Chapter 18
1.
Solve the initial-value problem.
y   16 y   233 y  0 , y (0)  1, y (0)  2
2.
Solve the initial-value problem.
y  16 y  64 y  0 , y (2)  0, y(2)  1
3.
Solve the initial-value problem.
 
 
y  49 y  0 , y   0, y   1
7
7
4.
Solve the boundary-value problem, if possible.
y  14 y  49 y  0 , y (0)  0, y (1)  6
Select the correct answer.
a. y  7 xe( 6 x  6)
no solution
5.
b.
y  6 xe( 7 x  7)
c.
Select the correct answer.
d.
21
c3
32
1
5
21
y( x)  c1e  4 x  c2e  x  x 2  x 
4
8
32
1 2 5
21
y ( x)  x  x 
4
8
32
4 x
x
y( x)  c1e  c2e
e.
y( x)  c1e 4 x  c2e x
a.
b.
c.
y( x)  c1e  4 x  c2e  x 
Use power series to solve the differential equation.
y  7 x 2 y
7.
Solve the differential equation.
y  14 y  85 y  0
8.
d.
Solve the differential equation using the method of undetermined coefficients.
y  5 y  4 y  x 2
6.
y  6 xe( 7 x  7)
Solve the differential equation.
y   y  cos x
y  6e( 7 x  7)
e.
9.
Solve the differential equation using the method of variation of parameters.
y  2 y  y  e3x
Select the correct answer.
a.
y( x)  e x  xe x
b.
y( x)  c1e x  c2 xe x 
c.
d.
e.
1 3x
e
4
1
y( x)  c1e x  c2e x  e3x
4
1 3x
x
y( x)  c1e  e
4
x
y( x)  c1e x  c2e x  e3x
4
10. Solve the differential equation using the method of variation of parameters.
y  3 y  e3x
11. Solve the differential equation using the method of variation of parameters.
y  3 y  2 y 
5
1  e x
12. Solve the differential equation using the method of variation of parameters.
y  4 y  4 y 
e 2 x
x3
Select the correct answer.
a.
b.
c.
d.
e.
1 

y( x)  e  2 x  c1  c2

2
x

1 

y( x)  e  2 x  c1  c2 x  c3

2x 

1 

y( x)  e  2 x  c1  c2 x  
2x 

1
y( x)  e  2 x c1  c2 x  
2x
2 x
y( x)  e c1  c2 
13. A spring with a 16-kg mass has natural length 0.8 m and is maintained stretched to a length of 1.2 m
by a force of 19.6 N. If the spring is compressed to a length of 0.6 m and then released with zero
velocity, find the position x(t) of the mass at any time t.
14. A spring with a mass of 2 kg has damping constant 14, and a force of 6 N is required to keep the spring
stretched 0.5 m beyond its natural length. The spring is stretched 1m beyond
its natural length and then released with zero velocity. Find the position x(t) of the mass at any time t.
15. A spring with a mass of 2 kg has damping constant 14, and a force of 4.8 N is required to keep the
spring stretched 0.4 m beyond its natural length. Find the mass that would produce critical damping.
16. A spring with a mass of 3 kg has damping constant 35 and spring constant 129. Find the damping
constant that would produce critical damping.
Select the correct answer.
a.
c  6 43
b.
c6 7
c9 7
c.
d.
c  9 43
e.
c  43 7
17. A series circuit consists of a resistor R  20  , an inductor with L = 1 H, a capacitor with C = 0.002F,
and a generator producing a voltage of E (t )  12 cos10t . If the initial charge and current are both 0,
find the charge Q(t) at time t.
18. Use power series to solve the differential equation.
y   xy   y  0, y (0)  4, y (0)  0
19. Use power series to solve the differential equation.
y  x 2 y  0, y (0)  2, y(0)  0
Select the correct answer.

