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Page 2
=-4(?? 2
)(-?? 2
)=4?? 2
?? >0
And hence (1) is hyperbolic everywhere except on the coordinate axis ?? =0,?? =0.
The quadratic equation is ?? ?? 2
+???? +?? =0, i.e.,
? ?? 2
?? 2
+0?? -?? 2
=0
? ?? 2
?? 2
=?? 2
? ?? =
±?? ?? 2 distinct roots ?? 1
=
?? ?? ,?? 2
=
-?? ??
2 positive real roots
????
????
+?? 1
?=0,
????
????
+?? 2
=0
????
????
+
?? ?? ?=0 and
????
????
-
?? ?? =0
???????? +?????? ?=0 and ?????? -?????? =0
On integrating, ?
?? 2
2
+
?? 2
2
=?? 1
,
?? 2
2
-
?? 2
2
=?? 2
So, ??? 2
+?? 2
=?? 1
and ?? 2
-?? 2
=?? 2
are the required family of characteristics. Hence, these are families of clearly circles and
hyperbola's respectively.
3.2 Solve the following partial differential equation
???? +???? =?? ,?? ?? (?? )=?? ,?? ?? (?? )=?? ,?? ?? (?? )=?? ??
by method of characteristics.
(2010 : 20 Marks)
Solution:
Given, the equation is
???? +???? ?=?? ???? +???? -?? ?=0=?? (?? ,?? 1
?? ,?? ,?? )
?? 0
?=?? ,?? 0
=1,?? 0
=2?? ?? 3
'
(?? )?=?? 0
?? 1
'
(?? )+?? 0
?? 2
(?? )
?? 2
?? ????
=?? 0
×
????
?? +?? 0
×
?? (1)
????
2=?? 0
+0??? 0
=2
Page 3
=-4(?? 2
)(-?? 2
)=4?? 2
?? >0
And hence (1) is hyperbolic everywhere except on the coordinate axis ?? =0,?? =0.
The quadratic equation is ?? ?? 2
+???? +?? =0, i.e.,
? ?? 2
?? 2
+0?? -?? 2
=0
? ?? 2
?? 2
=?? 2
? ?? =
±?? ?? 2 distinct roots ?? 1
=
?? ?? ,?? 2
=
-?? ??
2 positive real roots
????
????
+?? 1
?=0,
????
????
+?? 2
=0
????
????
+
?? ?? ?=0 and
????
????
-
?? ?? =0
???????? +?????? ?=0 and ?????? -?????? =0
On integrating, ?
?? 2
2
+
?? 2
2
=?? 1
,
?? 2
2
-
?? 2
2
=?? 2
So, ??? 2
+?? 2
=?? 1
and ?? 2
-?? 2
=?? 2
are the required family of characteristics. Hence, these are families of clearly circles and
hyperbola's respectively.
3.2 Solve the following partial differential equation
???? +???? =?? ,?? ?? (?? )=?? ,?? ?? (?? )=?? ,?? ?? (?? )=?? ??
by method of characteristics.
(2010 : 20 Marks)
Solution:
Given, the equation is
???? +???? ?=?? ???? +???? -?? ?=0=?? (?? ,?? 1
?? ,?? ,?? )
?? 0
?=?? ,?? 0
=1,?? 0
=2?? ?? 3
'
(?? )?=?? 0
?? 1
'
(?? )+?? 0
?? 2
(?? )
?? 2
?? ????
=?? 0
×
????
?? +?? 0
×
?? (1)
????
2=?? 0
+0??? 0
=2
By eqn.,
?? 0
?? 0
+?? 0
?? 0
=?? 0
?2?? ×2+1×?? 0
=??
??? 0
=-3??
?????????????????????????????????????????????????????????????????? 0
=?? ,?? 0
=1,?? 0
=2?? ,?? 0
=2,?? 0
=-3?? ???????????????????????(1)
Now, characteristics are:
????
????
=?? ?? =?? (2)
????
????
=?? ?? =?? (3)
????
????
=???? +???? =?? (4)
????
????
=?? ?? -?? ?? ?? =1-?? 2
(5)
????
????
=-?? ?? -?? ?? ?? =-?? -???? =-?? (1+?? ) (6)
From (3)
????
????
=?? ?
????
?? =????
Integrating both sides, we get
At ?? =0,?? ?????????=?? 1
?? ?? (?? 1
is a constant )
?1???????? ?=?? 1
????
????
=1-?? 2
?
????
1-?? 2
=????
??
1
2
(
1
1-?? +
1
1+?? )???? =????
From (5)
????
????
?=1-?? 2
?
????
1-?? 2
=????
1
2
(
1
1-?? +
1
1+?? )???? ?=????
Integrating both sides, we get
Page 4
=-4(?? 2
)(-?? 2
)=4?? 2
?? >0
And hence (1) is hyperbolic everywhere except on the coordinate axis ?? =0,?? =0.
