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Page 1
Edurev123
5. Canonical Form
5.1 Reduce the following ?? nd
order partial differential equation into canonical form
and find its general solution.
?? ?? ????
+?? ?? ?? ?? ????
-?? ?? =??
(2010 : 20 Marks)
Solution:
Let the characteristic variables be ?? and ?? .
Now, the given equation is
?? ?? ????
+2?? 2
?? ????
-4
?? =0
??? -characteristic equation is
?? ?? 2
+2?? 2
?? =0
????? (???? +2?? 2
)=0
????? =0 or ?? =-2?? For ?? =0:?
????
????
+0=0
????? = Constant =?? For ?? =-2?? :?
????
????
-2?? =0
??????? =2?????? ????? =?? 2
+?? ? ( ?? is a constant)
????? -?? 2
=?? =?? ????? =?? and ?? =?? -?? 2
Now,
?? ????
=
?
2
?? ??? 2
×(
??? ??? )
2
+(
?
2
?? ??? 2
)×(
??? ??? )
2
+
2?
2
?? ??? ??? ×
??? ??? ×
??? ??? +
??? ??? ×
?
2
?? ??? 2
+
??? ??? ×
?
2
?? ??? 2
?
2
?? ??? 2
=
?
2
?? ??? 2
×0+
?
2
?? ??? 2
×4?? 2
+
2?
2
?? ??? ??? ×0+
??? ??? ×0+
??? ??? ×(-2)
???
?
2
?? ??? 2
=4?? 2
?
2
?? ??? 2
-2
??? ??? =?? ????
?? ????
=
?
2
?? ??? 2
×
??? ??? ×
??? ??? +
?
2
?? ??? 2
×
??? ??? ×
??? ??? +
??? ??? ×
?
2
?? ??? ??? +
??? ??? ×
?
2
?? ??? ??? +
?
2
?? ??? ??? (
??? ??? ??? ??? +
??? ??? ??? ??? )
?=0+
?
2
?? ??? 2
×(-2?? )+0+0+
?
2
?? ??? ??? ×1×(-2?? )
Page 2
Edurev123
5. Canonical Form
5.1 Reduce the following ?? nd
order partial differential equation into canonical form
and find its general solution.
?? ?? ????
+?? ?? ?? ?? ????
-?? ?? =??
(2010 : 20 Marks)
Solution:
Let the characteristic variables be ?? and ?? .
Now, the given equation is
?? ?? ????
+2?? 2
?? ????
-4
?? =0
??? -characteristic equation is
?? ?? 2
+2?? 2
?? =0
????? (???? +2?? 2
)=0
????? =0 or ?? =-2?? For ?? =0:?
????
????
+0=0
????? = Constant =?? For ?? =-2?? :?
????
????
-2?? =0
??????? =2?????? ????? =?? 2
+?? ? ( ?? is a constant)
????? -?? 2
=?? =?? ????? =?? and ?? =?? -?? 2
Now,
?? ????
=
?
2
?? ??? 2
×(
??? ??? )
2
+(
?
2
?? ??? 2
)×(
??? ??? )
2
+
2?
2
?? ??? ??? ×
??? ??? ×
??? ??? +
??? ??? ×
?
2
?? ??? 2
+
??? ??? ×
?
2
?? ??? 2
?
2
?? ??? 2
=
?
2
?? ??? 2
×0+
?
2
?? ??? 2
×4?? 2
+
2?
2
?? ??? ??? ×0+
??? ??? ×0+
??? ??? ×(-2)
???
?
2
?? ??? 2
=4?? 2
?
2
?? ??? 2
-2
??? ??? =?? ????
?? ????
=
?
2
?? ??? 2
×
??? ??? ×
??? ??? +
?
2
?? ??? 2
×
??? ??? ×
??? ??? +
??? ??? ×
?
2
?? ??? ??? +
??? ??? ×
?
2
?? ??? ??? +
?
2
?? ??? ??? (
??? ??? ??? ??? +
??? ??? ??? ??? )
?=0+
?
2
?? ??? 2
×(-2?? )+0+0+
?
