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Page 1
Edurev123
Partial Differential Equations
1. Formulation of P.D.E.
1.1 Show that the differential equation of all cones which have their vertex at the
origin is ???? +???? =?? . Verify that this equation is satisfied by the surface ???? +?? ?? +
???? =?? .
(2009 : 12 Marks)
Solution:
The equation cone having vertex at origin
?? ?? 2
+?? ?? 2
+?? ?? 2
+2h???? +2?????? +2?????? =0 (1)
where ???,?? ,?? ,?? ,?? ,h are parameters.
Differentiating w.r.t. ?? and ?? , we get
2???? +2h?? +2???? +2?????? +2?????? +2?????? ?=0
2???? +2?????? +2h?? +2?????? +2???? +2?????? ?=0
???? +h?? +???? +?? (???? +???? +???? )?=0×?? ???? +h?? +???? +?? (???? +???? +???? )?=0×?? ?? ?? 2
+h???? +?????? +?? (?? ?? 2
+?????? +?????? )?=0
?? ?? 2
+h???? +?????? +?? (?????? +?? ?? 2
+?????? )?=0
On adding,
???????????????? ?? 2
+?? ?? 2
+2h???? +?????? +?????? +???? +???? [???? +???? +???? ]=0?????????
???????????????????????????????-(?? ?? 2
+?????? +?????? )+(???? +???? +???? )(???? +???? )=0?????????
????????????????????????????????????????????????????????????????????(???? +???? +???? )(???? +???? -?? )=0?????????
Clearly, ???? +???? -?? =0 is required differential equation.
Given surface is ???? +???? +???? =0
Differentiating (?
*
) w.r.t. ?? and ?? , we get
???? +?? +???? +?? =0 (2)
?? +???? +???? +?? =0 (3)
So, we get
Page 2
Edurev123
Partial Differential Equations
1. Formulation of P.D.E.
1.1 Show that the differential equation of all cones which have their vertex at the
origin is ???? +???? =?? . Verify that this equation is satisfied by the surface ???? +?? ?? +
???? =?? .
(2009 : 12 Marks)
Solution:
The equation cone having vertex at origin
?? ?? 2
+?? ?? 2
+?? ?? 2
+2h???? +2?????? +2?????? =0 (1)
where ???,?? ,?? ,?? ,?? ,h are parameters.
Differentiating w.r.t. ?? and ?? , we get
2???? +2h?? +2???? +2?????? +2?????? +2?????? ?=0
2???? +2?????? +2h?? +2?????? +2???? +2?????? ?=0
???? +h?? +???? +?? (???? +???? +???? )?=0×?? ???? +h?? +???? +?? (???? +???? +???? )?=0×?? ?? ?? 2
+h???? +?????? +?? (?? ?? 2
+?????? +?????? )?=0
?? ?? 2
+h???? +?????? +?? (?????? +?? ?? 2
+?????? )?=0
On adding,
???????????????? ?? 2
+?? ?? 2
+2h???? +?????? +?????? +???? +???? [???? +???? +???? ]=0?????????
???????????????????????????????-(?? ?? 2
+?????? +?????? )+(???? +???? +???? )(???? +???? )=0?????????
????????????????????????????????????????????????????????????????????(???? +???? +???? )(???? +???? -?? )=0?????????
Clearly, ???? +???? -?? =0 is required differential equation.
Given surface is ???? +???? +???? =0
Differentiating (?
*
) w.r.t. ?? and ?? , we get
???? +?? +???? +?? =0 (2)
?? +???? +???? +?? =0 (3)
So, we get
?? ?=
-(?? +?? )
(?? +?? )
,?? =
-(?? +?? )
(?? +?? )
???? +???? -?? ?=
-(?? +?? )?? (?? +?? )
-
(?? +?? )
(?? +?? )
?? -?? ?=
-(?? +?? )?? -(?? +?? )?? -?? (?? +?? )
(?? +?? )
?=
-???? -???? -???? -???? -???? -????
(?? +?? )
?=
-2(???? +???? +???? )
?? +?? =
-20
?? +?? =0
1.2 From the partial differential equation by eliminating the arbitrary function ??
given by:
?? (?? ?? +?? ?? ,?? -???? )=??