The quadratic equation is ?? ?? 2
+???? +?? =0, i.e.,
? ?? 2
?? 2
+0?? -?? 2
=0
? ?? 2
?? 2
=?? 2
? ?? =
±?? ?? 2 distinct roots ?? 1
=
?? ?? ,?? 2
=
-?? ??
2 positive real roots
????
????
+?? 1
?=0,
????
????
+?? 2
=0
????
????
+
?? ?? ?=0 and
????
????
-
?? ?? =0
???????? +?????? ?=0 and ?????? -?????? =0
On integrating, ?
?? 2
2
+
?? 2
2
=?? 1
,
?? 2
2
-
?? 2
2
=?? 2
So, ??? 2
+?? 2
=?? 1
and ?? 2
-?? 2
=?? 2
are the required family of characteristics. Hence, these are families of clearly circles and
hyperbola's respectively.
3.2 Solve the following partial differential equation
???? +???? =?? ,?? ?? (?? )=?? ,?? ?? (?? )=?? ,?? ?? (?? )=?? ??
by method of characteristics.
(2010 : 20 Marks)
Solution:
Given, the equation is
???? +???? ?=?? ???? +???? -?? ?=0=?? (?? ,?? 1
?? ,?? ,?? )
?? 0
?=?? ,?? 0
=1,?? 0
=2?? ?? 3
'
(?? )?=?? 0
?? 1
'
(?? )+?? 0
?? 2
(?? )
?? 2
?? ????
=?? 0
×
????
?? +?? 0
×
?? (1)
????
2=?? 0
+0??? 0
=2
By eqn.,
?? 0
?? 0
+?? 0
?? 0
=?? 0
?2?? ×2+1×?? 0
=??
??? 0
=-3??
?????????????????????????????????????????????????????????????????? 0
=?? ,?? 0
=1,?? 0
=2?? ,?? 0
=2,?? 0
=-3?? ???????????????????????(1)
Now, characteristics are:
????
????
=?? ?? =?? (2)
????
????
=?? ?? =?? (3)
????
????
=???? +???? =?? (4)
????
????
=?? ?? -?? ?? ?? =1-?? 2
(5)
????
????
=-?? ?? -?? ?? ?? =-?? -???? =-?? (1+?? ) (6)
From (3)
????
????
=?? ?
????
?? =????
Integrating both sides, we get
At ?? =0,?? ?????????=?? 1
?? ?? (?? 1
is a constant )
?1???????? ?=?? 1
????
????
=1-?? 2
?
????
1-?? 2
=????
??
1
2
(
1
1-?? +
1
1+?? )???? =????
From (5)
????
????
?=1-?? 2
?
????
1-?? 2
=????
1
2
(
1
1-?? +
1
1+?? )???? ?=????
Integrating both sides, we get
?
1
2
ln?(
1+?? 1-?? )???? ???=?? +?? 2
????? ?
?
1+?? 1-?? =?? 3
?? 2?? ???????? ?
At ?? =0,??????? 0
?=2 ????????
?
3
-1
=?? 3
??? 3
=-3 ???????? ?
1+?? 1-?? =-3?? 2?? ???????? ?
?? =
3?? 2?? +1
3?? 2?? -1
????
????
=-?? (1+?? )=-?? (1+
3?? 2?? +1
3?? 2?? -1
)
???
????
????
=-?? ×
6?? 2?? (3?? 2?? -1)
???
????
?? =-
6?? 2?? 3?? 2?? -1
????
Integrating both sides, we get
???? =
?? 4
3?? 2?? -1
(1)
Now at ?? =0,?? 0
=-3??
? -3?? =
?? 4
3-1
??? 4
=-6?? ? ?? =
-6?? 3?? 2?? -1
Now, from (2) and (4)
????
?? ???????? ?=
????
?? ??????? ?????????=??????
Integrating both sides, we get
?? 2
-?? 2
=?? 5
At ?? =0,?? 0
=5,?? 0
=2 s
??????????????????????????????????????????????????????????????????? 2
-4?? 2
=?? 5
???? 5
=-3?? 2
?? 2
-?? 2
=-3?? 2
(10)
Page 5
=-4(?? 2
)(-?? 2
)=4?? 2
?? >0
And hence (1) is hyperbolic everywhere except on the coordinate axis ?? =0,?? =0.
The quadratic equation is ?? ?? 2
+???? +?? =0, i.e.,
? ?? 2
?? 2
+0?? -?? 2
=0
? ?? 2
?? 2
=?? 2
? ?? =
±?? ?? 2 distinct roots ?? 1
=
?? ?? ,?? 2
=
-?? ??
2 positive real roots
????
????
+?? 1
?=0,
????
????
+?? 2
=0
????
????
+
?? ?? ?=0 and
????
????
-
?? ?? =0
???????? +?????? ?=0 and ?????? -?????? =0
On integrating, ?