2
?? ??? ??? ×1×(-2?? )
???? ????
?=-2?? ?
2
?? ?? ?? 2
-2?? ?
2
?? ??? ??? (???? )
?? ?? ?=
??? ??? ×
??? ??? +
??? ??? ×
??? ??? =0-2?? ??? ??? =-2?? ??? ??? (?????? )
Using values in equations (i), (ii) and (iii), we get
???
?
2
?? ??? ??? =0, which is the canonical form of the equation.
Now,
?
2
?? ??? ??? =0
??? ??? =?? (?? )
?? =???? (?? )????
?=???? ?
(?? -?? 2
)?? ''
(?? -?? 2
)
which is the required solution.
5.2 Reduce the equation:
?? ?? ?? ?? ?? ?? ?? +(?? +?? )
?? ?? ?? ?? ?? ?? ?? +?? ?? ?? ?? ?? ?? ?? =??
to its canonical form where ?? ??? :
(2013 : 10 Marks)
Solution:
Comparing the equation to
???? +???? +???? ?=0 (??)
?? ?=?? ,?? =(?? +?? ),?? =?? ?? 2
-4???? ?=(?? +?? )
2
-4???? =(?? -?? )
2
>0 for ?? ???
So, the equation is hyperbolic in form.
The ?? -quadratic is
?? 2
?? +???? +?? ?=0??? 2
?? +(?? +?? )?? +?? =0
??(???? +?? )(?? +1)?=0??? =-
?? ?? ,-1
So, the corresponding quadratic equations are :
Page 3
Edurev123
5. Canonical Form
5.1 Reduce the following ?? nd
order partial differential equation into canonical form
and find its general solution.
?? ?? ????
+?? ?? ?? ?? ????
-?? ?? =??
(2010 : 20 Marks)
Solution:
Let the characteristic variables be ?? and ?? .
Now, the given equation is
?? ?? ????
+2?? 2
?? ????
-4
?? =0
??? -characteristic equation is
?? ?? 2
+2?? 2
?? =0
????? (???? +2?? 2
)=0
????? =0 or ?? =-2?? For ?? =0:?
????
????
+0=0
????? = Constant =?? For ?? =-2?? :?
????
????
-2?? =0
??????? =2?????? ????? =?? 2
+?? ? ( ?? is a constant)
????? -?? 2
=?? =?? ????? =?? and ?? =?? -?? 2
Now,
?? ????
=
?
2
?? ??? 2
×(
??? ??? )
2
+(
?
2
?? ??? 2
)×(
??? ??? )
2
+
2?
2
?? ??? ??? ×
??? ??? ×
??? ??? +
??? ??? ×
?
2
?? ??? 2
+
??? ??? ×
?
2
?? ??? 2
?
2
?? ??? 2
=
?
2
?? ??? 2
×0+
?
2
?? ??? 2
×4?? 2
+
2?
2
?? ??? ??? ×0+
??? ??? ×0+
??? ??? ×(-2)
???
?
2
?? ??? 2
=4?? 2
?
2
?? ??? 2
-2
??? ??? =?? ????
?? ????
=
?
2
?? ??? 2
×
??? ??? ×
??? ??? +
?
2
?? ??? 2
×
??? ??? ×
??? ??? +
??? ??? ×
?
2
?? ??? ??? +
??? ??? ×
?
2
?? ??? ??? +
?
2
?? ??? ??? (
??? ??? ??? ??? +
??? ??? ??? ??? )
?=0+
?
2
?? ??? 2
×(-2?? )+0+0+
?
2
?? ??? ??? ×1×(-2?? )
???? ????
?=-2?? ?
2
?? ?? ?? 2
-2?? ?
2
?? ??? ??? (???? )
?? ?? ?=
??? ??? ×
??? ??? +
??? ??? ×
??? ??? =0-2?? ??? ??? =-2?? ??? ??? (?????? )
Using values in equations (i), (ii) and (iii), we get
???
?
2
?? ??? ??? =0, which is the canonical form of the equation.
Now,
?
