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ODE DREAMIN’
CHAPTER 2
ADDITIONAL INTRODUCTORY MATERIAL
(50407) by Doug Jones
ode_dreamin_ch2_50407[7].doc
Page 1 of 10
Here is some practice at the important skill of
verifying that a given function is a
solution of a ODE
. Verification is how you check your answers, and if you can’t check your
answers, how do you know if they are right?
And since I’ve had some extra time on my hands during the short break between semesters,
I’ve also included some review of important Calc II techniques we’ll be using in this course.
Enjoy.
I.
Verify that
()
2
t
te
φ
−
=
is a solution of
91
40
yy
y
′
′′
+
+=
.
Verification
:
[1]
1
.
() ( ) ( )
22
2
24
tt
t
t
e
t
e
φφ
−−
−
′
=⇒
=
−⇒
=
[2].
Subs. [1] into the LHS of the ODE and simplify:
( ) (
)()
2
2
9
14
9
14
4
9
2
14
41
81
4
0
subs
t
t
y
e
e
e
e
−
−
′
′
++
⎯
⎯
⎯
→+
+
−
+
=−+
=
[3].
Thus, we see that the given function is in fact a solution to the ODE.
▄
II.
Verify that
2
t
=
is a solution of
2
dy
ty
dt
=
on the halfplane
0
y
>
, given that y is a
function of t.
Verification
:
[1].
() ()
2
t
t
e
′
=
[2].
Subs. [1] into the LHS of the ODE and simplify:
2
subs
t
dy
d
te
t
dt
dt
⎯⎯⎯
→==
[3].
Therefore,
2
d
t
dt
=
, and it follows (by def) that
2
t
=
is a solution of the
given ODE.
▄
1
Let’s get this “arrow” notation straight right at the outset! Misuse of the “arrow” (
⇒
) is one of my petpeeves! Technically,
the arrow means “implies.” This means that the statement (or equation) that comes right before the arrow (the “antecedent”) is
a sufficient condition, inandof itself, to imply the statement (or equation) that comes right after the arrow (the “consequent”).
In writing math, we bend the rule just a little and use the arrow to mean that the antecedent
plus
possibly some other obvious
facts together imply the consequent…. But in
any case
, the arrow means
implies
. It
does not mean
“equals.” (
=
) means
“equals.” (
⇒
) means “implies.” Don’t confuse them!
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III.
Verify that
() ( )
3sinh 2
tt
φ
=
is a solution of
2
2
40
dx
x
dt
−
=
.
Verification
:
[1].
() ( ) ( ) ( ) (
)(
)
3sinh 2
6cosh 2
12sinh 2
t
t
t
t
φφ
′′
′
=⇒
=
⇒
=
[2].
Subs. [1] into the LHS of the ODE and simplify:
() ()
()
22
4
4
12sinh 2
4 3sinh 2
0
subs
d
xt
t
dt
dt
−⎯
⎯
⎯
→−
=
−
=
[3].
Therefore,
2
2
d
dt
−=
, and it follows (by def) that
(
)
=
is a solution
of the given ODE.
▄
IV.
Here is an “INSIGHT” for you to consider. For example, in the ODE above,
2
2
x
dt
−
=
,
think of the “zero” as a function
!
V.
Now let’s complicate things just a bit. Up until now, all the solutions we have verified
have been
explicit functions
. However, in many, many problems you will be solving
in this course, the solutions will be expressed as
implicit functions
. So I need to show
you what such solutions look like and the slightly different technique used for
verification.
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 Summer '11
 Jones

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