Math 528
Homework 2 Solutions
Assignment:
Section 2.1: #1-9 odd
Section 2.2: #1-35 odd
Section 2.4: #1-5, 11, 13, 15
Section 2.6: #1,5,9,10,13
Section 2.7: # 1-17 odd
Section 2.10: #1-5
Section 4.3: #1-15 odd
Graded:
Section 2.1: #3,5,9
Section 2.2: #1,7,13,19,25,31
Section 2.4: #1,2
Section 2.6: #9,10,13
Section 2.7: # 1,5,9,13
Section 2.10: #2,3,4
Section 4.3: #3,7,9,11
Section 2.1
3. Let u = y 0 then u0 = y 00 and the problem becomes
u0 + u = 0.
This differential equation is separable.
Z
Z
du
= − dx ⇒ u = c˜1 e−x .
u
u = y 0 = c˜1 e−x ⇒ y = −c˜1 e−x + c2 = c1 e−x + c2 .
5. Let u = y 0 then u0 = y 00 and the problem becomes
yu0 = 3u2 .
Now,
u0 =
du dy
du
du
=
=u
dx
dy dx
dy
so
du
y
= 3u ⇒
dy
Z
du
=3
u
Z
dy
⇒ ln u = 3 ln y + c˜1 ⇒ u = c˜2 y 3 ⇒ y 0 = c˜2 y 3 .
y
This is now separable,
Z
Z
dy
1
3
−3
= c˜2 y ⇒ y dy = c˜2 dx ⇒ − y −2 = c˜2 x + c˜3 ⇒ y = (c1 x + c2 )−1/2 .
dx
2
9. We can rewrite the differential equation in standard form as
9
5
y 00 − y 0 + 2 y = 0.
x
x
5
3
Since p(x) = − x and y1 = x
U=
1 R
e
x6
5
dx
x
=
1 5 ln x
1
1
e
= 6 · x5 = .
6
x
x
x
Thus,
Z
y2 = y1
U dx = x
3
Z
1
dx = x3 ln x ⇒ y = c1 x3 + c2 x3 ln x.
x
Section 2.2
1. The characteristic equation is
4λ2 − 25 = 0.
Its roots are
5
5
and λ2 = −
2
2
so that we obtain the general solution
λ1 =
5
5
y = c1 e 2 x + c2 e − 2 x .
3. The characteristic equation is
λ2 + 6λ + 8.96 = 0.
By the quadratic formula, its roots are
λ1 = −2.8 and λ2 = −3.2
so that we obtain the general solution
y = c1 e−2.8x + c2 e−3.2x .
5. The characteristic equation is
λ2 + 2πλ + π 2 = (λ + π)2 = 0.
Its root is
λ=π
so that we obtain the general solution
y = c1 e−πx + c2 xe−πx .
7. The characteristic equation is
λ2 + 4.5λ = λ(λ + 4.5) = 0.
Its roots are
λ1 = 0 and λ2 = −4.5
so that we obtain the general solution
y = c1 + c2 e−4.5x .
9. The characteristic equation is
λ2 + 1.8λ − 2.08 = 0.
By the quadratic formula, its roots are
λ1 = .8 and λ2 = −2.6
so that we obtain the general solution
y = c1 e.8x + c2 e−2.6x .
11. The characteristic equation is
4λ2 − 4λ − 3 = 0.
By the quadratic formula, its roots are
λ1 =
3
2
and λ2 = −
so that we obtain the general solution
3
1
y = c1 e 2 x + c2 e − 2 x .
1
2
13. The characteristic equation is
9λ2 − 30λ + 25 = 0
Its root is
λ=
5
3
so that we obtain the general solution
5
5
y = c1 e 3 x + c2 xe 3 x .
15. The characteristic equation is
λ2 + .54λ + (0.0729 + π) = 0
By the quadratic formula, its complex roots are
√
√
λ1 = −0.27 + i π and λ2 = −0.27 − i π
so that we obtain the general solution
y = e−0.27x (A cos
√
√
πx + B sin πx).
17. The ODE must have a characteristic equation with the root
√
λ=− 5
so the characteristic equation must be
√
√
(λ + 5)2 = λ2 + 2 5λ + 5 = 0.
The following ODE has the required characteristic equation:
√
y 00 + 2 5y 0 + 5y = 0.
19. The ODE must have a characteristic equation with complex roots
λ1 = −2 + i and λ2 = −2 − i
so the characteristic equation must be
(λ + 2 − i)(λ + 2 + i) = λ2 + 4λ + 5 = 0.
The following ODE has the required characteristic equation:
y 00 + 4y 0 + 5y = 0.
21. The characteristic equation is
λ2 + 25 = 0.
Its complex roots are
λ1 = 5i and λ2 = −5i
so that we obtain the general solution
y = A cos 5x + B sin 5x
and
y 0 = −5A sin 5x + 5B cos 5x.
