ODE
Homework 3 Solutions
Spring 2013
March 25, 2013
Problem 4.3 #38
Find the general solution of each of the differential equations.
y 00 + 7y 0 + 10y = 4xe−3x
(1)
with the initial condition y(0) = 0 and y 0 (0) = −1.
Solution. Note that a solution to the homogenous equation (1) will also be a solution to the particular differential equation. First we solve the following homogeneous equation
y 00 + 7y 0 + 10y = 0
(2)
Note that this ode has a characteristic polynomial of
r2 + 7r + 10 = (r + 2)(r + 5)
(3)
Which has zeros r = −2, r = −5. Thus, a solution to the homogeneous solution is
yc = c1 e−2x + c2 e−5x
(4)
−2x −5x . It’s easy to
In the language of the text, the homogenous equation
a UC set
has
of S1 = e , e
−3x
−3x
see that a UC set for the particular solution will be xe , e
. Therefore, yp = Cxe−3x +De−3x .
yp0 = −3Cxe−3x + Ce−3x − 3De−3x
yp00
−3x
= 9Cxe
Plugging the above computations of
yp0 ,
−3x
− 3Ce
yp00
−3x
− 3Ce
(5)
−3x
+ 9De
(6)
into (1) gives Grouping terms
(9C + 7(−3)C + 10C)xe−3x + (7(C − 3D) − 3C − 3C + 9D + 10D)e−3x = 4xe−3x + 0e−3x
(7)
Because of independence, the coefficients of the LHS must equal the coefficients of the RHS
9C − 21C + 10C = 4
7C − 21D − 6C + 19D = 0
(8)
Then C = −2, −2 − 2D = 0 and D = −1. The solution of the ODE is then y = yc + yp or
y = c1 e−2x + c2 e−5x − 2xe−3x − e−3x
(9)
Using the initial condition y(0) = 0,
0 = c1 + c2 − 1
(10)
Using the initial condition y 0 (0) = −1, we need y 0
y 0 = −2c1 e−2x − 5c2 e−5x + 6xe−3x − 2e−3x + 3e3x
(11)
−1 = −2c1 − 5c1 + 0 − 2 + 3
(12)
gives a system of equations
(
0
0
= c1 + c2 − 1
= −2c1 − 5c2 + 2
(13)
Has solution c1 = 1 and c2 = 0. Therefore,
y = e−2x + −2xe−3x − e−3x
is the unique solution to the initial value problem.
2
(14)
Problem 4.4 # 4
y 00 + y = sec3 x
(15)
yc is then y 00 + y = 0 which has the characteristic equation E
r2 + 1 = 0
(16)
Which has roots r = ±i and
1. yc = c1 cos x + c2 sin x
2. (a) y1 = cos x
(b) y2 = sin x
Solve the system
(
−c01 sin x + c02 cos x = sec3 x
c01 cos x + c2 sin x = 0
(17)
Multiplying the first equation by cos x and the second equation by sin x gives the equivalent system
(
−c01 sin x cos x + c02 cos2 x = sec2 x
(18)
c01 sin x cos x + c2 sin2 x = 0
Adding the two equations in (18)
c02 (cos2 x + sin2 x) = sec2 x
|
{z
}
(19)
=1
and
c02
=
sec2 x.
We can now solve for
c01
in the homogeneous equation of (17) via substitution
c01 cos x + sec2 x sin x = 0
(20)
⇒
− sec2 x sin x
cos x
− sin x
=
cos3 x
c01 =
(21)
(22)
Integrating (22)
Z
sin x
dx
cos3 x
use u = cos x, then du = − sin x and (23) becomes
Z
du
1
1
c1 = −
= u−2 /2 = cos−2 x = sec2 x
u3
2
2
R
Solving for c2 : c02 = sec2 x, i.e., c2 = sec2 x = tan x. Therefore
c1 = −
yp =
(23)
(24)
1
sec2 x cos x + tan x sin x
2
(25)
1
sec x + tan x sin x
2
(26)
Simplifying
yp =
3
Problem 4.5 # 27
Solve the initial-value problem in each of the exercises. In each case assume that x > 0.
x2 y 00 − 4xy 0 + 4y = 4x2 − 6x3
(27)
y(2) = 4 and y 0 (2) = −1.