2
?? ??? ??? =0
??? ??? =?? (?? )
?? =???? (?? )????
?=???? ?
(?? -?? 2
)?? ''
(?? -?? 2
)
which is the required solution.
5.2 Reduce the equation:
?? ?? ?? ?? ?? ?? ?? +(?? +?? )
?? ?? ?? ?? ?? ?? ?? +?? ?? ?? ?? ?? ?? ?? =??
to its canonical form where ?? ??? :
(2013 : 10 Marks)
Solution:
Comparing the equation to
???? +???? +???? ?=0 (??)
?? ?=?? ,?? =(?? +?? ),?? =?? ?? 2
-4???? ?=(?? +?? )
2
-4???? =(?? -?? )
2
>0 for ?? ???
So, the equation is hyperbolic in form.
The ?? -quadratic is
?? 2
?? +???? +?? ?=0??? 2
?? +(?? +?? )?? +?? =0
??(???? +?? )(?? +1)?=0??? =-
?? ?? ,-1
So, the corresponding quadratic equations are :
????
????
+?? ?=0 and
????
????
+?? 2
=0
????
????
-
?? ?? ?=0 and
????
????
-1=0
???? 2
2
-
?? 2
?? ?=?? 1
and ?? -?? =?? 2
?? ?=
1
2
(?? 2
-?? 2
);?? =?? -?? ?? ?=
??? ??? =
??? ??? ·
??? ??? +
??? ??? ·
??? ??? =-?? ??? ??? -
??? ??? ?? ?=
??? ??? =
??? ??? ·
??? ??? +
??? ??? ·
??? ??? =?? ??? ??? +
??? ???
?? ?=
?
2
?? ??? 2
=
?
??? (
??? ??? )=
?
??? (-?? ??? ??? -
??? ??? )
?=-
??? ??? -?? ?
??? ??? ??? -
?
??? (
??? ??? )
?=,-
??? ??? -?? [
?
2
?? ??? ·
??? ??? +
?
2
?? ??? ??? ·
??? ??? ]-[
?
2
?? ??? ??? ·
??? ??? +
?
2
?? ??? 2
·
??? ??? ]
?=-
??? ??? +?? 2
?
2
?? ??? 2
+2?? ?
2
?? ??? ??? +
?
2
?? ??? 2
?? ?=
?
2
?? ??? ??? =
?
??? (
??? ??? )=
?
??? (-?? ??? ??? -
??? ??? )
?=-?? [
?
2
?? ??? 2
·
??? ??? +
?
2
?? ??? ??? ·
??? ??? ]-[
?
2
?? ??? ??? ·
??? ??? +
?
2
?? ??? 2
·
??? ??? ]
?=-????
?
2
?? ??? 2
-(?? +?? )
?
2
?? ??? ??? -
?
2
?? ??? ?? ?=
?
2
?? ??? 2
=
?
??? (
??? ??? )=
?
??? [?? ??? ??? +
??? ??? ]
?=
??? ??? +?? 2
?
2
?? ??? 2
+2?? ?
2
?? ??? ??? +
?
2
?? ??? 2
Substituting in (i), we have
{4???? -(?? +?? )
2
}
?
2
?? ??? ??? -?? ??? ??? +?? ??? ??? =0
?(?? -?? )
2
?
2
?? ??? ??? +(?? -?? )
??? ??? =0
which is the canonical form.
[Note : In hyperbolic form only coefficient of
?
2
?? ??? ??? remains and this can be used].
5.3 Reduce the equation
?? ?? ?? ?? ?? ?? =?? ?? ?? ?? ?? ?? ?? ?? to Canonical form.
Page 4
Edurev123
5. Canonical Form
5.1 Reduce the following ?? nd
order partial differential equation into canonical form
and find its general solution.
?? ?? ????
+?? ?? ?? ?? ????
-?? ?? =??
(2010 : 20 Marks)
Solution:
Let the characteristic variables be ?? and ?? .
Now, the given equation is
?? ?? ????
+2?? 2
?? ????