Plugging in the initial conditions we find that
A
= 4.6
5B = −1.2
so A = 4.6 and B = −0.24 and the solution to the IVP is
y = 4.6 cos 5x − 0.24 sin 5x.
23. The characteristic equation is
λ2 + λ − 6 = (λ + 3)(λ − 2) = 0.
Its roots are
λ1 = −3 and λ2 = 2
so that we obtain the general solution
y = c1 e−3x + c2 e2x
and
y 0 = −3c1 e−3x + 2c2 e2x .
Plugging in the initial conditions we find that
c1 + c2 = 10
−3c1 + 2c2 = 0
so c1 = 4 and c2 = 6 and the solution to the IVP is
y = 4e−3x + 6e2x
25. The characteristic equation is
λ2 − 1 = 0.
Its roots are
λ1 = 1 and λ2 = −1
so that we obtain the general solution
y = c1 ex + c2 e−x
and
y 0 = c1 ex − c2 e−x .
Plugging in the initial conditions we find that
c1 + c2 =
2
c1 − c2 = −2
so c1 = 0 and c2 = 2 and the solution to the IVP is
y = 2e−x .
27. The characteristic equation is
λ2 + 2πλ + π 2 = (λ + π)2 = 0.
Its root is
λ=π
so that we obtain the general solution
y = c1 e−πx + c2 xe−πx .
and
y 0 = −πc1 e−πx + c2 e−πx (−πx + 1).
Plugging in the initial conditions we find that
c1
= 4.5
−πc1 + c2 = −4.5π − 1
so c1 = 4.5 and c2 = −1 so the solution to the IVP is
y = (4.5 − x)e−πx .
29. The characteristic equation is
λ2 + .54λ + (0.0729 + π) = 0
By the quadratic formula, its complex roots are
√
√
λ1 = −0.27 + i π and λ2 = −0.27 − i π
so that we obtain the general solution
y = e−0.27x (A cos
√
√
πx + B sin πx).
and
√
√
√
√ y 0 = e−0.27x (− πA − 0.27B) sin πx + ( πB − 0.27A) cos πx .
Plugging in the initial conditions we find that
A
= 0
√
−0.27A +
πB = 1
√
so A = 0 and B = 1/ π so the solution to the IVP is
√
1
y = √ e−0.27x sin πx.
π
31. We find the Wronskian of the given functions:
kx
e
xekx
kx
kx
W [e , xe ] = kx
ke
(kx + 1)ekx
= e2kx .
Since e2kx 6= 0 for any x (or any k), the given functions are linearly independent
on all real numbers.
33. We find the Wronskian of the given functions:
2
x
x2 ln x
2
2
W [x , x ln x] = 2x x(2 ln x + 1)
= x3 .
Since x3 6= 0 for any x > 1 the given functions are linearly independent on all
real numbers x > 1.
35. We find the Wronskian of the given functions:
sin 2x
cos x sin x
W [sin 2x, cos x sin x] = 2 cos 2x cos2 x − sin2 x
sin 2x
=
2 cos 2x
sin 2x =0
cos 2x 1
2
Since the Wronskian of the given functions is 0, they are linearly dependent
(on all intervals).
Section 2.4
1. The model for the given data is
y(t) = A cos ω0 t + B sin ω0 t
y(0) = y0
y 0 (0) = v0 .
Thus, A = y0 and B = v0 /ω0 so
y = y0 cos ω0 t + (v0 /ω0 ) sin ω0 t.
2. We can find the spring constant k with W = 0.02k ⇒ 20 = 0.02k ⇒ k = 1000
nt/m. The mass is m = W/g = 20/9.8 ≈ 2.041 kg. This gives a frequency of
p
k/m/(2π) ≈ 3.523 Hz. The period is 1/3.523 ≈ 0.2838 s.
Section 2.6
9. (a) We need the characteristic equation to have the complex roots ±5i. Thus,
the characteristic equation is λ2 + 25 = 0 so the ODE is
y 00 + 25y = 0.
(b) We find the Wronskian of the given functions:
cos 5x
sin 5x
W [cos 5x, sin 5x] = −5 sin 5x 5 cos 5x
=5
Since 5 6= 0 for any x the given functions are linearly independent on all
real numbers.
(c) The solution the given initial conditions is
y = 3 cos 5x − sin 5x.
10. (b) We find the Wronskian of the given functions:
m2
xm1
x
m1
m2
m1 +m2 −1
W [x , x ] = m1 −1
m2 −1 = (m2 − m1 )x
m1 x
m2 x