Solution:
The function is
???????????????????????????????????????????????????????????????????? =???? +?? (?? 2
+?? 2
)??????????????????????????????????????????????????????????????(1)
Now differentiating partially (1) w.r.t. ?? we get
??? ??? =?? +?? '
(?? 2
+?? 2
)2??
So,
?? -?? 2?? =?? '
(?? 2
+?? 2
) (2)
Now, differentiating partially (1) w.r.t. ?? , we get
??? ??? =?? +?? (?? 2
+?? 2
)·2??
So,
?? -?? 2?? =?? '
(?? 2
+?? 2
) (3)
Equating (2) and (3), we get
?? -?? 2?? =
?? -?? 2??
So, ???? -???? =?? 2
-?? 2
is linear PDE.
1.3 Find the surface satisfying the P.D.E. (?? ?? -?? ?? ?? '
+?? ?? )?? =?? and the
conditions that ???? =?? ?? when ?? =?? and ???? =?? ?? when ?? =?? .
Page 3
Edurev123
Partial Differential Equations
1. Formulation of P.D.E.
1.1 Show that the differential equation of all cones which have their vertex at the
origin is ???? +???? =?? . Verify that this equation is satisfied by the surface ???? +?? ?? +
???? =?? .
(2009 : 12 Marks)
Solution:
The equation cone having vertex at origin
?? ?? 2
+?? ?? 2
+?? ?? 2
+2h???? +2?????? +2?????? =0 (1)
where ???,?? ,?? ,?? ,?? ,h are parameters.
Differentiating w.r.t. ?? and ?? , we get
2???? +2h?? +2???? +2?????? +2?????? +2?????? ?=0
2???? +2?????? +2h?? +2?????? +2???? +2?????? ?=0
???? +h?? +???? +?? (???? +???? +???? )?=0×?? ???? +h?? +???? +?? (???? +???? +???? )?=0×?? ?? ?? 2
+h???? +?????? +?? (?? ?? 2
+?????? +?????? )?=0
?? ?? 2
+h???? +?????? +?? (?????? +?? ?? 2
+?????? )?=0
On adding,
???????????????? ?? 2
+?? ?? 2
+2h???? +?????? +?????? +???? +???? [???? +???? +???? ]=0?????????
???????????????????????????????-(?? ?? 2
+?????? +?????? )+(???? +???? +???? )(???? +???? )=0?????????
????????????????????????????????????????????????????????????????????(???? +???? +???? )(???? +???? -?? )=0?????????
Clearly, ???? +???? -?? =0 is required differential equation.
Given surface is ???? +???? +???? =0
Differentiating (?
*
) w.r.t. ?? and ?? , we get
???? +?? +???? +?? =0 (2)
?? +???? +???? +?? =0 (3)
So, we get
?? ?=
-(?? +?? )
(?? +?? )
,?? =
-(?? +?? )
(?? +?? )
???? +???? -?? ?=
-(?? +?? )?? (?? +?? )
-
(?? +?? )
(?? +?? )
?? -?? ?=
-(?? +?? )?? -(?? +?? )?? -?? (?? +?? )
(?? +?? )
?=
-???? -???? -???? -???? -???? -????
(?? +?? )
?=
-2(???? +???? +???? )
?? +?? =
-20
?? +?? =0
1.2 From the partial differential equation by eliminating the arbitrary function ??
given by:
?? (?? ?? +?? ?? ,?? -???? )=??
Solution:
The function is
???????????????????????????????????????????????????????????????????? =???? +?? (?? 2
+?? 2
)??????????????????????????????????????????????????????????????(1)
Now differentiating partially (1) w.r.t. ?? we get
??? ??? =?? +?? '
(?? 2
+?? 2
)2??
So,
?? -?? 2?? =?? '
(?? 2
+?? 2
) (2)
Now, differentiating partially (1) w.r.t. ?? , we get
??? ??? =?? +?? (?? 2
+?? 2
)·2??
So,
?? -?? 2?? =?? '
(?? 2
+?? 2
) (3)
Equating (2) and (3), we get
?? -?? 2?? =
?? -?? 2??
