Math 308, Sections 301, 302,
Summer 2008
Lecture 11.
06/23/2008
Chapter 4. Linear Second Order Equations
Section 4.9 Variation of Parameters
Consider the nonhomogeneous linear second order differential
equation
y ′′ + p(x)y ′ + q(x)y = g (x).
(1)
Let {y1 (x), y2 (x)} be a fundamental solution set to the
corresponding homogeneous equation
y ′′ + p(x)y ′ + q(x)y = 0.
The general solution to this homogeneous equation is
yh (x) = c1 y1 (x) + c2 y2 (x), where c1 and c2 are constants. To find
a particular solution to (1) we assume that c1 = c1 (x) and
c2 = c2 (x) are functions of x and we seek a particular solution
yp (x) in form
yp (x) = c1 (x)y1 (x) + c2 (x)y2 (x).
Let’s substitute yp (x), yp′ (x), and yp′′ (x) into (1).
yp′ (x) = c1′ (x)y1 (x) + c2′ (x)y2 (x) + c1 (x)y1′ (x) + c2 (x)y2′ (x).
Assume that c1′ (x)y1 (x) + c2′ (x)y2 (x) = 0 , then
yp′ (x) = c1 (x)y1′ (x) + c2 (x)y2′ (x).
yp′′ (x) = c1′ (x)y1′ (x) + c2′ (x)y2′ (x) + c1 (x)y1′′ (x) + c2 (x)y2′′ (x).
yp′′ + p(x)yp′ + q(x)yp = c1′ (x)y1′ (x) + c2′ (x)y2′ (x) = g (x).
We can find c1 (x) and c2 (x) solving the system
′
c1 (x)y1 (x) + c2′ (x)y2 (x) = 0
c1′ (x)y1′ (x) + c2′ (x)y2′ (x) = g (x)
for c1′ (x) and c2′ (x).
Z
Z
−g (x)y2 (x)
g (x)y1 (x)
c1 (x) =
dx , c2 (x) =
dx
W [y1 , y2 ](x)
W [y1 , y2 ](x)
Example 1. Find the general solution to the equation
1
y ′′ + y =
.
sin x
Section 4.11 A closer look at free mechanical vibrations
A damped mass-spring oscillator consists of a mass m attached to
a spring fixed at one end.
Model for the motion of the mass is expressed by the initial value
problem
my ′′ + by ′ + ky = Fexternal ,
y (0) = y0 , y ′ (0) = v0 ,
where m is a mass, b is the damping coefficient, k is the stiffness.
Let’s Fexternal = 0
Undamped free case: b = 0
The equation reduces to
my ′′ + ky = 0| : m
y ′′ + ω 2 y = 0
where ω =
q
k
m.
The solution of this equation is
y (t) = C1 cos ωt + C2 sin ωt
y (t) = A sin(ωt + φ),
where A =
q
C12 + C22 , tan φ =
C1
C2 .
The motion is periodic with
period 2π/ω
natural frequency ω/2π
angular frequency ω
amplitude A.
Example 2. A 2-kg mass is attached to a spring with stiffness
k = 50 N/m. The mass is displaced 1/4 m to the left of the
equilibrium point and given a velocity of 1 m/sec to the left.
Neglecting damping,
a) Set up an initial value problem for this system.
b) Find the equation of motion of the mass.
c) Find the amplitude, period and frequency of the motion.
d) How long after release does the mass pass through the
equilibrium position?
Underdamped or oscillatory motion (b 2 < 4mk)
The solution to the equation
my ′′ + by ′ + ky = 0
is
y (t) = e αt (C1 cos βt + C2 sin βt) = Ae αt sin(βt + φ),
where
q
b
C1
1 p
4mk − b 2 , A = C12 + C22 , tan φ =
α=−
,β=
.
2m
2m
C2
The solution y (t) varies between −Ae αt and Ae αt with
4πm
2π
=√
and quasifrequency 1/P.
quasiperiod P =
β
4mk − b 2
y (t) → 0 as t → ∞.
An exponential factor Ae αt is called a damping factor.
The system is called underdamped because there is not enough
damping present (b is too small) to prevent the system from
oscillating.
Overdamped motion (b 2 > 4mk)
The solution to the equation
my ′′ + by ′ + ky = 0
is
y (t) = c1 er1 t + c2 er2 t ,
b
b
1 p
1 p
where r1 = −
4mk − b 2 , r2 = −
4mk − b 2
+
−
2m 2m
2m 2m
r2 < 0, and since b 2 > b 2 − 4mk, r1 < 0. y (t) → 0 as t → ∞.
Critically damped motion (b 2 = 4mk) The solution to the
equation
my ′′ + by ′ + ky = 0
is
b
y (t) = (c1 + c2 t)e− 2m t .
y (t) → 0 as t → ∞.
Example 3. The motion of the mass-spring system with damping
is governed by
y ′′ (t) + by ′ (t) + 64y (t) = 0,
y (0) = 1, y ′ (0) = 0
Find the equation of the motion for b = 10, 16, 20.