Assuming m1 6= m2 , (m2 − m1 )xm1 +m2 −1 6= 0 for any x 6= 0 the given
functions are linearly independent on all real numbers.
(c) The solution the given initial conditions is
y = 2xm1 − 4xm2
13. (a) We need the characteristic equation to have the roots 0 and −2. Thus,
the characteristic equation is λ(λ + 2) = 0 so the ODE is
y 00 + 2y 0 = 0
(b) We find the Wronskian of the given functions:
1 e−2x −2x
= −2e−2x
W [1, e ] = 0 −2e−2x Since −2e−2x 6= 0 for any x the given functions are linearly independent
on all real numbers.
(c) The solution the given initial conditions is
1 1 −2x
+ e .
2 2
Section 2.7
1. The characteristic equation is
λ2 + 5λ + 4 = (λ + 4)(λ + 1) = 0
so the solution of the homogeneous ODE is
yh = c1 e−4x + c2 e−x
By the basic rule
yp = Ce−3x
yp0 = −3Ce−3x
yp00 = 9Ce−3x
Plugging into the ODE
9Ce−3x − 15Ce−3x + 4Ce−3x = 10e−3x
so C = −5. Therefore the general solution to the ODE is
y = c1 e−4x + c2 e−x − 5e−3x .
5. By similar methods as 1 we find the solution to the homogeneous ODE is
yh = c1 e−2x + c2 xe−2x .
By the basic rule
yp = e−x (K cos x + M sin x).
We find K and M by finding yp0 and yp00 and plugging into the ODE.
9. By similar methods as 1 we find the solution to the homogeneous ODE is
yh = c1 e4x + c2 e−4x .
By the modification rule and the sum rule
yp = Kxe4x + K̃ex
We find K and K̃ by finding yp0 and yp00 and plugging into the ODE.
13. By similar methods as 1 we find the solution to the homogeneous ODE is
1
1
y h = c1 e 2 x + c2 e 4 x .
By the basic rule
yp = K cosh x + M sinh x.
We find K and M by finding yp0 and yp00 and plugging into the ODE. We find
c1 and c2 by plugging in y(0) = 0.2 and y 0 (0) = 0.05. (We can also solve this
by replacing cosh x with 21 (ex + e−x ).
Section 2.10
1. First we find the homogeneous solution. The characteristic equation is
λ2 + 9 = 0.
Its complex roots are
λ1 = 3i and λ2 = −3i
so that we obtain the solution
yh = A cos 3x + B sin 3x.
A basis of the homogeneous ODE is thus y1 = cos 3x, y2 = sin 3x. This gives
the Wronskian
cos 3x
sin
3x
=3
W [sin 2x, cos x sin x] = −3 sin 3x 3 cos 3x Using the formula for variation of parameters, we get the particular solution
of the given ODE
Z
Z
sin 3x sec 3x
cos 3x sec 3x
yp = − cos 3x
dx + sin 3x
dx
3
3
Z
Z
cos 3x
sin 3x
sin 3x
=−
dx +
dx
3
cos 3x
3
1
1
= cos 3x ln | cos 3x| + x sin 3x.
9
3
Therefore, we obtain the solution to the ODE
y = A cos 3x + B sin 3x +
1
1
cos 3x ln | cos 3x| + x sin 3x.
9
3
2. First we find the homogeneous solution. From 1 we obtain the solution
yh = A cos 3x + B sin 3x
and the Wronskian W = 3. Using the formula for variation of parameters, we
get the particular solution of the given ODE
Z
Z
cos 3x csc 3x
sin 3x csc 3x
dx + sin 3x
dx
yp = − cos 3x
3
3
Z
Z
cos 3x
sin 3x
cos 3x
=−
dx +
dx
3
3
sin 3x
1
1
= − x cos 3x + sin 3x ln | sin 3x|.
3
9
Therefore, we obtain the solution to the ODE
1
1
y = A cos 3x + B sin 3x − x cos 3x + sin 3x ln | sin 3x|.
3
9
3. We can rewrite the problem in standard form
2
2
y 00 − y 0 + 2 y = x sin x.
x
x
By inspection we can find y1 = x, y2 = x2 is a basis of the homogeneous ODE.
This gives the Wronskian
x sin x2 2
= x2 .
W [x, x ] = 1
2x Using the formula for variation of parameters, we get the particular solution
of the given ODE
