Exam 1: MAP 2302
*
September 26, 2008
Name
:
Student ID
:
This is a
closed book
exam and the use of calculators is
not
allowed.
1. Solve the IVP:
dy
dx
=

xy, y
(1) =

1
.
By the existence and uniqueness theorem, we know there is a unique solution. Separable
equation:
1
y
dy
=

xdx,
ln

y

=

x
2
2
+
C, C
arbitrary constant
,

y

=
C
′
e
−
x
2
2
, C
′
positive constant (since
C
′
= e
C
)
,
y
=
±
C
′
e
−
x
2
2
=
C
′′
e
−
x
2
2
, C
′′
nonzero constant
Since
y
(1) =

1, it follows that

1 =
C
′′
e
−
1
2
or
C
′′
=

e
1
2
,
and thus
y
(
x
) =

e
1
2
−
x
2
2
Common mistake
:
After finding

y

=
C
′
e
−
x
2
2
,
one determines
C
′
, using the initial condition:
 
1

=
C
′
e
−
1
2
,
so that
C
′
= e
1
2
.
So far, so good, but then one drops the absolute value and concludes:
y
= e
1
2
−
x
2
2
.
This is wrong because
y
(1) = +1 instead of

1. The mistake made is that the absolute value
cannot be dropped. A careful argument is as follows:
y
=
±
e
1
2
−
x
2
2
.
These are two functions
y
(
x
), while we know there should only be one (by the existence and
uniqueness theorem). So which one is the solution to our problem? That’s easy: the one
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 Summer '06
 DeLeenheer
 Equations, Trigraph, Elementary algebra, The Mistake, initial condition

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