175
Answer: (a). a > 0 and b > 0, but c = 0
The characteristic equation is
ar2 + cr = 0
p
p
Thus the possible values of r are r1 = ac i and r2 = − ac i, and the
general solution of the equation is
r
r
c
c
y(t) = c1 cos
t + c2 sin
t.
a
a
Hence, all solutions are bounded as t → ∞.
(b): a > 0 and b > 0, but c = 0
The characteristic equation is
ar2 − br = 0
Thus the possible values of r are r1 = 0 and r2 = − ab , and the general
solution of the equation is
y(t) = c1 + c2 e
−bt
a
.
Hence, y(t) → c1 , as t → ∞, since a, b are positive constants. Obviously, c1 depends on the initial conditions.
16
0
From y (0) = y00 , we can get c2 = −
From y(0) = y0 , we get c1 = −
ay00
b
ay00
.
b
+ y0 .
Problem 39 on page 175
Answer:
(sin t)00 + (k sin2 t) sin0 t + (1 − k cos t sin t) sin t = 0
− sin t(k sin2 t) cos t + (1 − k cos t sin t) sin t = 0
1
1
On the other hand:1 − k cos t sin t = 1 − k sin 2t > 1 − k > 0 and
2
2
k sin2 t > 0 is naturally true.
Part 4 :Pages 184-186
Problem 2,3,5,7,8,11,13 on page 184
Answer: 2. The characteristic equation is
r2 + 2r + 5 = 0
Thus the possible values of r are r1 = −1 + 2i and r2 = −1 − 2i, and
the general solution of the homogeneous equation is
y(t) = e−t (c1 cos 2t + c2 sin 2t).
Let Y (t) = A cos 2t + B sin 2t where A, B are to be determined. On
substituting to the original equation, we get
(A + 4B) cos 2t + (−4A + B) sin 2t = 4 sin 2t.
12
So A = − 17
,B=
3
17
and
y(t) = e−t (c1 cos 2t + c2 sin 2t) −
16
4
cos 2t +
sin 2t
17
17
17
is the general solution of the original equation.
3.The characterisic equation is
r2 − r − 2 = 0
Thus the possible values of r are r1 = −1 andr2 = 2, and the general
solution of the homogeneous equation is
y(t) = c1 e−t + c2 e2t .
Let Y (t) = At2 + Bt + C where A, B, C are to be determined.On
substituting to the original equation, we get
−2At2 − 2(A + B)t + 2A − B − 2C = −3 + 4t2 .
So A = −2, B = 2, C = − 32 and
y(t) = c1 e−t + c2 e2t − 2t2 + 2t −
3
2
is the general solution of the original equation.
5. It is easy to see, the general solution of the homogeneous equation
is
y(t) = c1 e3t + c2 e−t .
Let Y (t) = t(At + B)e−t where A, B is a constant to be determined.
On substituting to the equation, we get
−8Ate−t + (2A − 4B)e−t = −6te−t .
So A = 43 , B = 83 . and
3
3
y(t) = c1 e3t + c2 e−t + t( t + )e−t
4
8
is the general solution of the original equation.
18
7.It is easy to see, the general solution of the homogeneous equation
is
y(t) = c1 cos 3t + c2 sin 3t.
Let Y (t) = D + (At2 + Bt + C)e3t where A, B, C, D are constants to
be determined. On substituting to the original equation, we get
[18At2 + (18B + 12A)t + 18C + 6B + A]e3t + 9D = 18 + t2 e3t .
So A =
1
,
18
1
D = 2, B = − 27
, C=
y(t) = c1 cos 3t + c2 sin 3t + (
1
162
and
1 2
1
1 3t
t − t+
)e + 2
18
27
162
is the general solution of the original equation.
8.It is easy to see,the general solution of the homogeneous equation
is
y(t) = c1 e−t + c2 te−t
Suppose Y1 (t) = At2 e−t is a particular solution and substitute it into the equation and we get A = 2,Thus the general solution to the
equation is:
y(t) = c1 e−t + c2 te−t + 2t2 e−t
11.The characteristic equation is
r2 + ω02 = 0
Thus the possible values of r are r1 = iω0 and r2 = −iω0 , and the
general solution of the homogeneous equation is
u(t) = c1 cos ω0 t + c2 sin ω0 t.