Z 3
Z 2
x sin x
x sin x
2
yp = −x
dx
+
x
dx
x2
x2
Z
Z
2
= −x x sin x dx + x
sin x dx
Z
= −x −x cos x + cos x dx − x2 cos x
= x2 cos x − x sin x − x2 cos x
= −x sin x.
Therefore, we obtain the solution to the ODE
y = c1 x + c2 x2 − x sin x.
4. First we find the homogeneous solution. The characteristic equation is
λ2 − 4λ + 5 = 0.
By the quadratic formula we find its complex roots are
λ1 = 2 + i and λ2 = 2 − i
so that we obtain the solution
yh = e2x (A cos x + B sin x).
A basis of the homogeneous ODE is thus y1 = e2x cos x, y2 = e2x sin x. This
gives the Wronskian
2x
2x
e
cos
x
e
sin
x
2x
2x
= e4x
W [e cos x, e sin x] = 2x
2x
2x
2x
−e sin x + 2e cos x e cos x + 2e sin x Using the formula for variation of parameters, we get the particular solution
of the given ODE
Z
Z
(e2x sin x) (e2x csc x)
(e2x cos x) (e2x csc x)
2x
2x
dx
+
e
sin
x
dx
yp = −e cos x
e4x
e4x
Z
Z
cos x
2x
2x
= −e cos x dx + e sin x
dx
sin x
= −xe2x cos x + e2x sin x ln | sin x|.
Therefore, we obtain the solution to the ODE
y = e2x (A cos x + B sin x) − xe2x cos x + e2x sin x ln | sin x|.
5. First we find the homogeneous solution. The characteristic equation is
λ2 + λ = 0.
We find its complex roots are
λ1 = i and λ2 = −i
so that we obtain the solution
yh = A cos x + B sin x.
Using the method of undetermined coefficients and the Modification Rule we
choose
yp = Kx cos x + M x sin x.
yp0 = −Kx sin x + K cos x + M x cos x + M sin x
= (M x + K) cos x + (−Kx + M ) sin x
yp00 = −(M x + K) sin x + M cos x + (−Kx + M ) cos x − K sin x
= (−Kx + 2M ) cos x + (−M x − 2K) sin x
Plugging into the ODE, we find
(−Kx + 2M ) cos x + (−M x − 2K) sin x + Kx cos x + M x sin x = cos x − sin x
2M cos x − 2K sin x = cos x − sin x
Thus, M = K = 21 . Therefore, we obtain the solution to the ODE
1
1
y = A cos x + B sin x + x cos x + x sin x.
2
2
Section 4.3
1. We must find solutions of the system
0
y = Ay =
1
1
3 −1
y.
The characteristic equation is
1−λ
1
det(A − λI) = 3
−1 − λ
= λ2 − 4 = 0.
This gives eigenvalues λ1 = 2 and λ2 = −2. Eigenvectors are obtained from
(1 − λ)x1 + x2 = 0.
For λ1 = 2 this is −x1 + x2 = 0. Hence we can take x
(1)
this becomes 3x1 + x2 = 0, and an eigenvector is x(2)
1
=
. For λ2 = −2
1
1
=
. This gives
−3
the general solution
y=
y1
y2
= c1
1
1
2t
e + c2
1
−3
e−2t .
3. We must find solutions of the system
0
y = Ay =
1 2
1
1
2
y.
The characteristic equation is
1−λ
2
det(A − λI) = 1
1
−
λ
2
= λ2 − 2λ = 0.
This gives eigenvalues λ1 = 0 and λ2 = 2. Eigenvectors are obtained from
(1 − λ)x1 + 2x2 = 0.
2
For λ1 = 0 this is x1 + 2x2 = 0. Hence we can take x =
. For λ2 = 2
−1
2
(2)
this becomes −x1 + 2x2 = 0, and an eigenvector is x =
. This gives
1
the general solution
y1
2
2
= c1
y=
+ c2
e2t .
y2
−1
1
(1)
5. We must find solutions of the system
0
y = Ay =
2
5
5 12.5
The characteristic equation is
2−λ
5
det(A − λI) = 5
12.5 − λ
y.
= λ2 − 14.5λ = 0.
This gives eigenvalues λ1 = 0 and λ2 = 14.5. Eigenvectors are obtained from
(2 − λ)x1 + 5x2 = 0.
5
For λ1 = 0 this is 2x1 + 5x2 = 0. Hence we can take x =
. For
−2 2
λ2 = 14.5 this becomes −12.5x1 + 5x2 = 0, and an eigenvector is x(2) =
.
5
This gives the general solution
y1
5
2
= c1
y=
+ c2
e14.5t .
y2
−2
5
(1)
7. We must find solutions of the system