Solution. Using the method outlined in the book according to Theorem 4.14 p.172. reduces to
dy
d2 y
+ A1
+ A2 y = G(t)
(28)
2
dt
dt
where A0 = a0 , A1 = a1 − a0 , A2 = a2 and G(t) = F (et ). In (27) we have a0 = 1, a1 = −4,
a2 = 4 and F (x) = 4x2 − 6x3 . Simple substitution gives A0 = 1, A1 = −4 − 1 = −5, A2 = 4, and
G(t) = F (et ) = 4e2t − 6e3t and (28) becomes
A0
d2 y
dy
− 5 + 4y = 4e2t − 6e3t
(29)
2
dt
dt
To solve 29, first solve the homogeneous equation y 00 − 5y 0 + 4y = 0 has a characteristic equation
r2 − 5r + 4 = 0 or (r − 4)(r − 1) = 0, i.e.,
yc = c1 et + c2 e4t
(30)
2t 3t To solve the particular equation, note that a UC set is e , e , i.e., yp = Ae2t + Be3t . Compute
the derivatives of yp :
1. yp0 = 2Ae2t + 3Be3t
2. yp00 = 4Ae2t + 9Be3t
Plugging yp , yp0 , yp00 into (29):
4Ae2t + 9Be3t − 5 2Ae2t + 3Be3t + 4 Ae2t + Be3t = 4e2t − 6e3t
(31)
Refactor (31) in the form αe2t + βe3t :
(4A − 10A + 4A)e2t + (9B − 15B + 4B)e3t = 4e2t − 6e3t
(32)
Since the UC set is a set of linearly independent functions, we must have matching coefficients, i.e.,
4A − 10A + 4A = 4
9B − 15B + 4B = −6
(33)
or −2A = 4, A = −2, and −2B = −6 and B = 3. Therefore
yp (t) = −2e2t + 3e3t
(34)
y(t) = c1 et + c2 e4t − 2e2t + 3e3t
(35)
and y(t) = yc (t) + yp (t) or
back substituting x =
et
into (35)
y(x) = c1 x + c2 x4 − 2x2 + 3x3
(36)
with initial value conditions gives
y(x) =
−23 4
5
x + 3x3 − 2x2 + x
24
3
4
(37)
Problem 5.3 #2
To establish a spring constant mg = k` or 16 = 8/12k, i.e., k = 24. To establish the mass w = mg
⇒ 16 = m32 ⇒ m = 12 . Therefore, obtain the equation
1 00
x + 6x0 + 24x = 0
2
(38)
with the initial conditions x0 (0) = 2 and x(0) = 0. From (38), solve the characteristic polynomial
1 2
r + 6r + 24 = 0
2
(39)
√
r = −6 ± 2 3i
(40)
√
√
x(t) = c1 e−6t sin 2 3t + c2 e−6t cos 2 3t
(41)
has solutions
So that
With the initial condition x(0) = 0 plugged into (41) we get
0 = x(0) = c1 · 1 · 0 + c2 · 1 · 1 = c2
(42)
√
x = c1 e−6t sin 2 3t
(43)
i.e., 0 = c2 . Hence
Differentiate (43) and use the initial condition x0 (0) = 2
√
√
√
x0 = c1 (−6)e−6t sin 2 3t + 2 3c1 e−6t cos 2 3t
plug in t = 0 and x0 = 2
√
or 2 = 2 3c1 or c1 =
√1 .
3
(44)
√
2 = c1 (−6) · 1 · 0 + 2 3c1 · 1·
(45)
√
1
x(t) = √ e−6t sin 2 3t
3
(46)
Hence,
Problem 5.4 #2
A 16-lb weight is attached to the lower end of a coil spring suspended from the ceiling. The weight
comes to rest in its equilibrium position, thereby stretching the spring 0.4ft. Then, beginning at
t = 0, an external force given by F (t) = 40 cos 16t is applied to the system. The medium offers
resistance in pounds numerically equal to 4x0 , where x0 is the instantaneous velocity in feet per
second.
1. Find the displacement of the weight as a function of time.
2. Graph separately the transient and steady-state terms of the motion found in step 1 and then
use the curves so obtained to graph the entire displacement itself. See (1)
5
Solution. Hooke’s law, we have F = ks or k = 16/0.4 = 16/(4/10) = 16(10/4) = 160/4 = 40. To
solve for the mass of the weight, note that F = gm or m = 16/(32) = 1/2 slugs. The resistance of
the medium is 4x0 , i.e., a = 4 in the following equation:
mx00 + ax0 + kx = F1 (t)
(47)
The acted on force is F1 (t) = 40 cos 16t and (47) becomes
1 00
x + 4x0 + 40x = 40 cos 16t
2
(48)
with the initial conditions x(0) = 0, x0 (0) = 0
Has a homogeneous solution
xh (t) = c1 e−4t sin(8t) + c2 e−4t cos(8t)
(49)
and a particular solution
xp (t) =
1
(8 sin(16t) − 11 cos(16t))
37
(50)
Therefore, x(t) = xh + xp or
x(t) = c1 e−4t sin(8t) + c2 e−4t cos(8t) +
1
(8 sin(16t) − 11 cos(16t))
37
(51)
With initial conditions, x(0) = 0 and x0 (0) = 0 (47) Let x(0) = 0 and notice that only cos coefficients
survive, i.e.,
0 = c2 − 11/37
(52)
and c2 = 11/37. With the same observation, we only compute derivaties of summands in (??)
which will result in a factor of cos:
0 = x0 (0) = 8c1 − 4c2 + (8 · 16)/37
(53)
Or 0 = 8c1 − 4c2 + 128/37 or 0 = 8c1 − 44/37 + 128/37, i.e., c1 = − 21
74 . Therefore,
x(t) = −
11
1
21 −4t
e sin(8t) + e−4t cos(8t) +
(8 sin(16t) − 11 cos(16t))
74
37
37
is a solution to the initial value problem.
6
(54)
Figure 1: Part (b) Graph of the displacement x(t)
7