Unformatted text preview: 2 rt
L[y (r, t)] = (r2 + pr + q )aert = r +
ae .
2
p Clearly L[y (r = −p/2, t)] = 0. And therefore y1 (t) = ae− 2 t is a solution. But, note
also that
∂
∂y
L[y (r, t)] = L
(r, t)
∂r
∂r
= (r + p/2)2 atert + 2(r + p/2)aert = 0 also when r = − 2. Therefore, at a double root, both y1 (t) = y (r = −p/2, t) and
p/
y2 (t) = ∂ y (r, t)
are solutions.
∂r
r =−p/2
p
∂y
= atert
= ate− 2 t .
∂ r r=−p/2
r =−p/2 p p Thus y1 (t) = e− 2 t and y2 (t) = te− 2 t are two essentially diﬀerent solutions of (6.7)
and the general solution is
p p y (t) = Ae− 2 t + Bte− 2 t
when p2 = 4q .
We will come to the precise meaning of essentially diﬀerent shortly.
Example:
y (0) = 2, y (0) = −1. 9y − 12y + 4y = 0,
y − 4 y + 4 y = 0.
3
9
y = ert :⇒ r2 − 4 r +
3
2 4
9 =0⇒ r= 2
twice.
3 2 Solutions: y1 (t) = e 3 t , y2 (t) = te 3 t .
2 2 General solution: y (t) = Ae 3 t + Bte 3 t .
Initial conditions: y (0) = 2 = A
y (0) = −1 = 2 A + B
3 ⇒ A = 2, B = −7/3. Thus
y ( t) = 2 e 2
3t 72
− te 3 t =
3
8
2
7
2 − t e 3 t.
3 Reduction of order.
We interrupt the discussion at this point to introduce a very useful new idea,
reduction of order, discussed in NSS §4.7 pg. 198. The idea is this: If we know one
solution y = y1 (t) of (6.2)
d2 y
dy
+ p( t )
+ q (t)y = 0,
dt2
dt (6.2) we can ﬁnd another one y2 (t) by setting
y 2 ( t ) = y 1 ( t) v ( t ) . (6.11) Diﬀerentiate (6.11) twice
dy2
dy1
dv
=
v + y1
dt
dt
dt (6.12a) d 2 y2
d 2 y1
dy1 dv
d2 v
=
v+2
+ y1 2
2
2
dt
dt
dt dt
dt (6.12b) and substitute (6.12) into (6.2). We ﬁnd, on collecting terms
2
d2 v
dy1
dv
d y1
dy1
y1 2 + 2
+ p( t) y 1
+
+ p( t )
+ q (t)y1 v = 0.
dt
dt
dt
dt2
dt
But the last bracket is zero because y1 (t) solves (6.2). Therefore we are left with a
ﬁrst order ode for u(t) = dv
dt
du
dy1
y1
+2
+ p( t ) y 1 u = 0
(6.13)
dt
dt
(6.13) is a linear, ﬁrst order ode for u(t) which we solve using what we learned in
Lectures 56. Divide across by y1 . Then the integrating factor for (6.13) is
2 dy1
IF = exp
+ p(t) dt
y1 dt
t
= exp 2 ln y1  +
p(t )dt
2
y1 exp =
Therefore
d
dt
or 2
uy1 exp t p(t )dt .
t p(t )dt =0 t
dv
1
= 2 exp −
p(t )dt .
dt
y1
Therefore a second solution y2 (t) of (6.2) is
t
dt
y 2 ( t ) = y 1 ( t)
exp −
p(t )dt .
2
y 1 ( t)
u= 9 Furthermore, we will ﬁnd when we get to deﬁne what linearly independent means,
y1 (t) and y2 (t) are linearly independent (essentially diﬀerent) on any interval on
which p(t), q (t) are continuous.
Application to double root situation.
Let q = p2 /4. We know one solution to
d2 y
dy p2
+p
+ y=0
dt2
dt
4
is y1 (t) = e−p/2 t.
Let y2 (t) = ve−p/2 t . Then
y2 (t) = v e−p/2 t − p/2 ve−p/2 t , y2 = v e−p/2 t − p...
View
Full
Document
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

Click to edit the document details