LECTURE 13: NONHOMOGENEOUS EQUATIONS
MINGFENG ZHAO
October 03, 2014
Mass-Spring System:
Figure 1. Mass-Spring System
Let x(t) be the displacement of the mass, then
mx00 + cx0 + kx = F (t) .
For free motion (that is, F (t) = 0), rewrite the equation, we have
x00 + 2px0 + ω02 = 0,
where
c
p=
,
2m
r
and ω0 =
k
.
m
The characteristic equation of x00 + 2px0 + ω02 = 0 is:
r2 + 2pr + ω02 = 0.
Solve r2 + 2pr + ω02 = 0, we get
r1,2 = −p ±
Then
1
q
p2 − ω02 .
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MINGFENG ZHAO
I. Free undamped motion: c = 0 (that is, p = 0)
The general solution to x00 + ω02 x = 0 is:
x(t) = A cos(ω0 t) + B sin(ω0 t) .
II. Overdamping: c2 − 4km > 0 (that is, p2 − ω02 > 0)
The general solution to x00 + 2px0 + ω02 x = 0 is:
y(t) = Aer1 t + Ber2 t ,
where
r1 = −p −
q
p2 − ω02 < 0,
and r2 = −p +
q
p2 − ω02 < 0.
III. Critical damping: c2 − 4km = 0 (that is, p2 − ω02 = 0)
The general solution to x00 + 2px0 + ω02 x = 0 is:
y(t) = Ae−pt + Bte−pt .
IV. Underdamping: c2 − 4km < 0 (that is, p2 − ω02 < 0)
The general solution to x00 + 2px0 + ω02 x = 0 is:
y(t) = Ae−pt cos
q
q
ω02 − p2 t + Be−pt sin
ω02 − p2 t ,
Example 1. Suppose that m = 2 kg and k = 8 N/m. The whole mass and spring setup is sitting on a truck that was
traveling at 1 m/s. The truck crashes and hence stops. The mass was held in place 0.5 meters forward from the rest
position. During the crash the mass gets loose. That is, the mass is now moving forward at 1 m/s, while the other end
of the spring is held in place. The mass therefore starts oscillating. What’s is the frequency of the resulting oscillation
and what is the amplitude.
Let x(t) be the displacement of the mass at time t (the moving forward is in the positive direction), them the
differential equation is:
mx00 + kx = 2x00 + 8x = 0.
By the assumption, we have
x(0) = 0.5,
and x0 (0) = 1.
Rewrite the equation, we have
x00 + 4x = 0.
LECTURE 13: NONHOMOGENEOUS EQUATIONS
3
The characteristic equation of x00 + 4x = 0 is:
r2 + 4 = 0.
Solve r2 + 4 = 0, we have
r1 = 2i,
and r2 = −2i.
So the general solution to 2x00 + 8x = 0 is:
x(t) = A cos(2x) + B sin(2x).
Then
x0 (t) = −2A sin(2x) + 2B cos(2x).
Since x(0) = 0.5 and x0 (0) = 1, then
A = 0.5,
and
2B = 1.
A = 0.5,
and B = 0.5
Then
Therefore, the solution to 2x00 + 8x = 0, x(0) = 0.5, x0 (0) = 1 is:
√
2
π
cos 2x +
.
2
4
√
p
2
2
0.52 + 0.52 =
.
Hence the frequency is 2 =
Hz , and the amplitude is
2π
2
x(t) = 0.5 cos(2x) + 0.5 sin(2x) =
Example 2. Suppose you want to use a spring to weigh items. You place the mass on the spring and put it in motion.
You have two reference weight 1 kg and 2 kg to calibrate your setup. You put each in motion on your spring an measure
the frequency. For the 1 kg weight you measured 1.1 Hz, for the 2 kg weight you measured 0.8 Hz. a) Find the spring
constant k and the damping constant c. b) Find a formula for the mass in terms of the frequency in Hz. c) For an
unknown object you measure 0.2 Hz, what is the mass of the object? Suppose that you know that the mass of the
unknown object is more than a kilogram.
The differential equation for the mass spring system is:
mx00 + cx0 + kx = 0.
Rewrite the equation, we have
x00 + 2px0 + ω02 x = 0,
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MINGFENG ZHAO
where
c
p=
,
2m
r
and ω0 =
k
.
m
a) By the assumption, we know that the motion is undamped, and the frequency is
r
p
q
4ω02 − 4p2
1
1
k
c2
1
.
·
=
· ω02 − p2 =
·
−
2π
2
2π
2π
m 4m2
By the assumption, we know that
r
1
k
c2
= 1.1,
·
−
2π
1 4 · 12
and
1
·
2π
r
k
c2
= 0.8.