Let Y (t) = A cos ωt + B sin ωt where A and B are constants to be
determined. On substituting to the original equation, we get
(ω02 − ω 2 )B sin ωt + (ω02 − ω 2 )A cos ωt = cos ωt.
19
So A =
1
,
ω02 −ω 2
B = 0 and
u(t) = c1 cos ω0 t + c2 sin ω0 t +
ω02
1
cos ωt
− ω2
is the general solution of the original equation.
13. The characteristic equation of the homogeneous equation is
r2 + r + 4 = 0
√
√
−1 + 15i
−1 − 15i
Thus the possible values of r are r1 =
and r2 =
,
2
2
and the general solution of the homogeneous equation is
√
√
1
1
− t
− t
15
15
y(t) = c1 e 2 cos
+ c2 e 2 sin
.
2
2
Since f (t) = 2 sinh t, we can assume a particular solution of ODE is
1
1
c1 et + c2 e−t , then we have c1 = , c2 = − . so, the general solution is
3
2
√
√
1
1
− t
− t
15
15 1 t 1 −t
+ c2 e 2 sin
+ e − e .
y(t) = c1 e 2 cos
2
2
3
2
Problem 16 on page 184
Answer: Obviously the general solution to the homogeneous equation
is y = c1 cos 2t + c2 sin 2t ,suppose the particular solution is Y (t) =
1
At2 + Bt + C + Dtet ,substitute it into the equation we get:A = , B =
4
1
3
0, C = − , D = .So the general solution is
8
5
1
1 3
y(t) = c1 cos 2t + c2 sin 2t + t2 − + tet .
4
8 5
According to the initial condition we get the specific solution;
y=−
19
1
1
1 3
cos 2t + sin 2t + t2 − + tet .
40
5
4
8 5
20
Problem 18 on page 184
Answer: Obviously the general solution to the homogeneous equation
is y = c1 e−t + c2 e3t ,suppose the particular solution is Y (t) = Ae2t +
2
Bte2t ,substitute it into the equation we get:A = − , B = −1.So the
3
general solution is
2
y(t) = c1 e−t + c2 e3t − e2t − te2t .
3
According to the initial condition we get the specific solution;
3
1
2
y = e3t + 6e−t − e2t − te6t
2
3
3
Problem 20 on page 184
Answer: The characteristic equation is r2 + 2r + 5 = 0, with roots
r = −1 ± 2i. The general solution of the homogeneous equation is
y = (c1 cos 2t + c2 sin 2t)e−t .
Let Y (t) = Ate−t cos 2t + Bte−t sin 2t. Substituting, we find Y =
te−t sin 2t.
Hence the general solution is
y(t) = (c1 cos 2t + c2 sin 2t)e−t + te−t sin 2t.
Invoking the initial conditions, y(t) = te−t sin 2t.
Problem 29 on page 185
Answer: (a) Since Y = ve−t ,we have Y 0 = v 0 e−t − ve−t ,Y 00 = v 00 e−t −
2v 0 e−t + ve−t .
Then Y 00 − 3Y 0 − 4Y = 2e−t ⇒ v 00 − 5v 0 = 2.
(b) Let ω = v 0 ,v 00 − 5v 0 = 2 ⇒ ω 0 − 5ω = 2
2
Solve this ODE ,we find ω = Ce5t −
5
21
2
(c) Integrate ω,we find v = C1 e5t − t + C2
5
2 −t
−t
4t
Thus Y = C1 e − te + C2 e
5
Problem 33 on page 185
Answer: Let u = Y1 (t) − Y2 (t), clearly, u is the solution of the homogeneous equation
ay 00 + by 0 + cy = 0.
Hence by the problem 38 of section 3.5, we know that
Y1 (t) − Y2 (t) = u(t) → 0.
By (a) part of problem 39 of section 3.5, we know that the result is
false for b = 0.
