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ODE A2 - 1
ODE Chapter 2 AAssignment
2-1 For each of the following initial value problems, (i) state the largest interval in whichthere will be a unique solution, (ii) solve it, if you can.
10
2
=
=+
)(
sin)(tan)(
ywhere
xyxya
30
1
=
=+
)(
)(
ywhere
xx
yyf
10
2 2
=
=+
)(
cossin)(sin)(cos)(
ywhere
xxyxyxb
10
2
=
=+
)(
sin)(cos)(
ywhere
xyeyxg x
21
2
)1(
12)(
=
+=+
ywhere
xxyyxc
10 =
=+
)(
sin)(sin)(
ywhere
xyxyxh
30
41
=
=
)(
)(
ywhere
x
yyd
42
2
eeywhere
xxyyxxi
=
=+
)(
ln)ln()(
30
141
=
=
)(
),()()(
ywhere
xyyxe
2-2 For each of the following initial value problems determine whether or not the
Fundamental Existence and Uniqueness Theorem (for 1storder ordinary differential
equations) guarantees that the problem has a unique solution in a neighbourhood of
the point indicated.
0151
== )(,)( ywhereyxyi 10
2
== )(,)( ywherex
y
yiii
0)1(,)( 51 == ywherexyyii 1)0(,)(2
== ywherey
xyiii
2-3 Does the initial value problem ( ) 02 == ywherexyy ,sin have a unique solution
in any neighbourhood of2=x ? If so
(a) find the unique solution, and
(b) show how the existence follows from the Fundamental Existence and
Uniqueness Theorem.If not
(a) write down at least two solutions, and
(b) show why the Fundamental Existence and Uniqueness Theorem fails to apply
to this initial value problem.
2-4 Show how the Existence and Uniqueness Theorem guarantees the existence of a
unique solution of ),( yxxy += where y(2) = 5 near x= 2.
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ODE A2 - 2
2-5 (a) State the largest open interval in which you can be sure that the initial value
problem 10312
1==
+ )(,
)(ywherey
xy
will have a unique solution and,
(b) Solve the initial value problem.(c) What is the largest x-interval in which you have found a continuous solution
2-6 (a) Find all constant solutions and the general solution of
1
2 2
+=
x
y
dx
dy )(
(b) Solve1
2 2
+=
x
y
dx
dy )(, where y(0) = 0.
2-7 Verify that the solution of the initial value problem ( )311 +++= )ln()( xxxy iscontinuous on ( ) ,1 .
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ODE A2 - 3
Answers to ODE Chapter 2 AAssignment
2-1 (a) xxy coscos,, 3222
2+=
(b) xxy coscos,, 3222
2
+=
(c) ( )2
1
12
1
340
2
2
++=x
xxy,,
(d) ( ) )()ln()(,, xxxy += 131141 (e) same as (d)
(f) ( ) ( )3111 +++= )ln()(,, xxxy
(g) lyanalyticalevaluatedbecannotdxx
eFI
x
=
cos..,,
2
22
(h) no unique solution at y(0) = 1
(i) ( )x
xxy
ln,,
421
22
=
2-2 (i) Yes, (ii) No, (iii) No, (iv) Yes.
2-3 No.
2-5 (a)
( )1,
(b)x
xy
+=1
312 )(
(c) ( )1,
2-6 (a) 2=ySolutionntConsta :
Cxy
SolutionGeneral +=+
)ln(: 12
1
(b)2
11
2
1=
+
)ln( xy
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