So, ???? -???? =?? 2
-?? 2
is linear PDE.
1.3 Find the surface satisfying the P.D.E. (?? ?? -?? ?? ?? '
+?? ?? )?? =?? and the
conditions that ???? =?? ?? when ?? =?? and ???? =?? ?? when ?? =?? .
(2010 : 12 Marks)
Solution:
Given, the equation is
(?? 2
-2???? +?? 2
)?? =0
?????????????????????????????????????????????????????????????????????????????(?? -?? '
)
2
?? =0
The auxiliary eqn. for above eqn. is
(?? -1)
2
=0
??????????????????????????????????????????????????????????????????????????????????????? =1,1
? The solution of above eqn. is
?? =?? 1
(?? +?? )+?? ?? 2
(?? +?? )
Given, at ?? =0,???? =?? 2
?
?? =
?? 2
?? i.e., ?
?? 2
?? =?? 1
(?? )+0??? 1
(?? )=
?? 2
?? ????? 1
(?? +?? )=
(?? +?? )
2
?? at ?? =0,???? =?? 2
???? =
?? 2
?? i.e., ?
?? 2
?? =?? 1
(?? )+?? ?? 2
(?? )
???
?? 2
?? =
?? 2
?? +?? ?? 2
(?? )
????? ?? 2
(?? )=?? 2
(
1
?? -
1
?? )??? 2
(?? )=?? (
1
?? -
1
?? )
???? 2
(?? +?? )=(?? +?? )(
1
?? -
1
?? )
? Putting these values of ?? 1
and ?? 2
in the solution, we get
?? =
(?? +?? )
2
?? +?? (?? +?? )(
1
?? -
1
?? )
1.4 Find the surface satisfying
?? ?? ?? ?? ?? ?? =?? ?? +?? and touching ?? =?? ?? +?? ?? along its
section by the plane ?? +?? +?? =?? .
(2011 : 20 Marks)
Page 4
Edurev123
Partial Differential Equations
1. Formulation of P.D.E.
1.1 Show that the differential equation of all cones which have their vertex at the
origin is ???? +???? =?? . Verify that this equation is satisfied by the surface ???? +?? ?? +
???? =?? .
(2009 : 12 Marks)
Solution:
The equation cone having vertex at origin
?? ?? 2
+?? ?? 2
+?? ?? 2
+2h???? +2?????? +2?????? =0 (1)
where ???,?? ,?? ,?? ,?? ,h are parameters.
Differentiating w.r.t. ?? and ?? , we get
2???? +2h?? +2???? +2?????? +2?????? +2?????? ?=0
2???? +2?????? +2h?? +2?????? +2???? +2?????? ?=0
???? +h?? +???? +?? (???? +???? +???? )?=0×?? ???? +h?? +???? +?? (???? +???? +???? )?=0×?? ?? ?? 2
+h???? +?????? +?? (?? ?? 2
+?????? +?????? )?=0
?? ?? 2
+h???? +?????? +?? (?????? +?? ?? 2
+?????? )?=0
On adding,
???????????????? ?? 2
+?? ?? 2
+2h???? +?????? +?????? +???? +???? [???? +???? +???? ]=0?????????
???????????????????????????????-(?? ?? 2
+?????? +?????? )+(???? +???? +???? )(???? +???? )=0?????????
????????????????????????????????????????????????????????????????????(???? +???? +???? )(???? +???? -?? )=0?????????
Clearly, ???? +???? -?? =0 is required differential equation.
Given surface is ???? +???? +???? =0
Differentiating (?
*
) w.r.t. ?? and ?? , we get
???? +?? +???? +?? =0 (2)
?? +???? +???? +?? =0 (3)
So, we get
?? ?=
-(?? +?? )
(?? +?? )
,?? =
-(?? +?? )
(?? +?? )
???? +???? -?? ?=
-(?? +?? )?? (?? +?? )
-
(?? +?? )
(?? +?? )
?? -?? ?=
-(?? +?? )?? -(?? +?? )?? -?? (?? +?? )
(?? +?? )
?=
-???? -???? -???? -???? -???? -????