Problem 34 on page 185
Answer: Assume that c 6= 0. Then, Y (t) = d/c is a particular solution
to Eq. (i). Therefore the general solution of the Eq. (i) is
y(t) = d/c + u(t)
where u(t) is the general solution of the corresponding homogeneous
equation
ay 00 + by 0 + cy = 0.
By problem 31 of this section, one has u(t) → 0 as t → ∞. Therefore,
y(t) → d/c as t → ∞.
Section 3.6
22
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
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
23
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
Z
Z
Y (t) = − cos t 2 sin t tan tdt + sin t 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 <
π
2
are the general solutions of the original equation.
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 .
24
W (y1 , y2 ) = e−4t then a particular solution of the original equation
is
Z
y2 (t)g(t)
Y (t) = −y1 (t)
dt + y2 (t)
W (y1 , y2 )
= −2e−2t ln t − e−2t
Z
y1 (t)g(t)
dt
W (y1 , y2 )
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
1
1
particular solution is u(t) = u1 (t) cos t + u2 (t) sin t,then
2
2

1
1


u01 cos t + u02 sin t = 0
2
2
1 0
1
1 0
1

 − u1 sin t + u2 cos t = 8 sec(t/2)
2
2
2
2
Hence


 u01 = −4 sin 1 t sec 1 t
2
2
1
1

 u02 = 4 cos t sec t
2
2
Solve this equation,
1
u1 = 8 ln cos t
2
u2 =
4t
1
1
1
Then u(t) = 8(ln cos t) cos t + 4t sin t. The general solution is
2
2
2
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
Answer: 11. The characteristic equation is
r2 − 7r + 10 = 0
25
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 te5t .
W (y1 , y2 ) = 3e7t then a particular solution of the original equation
is
Z
Y (t) = −y1 (t)
Z
=
y2 (t)g(t)
dt + y2 (t)
W (y1 , y2 )
Z
y1 (t)g(t)
dt
W (y1 , y2 )
[e5(t−s) − e2(t−s) ]g(s)ds
Hence,
2t
5t
y(t) = c1 e + c2 e +
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
3
0
2 00
2
18. x y + xy + (x − 0.25)y = 3x 2 sin x, x > 0; y1 =
1
1
−
−
x 2 sin x, y2 = x 2 cos x
20.
1
1
−
x y +xy +(x −0.25)y = g(x), x > 0; y1 = x 2 sin x, y2 = x 2 cos x
2
00
0
−
2
Answer: 13. It is easy to check that y1 and y2 satisfy
00
t2 y − 2y = 0, t > 0.
26
W (y1 , y2 ) = −3 and let
Z −1 2
Z 2 2
t (4t − 3)
t (4t − 3)
2
−1
Y (t) = −t
+t
−3
−3
2
1
= t4 − t2 ln t + t2 , t > 0.
5
3
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
Y (t) = −(1 + t)
+
e
tet
tet
5
5
1
= ( t2 − t + )e2t .
2
4
4
then a particular solution of the original equation is
5
5
1
y(t) = ( t2 − t + )e2t .
2
4
4
18. It is easy to check that y1 and y2 satisfy
Z
00
0
x2 y + xy + (x2 − 0.25)y = 0
1
and let
x
Z
Z
y1 (t)g(t)
y2 (t)g(t)
dt + y2 (t)
dt
Y (t) = −y1 (t)
W (y1 , y2 )
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
27
1
and let
x
Z
Z
y1 (t)g(t)
y2 (t)g(t)
dt + y2 (t)
dt
Y (t) = −y1 (t)
W (y1 , y2 )
W (y1 , y2 )
W (y1 , y2 ) = −
1Z 1
= x 2 t 2 sin(x − t)g(t)dt.
−
then a particular solution of the original equation is
1Z 1
−
y(t) = x 2 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.
28
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
Z t
y1 (s)y2 (t) − y1 (t)y2 (s)
Y (t) =
g(s)ds.
0
0
t0 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.