−
2 4 · 22
Then we have
k = 4π 2 · 1.35,
and c = 2π ·
√
0.56 .
b) Let f be the frequency corresponding to the weight m, then we have
r
4π 2 · 1.35 4π 2 · 0.56
1
·
−
= f.
2π
m
4m2
That is, we have
f2 =
1.35 0.14
− 2 .
m
m
c) If f = 0.2, then
1.35 0.14
− 2 = 0.22
m
m
That is, we have
0.04m2 − 1.35m + 0.14 = 0.
Then
m=
1.35 +
√
1.352 − 4 · 0.04 · 0.14
≈ 33.645 .
2 · 0.04
Nonhomogeneous equations
Question 1. How to solve the equation y 00 + 5y 0 + 6y = 2x + 1? Is there any relation between the nonhomogeneous
equation y 00 + 5y 0 + 6y = 2x + 1 and the homogeneous equation y 00 + 5y 0 + 6y = 0?
Assume we already have a particular solution yp to y 00 + 5y 0 + 6y = 2x + 1, then
I. For any other solution y(x) to y 00 + 5y 0 + 6y = 2x + 1, it’s easy to see that
y(x) − yp (x) is a solution to y 00 + 5y 0 + 6y = 0.
LECTURE 13: NONHOMOGENEOUS EQUATIONS
5
II. On the other hand, for any solution yc (x) to y 00 + 5y 0 + 6y = 0, it’s easy to see that
yc (x) + yp (x) is a solution to y 00 + 5y 0 + 6y = 2x + 1.
Theorem 1. Consider the nonhomogeneous equation y 00 + p(x)y 0 + q(x)y = f (x). Let yc (x) be a complementary solution
(that is, yc (x) is a solution to the homogeneous equation y 00 + p(x)y 0 + q(x)y = 0), 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).
Undetermined coefficients
In this course, we only consider the equation:
ay 00 + by 0 + cy = f (x),
where
• a, b and c are constants.
• f (x) is a polynomial, exponential functions, sines, cosines, and the multiplies of polynomial, exponential functions, sines and cosines.
We already know how to solve the homogeneous equation ay 00 + by 0 + cy = 0, that is, we can find all complementary
solutions to ay 00 + by 0 + cy = f (x), so now we only need to find a particular solution to the nonhomogeneous equation
ay 00 + by 0 + cy = f (x), then we can find all general solution to the nonhomogeneous equation ay 00 + by 0 + cy = f (x).
Let a, b, c be constants, consider
ay 00 + by 0 + cy = f (x).
The characteristic equation is:
ar2 + br + c = 0.
Then
r1,2 =
−b ±
√
b2 − 4ac
.
2a
We have the following cases:
I. When c 6= 0 and f (x) is a polynomial, a particular solution can be also a polynomial with the same degree as
f (x).
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MINGFENG ZHAO
Example 3. Find the general solution to y 00 + 5y 0 + 6y = 2x + 1.
The characteristic equation to y 00 + 5y 0 + 6y = 0 is :
r2 + 5r + 6 = 0.
Solve r2 + 5r + 6 = 0, we get
r1 = −2,
and r2 = −3.
So the general solution to the homogeneous equation y 00 + 5y 0 + 6y = 0 is:
y(x) = C1 e−2x + C2 e−3x .
Let yp (x) = Ax + B for some constants A and B, then
yp00 + 5yp0 + 6yp = 0 + 5A + 6(Ax + B) = 2x + 1.
That is, we have
(6A − 2)x + 5A + 6B − 1 = 0.
So
6A − 2 = 0,
and
5A + 6B − 1 = 0.
So
A=
1
,
3
1
and B = − .
9
1
x
− is a particular solution to the nonhomogeneous equation y 00 + 5y 0 + 6y = 2x + 1.
3
9
Therefore, he general solution to the nonhomogeneous equation y 00 + 5y 0 + 6y = 2x + 1 is:
Hence yp (x) =
y(x) =
x 1
− + C1 e−2x + C2 e−3x .
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II. When f (x) = menx is an exponential function with n 6= r, a particular solution can be also an exponential
function like yp (x) = Aenx for some constant A.
Example 4. Find a particular solution to y 00 + 2y 0 + 3y = e5x .
Let yp (x) = Ae5x for some constant A, then
yp0 (x)
=
5Ae5x
yp00 (x)
=
25Ae5x
yp00 + 2yp0 + 3yp
=
25Ae5x + 10Ae5x + 3Ae5x
LECTURE 13: NONHOMOGENEOUS EQUATIONS
=
7
e5x
That is,
e5x (25A + 10A + 3A) = e5x .
So 38A = 1, that is,
A=
1
.
38
So a particular solution to y 00 + 2y 0 + 3y = e5x is:
yp (x) =
1 5x
e .
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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