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11
Exam 2 –Review
This page gives a summary of the material required for Exam 2.
Subject Matter:
1. Special Second Order Equations:
(a)
y
00
=
f
(
t,y
0
)
Set
y
0
=
v
, get
v
0
=
f
(
t,v
)
Solve:
v
=
v
(
t,C
1
),
Then, integrate to get
y
=
R
v
(
t,C
1
) +
C
2
(b)
y
00
=
f
(
y,y
0
)
Set
y
0
=
v
,
y
00
=
dv
dt
=
dv
dy
dy
dt
=
dv
dy
v
=
f
(
y,v
)
Solve for
v
=
v
(
y,C
1
)
Then,
y
0
=
v
(
y,C
1
) – separable
Solve for
y
=
y
(
t,C
1
,C
2
) — May get implicit solution
2. Reduction of order
L
(
y
) =
y
00
+
p
(
t
)
y
0
+
q
(
t
)
y
= 0,
Have one nonzero solution
y
1
.
Get second independent solution by
y
2
(
t
) =
y
1
(
t
)
v
(
t
) for some non
constant
v
(
t
)
Plug into equation and get
v
0
(2
y
0
1
+
py
1
) +
y
1
v
00
= 0
Solve for
v
0
, then integrate to get
v
– Do not need to keep the constants
3. Linear Second Order equations – constant coeﬃcients
L
(
y
) =
ay
00
+
by
0
+
cy
=
h
(
t
)
(a) homogeneous:
L
(
y
) = 0:
Get characteristic polynomial:
z
(
r
) =
ar
2
+
br
+
c
,
Get roots
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This note was uploaded on 02/03/2012 for the course MTH 235 taught by Professor Staff during the Fall '11 term at Michigan State University.
 Fall '11
 STAFF
 Equations

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