Section 2.3
Existence and Uniqueness of Solutions of Nonlinear Equation s
11
2.2.38. y
1
D e
2x
is a solution of y
0
2y D 0. Look for solutions of the nonlinear equation of t he
form y D ue
2x
. Then u
0
e
2x
D
xe
2x
1 u
; u
0
D
x
1 u
; .1 u/u
0
D x;
.1 u/
2
2
D
1
2
.x
2
c/;
u D 1 ˙
p
c x
2
; y D e
2x
1 ˙
p
c x
2
.
2.3 EXISTENCE AND UNIQUENESS OF SOLUTIONS OF NONLINEAR EQUATIONS
2.3.2. f .x; y/ D
e
x
Cy
x
2
C y
2
and f
y
.x; y/ D
1
x
2
C y
2
2y.e
x
C y/
.x
2
C y
2
/
2
are both continuous at all .x; y/ ¤
.0; 0/. Hence, Theorem 2.3.1 implies that if .x
0
; y
0
/ ¤ .0; 0/, then the initial value problem h as a a
unique solution on some open interval containing x
0
. Theorem 2.3.1 does no t apply if .x
0
; y
0
/ D .0; 0/.
2.3.4. f .x; y/ D
x
2
Cy
2
ln xy
and f
y
.x; y/ D
2y
ln xy
x
2
C y
2
x.ln xy/
2
are bo th continuous at all .x; y/ such
that xy > 0 and xy ¤ 1. Hence, Theorem 2.3.1 implies that if x
0
y
0
> 0 and x
0
y
0
¤ 1, then the initial
value problem has unique solution on an open interval containing x
0
. Theorem 2.3.1 does not appl y if
x
0
y
0
0 or x
0
y
0
D 1.
2.3.6. f .x; y/ D 2xy and f
y
.x; y/ D 2x are both continuous at all .x; y/. Hence, Theorem 2.3.1
implies that if .x
0
; y
0
/ is arbitrary, then t he initial value problem has a unique solution on some open
inter val contai ning x
0
.
2.3.8. f .x; y/ D
2x C 3y
x 4y
and f
y
.x; y/ D
3
x 4y
C 4
2x C 3y
.x 4y/
2
are both continuous at all .x; y/ such
that x ¤ 4y. Hence, Theorem 2.3.1 impli es that if x
0
¤ 4y
0
, then the ini tial value problem has a unique
solution on some open interval containing x
0
. Theorem 2.3.1 does not apply if x
0
D 4y
0
.
2.3.10. f .x; y/ D x.y
2
1/
2=3
is continuous at all .x; y/, but f
y
.x; y/ D
4
3
xy.y
2
1/
1=3
is continuous
at .x; y/ if and only if y ¤ ˙1. Hence, Theorem 2.3.1 implies that if y
0
¤ ˙1, t hen the initial value
problem has a unique solution on some o pen interval containing x
0
, while if y
0
D ˙1, then the initial
value prob lem has at least one so lution (possibly not unique on any open inter val containing x
0
).
2.3.12. f .x; y/ D .x C y/
1=2
and f
y
.x; y/ D
1
2.x C y/
1=2
are b oth continuous at all .x; y/ such th at
x Cy > 0 Hence, Theorem 2.3.1 implies that if x
0
Cy
0
> 0, then the initial value problem has a uniq ue
solution on some open interval containing x
0
. Theorem 2.3.1 does not apply if x
0
C y
0
0.
2.3.14. To apply Theorem 2.3.1, rewrite the given initial value problem as (A) y
0
D f .x; y/; y.x
0
/ D y
0
,
where f .x; y/ D p.x/y C q.x/ and f
y
.x; y/ D p.x/. If p and f are continuous on some open
inter val .a; b/ containing x
0
, then f and f
y
are continuous on some open rectangle containing .x
0
; y
0
/,
so Theorem 2.3.1 implies that (A) has a un ique sol ution on so m e open i nterval containing x
0
. The
conclusion of Theorem 2.1.2 i s more specific: the solut ion of (A ) exists and is unique on .a; b/. For
example, in the extreme case where .a; b/ D .1; 1/, Theorem 2.3.1 stil l implies only existence and
uniqueness on some open interval contain ing x
0
, while Theorem 2.1.2 implies that the solu tion exists and
is unique on .1; 1/.
2.3.16. First find solutions of (A) y
0
D y
2=5
. Obviously y 0 is a so lution. If y 6 0, then we
can separate variables on any open interval wh ere y has no zeros: y
2=5
y
0
D 1;
5
3
y
3=5
D x C c;
y D
3
5
.x C c/
5=3
. (Note that this solution is also defined at x D c, even though y.c/ D 0.