Answer:
Z
Y (t) = −y1 (t)
Z
t
=
t0
Z
y2 (s)g(s)
ds + y2 (t)
W (y1 , y2 )(s)
−y1 (t)y2 (s)g(s)
ds +
0
0
y1 (s)y2 (s) − y2 (s)y1 (s)
t
Z
=
t0
Z
t
t0
y1 (s)g(s)
ds
W (y1 , y2 )(s)
y2 (t)y1 (s)g(s)
ds
0
0
y1 (s)y2 (s) − y2 (s)y1 (s)
y1 (s)y2 (t) − y2 (s)y1 (t)
g(s)ds
0
0
y1 (s)y2 (s) − y2 (s)y1 (s)
Hence,
t0
Z
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
t
Z
y2 (s)g(s)
ds + y2 (t)
W (y1 , y2 )(s)
Y (t) = −y1 (t)
t0
t
Z
y1 (s)g(s)
ds,
W (y1 , y2 )(s)
t0
we get
0
Z
0
t
Y (t) = −y1 (t)
t0
y2 (s)g(s)
0
ds + y2 (t)
W (y1 , y2 )(s)
Z
t
t0
y1 (s)g(s)
ds
W (y1 , y2 )(s)
then
Z t0
y2 (s)g(s)
y1 (s)g(s)
0
Y (t0 ) = −y1 (t0 )
ds + y2 (t0 )
ds = 0
t0 W (y1 , y2 )(s)
t0 W (y1 , y2 )(s)
Z t
Z t
y2 (s)g(s)
y1 (s)g(s)
00
00
00
Y (t) = −y1 (t)
ds + y2 (t)
ds + g(t)
t0 W (y1 , y2 )(s)
t0 W (y1 , y2 )(s)
0
0
Z
t0
29
00
0
Hence, Y (t) + p(t)Y (t) + q(t)Y (t)
00
t
Z
0
y1 (s)g(s)
ds
W (y1 , y2 )(s)
= [y2 (t) + p(t)y2 (t) + q(t)y2 (t)]
t0
Z
0
00
t
−[y1 (t) + p(t)y1 (t) + q(t)y1 (t)]
t0
y2 (s)g(s)
ds + g(t) ≡ g(t).
W (y1 , y2 )(s)
Therefore, Y (t) is a solution of the initial value problem
0
L[y] = g(t), y(t0 ) = 0, y (t0 ) = 0.
28. The method of reduction of order (Section 3.5) 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
0
00
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)
30
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
5
3
7 0
00
y (t) − y (t) + 2 y(t) = .
t
t
t
Let v satisfies
00
0
0
y1 v + [2y1 (t) + p(t)y1 (t)]v = g(t)
00
0
then tv + 5t v = 3 and v = 14 t2 + c1 t−4 .
00
0
So y2 = y1 v = 41 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
00
y (t) +
t
1
−
= 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)
0
t
)v = 2(1 − t)e−2t and v = ( 12 − t)e−2t + c1 te−t .
1−t
00
0
So y2 = y1 v = ( 12 − t)e−t + c1 t is the solution of (1 − t)y + ty −
2(t − 1)2 e−t , 0 < t < 1; y1 (t) = et .
00
then v + (2 +
Hence, the general solution are
1
y2 = c1 t + c2 et + ( − t)e−t .
2
y=
31
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
√
√
2π
1 + 3 = 2 and ω0 = 1, δ = arctan(− 3) = −
3
2π
Hence u = 2 cos(t −
).
3√
√
3. R = 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
Answer: 2. 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
32
√
√
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
1
Q = 0.
C
for the charge Q. The initial conditions are
LQ00 +
Q(0) = Q0 , Q0 (0) = I(0) = 0
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
33
So we have
1
Q = 0.
C
for the charge Q. The initial conditions are
LQ00 + RQ0 +
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 thep
characteristic equation is
−γ
±
γ2 − 4
r2 + γr + 1 = 0, the solution is r =
. For γ 2 − 4 < 0,
2
the general solution is
p
p
γ
γ
−
−
4 − γ2
4 − γ2
u(t) = c1 e 2 cos
t + c2 e 2 sin
t.
2
2
p
4 − γ2
The quasi-period is 2π/(
).
2
For the undamped motion
u00 + u = 0,
√
4
) = 2π.
