LECTURE 11: HOMOGENEOUS SECOND ORDER LINEAR ODES WITH CONSTANT
COEFFICIENTS AND MECHANICAL VIBRATIONS
MINGFENG ZHAO
October 02, 2015
Theorem 1. Let p(x) and q(x) be continuous functions, y1 and y2 are two linearly independent solutions to a homogeneous equation y 00 + p(x)y 0 + q(x)y = 0. Then the general solution to y 00 + p(x)y 0 + q(x)y = 0 is:
y = C1 y1 (x) + C2 y2 (x).
Definition 1. Any two linearly independent solutions y1 and y2 to a homogeneous equation y 00 + p(x)y 0 + q(x)y = 0 is
called a fundamental set of solutions to this homogeneous equation y 00 + p(x)y 0 + q(x)y = 0.
Homogeneous second order linear ODEs with constant coefficients
Conside the following constant coefficients second order liner differential equation has the form of:
ay 00 + by 0 + cy = 0.
(1)
To find the general solution to the above differential equation:
1) Try y = erx to be a solution to (1, then
y 0 = rerx ,
and y 00 = r2 erx .
Then
0
= ay 00 + by 0 + cy
= ar2 erx + brerx + cerx
= erx [ar2 + br + c].
So r is a root of the characteristic equation of ay 00 + by 0 + cy = 0, that is,
(2)
ar2 + br + c = 0.
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2
MINGFENG ZHAO
2) Find the solutions to the characteristic equation ar2 + br + c = 0, that is,
√
−b ± b2 − 4ac
r1,2 =
.
2a
3) Write down the general solution to ay 00 + by 0 + cy = 0:
– If b2 − 4ac > 0, that is, (2) has two different real roots r1 and r2 , then the general solution to (1) is:
y = C1 er1 x + C2 er2 x .
– If b2 − 4ac = 0, that is, (2) has only one real root r1 = r2 = −
−b
b
, then the general solution to (1) is:
2a
−b
y = C1 e 2a ·x + C2 xe 2a ·x .
– If b2 − 4ac < 0, that is, (2) has two different complex roots, that is,
√
4ac − b2
b
.
r1,2 = − ± i
2a
2a
Notice that
e r1 x
= e
b
− 2a
+i
b
= e− 2a x+i
√√
√√
4ac−b2 2a x
4ac−b2 2a
√√
b
2
= e− 2a x · ei 4ac−b 2ax
qp
qp
b
x
− 2a
2
2
cos
4ac − b 2ax + i sin
4ac − b 2a
= e
b
= e− 2a x cos
qp
Then the general solution to (1) is:
√
y = C1 e
b
·x
− 2a
cos
qp
b
4ac − b2 2ax + ie− 2a x sin
4ac − b2 2ax .
4ac − b2
x
2a
√
!
b
− 2a
·x
+ C2 e
sin
!
4ac − b2
x .
2a
Example 1. Let α be a constant. Solve y 00 − 2y 0 + α(2 − α)y = 0 for y(0) = 0 and y 0 (0) = 2.
Let y = erx be a solution to y 00 − 2y 0 + α(2 − α)y = 0, then
y 0 = rerx ,
and y 00 = r2 erx .
So we get
0
= y 00 − 2y 0 + α(2 − α)y
= r2 erx − 2rerx + α(2 − α)erx
LECTURE 11: HOMOGENEOUS SECOND ORDER LINEAR ODES WITH CONSTANT COEFFICIENTS AND MECHANICAL VIBRATIONS
3
=
erx [r2 − 2r + α(2 − α)].
So we get r2 − 2r + α(2 − α) = 0. Then
r1,2
p
4 − 4α(2 − α)
=
2
p
= 1 ± 1 − α(2 − α)
p
= 1 ± 1 − 2α + α2
2±
p
(1 − α)2
=
1±
=
1 ± |1 − α|
=
1 ± (1 − α).
which implies that
r1 = α,
and r2 = 2 − α.