-4
?? =0
??? -characteristic equation is
?? ?? 2
+2?? 2
?? =0
????? (???? +2?? 2
)=0
????? =0 or ?? =-2?? For ?? =0:?
????
????
+0=0
????? = Constant =?? For ?? =-2?? :?
????
????
-2?? =0
??????? =2?????? ????? =?? 2
+?? ? ( ?? is a constant)
????? -?? 2
=?? =?? ????? =?? and ?? =?? -?? 2
Now,
?? ????
=
?
2
?? ??? 2
×(
??? ??? )
2
+(
?
2
?? ??? 2
)×(
??? ??? )
2
+
2?
2
?? ??? ??? ×
??? ??? ×
??? ??? +
??? ??? ×
?
2
?? ??? 2
+
??? ??? ×
?
2
?? ??? 2
?
2
?? ??? 2
=
?
2
?? ??? 2
×0+
?
2
?? ??? 2
×4?? 2
+
2?
2
?? ??? ??? ×0+
??? ??? ×0+
??? ??? ×(-2)
???
?
2
?? ??? 2
=4?? 2
?
2
?? ??? 2
-2
??? ??? =?? ????
?? ????
=
?
2
?? ??? 2
×
??? ??? ×
??? ??? +
?
2
?? ??? 2
×
??? ??? ×
??? ??? +
??? ??? ×
?
2
?? ??? ??? +
??? ??? ×
?
2
?? ??? ??? +
?
2
?? ??? ??? (
??? ??? ??? ??? +
??? ??? ??? ??? )
?=0+
?
2
?? ??? 2
×(-2?? )+0+0+
?
2
?? ??? ??? ×1×(-2?? )
???? ????
?=-2?? ?
2
?? ?? ?? 2
-2?? ?
2
?? ??? ??? (???? )
?? ?? ?=
??? ??? ×
??? ??? +
??? ??? ×
??? ??? =0-2?? ??? ??? =-2?? ??? ??? (?????? )
Using values in equations (i), (ii) and (iii), we get
???
?
2
?? ??? ??? =0, which is the canonical form of the equation.
Now,
?
2
?? ??? ??? =0
??? ??? =?? (?? )
?? =???? (?? )????
?=???? ?
(?? -?? 2
)?? ''
(?? -?? 2
)
which is the required solution.
5.2 Reduce the equation:
?? ?? ?? ?? ?? ?? ?? +(?? +?? )
?? ?? ?? ?? ?? ?? ?? +?? ?? ?? ?? ?? ?? ?? =??
to its canonical form where ?? ??? :
(2013 : 10 Marks)
Solution:
Comparing the equation to
???? +???? +???? ?=0 (??)
?? ?=?? ,?? =(?? +?? ),?? =?? ?? 2
-4???? ?=(?? +?? )
2
-4???? =(?? -?? )
2
>0 for ?? ???
So, the equation is hyperbolic in form.
The ?? -quadratic is
?? 2
?? +???? +?? ?=0??? 2
?? +(?? +?? )?? +?? =0
??(???? +?? )(?? +1)?=0??? =-
?? ?? ,-1
So, the corresponding quadratic equations are :
????
????
+?? ?=0 and
????
????
+?? 2
=0
????
????
-
?? ?? ?=0 and
????
????
-1=0
???? 2
2
-
?? 2
?? ?=?? 1
and ?? -?? =?? 2
?? ?=
1
2
(?? 2
-?? 2
);?? =?? -?? ?? ?=
??? ??? =
??? ??? ·
??? ??? +
??? ??? ·
??? ??? =-?? ??? ??? -
??? ??? ?? ?=
??? ??? =
??? ??? ·
??? ??? +
??? ??? ·
??? ??? =?? ??? ??? +
??? ???
?? ?=
?
2
?? ??? 2
=
?
??? (
??? ??? )=
?
??? (-?? ??? ??? -
??? ??? )
?=-
??? ??? -?? ?
??? ??? ??? -
?
??? (
??? ??? )
?=,-
??? ??? -?? [
?
2
?? ??? ·
??? ??? +
?
2
?? ??? ??? ·
??? ??? ]-[
?
