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4. SECOND-ORDER LINEAR EQUATIONS
Example.
(1) Solve
y 00 + 2y 0 + 5y = 0, y(0) = 1,
The associated characteristic equation
y 0(0) = 1.
r2 + 2r + 5 = 0
has roots
r=
Let ↵ =
1,
2±
= 2. Then
p
4
2
20
=
2 ± 4i
=
2
1 ± 2i.
y(t) = c1e t cos 2t + c2e t sin 2t
is the general solution and
y 0(t) =
2c1e t sin 2t
c1e t cos 2t + 2c2e t cos 2t
c2e t sin 2t.
Then
y(0) = c1 = 1 and
y 0(0) = c1 + 2c2 = 1 + 2c2 = 1 =) c2 = 1.
Thus the solution to the IVP is
x(t) = e t cos 2t + e t sin 2t.
Referring back to the unforced mass-spring oscillator
or
[inertia] ⇥ y 00 + [damping] ⇥ y 0 + [sti↵ness] ⇥ y = 0
my 00(t) + by 0(t) + ky(t) = 0,
b
the exponential decay rate is ↵ =
and the angular frequency is =
2m
p
p
p
4mk b2
. Since b = 2 < 4mk = 20, this is an example of an under2m
damped oscillator.