Unformatted text preview: v e−p/2 t + p2 /4 ve−p/2 t . Substitute into d 2 y2
dy2
+p
+ p2 /4 y2 = 0
dt2
dt and ﬁnd
p2 −p/2 t
ve
4
p2
+p(v e−p/2 t − p/2ve−p/2 t + q ve−p/2 t = 0,
4
e−p/2 t v − pv e−p/2 t + or, cleaning up,
v = 0
or
v = At + B.
Then y2 (t) = (At + B )e−p/2 t is a second solution. In particular, the solution with
A = 1, B = 0, namely y2 (t) = te−p/2 t is “essentially diﬀerent”.
Please do examples #1,3,5 on NSS 165.
We now return to Case 3, p2 < 4q .
Case 3. p2 < 4q , complex roots. See Notes on Complex Arithmetic.
√
1
r1 ,2 =
−p ± i 4q − p2 , i = −1.
2
Solution:
√
√
1
1
2
2
y (t) = Ae 2 (−p+i 4q−p )t + Be 2 (−p−i 4q−p )t
There are questions. To discuss things we must know a little complex arithmetic.
A complex number z is an ordered pair (x, y ) of real numbers and can be written
as x + iy . It can also be written in polar form as reiθ where x = r cos θ and
y = r sin θ. Here we have used the identity eiθ = cos θ + i sin θ.
10 y✻ ✻
y
❄
✚
✛ r✚
✚
✚θ ✚ x ✚
✚ ✚ θ
z = x + iy = re ✲
x ✲ Also prove cos θ = (eiθ + e−iθ )/2 and sin θ = (eiθ − e−iθ )/2i.
The complex conjugate of a complex number z is denoted z ∗ or z and is deﬁned as
¯
z ∗ = z = x − iy . Show z ∗ = re−iθ .
¯
Given z , to ﬁnd z ∗ , simply change i to −i in the expression for z , e.g. (sin(1+ i))∗ =
sin(1 − i).
The sum of z and its complex conjugate z ∗ is real and equals 2x = 2 real z .
√
√
1
1
2)
2
If y (t) is real, and y (t) = Ae 2 (−p+i 4q−p√ t + Be 2 (−p−i 4q−p )t , then, because
√
i
i
2
2
e− 2 4q−p t is the complex conjugate of e 2 4q−p t , B must be the complex conju∗
gate A of A.
Therefore, for p2 < 4q , the general solution of (6.7) is
√
√
1
1
2
2
y (t) = Ae 2 (−p+i 4q−p )t + A∗ e 2 (−p−i 4q−p )t (6.14) where A is in general complex.
It is often simpler to write (6.14) in another form. Let A = C −iD . Then A∗ = C +iD ,
2
2
when C and D are real. Then
p
t
t
y (t) = e− 2 t (C cos 4q − p2 + D sin 4q − p2 ).
2
2
There is also another convenient way to write this expression which gives one a
better picture of what the solution does.
A little trigonometry:
C cos θ + D sin θ can be written
C
D
C 2 + D2 √
cos θ + √
sin θ .
C 2 + D2
C 2 + D2
Now deﬁne the angle φ as
cos φ = √ C
,
+ D2 sin φ = √ C2 11 D
+ D2 C2 and then
C cos θ + D sin θ = C 2 + D2 cos(θ − φ). Therefore we can rewrite (6.14) as
p
t
2 + D 2 e− 2 t cos
2 −φ
y ( t) = C
4q − p
2
where cos φ =
= −2 Im A. √ C
C 2 +D 2 D
and sin φ √C 2 +D2 , C = A + A∗ = 2 Rl A, D = i(A − A∗ ) We see that y (t) is the product of an
amplitude or envelope
p
C 2 + D2 exp(− t)
2
which decays, and an oscillatory sinusoidal part
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This note was uploaded on 01/22/2014 for the course MATH 254 taught by Professor Indik during the Fall '08 term at Arizona.
 Fall '08
 INDIK

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