Math 334 Lecture #4
§
2.2: Separable First Order ODE’s
The ODE’s
dy
dx
=

1
5
y
+ 9
.
8
and
y
=
xy
2
1 +
x
2
can be rewritten respectively as

1 +
1

y/
5 + 9
.
8
dy
dx
= 0
and

x
1 +
x
2
+
1
y
2
y
= 0
.
[The first ODE is linear; the second ODE is nonlinear.]
Each of these is an example of a
separable first order
ODE, the general form of which
is
M
(
x
) +
N
(
y
)
dy
dx
= 0
for functions
M
(
x
) and
N
(
y
).
Method of Solution.
Let
H
(
x
) and
G
(
y
) be antiderivatives of
M
(
x
) and
N
(
y
):
d
dx
H
(
x
)
=
M
(
x
)
and
d
dy
G
(
y
)
=
N
(
y
)
.
The general form of the separable first order ODE becomes
d
dx
H
(
x
) +
d
dy
G
(
y
)
dy
dx
= 0
.
By the Chain Rule, the second term is
d
dx
G
(
y
)
=
d
dy
G
(
y
)
dy
dx
,
so that the ODE becomes
d
dx
H
(
x
) +
d
dx
G
(
y
)
= 0
⇒
d
dx
H
(
x
) +
G
(
y
)
= 0
.
Integration gives
H
(
x
) +
G
(
y
) =
C
as a one parameter family of solutions of the separable first order ODE.
[It may not be a general solution, i.e., it may not contain all possible solutions of the
ODE, but it can be useful for solving IVP’s.]
[We will consider in a moment an example where the one parameter family of solutions
does not contain all possible solutions!]
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 Spring '11
 Smith
 Math, Calculus, Derivative, Ode, IVP, dx dy dx

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