2
?? ??? ??? ·
??? ??? +
?
2
?? ??? 2
·
??? ??? ]
?=-
??? ??? +?? 2
?
2
?? ??? 2
+2?? ?
2
?? ??? ??? +
?
2
?? ??? 2
?? ?=
?
2
?? ??? ??? =
?
??? (
??? ??? )=
?
??? (-?? ??? ??? -
??? ??? )
?=-?? [
?
2
?? ??? 2
·
??? ??? +
?
2
?? ??? ??? ·
??? ??? ]-[
?
2
?? ??? ??? ·
??? ??? +
?
2
?? ??? 2
·
??? ??? ]
?=-????
?
2
?? ??? 2
-(?? +?? )
?
2
?? ??? ??? -
?
2
?? ??? ?? ?=
?
2
?? ??? 2
=
?
??? (
??? ??? )=
?
??? [?? ??? ??? +
??? ??? ]
?=
??? ??? +?? 2
?
2
?? ??? 2
+2?? ?
2
?? ??? ??? +
?
2
?? ??? 2
Substituting in (i), we have
{4???? -(?? +?? )
2
}
?
2
?? ??? ??? -?? ??? ??? +?? ??? ??? =0
?(?? -?? )
2
?
2
?? ??? ??? +(?? -?? )
??? ??? =0
which is the canonical form.
[Note : In hyperbolic form only coefficient of
?
2
?? ??? ??? remains and this can be used].
5.3 Reduce the equation
?? ?? ?? ?? ?? ?? =?? ?? ?? ?? ?? ?? ?? ?? to Canonical form.
(2014: 15 marks)
Solution:
Let, the given equation,
?
2
?? ??? 2
=
?? 2
?
2
?? ??? 2
(??)
Re-writing the given equation becomes ?? -?? 2
?? =0
comparing (i) with
???? +5?? +?? ?? +?? (?? ,?? ,?? ,?? ,?? )=0
We have,
Now, the ?? -quadratic equation,
?? =1,?? =0,?? =-?? 2
?? ?? 2
+5?? +7?=0
?? 2
-?? 1
?=0
???? ?=????
Here, ?? 1
=?? and ?? 2
=-?? (real & distinct roots)
Hence, the characteristic equations.
????
????
+?? 1
=0 and
????
????
+?? 2
=0
????
????
+?? =0 and
????
????
-?? =0
Integrating these two equations, we get
?? +(
?? 2
2
)=?? 1
?? -(
?? 2
2
)=?? 2
Hence in order to reduce (i) to canonical form, we change ?? ,?? to ?? ,?? by taking.
Page 5
Edurev123
5. Canonical Form
5.1 Reduce the following ?? nd
order partial differential equation into canonical form
and find its general solution.
?? ?? ????
+?? ?? ?? ?? ????
-?? ?? =??
(2010 : 20 Marks)
Solution:
Let the characteristic variables be ?? and ?? .
Now, the given equation is
?? ?? ????
+2?? 2
?? ????
-4
?? =0
??? -characteristic equation is
?? ?? 2
+2?? 2
?? =0
????? (???? +2?? 2
)=0
????? =0 or ?? =-2?? For ?? =0:?
????
????
+0=0
????? = Constant =?? For ?? =-2?? :?
????
????
-2?? =0
??????? =2?????? ????? =?? 2
+?? ? ( ?? is a constant)
????? -?? 2
=?? =?? ????? =?? and ?? =?? -?? 2
Now,
?? ????
=
?
2
?? ??? 2
×(
??? ??? )
2
+(
?
2
?? ??? 2
)×(
??? ??? )
2
+
2?
2
?? ??? ??? ×
??? ??? ×
??? ??? +
??? ??? ×
?
2
?? ??? 2
+
??? ??? ×
?
2
?? ??? 2
?
2
?? ??? 2
=
?
2
?? ??? 2
×0+
?
2
?? ??? 2
×4?? 2
+
2?
2
?? ??? ??? ×0+
??? ??? ×0+
??? ??? ×(-2)
???
?
2
?? ??? 2
=4?? 2
?