0
1 0
0 1  y.
y0 = Ay =  −1
0 −1 0
The characteristic equation is
−λ
1
0
1
det(A − λI) = −1 −λ
0 −1 −λ
= −λ3 − 2λ = 0.
√
√
This gives eigenvalues λ1 = 0, λ2 = 2i, and λ3 = − 2i. Eigenvectors are
obtained from
−λx1 + x2
= 0
−x1 − λx2 + x3 = 0.
For λ1 = 0 this is
x2
−x1
For λ2 =
√
x3
= 0
= 0.
2i this is
√
− 2ix1 + √ x2
= 0
2ix2 + x3 = 0.
−x1 −
√
For λ3 = − 2i this is
√
2ix1 + √ x2
= 0
−x1 +
2ix2 + x3 = 0.
Hence we can take

x(1)

1
=  0 ,
−1

x(2)

√1
=  2i  ,
−1

x(3)

√1
=  − 2i 
−1
This gives the general solution








1
y1
1
√
√
√1
√
y =  y2  = c1  0  + c˜2  2i  e 2it + c˜3  − 2i  e− 2it .
y3
−1
−1
−1
Expanding this solution, using Euler’s formula (eix = cos x+i sin x), and letting
c2 = c˜2 + c˜3 and c3 = i(c˜2 − c˜3 ) we get the solution
√
√
y1 = c1 + c2 cos 2t + c3 sin 2t
√
√
√
√
y2 = 2c3 cos 2t − 2c2 sin 2t
√
√
y3 = c1 − c2 cos 2t − c3 sin 2t
9. We must find solutions of the system