?? 2
2
+
?? 2
2
=?? 1
,
?? 2
2
-
?? 2
2
=?? 2
So, ??? 2
+?? 2
=?? 1
and ?? 2
-?? 2
=?? 2
are the required family of characteristics. Hence, these are families of clearly circles and
hyperbola's respectively.
3.2 Solve the following partial differential equation
???? +???? =?? ,?? ?? (?? )=?? ,?? ?? (?? )=?? ,?? ?? (?? )=?? ??
by method of characteristics.
(2010 : 20 Marks)
Solution:
Given, the equation is
???? +???? ?=?? ???? +???? -?? ?=0=?? (?? ,?? 1
?? ,?? ,?? )
?? 0
?=?? ,?? 0
=1,?? 0
=2?? ?? 3
'
(?? )?=?? 0
?? 1
'
(?? )+?? 0
?? 2
(?? )
?? 2
?? ????
=?? 0
×
????
?? +?? 0
×
?? (1)
????
2=?? 0
+0??? 0
=2
By eqn.,
?? 0
?? 0
+?? 0
?? 0
=?? 0
?2?? ×2+1×?? 0
=??
??? 0
=-3??
?????????????????????????????????????????????????????????????????? 0
=?? ,?? 0
=1,?? 0
=2?? ,?? 0
=2,?? 0
=-3?? ???????????????????????(1)
Now, characteristics are:
????
????
=?? ?? =?? (2)
????
????
=?? ?? =?? (3)
????
????
=???? +???? =?? (4)
????
????
=?? ?? -?? ?? ?? =1-?? 2
(5)
????
????
=-?? ?? -?? ?? ?? =-?? -???? =-?? (1+?? ) (6)
From (3)
????
????
=?? ?
????
?? =????
Integrating both sides, we get
At ?? =0,?? ?????????=?? 1
?? ?? (?? 1
is a constant )
?1???????? ?=?? 1
????
????
=1-?? 2
?
????
1-?? 2
=????
??
1
2
(
1
1-?? +
1
1+?? )???? =????
From (5)
????
????
?=1-?? 2
?
????
1-?? 2
=????
1
2
(
1
1-?? +
1
1+?? )???? ?=????
Integrating both sides, we get
?
1
2
ln?(
1+?? 1-?? )???? ???=?? +?? 2
????? ?
?
1+?? 1-?? =?? 3
?? 2?? ???????? ?
At ?? =0,??????? 0
?=2 ????????
?
3
-1
=?? 3
??? 3
=-3 ???????? ?
1+?? 1-?? =-3?? 2?? ???????? ?
?? =
3?? 2?? +1
3?? 2?? -1
????
????
=-?? (1+?? )=-?? (1+
3?? 2?? +1
3?? 2?? -1
)
???
????
????
=-?? ×
6?? 2?? (3?? 2?? -1)
???
????
?? =-
6?? 2?? 3?? 2?? -1
????
Integrating both sides, we get
???? =
?? 4
3?? 2?? -1
(1)
Now at ?? =0,?? 0
=-3??
? -3?? =
?? 4
3-1
??? 4
=-6?? ? ?? =
-6?? 3?? 2?? -1
Now, from (2) and (4)
????
?? ???????? ?=
????
?? ??????? ?????????=??????
Integrating both sides, we get
?? 2
-?? 2
=?? 5
At ?? =0,?? 0
=5,?? 0
=2 s
??????????????????????????????????????????????????????????????????? 2
-4?? 2
=?? 5
???? 5
=-3?? 2
?? 2
-?? 2
=-3?? 2
(10)
Now, by eqn.,
???? +???? =??
Using values from (7), (8), (9), we get
?? ·(
3?? 2?? +1
3?? 2?? -1
)+?? ·(
-6?? 3?? 2?? -1
)?????????=?? ???? ·
(3?? 2
+1)
3?? 2
-1
+?? ·
-6?? 3?? 2
-1
?????????=??
?
-6????
3?? 2
-1
=
?? -?? (3?? 2
+1)
3?? 2
-1
? -6???? =?? (3?? 2
-1)-?? (3?? 2
+1)
? ?? =
3?? 2
?? -3?? 2
?? -?? -?? -6?? (11)
Putting this value of ?? in (10), we get
?? 2
-?? 2
?=
-3×(3?? 2
?? -3?? 2
?? -?? -?? )
3
36?? 2
???? 2
-?? 2
?=
-{3?? 2
(?? -?? )-(?? +?? )}
2
12?? 2
which is the required solution.
3.3 Determine the characteristics of the equation ?? =?? ?? -?? ?? and find the integral
surface which passes through the parabola ?? ?? +?? ?? =?? ,?? =?? .
(2016 : 15 Marks)
Solution:
Given equation is
?? =?? 2
-?? 2
or ?? 2
-?? 2
-?? =0=?? (??)
The curve is 4?? +?? 2
=0 and ?? =0.
Let characteristic variables are ?? and ?? .
Now, let
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Nov 21, 2024 Last updated |
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