LECTURE 16: NONHOMOGENEOUS EQUATIONS AND FORCED OSCILLATIONS AND
RESONANCE
MINGFENG ZHAO
October 19, 2015
Theorem 1. Let yc (x) be the general solution to the homogeneous equation y 00 + p(x)y 0 + q(x)y = 0 ( which is call the
complementary solution), and yp (x) be any particular solution to the nonhomogeneous equation y 00 +p(x)y 0 +q(x)y = f (x),
then the general solution to the nonhomogeneous equation y 00 + p(x)y 0 + q(x)y = f (x) is:
y(x) = yp (x) + yc (x).
Variation of Parameters
Definition 1. Let y1 (x) and y2 (x) be two solutions, the Wronskian of y1 and y2 is defined as:

W (y1 , y2 ) = det 
y1 (x) y2 (x)
y10 (x)
y20 (x)

 = y1 (x)y20 (x) − y10 (x)y2 (x).
To find a particular solution to the nonhomogeneous equation y 00 + p(x)y 0 + q(x)y = f (x):
I. Find a fundamental set of solutions y1 (x) and y2 (x) to the homogeneous equation y 00 + p(x)y 0 + q(x)y = 0, that
is, the general solution to y 00 + p(x)y 0 + q(x)y = 0 is:
yc (x) = C1 y1 (x) + C2 y2 (x).
II. Let yp (x) = C1 (x)y1 (x) + C2 (x)y2 (x) be a particular solution to y 00 + p(x)y 0 + q(x)y = f (x)
III. Compute yp0 (x), we get
yp0 (x) = C10 (x)y1 (x) + C20 (x)y2 (x) + C1 (x)y10 (x) + C2 (x)y20 (x)
Take
C10 (x)y1 (x) + C20 (x)y2 (x) = 0.
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2
MINGFENG ZHAO
Then
yp0 (x)
=
C1 (x)y10 (x) + C2 (x)y20 (x)
yp00 (x)
=
C10 (x)y10 (x) + C1 (x)y100 (x) + C20 (x)y20 (x) + C2 (x)y200 (x).
IV. Plug yp (x), yp0 (x) and yp00 (x) into y 00 + p(x)y 0 + q(x)y = f (x), we get
C10 (x)y10 (x) + C20 (x)y20 (x) = f (x).
V. Solve C10 (x) and C20 (x) from the system:


 C10 (x)y1 (x) + C20 (x)y2 (x) = 0

 C 0 (x)y 0 (x) + C 0 (x)y 0 (x) = f (x).
1
1
2
2
Then
C10 (x) =
−y2 (x)f (x)
,
W (y1 , y2 )
and C20 (x) =
y1 (x)f (x)
.
W (y1 , y2 )
VI. Solve C1 (x) and C2 (x), then
Z
C1 (x) =
−y2 (x)f (x)
dx,
W (y1 , y2 )
Z
and C2 (x) =
y1 (x)f (x)
dx.
W (y1 , y2 )
VII. Write down the solution:
Z
yp (x) = −y1 (x)
y2 (x)f (x)
dx + y2 (x)
W (y1 , y2 )
Z
y1 (x)f (x)
dx .
W (y1 , y2 )
Example 1. Find a particular solution of y 00 − y = 2 sin(x2 ).
The characteristic equation of y 00 − y = 0 is:
r2 − 1 = 0.
Solve r2 − 1 = 0, then
r1 = 1,
and r2 = −1.
Then the general solution to y 00 − y = 0 is:
y(x) = C1 ex + C2 e−x .
Let yp (x) = C1 (x)ex + C2 (x)e−x be a particular solution to y 00 − y = 2 sin(x2 ), then
yp0 (x) = C10 (x)ex + C1 (x)ex + C20 (x)e−x − C2 (x)e−x
LECTURE 16: NONHOMOGENEOUS EQUATIONS AND FORCED OSCILLATIONS AND RESONANCE
Take
C10 (x)ex + C20 (x)e−x = 0.
Then we have
yp0 (x)
= C1 (x)ex − C2 (x)e−x
yp00 (x)
= C10 (x)ex + C1 (x)ex − C20 (x)e−x + C2 (x)e−x
So we have
yp00 − yp
= C10 (x)ex + C1 (x)ex − C20 (x)e−x + C2 (x)e−x − C1 (x)ex + C2 (x)e−x
= C10 (x)ex − C20 (x)e−x
=
sin(x2 )
In summary, we know that C10 (x) and C20 (x) satisfy


