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
=
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 Spring '10
 Hald,OH
 Math, Differential Equations, Linear Algebra, Algebra, Equations, Trigraph, dx, ty, constant solutions, ln V

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