10 −10 −4
1 −14  y.
y0 = Ay =  −10
−4 −14 −2
The characteristic equation is
10 − λ −10
−4
det(A − λI) = −10 1 − λ −14
−4
−14 −2 − λ
= −(λ − 18)(λ + 18)(λ − 9) = 0.
This gives eigenvalues λ1 = 18, λ2 = −18, and λ3 = 9. Eigenvectors are
obtained from
(10 − λ)x1 −
10x2 − 4x3 = 0
−10x1 + (1 − λ)x2 − 14x3 = 0.
For λ1 = 18 this is
−8x1 − 10x2 − 4x3 = 0
−10x1 − 17x2 − 14x3 = 0.
For λ2 = −18 this is
28x1 − 10x2 − 4x3 = 0
−10x1 + 19x2 − 14x3 = 0.
For λ3 = 9 this is
x1 − 10x2 − 4x3 = 0
−10x1 + −8x2 − 14x3 = 0.
Hence we can take

x(1)

2
=  −2  ,
1

x(2)

1
=  2 ,
2

x(3)

−2
=  −1 
2
This gives the general solution




 


y1
2
1
−2
y =  y2  = c1  −2  e18t + c2  2  e−18t + c3  −1  e9t .
y3
1
2
2
11. We must find solutions of the system
0
y = Ay =
2
5
y.
− 21 − 32
The characteristic equation is
2−λ
5
det(A − λI) = 3
1
−2 −2 − λ
= (2λ + 1)(λ − 1) = 0.
This gives eigenvalues λ1 = 1 and λ2 = − 12 . Eigenvectors are obtained from
(2 − λ)x1 + 5x2 = 0.
For λ1 = 1 this is x1 + 5x2 = 0. Hence we can take x
(1)
=
λ2 = − 12 this becomes 25 x1 + 5x2 = 0, and an eigenvector is x(2)
5
. For
−1
2
=
.
−1
This gives the general solution
y1
5
2
t
y=
= c1
e + c2
e−0.5t .
y2
−1
−1
Plugging in the initial values we get
5
2
c1
−12
=
.
−1 −1
c2
0
Solving this we find c1 = −4 and c2 = 4 giving the solution to the IVP
y1
5
2
t
y=
= −4
e +4
e−0.5t .
y2
−1
−1
13. We must find solutions of the system
0
y = Ay =
0 1
1 0
y.
The characteristic equation is
−λ 1
det(A − λI) = 1 −λ
= λ2 − 1 = 0.
This gives eigenvalues λ1 = 1 and λ2 = −1. Eigenvectors are obtained from
−λx1 + x2 = 0.
1
. For λ2 = − 12
1
For λ1 = 1 this is −x1 + x2 = 0. Hence we can take x(1) =
1
(2)
this becomes x1 + x2 = 0, and an eigenvector is x =
. This gives the
−1
general solution
y1
1
1
t
= c1
y=
e + c2
e−t .
y2
1
−1
Plugging in the initial values we get
1
1
c1
0
=
.
1 −1
c2
2
Solving this we find c1 = 1 and c2 = −1 giving the solution to the IVP
y1
1
1
t
y=
=−
e +
e−t .
y2
1
−1
Notice that y1 = 2 sinh t and y2 = 2 cosh t.
15. We must find solutions of the system
0
y = Ay =
3 2
2 3
The characteristic equation is
3−λ
2
det(A − λI) = 2
3−λ
y.
= (λ − 1)(λ − 5) = 0.
This gives eigenvalues λ1 = 1 and λ2 = 5. Eigenvectors are obtained from
(3 − λ)x1 + 2x2 = 0.
1
For λ1 = 1 this is 2x1 + 2x2 = 0. Hence we can take x(1) =
. For
−1 1
λ2 = 5 this becomes −2x1 + 2x2 = 0, and an eigenvector is x(2) =
. This
1
gives the general solution
y1
1
1
t
y=
= c1
e + c2
e5t .
y2
−1
1
Plugging in the initial values we get
1 1
c1
0.5
=
.
−1 1
c2
−0.5
Solving this we find c1 =
1
2
and c2 = 0 giving the solution to the IVP
1
y1
1
y=
=
et
y2
2 −1