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t
cos
4q − p2 − φ .
2
We now look at several examples. Please use JOde to plot solutions in each case!
Summary:
General solution and properties of (6.7)
L[ y ] = d2 y
dy
+p
+ qy = 0.
2
dt
dt √2
√2
Case 1. p2 > 4q. y (t) = Ce(−p+ p −4q)t/2 + De(−p− p −4q)t/2 with C and D real.
For nonzero C, D;
y (t) → ∞ if
 y ( t)  → 0
if q < 0 or if p < 0, q > 0
p > 0, q > 0.
p Case 2. p2 = 4q. y (t) = (C + Dt)e− 2 t , C, D real. Again for nonzero C, D;
y (t) → ∞ if
 y ( t)  → 0
if 12 p0
p > 0. Case 3. p2 < 4q .
p
y (t) = e− 2 t C cos 4q − p2 t/2 + D sin 4q − p2 t/2
p
or C 2 + D2 e− 2 t cos
4q − p2 t/2 − φ √
√
where cos φ = C/ C 2 + D2 and sin φ = D/ C 2 + D2 . Again for nonzero C, D;
 y ( t)  → ∞
 y ( t)  → 0
y ( t) if
if
remains ﬁnite and bounded if 13 p<0
p>0
p = 0. Application to MassSpring systems.
Units: The units of mass M is kg., the unit of distance is meter (m), the unit of
time is second (sec). A mass M kg. under gravity g exerts a force of M g newtons.
Interpretation of results of example for damped mass spring system
d2 x
dx
+p
+ qx = 0,
dt2
dt
where p is the friction divided by the mass and is the spring constant divided by
the mass.
Case 1: p = 4, q = 1, overdamped. The mass is pulled down by one unit of length
beyond equilibrium
Case 1.
>
>
>
>
>
>
>
>
>equilibrium>
②
         > ✻>
>
x(0) >
>
❄② and then let go. The damping is so strong that it dominates the spring force so
that there are no oscillations. Instead x(t) simply decays exponentially in time
from x(0) = 1 to zero.
Case 2: p = 2, q = 1, critically damped. Similar behavior. However it is borderline
in that the mass is very close to making small oscillations before damping to zero.
Case 3: p = 1, q = 1, p2 < 4q underdamped. In this case, the friction is not enough
to prevent oscillations, but merely dampens them. The motion is not periodic
unless p = 0.
Periodic means there exists T > 0 such that x(t + T ) = x(√ for all t. For p = 0, x(t)
t)
2
is periodic with period √π . On the other hand, even for 4q > p > 0, the distance
q
between zeros t0 at which x (t0 ) > 0 is constant and equal to 2π
2
q− p
4 . (Underdamped massspring systems are a very important paradigm for understanding many physical, engineering and chemical processes.) 14 Find the solution of the initial value problem
d2 y
dy
+p
+ 4y = 0,
dt2
dt
y (0) = 0,
dy
(0) = 1,
dt
in the three cases (i) p = 5, (ii) p = 4, (iii) p = 2.
In each case:
a) ﬁnd the behavior of the solutions near t = 0 and the dominant behavior for large
t,
(b) use JOde to draw the graphs of y (t),
(c) ﬁnd ymax and the value of t where it occurs. 15 Properties of solutions of linear, homogeneous odes L[y ] ≡ (D2 + p(t)D + q (t))y ≡ d2 y
dy
+ p( t )
+ q ( t) y = 0
dt2
dt (6.15) with initial conditions
dy...
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 Fall '08
 INDIK

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