2
?? ??? 2
-2
??? ??? =?? ????
?? ????
=
?
2
?? ??? 2
×
??? ??? ×
??? ??? +
?
2
?? ??? 2
×
??? ??? ×
??? ??? +
??? ??? ×
?
2
?? ??? ??? +
??? ??? ×
?
2
?? ??? ??? +
?
2
?? ??? ??? (
??? ??? ??? ??? +
??? ??? ??? ??? )
?=0+
?
2
?? ??? 2
×(-2?? )+0+0+
?
2
?? ??? ??? ×1×(-2?? )
???? ????
?=-2?? ?
2
?? ?? ?? 2
-2?? ?
2
?? ??? ??? (???? )
?? ?? ?=
??? ??? ×
??? ??? +
??? ??? ×
??? ??? =0-2?? ??? ??? =-2?? ??? ??? (?????? )
Using values in equations (i), (ii) and (iii), we get
???
?
2
?? ??? ??? =0, which is the canonical form of the equation.
Now,
?
2
?? ??? ??? =0
??? ??? =?? (?? )
?? =???? (?? )????
?=???? ?
(?? -?? 2
)?? ''
(?? -?? 2
)
which is the required solution.
5.2 Reduce the equation:
?? ?? ?? ?? ?? ?? ?? +(?? +?? )
?? ?? ?? ?? ?? ?? ?? +?? ?? ?? ?? ?? ?? ?? =??
to its canonical form where ?? ??? :
(2013 : 10 Marks)
Solution:
Comparing the equation to
???? +???? +???? ?=0 (??)
?? ?=?? ,?? =(?? +?? ),?? =?? ?? 2
-4???? ?=(?? +?? )
2
-4???? =(?? -?? )
2
>0 for ?? ???
So, the equation is hyperbolic in form.
The ?? -quadratic is
?? 2
?? +???? +?? ?=0??? 2
?? +(?? +?? )?? +?? =0
??(???? +?? )(?? +1)?=0??? =-
?? ?? ,-1
So, the corresponding quadratic equations are :
????
????
+?? ?=0 and
????
????
+?? 2
=0
????
????
-
?? ?? ?=0 and
????
????
-1=0
???? 2
2
-
?? 2
?? ?=?? 1
and ?? -?? =?? 2
?? ?=
1
2
(?? 2
-?? 2
);?? =?? -?? ?? ?=
??? ??? =
??? ??? ·
??? ??? +
??? ??? ·
??? ??? =-?? ??? ??? -
??? ??? ?? ?=
??? ??? =
??? ??? ·
??? ??? +
??? ??? ·
??? ??? =?? ??? ??? +
??? ???
?? ?=
?
2
?? ??? 2
=
?
??? (
??? ??? )=
?
??? (-?? ??? ??? -
??? ??? )
?=-
??? ??? -?? ?
??? ??? ??? -
?
??? (
??? ??? )
?=,-
??? ??? -?? [
?
2
?? ??? ·
??? ??? +
?
2
?? ??? ??? ·
??? ??? ]-[
?
2
?? ??? ??? ·
??? ??? +
?
2
?? ??? 2
·
??? ??? ]
?=-
??? ??? +?? 2
?
2
?? ??? 2
+2?? ?
2
?? ??? ??? +
?
2
?? ??? 2
?? ?=
?
2
?? ??? ??? =
?
??? (
??? ??? )=
?
??? (-?? ??? ??? -
??? ??? )
?=-?? [
?
2
?? ??? 2
·
??? ??? +
?
2
?? ??? ??? ·
??? ??? ]-[
?
2
?? ??? ??? ·
??? ??? +
?
2
?? ??? 2
·
??? ??? ]
?=-????
?
2
?? ??? 2
-(?? +?? )
?
2
?? ??? ??? -
?
2
?? ??? ?? ?=
?
2
?? ??? 2
=
?
??? (
??? ??? )=
?
??? [?? ??? ??? +
??? ??? ]
?=
??? ??? +?? 2
?
2
?? ??? 2
+2?? ?
2
?? ??? ??? +
?