(?? +?? )
?=
-2(???? +???? +???? )
?? +?? =
-20
?? +?? =0
1.2 From the partial differential equation by eliminating the arbitrary function ??
given by:
?? (?? ?? +?? ?? ,?? -???? )=??
Solution:
The function is
???????????????????????????????????????????????????????????????????? =???? +?? (?? 2
+?? 2
)??????????????????????????????????????????????????????????????(1)
Now differentiating partially (1) w.r.t. ?? we get
??? ??? =?? +?? '
(?? 2
+?? 2
)2??
So,
?? -?? 2?? =?? '
(?? 2
+?? 2
) (2)
Now, differentiating partially (1) w.r.t. ?? , we get
??? ??? =?? +?? (?? 2
+?? 2
)·2??
So,
?? -?? 2?? =?? '
(?? 2
+?? 2
) (3)
Equating (2) and (3), we get
?? -?? 2?? =
?? -?? 2??
So, ???? -???? =?? 2
-?? 2
is linear PDE.
1.3 Find the surface satisfying the P.D.E. (?? ?? -?? ?? ?? '
+?? ?? )?? =?? and the
conditions that ???? =?? ?? when ?? =?? and ???? =?? ?? when ?? =?? .
(2010 : 12 Marks)
Solution:
Given, the equation is
(?? 2
-2???? +?? 2
)?? =0
?????????????????????????????????????????????????????????????????????????????(?? -?? '
)
2
?? =0
The auxiliary eqn. for above eqn. is
(?? -1)
2
=0
??????????????????????????????????????????????????????????????????????????????????????? =1,1
? The solution of above eqn. is
?? =?? 1
(?? +?? )+?? ?? 2
(?? +?? )
Given, at ?? =0,???? =?? 2
?
?? =
?? 2
?? i.e., ?
?? 2
?? =?? 1
(?? )+0??? 1
(?? )=
?? 2
?? ????? 1
(?? +?? )=
(?? +?? )
2
?? at ?? =0,???? =?? 2
???? =
?? 2
?? i.e., ?
?? 2
?? =?? 1
(?? )+?? ?? 2
(?? )
???
?? 2
?? =
?? 2
?? +?? ?? 2
(?? )
????? ?? 2
(?? )=?? 2
(
1
?? -
1
?? )??? 2
(?? )=?? (
1
?? -
1
?? )
???? 2
(?? +?? )=(?? +?? )(
1
?? -
1
?? )
? Putting these values of ?? 1
and ?? 2
in the solution, we get
?? =
(?? +?? )
2
?? +?? (?? +?? )(
1
?? -
1
?? )
1.4 Find the surface satisfying
?? ?? ?? ?? ?? ?? =?? ?? +?? and touching ?? =?? ?? +?? ?? along its
section by the plane ?? +?? +?? =?? .
(2011 : 20 Marks)
Solution:
Given :
?
2
?? ??? 2
=6?? +2
??
??? ??? =6?? +2 where ?? =
??? ??? (??)
Integrating (i) w.r.t. ?? ,
?? ?=3?? 2
+2?? +?? (?? )
??????????????????????????????????????????????????????????????
??? ??? =3?? 2
+2?? +?? (?? )????????????????????????????????????????????????????????????(???? )
Integrating (ii) w.r.t. ?? ,
?? =?? 3
+?? 2
+???? (?? )+?? (?? ) (?????? )
where ?? (?? ) and ?? (?? ) are arbitrary functions.
The given surface is
?? =?? 3
+?? 3
(???? )
and the given plane is
?? +?? +1=0 (?? )
Since (iii) and (iv) touch each other, along their section by ( ?? ), the values of ?? and ?? at
any point on ( ?? ) must be equal. Thus, we must have
and ?????????????????????????????????????????????????????
3?? 2
+2?? +?? (?? )?=3?? 2
??????????????????????????????????????????????????????????????(???? )
???? (?? )+?? (?? )?=3?? 2
?????????????????????????????????????????????????????????????(?????? )
From (v) and (vi),
?? (?? ) ?=-2?? =2(?? +1)???????????????????????????????????????????????(???????? )
?? (?? ) ?=2
from?(vii)????????????????????????????????????????2?? +?? (?? ) ?=3?? 2
?? (?? ) ?=3?? 2
-2?? ?=3?? 2
+2(?? +1)??????????????????????????????????????????????????(???? )
?? (?? ) ?=?? 3
+?? 2
+2?? +??
where ?? is an arbitrary constant.
