Polytechnic Institute of NYU
MA 2132
Final Practice Answers
Fall 2008B
Studying from past or sample exams is NOT recommended. If you do, it should be only AFTER
you know how to do all of the homework and worksheet problems. These are additional practice
problems designed to cover the material, but not necessarily specific to the exam.
The final is cumulative, covering all material in the course.
In general, 7580% of the exam is
new material.
(1) Consider the initial value problem
P
0
=
te
t
2

P
,
y
(0) = 1
.
(a) Use the Euler Method with stepsize
h
= 0
.
25 to find the approximate value for the
solution of the above problem at
x
= 0
.
5.
We will need (0
.
5

0)(
.
25) = 2 iterations to get the answer.
n
t
n
=
t
0
+
n
*
h
y
n
f
(
t
n
, P
n
) =
te
t
2

P
P
n
+1
=
P
n
+
h
*
f
(
t
n
, P
n
)
0
0
0
0
*
e
0
2

0
= 0
P
1
= 0 + (0
.
25)(0) = 0
1
0
.
25
0
0
.
25
*
e
(0
.
25)
2

0
= 0
.
2661
P
2
= 0 + 0
.
25(0
.
2661) = 0
.
0665
2
0
.
5
0
.
0665
n/a
n/a
(b) Find the exact solution and determine its value at
x
= 0
.
5.
Separable:
e
P
dP
=
te
t
2
dt
so
P
= ln(0
.
5
*
e
t
2
+
C
)
Using the IVP (0
,
0), 0 = ln(0
.
5
*
1 +
C
)
⇒
1 = 0
.
5 +
C
⇒
C
= 0
.
5
.
So,
P
= ln(0
.
5
*
e
t
2
+ 0
.
5) and
P
(0
.
5) = 0
.
1328. The approximation is poor due to the
relatively large stepsize.
(2) Consider the initial value problem
x
2
y
0
= 2
xy

x
3
,
y
(1) = 3
.
(a) Use the Euler Method with stepsize
h
= 0
.
1 to find the approximate value for the
solution of the above problem at
x
= 1
.
3.
n
x
n
y
n
f
(
x
n
, y
n
) = 2
*
(
y/x
)

x
y
n
+1
=
y
n
+
h
*
f
(
x
n
, y
n
)
0
1
3
2
*
3
/
1

1 = 5
y
1
= 3 + (0
.
1)(5) = 3
.
5
1
1 +
.
1 = 1
.
1
3
.
5
5
.
2636
y
2
= 3
.
5 + 0
.
1(5
.
2636) = 4
.
02636
2
1
.
2
4
.
02636
5
.
5106
y
3
= 4
.
02636 + 0
.
1(5
.
5106) = 4
.
577
3
1
.
3
4
.
577
n/a
n/a
1
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2
(b) Find the exact solution and determine its value at
x
= 1
.
3.
First order linear:
You must reformulate the problem as
y
0
= 2
y/x

x
as in the
f
(
x
n
, y
n
) column above.
The integrating factor is
e

2
/x dx
=
e

2
ln

x

=
x

2
.
Then
x

2
y
0

2
x

3
y
=

x

1
.
(
x

2
y
) =

ln

x

+
C
⇒
y
=

x
2
ln

x

+
Cx
2
.
Using the IVP (1
,
3),
C
= 3, so
y
=

x
2
ln

x

+ 3
x
2
. Then
y
(1
.
3) = 5
.
5134
.
(3) Consider the initial value problem
xy
0
=
y
(ln(
y/x
) + 1)
,
y
(1) =
e.
(a) Use the Euler Method with stepsize
h
= 0
.
1 to find the approximate value for the
solution of the above problem at
x
= 1
.
2.
n
x
n
y
n
f
(
x
n
, y
n
) = (
y/x
)
*
(ln(
y/x
) + 1)
y
n
+1
=
y
n
+
h
*
f
(
x
n
, y
n
)
0
1
e
e
*
(ln(
e
) + 1) = 2
e
y
1
=
e
+ 0
.
1
*
(2
e
) = 1
.
2
*
e
= 3
.
2619
1
1
.
1
3
.
2619
6
.
1888
y
2
= 3
.
8808
2
1
.
2
3
.
8808
n/a
n/a
(b) Find the exact solution and determine its value at
x
= 1
.
2.
Homogeneous: Use
y
=
vx
substitution to get
v
0
x
+
v
=
v
(ln(
v
) + 1). Note that with
the given IVP,
x
≥
1 in this problem.
Then
v
0
=
v
ln(
v
)
*
x

1
⇒
dv
v
ln(
v
)
=
dx
x
⇒
ln(ln(
v
)) = ln(
x
) +
C
⇒
ln(
v
) =
xe
C
=
kx
for
k
=
e
C
. So,
v
=
e
kx
⇒
y
=
xe
kx
.
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 Spring '07
 King
 Differential Equations, Addition, Equations, Sin, Cos, Trigraph

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