2
Since the quasi-period of the damped motion is 50% greater than the
the period is 2π/(
period of the corresponding undamped motion, so
p
4 − γ2
5
2π/(
) = × 2π,
2
4
r
6
hence γ =
.
5
34
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).
√
B
B
Hence r = A2 + B 2 and θ = arctan or θ = arctan + 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.
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
35
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
v
u
uk
t
−
t
the general solution is u(t) = (c1 + c2 t)e m , and we always have
u(t) → 0 as t → ∞.
From u(0) = u0 , u0 (0) = v0 , we can easily get
r
u(t) = (u0 + (v0 + u0
r
If v0 = 0, we get u(t) = u0 (1+
k
)t)e
m
k
)t)e
m
v
u
u
t
−
v
u
u
t
−
k
t
m
k
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 +
r
k
u0
)t0 = 0. then the necessary and sufficient condition for such t0
m
r
k
exists is v0 + u0
< 0, i.e.
m
r
k
.
v0 < −u0
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
36
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 =
√
and then B = ± 5.
Hence,
√
A2 + B 2 = 3
√
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) = 0,
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
p
p
2
γt
4km
−
γ
4km − γ 2
u = e− 2m [A cos
t + B sin
t]
2m
2m
By u(0) = u0 and u0 (0) = v0 we obtain that
2m
γu0
A = u0 , B = p
(v0 +
)
2
2m
4km − γ
(b)
u(t) = R exp(−γt/2m) cos(µt − δ),
37
where
√
s
4m2 v02 + 4mv0 γu0 + 4kmu20
4km − γ 2
p
4km − γ 2
µ=
2m
B
2m
v0
γ
δ = arctan = arctan p
( +
)
A
4km − γ 2 u0 2m
R=
A2 + B 2 =
(c) R increases when γ increases.
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
and that the potential energy initially stored in the spring is
that initially the total energy in the system is
mb2
,
2
ka2
,
2
so
(ka2 +mb2 )
.
2
(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
Answer: (a)u (0) = b ⇒ the kinetic energy initially imparted to the
0
mass is
m(u (0))2
2
stored in the
the system is
mb2
; u(0) = a ⇒ the potential energy initially
2
2
2
spring is k(u(0))
= ka2 , then the initially total energy in
2
(ka2 +mb2 )
.
2
=
(b) Obviously, the general solution of the equation is
u(t) = Acosω0 t + Bsinω0 t
ω02 =
√b
k
m
k
m
⇒ ω0 =
q
k
,
m
0
u(0) = a ⇒ A = a and u (0) = b ⇒ B =
. Hence,
p
p
p
u(t) = acos( k/mt) + b m/ksin( k/mt)
b
ω0
=
38
(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 .
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.sin9t − sin 6t
4.sin4t + sin5t
Answer:
2.
2 sin
3t
15t
cos
.
2
2
4.
2 sin(9t/2) cos(t/2)
39
6. A mass of 5kg stretches a spring 10cm. The mass is acted on by an
t
external force of 10 sin( )N , and moves in a medium that imparts a
2
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),
u(0) = 0,
u0 (0) = 0.03,
where u in m, t in sec.
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
cos 7t. The initial conditions are u(0) =
3
0, u0 (0) = 0. The general solution is u(t) = A cos 8t + B sin 8t +
128
45
128
45
cos 7t. Invoking the initial conditions, we have u(t) = − 128
cos 8t +
45
cos 7t =
256
45
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.
40
Answer: (a) m =
8lb
8lb
lb
1 lb − sec2
, k =
= 16 , γ =
=
2
1
32f t/sec
4
ft
ft
ft
2
lb − sec2
.
ft
From the equation
0.25
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
8
Let U (t) = A cos 2t + B sin 2t, we have A = 4mB, B =
,
1 − 16m2
√
hence we have R = A2 + B 2 , we want the steady-state response is
1
maximum, by computation, we need m = lb.
4
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
q
Further, we find that R = 13
and δ = arctan(−7/4). Hence we can
45
q
write uss (t) = 13
cos(3t − π + arctan(−7/4)).
45
41
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 −→ ∞.