From (viii) and (ix), and using (iii), we get,
Page 5
Edurev123
Partial Differential Equations
1. Formulation of P.D.E.
1.1 Show that the differential equation of all cones which have their vertex at the
origin is ???? +???? =?? . Verify that this equation is satisfied by the surface ???? +?? ?? +
???? =?? .
(2009 : 12 Marks)
Solution:
The equation cone having vertex at origin
?? ?? 2
+?? ?? 2
+?? ?? 2
+2h???? +2?????? +2?????? =0 (1)
where ???,?? ,?? ,?? ,?? ,h are parameters.
Differentiating w.r.t. ?? and ?? , we get
2???? +2h?? +2???? +2?????? +2?????? +2?????? ?=0
2???? +2?????? +2h?? +2?????? +2???? +2?????? ?=0
???? +h?? +???? +?? (???? +???? +???? )?=0×?? ???? +h?? +???? +?? (???? +???? +???? )?=0×?? ?? ?? 2
+h???? +?????? +?? (?? ?? 2
+?????? +?????? )?=0
?? ?? 2
+h???? +?????? +?? (?????? +?? ?? 2
+?????? )?=0
On adding,
???????????????? ?? 2
+?? ?? 2
+2h???? +?????? +?????? +???? +???? [???? +???? +???? ]=0?????????
???????????????????????????????-(?? ?? 2
+?????? +?????? )+(???? +???? +???? )(???? +???? )=0?????????
????????????????????????????????????????????????????????????????????(???? +???? +???? )(???? +???? -?? )=0?????????
Clearly, ???? +???? -?? =0 is required differential equation.
Given surface is ???? +???? +???? =0
Differentiating (?
*
) w.r.t. ?? and ?? , we get
???? +?? +???? +?? =0 (2)
?? +???? +???? +?? =0 (3)
So, we get
?? ?=
-(?? +?? )
(?? +?? )
,?? =
-(?? +?? )
(?? +?? )
???? +???? -?? ?=
-(?? +?? )?? (?? +?? )
-
(?? +?? )
(?? +?? )
?? -?? ?=
-(?? +?? )?? -(?? +?? )?? -?? (?? +?? )
(?? +?? )
?=
-???? -???? -???? -???? -???? -????
(?? +?? )
?=
-2(???? +???? +???? )
?? +?? =
-20
?? +?? =0
1.2 From the partial differential equation by eliminating the arbitrary function ??
given by:
?? (?? ?? +?? ?? ,?? -???? )=??
Solution:
The function is
???????????????????????????????????????????????????????????????????? =???? +?? (?? 2
+?? 2
)??????????????????????????????????????????????????????????????(1)
Now differentiating partially (1) w.r.t. ?? we get
??? ??? =?? +?? '
(?? 2
+?? 2
)2??
So,
?? -?? 2?? =?? '
(?? 2
+?? 2
) (2)
Now, differentiating partially (1) w.r.t. ?? , we get
??? ??? =?? +?? (?? 2
+?? 2
)·2??
So,
?? -?? 2?? =?? '
(?? 2
+?? 2
) (3)
Equating (2) and (3), we get
?? -?? 2?? =
?? -?? 2??
So, ???? -???? =?? 2
-?? 2
is linear PDE.
1.3 Find the surface satisfying the P.D.E. (?? ?? -?? ?? ?? '
+?? ?? )?? =?? and the
conditions that ???? =?? ?? when ?? =?? and ???? =?? ?? when ?? =?? .
(2010 : 12 Marks)
Solution:
Given, the equation is
(?? 2
-2???? +?? 2
)?? =0
?????????????????????????????????????????????????????????????????????????????(?? -?? '
)
2
?? =0
The auxiliary eqn. for above eqn. is
(?? -1)
2
=0
??????????????????????????????????????????????????????????????????????????????????????? =1,1
? The solution of above eqn. is
?? =?? 1
(?? +?? )+?? ?? 2
(?? +?? )
Given, at ?? =0,???? =?? 2
?
