Math 3301
Homework Set 2 – Solution
10 Points
1. (3 pts)
( )
( )
(
)
( )
(
)
2
2
2
5
2
2
5
2
2
2
2
5
2
8
3
2
5
4
3
2
4 1
3 1
7
3
9
4 4
7
2
3
121
32
40
4
3
7
2
0
8
8
y
dy
x dx
y
y
x
x
c
c
x
x
x
x
y
y
x
x
y
+
=

+
=

+
+
=
=
 –
+
+

 –
+

+

+

=
=
=
°
°
Reapplying the initial condition shows that “+” is the correct sign.
The solution is then,
(
)
2
3
121
32
40
8
x
x
y x
 +
+

=
For the interval of validity we need :
2
40
32
121
0
x
x

+
‡
.
The quadratic is zero at 1.3847 and
2.1847.
So, the possible interval of validity is,
1.3847
.13847
2.1847
2.1847
x
x
x
¥ <
£ 

£
£
£
< ¥
The interval of validity is
.13847
2.1847
x

£
£
because it contains the initial condition and the
quadratic is positive in this range.
2.
(3 pts)
( )
(
)
1
1
2
2
1
2
3
4
2
7
4
2
4
2
4
2
7
7
8
8
7
8
2
4
2
8
2
8
2
8
y
dy
x
x dx
y
x
x
c
c
y
x
x
y x
x
x

=
+
=
+
+
=
=
+
+
=
+
+
°
°
In the right (implicit) solution the polynomial is always positive and so we won’t have any issues with the
root being negative out of it.
Also, the explicit solution is a polynomial and so will exist for for all
x
’s.
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 Spring '08
 Deng
 Math, Differential Equations, Equations, Derivative, IVP

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