2
?? ??? 2
Substituting in (i), we have
{4???? -(?? +?? )
2
}
?
2
?? ??? ??? -?? ??? ??? +?? ??? ??? =0
?(?? -?? )
2
?
2
?? ??? ??? +(?? -?? )
??? ??? =0
which is the canonical form.
[Note : In hyperbolic form only coefficient of
?
2
?? ??? ??? remains and this can be used].
5.3 Reduce the equation
?? ?? ?? ?? ?? ?? =?? ?? ?? ?? ?? ?? ?? ?? to Canonical form.
(2014: 15 marks)
Solution:
Let, the given equation,
?
2
?? ??? 2
=
?? 2
?
2
?? ??? 2
(??)
Re-writing the given equation becomes ?? -?? 2
?? =0
comparing (i) with
???? +5?? +?? ?? +?? (?? ,?? ,?? ,?? ,?? )=0
We have,
Now, the ?? -quadratic equation,
?? =1,?? =0,?? =-?? 2
?? ?? 2
+5?? +7?=0
?? 2
-?? 1
?=0
???? ?=????
Here, ?? 1
=?? and ?? 2
=-?? (real & distinct roots)
Hence, the characteristic equations.
????
????
+?? 1
=0 and
????
????
+?? 2
=0
????
????
+?? =0 and
????
????
-?? =0
Integrating these two equations, we get
?? +(
?? 2
2
)=?? 1
?? -(
?? 2
2
)=?? 2
Hence in order to reduce (i) to canonical form, we change ?? ,?? to ?? ,?? by taking.
?? ?=?? +
?? 2
2
and ?? =?? -
?? 2
2
(???? )
?? ?=
??? ??? =
??? ??? ·
??? ??? +
??? ??? ·
??? ??? =
?? ??? ??? -?? ??? ??? ?? ?=?? (
??? ??? -
??? ??? ) (?????? )
?? ?=
??? ??? =
??? ??? ·
??? ??? +
??? ??? ·
??? ??? =
??? ??? +
??? ??? ???? ?=
?
2
?? ??? 2
=
?
??? (
??? ??? )=
?
??? {?? (
??? ??? -
??? ??? )}
???? ?=?? ?
??? (
??? ??? -
??? ??? )+1(
??? ??? -
??? ??? )
?=?? 2
[
?
2
?? ??? 2
-
2?
2
?? ??? ??? +
?
2
?? ??? 2
]+
??? ??? -
??? ??? =?? ?? ?=
?
2
?? ??? 2
=
?
??? (
??? ??? )=
?
??? (
??? ??? +
??? ??? )
?? ?=
?
??? (
?
??? )=(
?
??? +
?
??? )(
??? ??? +
??? ??? )
?=
?
2
?? ??? 2
+2
?
2
?? ??? ??? +
?
2
?? ??? 2
Put, the values of ?? &?? in equation (i)
?? 2
[
?
2
?? ??? 2
-2
?
2
?? ??? ??? +
?
2
?? ??? 2
]+
??? ??? -
??? ??? -?? 2
[
?
2
?? ??? 2
+2
?
2
?? ??? ??? +
?
2
?? ??? 2
]=0
or
?
2
?? ??? ??? ?=
1
4?? 2
(
??? ??? -
??? ??? )
?
2
?? ??? ??? ?=
1
4(?? -?? )
(
??? ??? -
??? ??? )
Which is the required canonical form of the given equation.
5.4 Reduce the second order partial differential equation :
?? ?? ?? ?? ?? ?? ?? ?? -?? ????
?? ?? ?? ?? ?? ?? ?? +?? ?? ?? ?? ?? ?? ?? ?? +?? ?? ?? ?? ?? +?? ?? ?? ?? ?? =??
into canonical form. Hence, find its general solution.
(2015 : 15 Marks)
Solution:
Given : ?? 2
?
2
?? ??? 2
-2????
?
2
?? ??? ??? +?? 2
?
2
?? ??? 2
+?? ??? ??? +?? ??? ??? =0
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Dec 22, 2024 Last updated |
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