Section 3.6: Second-Order Linear Equations
Math 224
CWRU
Math 224 (CWRU)
Section 3.6
1 / 16
Second-Order Linear Equations
In this section, we consider equations of the form
d 2y
dy
a 2 +b
+ cy = 0,
dt
dt
where a, b, and c are constants. Such an equation is called a
homogeneous, constant-coefficient, linear, second-order equation. This
includes equations for harmonic oscillators.
Solution Methods
Method 1: “Guess” y = e st and solve for s.
Method 2: Solve the associated linear system using eigenvalues and
eigenvectors.
Math 224 (CWRU)
Section 3.6
2 / 16
Solution Method 1: Educated Guessing
These equations can usually be solved quickly by “guessing” a solution of
the form e st and determining what value(s) s would have to be.
d 2y
dy
Example 1.
+2
2
dt
dt
3y = 0
When using this method, we’ll always have to solve the equation
as 2 + bs + c = 0. This is called the characteristic polynomial for the
equation.
Math 224 (CWRU)
Section 3.6
3 / 16
Solution Method 2: Solve the Associated Linear System
d 2y
dy
Example 2. Convert
+2
3y = 0 to a linear system and solve it
2
dt
dt
using eigenvalues, eigenvectors, etc.
Math 224 (CWRU)
Section 3.6
4 / 16
v
10
5
- 10
-5
5
10
y
-5
- 10
Math 224 (CWRU)
Section 3.6
5 / 16
Which Method is Better?
If you just need the analytic solution (formula), Method 1 is quicker.
If you need to understand the behavior of the solution, the phase
portrait from Method 2 can be more helpful.
Math 224 (CWRU)
Section 3.6
6 / 16
Example 3
d 2y
dy
Solve the initial value problem
+2
2
dt
dt
0
y (0) = 2.
3y = 0, y (0) = 6,
y
100
80
60
40
20
-2
-1
0
1
Math 224 (CWRU)
2
3
4
5
t
Section 3.6
7 / 16
Example 4 (Complex roots)
d 2y
dy
Solve the initial value problem
+4
+ 20y = 0, y (0) = 2,
2
dt
dt
0
y (0) = 8.
Math 224 (CWRU)
Section 3.6
8 / 16
v
y
10
400
200
-6
-4
-2
5
2
- 200
4
6
t
- 10
- 400
-5
5
10
y
-5
- 10
Math 224 (CWRU)
Section 3.6
9 / 16
Classification of Harmonic Oscillators
Recall the equation for a harmonic oscillator:
d 2y
dy
m 2 +b
+ ky = 0,
dt
dt
where m > 0 is the mass, k > 0 is the spring constant, and b
damping coefficient.
0 is the
Sometimes we write this as
d 2y
dy
+p
+ qy = 0,
2
dt
dt
where p = b/m and q = k/m.
✓
◆
#»
dY
#»
#»
y (t)
This is equivalent to the system
= AY , where Y (t) =
and
v (t)
dt
✓
◆
0
1
A=
.
q
p
Math 224 (CWRU)
Section 3.6
10 / 16
Undamped (b = 0)
The general solution is of the form y (t) = k1 cos !t + k2 sin !t. It is
periodic with period 2⇡/!. The origin is a center in the phase plane. The
mass oscillates about its rest position forever.
Damped (b > 0)
d 2y
dy
Here we have m 2 + b
+ ky = 0. The characteristic polynomial is
dt
dt
2
ms + bs + k = 0, which has roots
p
b ± b 2 4km
s=
.
2m
Math 224 (CWRU)
Section 3.6
11 / 16
e(i).
If b 2 4km < 0, the roots/eigenvalues are complex. Since the real
part of s is b/2m, which is negative, the origin is a spiral sink. The
mass oscillates about its rest position, but the amplitude decreases
over time, so the mass tends toward its rest position. In this case, the
oscillator is called underdamped.
e(ii).
If b 2 4km > 0, there are two distinct real roots. In fact, they both
must be negative. Thus the origin is a real sink. All solutions tend
toward the origin, and most solutions tend toward the origin tangent
to the line through the eigenvector corresponding to the larger
eigenvalue. They do not oscillate about the rest position. The
oscillator is called overdamped.
b
e(iii). If b 2 4km = 0, there is a repeated root, s = 2m
. The equilibrium
point at the origin is a sink. The solutions tend to the origin in a
direction that is tangent to the single line of eigenvectors. Again, the
solutions do not oscillate about the rest position. The oscillator is
called critically damped. That is because a small change in the
damping coefficient b results in a major change in the behavior of
solutions.
Math 224 (CWRU)
Section 3.6
12 / 16
Example 5
m = 9, k = 1, b = 6
Math 224 (CWRU)
Section 3.6
13 / 16
v
y
10
3
2
1
5
5
-1
10
15
20
t
- 10
-5
5
10
y
-2
-3
-5
- 10
Math 224 (CWRU)
Section 3.6
14 / 16
Example 6
m = 1, k = 7, b = 8
Math 224 (CWRU)
Section 3.6
15 / 16
v
y
10
1.4
1.2
1.0
5
0.8
0.6
- 10
-5
5
10
y
0.4
0.2
-5
0
1
2
3
4
5
t
- 10
Math 224 (CWRU)
Section 3.6
16 / 16