 C10 (x)ex + C20 (x)e−x = 0

 C 0 (x)ex − C 0 (x)e−x = 2 sin(x2 )
1
2
Solve C10 (x) and C20 (x), we get
C10 (x) = e−x sin(x2 ),
and C20 (x) = −ex sin(x2 ).
So
Z
C1 (x) =
e
−x
2
sin(x ) dx,
Z
and C2 (x) = −
ex sin(x2 ) dx.
Therefore, a particular solution to y 00 − y = 2 sin(x2 ) can be:
yp (x) = ex
Z
e−x sin(x2 ) dx − e−x
Z
ex sin(x2 ) dx .
Forced oscillations and resonance
Mass-Spring System:
Let x(t) be the displacement of the mass, then
mx00 + cx0 + kx = F (t) .
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MINGFENG ZHAO
Figure 1. Mass-Spring System
We are interested in periodic forcing, F (t) = F0 cos(ωt) with ω > 0. So we have
mx00 + cx0 + kx = F0 cos(ωt) .
Rewrite the equation, we have
x00 + 2px0 + ω02 x =
F0
cos(ωt) ,
m
where
c
p=
≥ 0,
2m
r
and ω0 =
k
> 0.
m
Undamped forced motion and resonance
The differential equation for the undamped forced motion (c = 0) is:
x00 + ω02 x =
F0
cos(ωt),
m
ω > 0.
The characteristic equation of x00 + ω02 x = 0 is:
r2 + ω02 = 0.
Solve r2 + ω02 = 0, we get
r1 = ω0 i,
and r2 = −ω0 i.
The general solution to x00 + ω02 x = 0 is:
xc (t) = A cos(ω0 t) + B sin(ω0 t) .
Since the nonhomogeneous function is
F0
cos(ωt), we have two cases:
m
LECTURE 16: NONHOMOGENEOUS EQUATIONS AND FORCED OSCILLATIONS AND RESONANCE
I. If ω 6= ω0 , let xp (t) = D cos(ωt) + E sin(ωt) be a particular solution to x00 + ω02 x =
x0p (t)
= −Dω sin(ωt) + Eω cos(ωt)
x00p (t)
= −Dω 2 cos(ωt) − Eω 2 sin(ωt)
x00 + ω02 x
5
F0
cos(ωt), then
m
= −Dω 2 cos(ωt) − Eω 2 sin(ωt) + Dω02 cos(ωt)
= D[ω02 − ω 2 ] cos(ωt) − Eω 2 sin(ωt)
=
F0
cos(ωt).
m
So we get
F0
, and E = 0.
m[ω02 − ω 2 ]
F0
cos(ωt) can be:
So a particular solution to x00 + ω02 x =
m
D=
xp (t) =
F0
cos(ωt) .
m[ω02 − ω 2 ]
II. If ω = ω0 . Let xp (t) = t[D cos(ω0 t) + E sin(ω0 t)] = Dt cos(ω0 t) + Et sin(ω0 t) be a particular solution to
F0
x00 + ω02 x =
cos(ω0 t), then
m
x0p (t)
= D cos(ω0 t) − Dω0 t sin(ω0 t) + E sin(ω0 t) + Eω0 t cos(ω0 t)
x00p (t)
= −Dω0 sin(ω0 t) − Dω0 sin(ω0 t) − Dω02 t cos(ω0 t) + Eω0 cos(ω0 t) + Eω0 cos(ω0 t) − Eω02 t sin(ω0 t)
= −2Dω0 sin(ω0 t) − Dω02 t cos(ω0 t) + 2Eω0 cos(ω0 t) − Eω02 t sin(ω0 t)
x00 + ω02 x
= −2Dω0 sin(ω0 t) − Dω02 t cos(ω0 t) + 2Eω0 cos(ω0 t) − Eω02 t sin(ω0 t)
+ω02 [Dt cos(ω0 t) + Et sin(ω0 t)]
= −2Dω0 sin(ω0 t) + 2Eω0 cos(ω0 t)
=
F0
cos(ω0 t).
m
Then we have
D = 0,
So a particular solution to x00 + ω02 x =
and E =
F0
.
2mω0
F0
cos(ω0 t) can be:
m
xp (t) =
F0
t sin(ω0 t) .
2mω0
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MINGFENG ZHAO
In summary, the general solution to x00 + ω02 x =
x(t) =
F0
cos(ωt) is:
m



 A cos(ω0 t) + B sin(ω0 t) +
F0
cos(ωt), if ω0 6= ω,
m[ω02 − ω 2 ]


 A cos(ω0 t) + B sin(ω0 t) +
F0
t sin(ω0 t),
2mω0
.
if ω0 = ω
F0
F0
t and
t, in particular, the
2mω0
2mω0
amplitude of x(t) will go to infinity. This kind of behavior is called resonance or perhaps pure resonance.
In the case that ω0 = ω, when t → ∞, the graph of x(t) oscillates between −
Figure 2. Graph of
1
t sin(πt)
π
Example 2. Take 0.5x00 + 8x = 10 cos(πt), x(0) = 0, x0 (0) = 0.
Then
r
m = 0.5,
ω0 =
√
8
= 16 = 4,
0.5
F = 10,
and ω = π.
The general solution to 0.5x00 + 8x = 10 cos(πt) is:
x(t) = A cos(4t) + B sin(4t) +
10
20
cos(πt) = A cos(4t) + B sin(4t) +
cos(πt).
0.5(42 − π 2 )
16 − π 2
Then
x0 (t) = −4A sin(4t) + 4B cos(4t) −
20π
sin(πt).
16 − π 2
Since x(0) = 0 and x0 (0) = 0, then
A+
20
= 0,
16 − π 2
and
4B = 0.
So we get
A=−
20
,
16 − π 2
and B = 0.
LECTURE 16: NONHOMOGENEOUS EQUATIONS AND FORCED OSCILLATIONS AND RESONANCE
7
So the solution to 0.5x00 + 8x = 10 cos(πt), x(0) = 0, x0 (0) = 0 is:
x(t) =
20
[cos(πt) − cos(4t)] .
16 − π 2
Figure 3. Graph of
20
[cos(πt) − cos(4t)]
16 − π 2
Problems you can do:
Lebl’s Book [2]: All exercises on Page 83.
Braun’s Book [1]: All exercises on Page 172 and Page 173. Read all materials in Section 2.6.
References
[1] Martin Braun. Differential Equations and Their Applications: An Introduction to Applied Mathematics. Springer, 1992.
[2] Jiri Lebl. Notes on Diffy Qs: Differential Equations for Engineers. Createspace, 2014.
Department of Mathematics, The University of British Columbia, Room 121, 1984 Mathematics Road, Vancouver, B.C.
Canada V6T 1Z2
E-mail address: mingfeng@math.ubc.ca