a.
y ( x) 
 (1)
n
n 1

c.
y( x)  2 
 (1)
n
n 1

e.
y ( x)  2 

2 x 4n
4n(4n  1)... 4  3
 (1)
n
n 1
n
x
4n(4n  1)... 4  3
y  11 y  24 y  0
Select the correct answer.
a.
y( x)  c1e8 x  c2 xe8 x
b.
y( x)  c1e3x  c2e8 x
c.
y( x)  ce3x
d.
y( x)  c1e3x  c2 xe3x
e.
y( x)  c1e3x  c2 xe x
y ( x)  2 
 (1)
n 1

4n
2x
4n(4n  1)... 4  3
20. Solve the differential equation.
b.
d.
y ( x)  2 
 (1)
n 1
n
n
x 4n
4n(4n  1)... 4  3
2 x 4n
4n(4n  1)... 4  3
1.
Solve the differential equation.
y   2 y   e x
2.
Solve the differential equation.
y  8 y  41 y  0
Select the correct answer.
3.
a.
y( x)  e4 x (c1 cos(5x)  c2 sin(5x))
b.
y( x)  ce4 x cos(5x)
c.
y( x)  e5 x (c1 cos(4x)  c2 sin(4x))
d.
y( x)  e4 x (c1 cos(5x)  c2 x sin(5x))
e.
y( x)  e4 x (c1 cos(4x)  c2 x sin(4x))
Solve the differential equation.
y  14 y  49 y  0
4.
Solve the differential equation.
11 y  2 y
5.
Solve the differential equation.
9 y  y  0
6.
Solve the differential equation.
7 y  y  0
7.
Solve the differential equation.
d 2 y dy

 10 y  0
dt 2 dt
8.
Solve the initial-value problem.
2 y  19 y  24 y  0, y (0)  0, y(0)  1
9.
Solve the initial-value problem.
y  16 y  64 y  0, y (2)  0, y(2)  1
10. Solve the differential equation using the method of undetermined coefficients.
y  4 y  e 2 x
11. Solve the differential equation using the method of variation of parameters.
y  5 y  4 y  sin x
12. Solve the initial-value problem.
y  81y  0, y  / 9  0, y  / 9  1
Select the correct answer.
a.
b.
c.
d.
e.
1
x sin(9 x)
9
1
y( x)   sin(9 x)
9
1
y( x)  sin(9 x)
9
1
y( x)   x sin(9 x)
9
y( x)   x sin(9 x)
y ( x) 
13. Solve the boundary-value problem, if possible.
y  10 y  25 y  0, y 0  0, y 1  8
14. Solve the boundary-value problem, if possible.
y  3 y  54 y  0, y 0  0, y 2  1
15. Solve the boundary-value problem, if possible.
y  y  2 y  0, y  1  1, y 1  0
16. Solve the differential equation using the method of undetermined coefficients.
y  3 y  sin 9 x
17. Graph the particular solution and several other solutions.
2 y  3 y  y  2  cos 2 x
Select the correct answer.
a.
b.
c.
18. Solve the differential equation using the method of undetermined coefficients.
y  8 y  17 y  e x
19. Find a trial solution for the method of undetermined coefficients. Do not determine the coefficients.
y  3 y  5 y  x 4e2 x
Select the correct answer.

( x)  Ax

a.
y p ( x)  Ax 3  Bx 2  Cx  D e 2 x
b.
yp
c.
y p ( x )  Ax 4
d.
y p ( x)  Ax 4  Bx 3  Cx 2  Dx  E e 2 x
e.
yp

( x)  Ax
4
4

 B e2x


 Bx 2  C e  2 x
20. Find a trial solution for the method of undetermined coefficients. Do not determine the coefficients.
y  2 y  1  xe2 x
Select the correct answer.
a.
b.
c.
d.
e.
y p1 ( x)  A
y p 2 ( x)  x( A  B)e  2 x
y p1 ( x)  Ax
y p 2 ( x)  x 2 ( Ax  B)e  2 x
y p1 ( x)  Ax
y p 2 ( x)  x( Ax  B)e  2 x
y p1 ( x)  A
y p 2 ( x)  x( A  Bx)e  2 x
y p1 ( x)  A
y p 2 ( x)  Bxe 2 x