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13
13
Boundary-Value Problems
in Rectangular Coordinates
EXERCISES 13.1
Separable Partial Differential Equations
1.
Substituting
u
(
x, y
)=
X
(
x
)
Y
(
y
) into the partial diferential equation yields
X
0
Y
=
XY
0
. Separating variables
and using the separation constant
−
λ
, where
λ
6
= 0, we obtain
X
0
X
=
Y
0
Y
=
−
λ.
When
λ
6
=0
X
0
+
λX
= 0
and
Y
0
+
λY
so that
X
=
c
1
e
−
λx
and
Y
=
c
2
e
−
λy
.
A particular product solution oF the partial diferential equation is
u
=
=
c
3
e
−
λ
(
x
+
y
)
,λ
6
.
When
λ
= 0 the diferential equations become
X
0
= 0 and
Y
0
= 0, so in this case
X
=
c
4
,
Y
=
c
5
, and
u
=
=
c
6
.
2.
Substituting
u
(
x, y
X
(
x
)
Y
(
y
) into the partial diferential equation yields
X
0
Y
+3
0
= 0. Separating
variables and using the separation constant
−
λ
we obtain
X
0
−
3
X
=
Y
0
Y
=
−
λ.
When
λ
6
X
0
−
3
λX
= 0
and
Y
0
+
λY
so that
X
=
c
1
e
3
λx
and
Y
=
c
2
e
−
λy
.
A particular product solution oF the partial diferential equation is
u
=
=
c
3
e
λ
(3
x
−
y
)
.
When
λ
= 0 the diferential equations become
X
0
= 0 and
Y
0
= 0, so in this case
X
=
c
4
,
Y
=
c
5
, and
u
=
=
c
6
.
3.
Substituting
u
(
x, y
X
(
x
)
Y
(
y
) into the partial diferential equation yields
X
0
Y
+
0
=
. Separating
variables and using the separation constant
−
λ
we obtain
X
0
X
=
Y
−
Y
0
Y
=
−
λ.
680

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*Sign up* 13.1
Separable Partial Diferential Equations
Then
X
0
+
λX
= 0
and
Y
0
−
(1 +
λ
)
Y
=0
so that
X
=
c
1
e
−
λx
and
Y
=
c
2
e
(1+
λ
)
y
.
A particular product solution of the partial diFerential equation is
u
=
XY
=
c
3
e
y
+
λ
(
y
−
x
)
.
4.
Substituting
u
(
x, y
)=
X
(
x
)
Y
(
y
) into the partial diFerential equation yields
X
0
Y
=
0
+
. Separating
variables and using the separation constant
−
λ
we obtain
X
0
X
=
Y
+
Y
0
Y
=
−
λ.
Then
X
0
+
λX
= 0
and
y
0
+(1+
λ
)
Y
so that
X
=
c
1
e
−
λx
and
Y
=
c
2
e
−
(1+
λ
)
y
.
A particular product solution of the partial diFerential equation is
u
=
=
c
3
e
−
y
−
λ
(
x
+
y
)
.
5.
Substituting
u
(
x, y
X
(
x
)
Y
(
y
) into the partial diFerential equation yields
xX
0
Y
=
yXY
0
. Separating vari-
ables and using the separation constant
−
λ
we obtain
xX
0
X
=
yY
0
Y
=
−
λ.
When
λ
6
xX
0
+
λX
= 0
and
0
+
λY
so that
X
=
c
1
x
−
λ
and
Y
=
c
2
y
−
λ
.
A particular product solution of the partial diFerential equation is
u
=
=
c
3
(
xy
)
−
λ
.
When
λ
= 0 the diFerential equations become
X
0
= 0 and
Y
0
= 0, so in this case
X
=
c
4
,
Y
=
c
5
, and
u
=
=
c
6
.
6.
Substituting
u
(
x, y
X
(
x
)
Y
(
y
) into the partial diFerential equation yields
yX
0
Y
+
xXY
0
= 0. Separating
variables and using the separation constant
−
λ
we obtain
X
0
xX
=
−
Y
0
=
−
λ.
When
λ
6
X
0
+
λxX
= 0
and
Y
0
−
λyY
so that
X
=
c
1
e
λx
2
/
2
and
Y
=
c
2
e
−
λy
2
/
2
.
A particular product solution of the partial diFerential equation is
u
=
=
c
3
e
λ
(
x
2
−
y
2
)
/
2
.
681

13.1
Separable Partial Diferential Equations
When
λ
= 0 the diferential equations become
X
0
= 0 and
Y
0
= 0, so in this case
X
=
c
4
,
Y
=
c
5
, and
u
=
XY
=
c
6
.
7.
Substituting
u
(
x, y
)=
X
(
x
)
Y
(
y
) into the partial diferential equation yields
X
0
Y
+
X
0
Y
0
+
0
=0
0
, which
is not separable.

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