This preview shows page 1. Sign up to view the full content.
Exercises 2.6
When
C
= 0, the above formula gives
θ
≡
0or
y
≡
0, which were lost in separating variables.
Also, we lost another solution,
u
+1
≡
0or
y
=
−
θ
.Thu
s
,th
ean
sw
e
ri
s
θy
2
=
C
(
y
+
θ
)
2
and
y
=
−
θ,
where
C
is an arbitrary constant.
41.
The righthand side of (8) from Example 2 of the text can be written as
y
−
x
−
1+(
x
−
y
+2)
−
1
=
−
(
x
−
y
+2)+1+(
x
−
y
+2)
−
1
=
G
(
x
−
y
+2)
with
G
(
v
)=
−
v
+
v
−
1
+1. With
v
=
x
−
y
+2, we have
y
0
=1
−
v
0
, and the equation becomes
1
−
dv
dx
=
−
v
+
v
−
1
+1
⇒
dv
dx
=
v
2
−
1
v
⇒
v
v
2
−
1
dv
=
dx
⇒
ln

v
2
−
1

=2
x
+
C
1
⇒
v
2
−
1=
Ce
2
x
,C
6
=0
.
Dividing by
v
2
−
1, we lost constant solutions
v
=
±
1, which can be obtained by taking
C
=0
in the above formula. Therefore, a general solution to the given equation is
(
x
−
y
+2)
2
=
This is the end of the preview. Sign up
to
access the rest of the document.
This note was uploaded on 03/29/2010 for the course MATH 54257 taught by Professor Hald,oh during the Spring '10 term at University of California, Berkeley.
 Spring '10
 Hald,OH
 Math, Differential Equations, Linear Algebra, Algebra, Equations

Click to edit the document details