HOMEWORK 4: MATH 265 Due in class on Oct 20
PARTIAL SOLUTIONS
Problem 1:
(a) Solve the nonhomogeneous ODE
y
′′
+16
y
= cos(
ωt
) with
y
(0) = 0,
y
′
(0) = 0 and express your solution
in terms of the frequency of the forcing term
ω >
0. Sketch the solution when
ω
= 3 and when
ω
= 4.
(b)
NEW (Can be added at end of your HW pages, if you have already done other problems)
In Part (a), you may have gotten a solution in the form of a sum or difference of two trig functions
such as sines or cosines. Use a trigonometric identity to reexpress that solution as a product of two
trig functions (such as
K
sin(
w
1
t
) sin(
w
2
t
)  but you must find the values of
K, w
1
, w
2
in terms of other
quantities in the problem).
(c) Based on part (b) explain the behaviour of the solution when
ω
is close to but not equal to 4. That is,
explain the phenomenon of
beats
. (You may find it useful to read about this in your book.)
(d) Use your favorite graphics software (or calculator) to plot the solution in two cases,
ω
= 3
.
5 and
ω
= 3
.
8. What is the period of the
envelope
of the oscillations in each case? What is the frequency of
the oscillations within that envelope in each case?
Solution to Problem 1:
First solve the homogeneous problem. Let
y
=
e
rt
. Then, the characteristic polynomial is
r
2
+ 16 = 0,
so that
r
=
±
4
i
. Thus,
y
= sin(4
t
) and
y
= cos(4
t
) are solutions to the homogeneous problem.
•
Find the particular solution when
ω
0
negationslash
= 4. The particular solution is easily found to be
y
p
=
cos(
ω
0
t
)
16
−
ω
2
0
.
The general solution, when
ω
0
negationslash
= 4, is
y
=
c
1
cos(4
t
) +
c
2
sin(4
t
) +
cos(
ω
0
t
)
16
−
ω
2
0
.
Satisfying the initial conditions
y
(0) =
y
′
(0) = 0, we get
c
1
= 0,
c
2
=
−
1
/
(16
−
ω
2
0
). Thus, the solution to
the IVP is
y
(
t
) =
1
(16
−
ω
2
0
)
[cos(
ω
0
t
)
−
cos(4
t
)]
,
when
ω
0
negationslash
= 4
.
(2)
•
When
ω
0
= 4 we have resonance and
y
p
= (
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 Winter '10
 LEAHKESHET
 Math, Differential Equations, Trigonometry, Equations, Cos, yp

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