We have the following two cases:
I. If r1 = r2 , that is, α = 2 − α, then α = 1. So the solution is like:
y(x) = C1 ex + C2 xex .
Since y(0) = 0, then C1 = 0, that is, y(x) = C2 xex . Then
y 0 (x) = C2 ex + C2 xex .
Since y 0 (0) = 2, that is, C2 = 2. So the solution is given by:
y(x) = 2xex .
II. If r1 6= r2 , that is, α 6= 2 − α, then α 6= 1. So the solution is like:
y(x) = C1 eαx + C2 e(2−α)x .
Then
y 0 (x) = C1 αeαx + C2 (2 − α)e(2−α)x .
Since y(0) = 0 and y 0 (0) = 2, then
C1 + C2 = 0,
and C1 α + C2 (2 − α) = 2.
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MINGFENG ZHAO
Since α 6= 1, then we get
C1 = −
1
,
1−α
and C2 =
1
.
1−α
Therefore, the solution is given by:
y(x) =
i
h
1
· −eαx + e(2−α)x .
1−α
In summary, we have


 2xex ,
if α = 1,
i
h
y(x) =
1


· −eαx + e(2−α)x , if α 6= 1.
1−α
Example 2. Find the general solution to y 00 + 2y 0 + 5y = 0.
Let y = erx be a solution to y 00 + 2y 0 + 5y = 0, then
y 0 = rerx ,
and y 00 = r2 erx .
So we get
0
= y 00 + 2y 0 + 5y
= r2 erx + 2rerx + 5erx
= erx [r2 + 2r + 5].
So we get r2 + 2r + 5 = 0, which implies that
√
√
−2 + −16
−2 + 4 − 4 · 5
=
= −1 + 2i,
r1 =
2
2
and r2 = −1 − 2i.
Notice that
e r1 x
= e(−1+2i)x
= e−x+i2x
= e−x · ei(2x)
= e−x [cos(2x) + i sin(2x)]
By Euler’s identity
= e−x cos(2x) + ie−x sin(2x).
So the general solution of y 00 + 2y 0 + 5y = 0 is:
y(x) = C1 e−x cos(2x) + C2 e−x sin(2x) .
LECTURE 11: HOMOGENEOUS SECOND ORDER LINEAR ODES WITH CONSTANT COEFFICIENTS AND MECHANICAL VIBRATIONS
5
Mass-Spring System
Suppose we have a mass m > 0 (in kilograms) connected by a spring with spring constant k > 0 (in Newtons per meter)
to a fixed wall. There may be some external force F (t) (in Newtons) acting on the mass at time t (in seconds). Finally,
there is some friction measure by c ≥ 0 (in Newton-sections per meter) as the mass slides along the floor.
Figure 1. Mass-Spring System
Let x(t) be the displacement of the mass (x = 0 is the rest position, that is, the spring is neither compressed nor
stretched), by Newton’s law, we have
mx00 (t) = all forces acting on the mass.
There are three kinds of forces acting on the mass:
• External force: F (t)
• Friction or Damping force: −cx0 (t) (negative means negative direction)
• Spring Compression force: −kx(t) (negative means negative direction)
So mx00 = F (t) − cx0 − kx, that is,
mx00 + cx0 + kx = F (t) .
Sometimes, we write it as:
x00 + 2px0 + ω02 x =
F (t)
,
m
where
c
p=
2m
r
(damping coefficient),
ω0 =
k
m
(frequency).
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MINGFENG ZHAO
Here are some terminology:
I. The motion is forced if F (t) 6≡ 0.
II. The motion is unforced/free if F (t) ≡ 0.
III. The motion is damped if c > 0.
IV. The motion is undamped if c = 0.
Problems you can do:
Lebl’s Book [2]: All exercises on Page 55 and Page 56.
Braun’s Book [1]: All Exercises on Page 140, Page 144, Page 145, Page 149 and Page 150. Read all materials
in Section 2.2.
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