?? =
?? 2
?? i.e., ?
?? 2
?? =?? 1
(?? )+0??? 1
(?? )=
?? 2
?? ????? 1
(?? +?? )=
(?? +?? )
2
?? at ?? =0,???? =?? 2
???? =
?? 2
?? i.e., ?
?? 2
?? =?? 1
(?? )+?? ?? 2
(?? )
???
?? 2
?? =
?? 2
?? +?? ?? 2
(?? )
????? ?? 2
(?? )=?? 2
(
1
?? -
1
?? )??? 2
(?? )=?? (
1
?? -
1
?? )
???? 2
(?? +?? )=(?? +?? )(
1
?? -
1
?? )
? Putting these values of ?? 1
and ?? 2
in the solution, we get
?? =
(?? +?? )
2
?? +?? (?? +?? )(
1
?? -
1
?? )
1.4 Find the surface satisfying
?? ?? ?? ?? ?? ?? =?? ?? +?? and touching ?? =?? ?? +?? ?? along its
section by the plane ?? +?? +?? =?? .
(2011 : 20 Marks)
Solution:
Given :
?
2
?? ??? 2
=6?? +2
??
??? ??? =6?? +2 where ?? =
??? ??? (??)
Integrating (i) w.r.t. ?? ,
?? ?=3?? 2
+2?? +?? (?? )
??????????????????????????????????????????????????????????????
??? ??? =3?? 2
+2?? +?? (?? )????????????????????????????????????????????????????????????(???? )
Integrating (ii) w.r.t. ?? ,
?? =?? 3
+?? 2
+???? (?? )+?? (?? ) (?????? )
where ?? (?? ) and ?? (?? ) are arbitrary functions.
The given surface is
?? =?? 3
+?? 3
(???? )
and the given plane is
?? +?? +1=0 (?? )
Since (iii) and (iv) touch each other, along their section by ( ?? ), the values of ?? and ?? at
any point on ( ?? ) must be equal. Thus, we must have
and ?????????????????????????????????????????????????????
3?? 2
+2?? +?? (?? )?=3?? 2
??????????????????????????????????????????????????????????????(???? )
???? (?? )+?? (?? )?=3?? 2
?????????????????????????????????????????????????????????????(?????? )
From (v) and (vi),
?? (?? ) ?=-2?? =2(?? +1)???????????????????????????????????????????????(???????? )
?? (?? ) ?=2
from?(vii)????????????????????????????????????????2?? +?? (?? ) ?=3?? 2
?? (?? ) ?=3?? 2
-2?? ?=3?? 2
+2(?? +1)??????????????????????????????????????????????????(???? )
?? (?? ) ?=?? 3
+?? 2
+2?? +??
where ?? is an arbitrary constant.
From (viii) and (ix), and using (iii), we get,
?? =?? 3
+?? 2
+2?? (?? +1)+?? 3
+?? 2
+2?? +?? (?? )
At the point of contact of (iv) and ( x ) values of ?? must be the same and hence, we have
?? 3
+?? 2
+2?? (?? +1)+?? 3
+?? 2
+2?? +?? =?? 3
+?? 3
(xi)
Using ?? =-?? -1 from (v), (xi) gives
?? =1
Putting ?? =1 in (?? ) , the required surface is
?? ?=?? 3
+?? 3
+2?? (?? +1)+?? 3
+?? 2
+2?? +1
?=?? 3
+?? 3
+(?? +?? +1)
2
1.5 Form a partial differential equation by eliminating the arbitrary functions ?? and
?? from ?? =???? (?? )+???? (?? ) .
(2013: 10 marks)
Solution:
Differentiating partially with respect to ?? and ??
?? =???? (?? )+???? (?? )
and
??? ??? ?=?? ?? '
(?? )+?? (?? );
??? ??? =?? (?? )+?? ?? '
(?? )
?
2
?? ??? ??? ?=?? (?? )+?? '
(?? )
?? ??? ??? +?? ??? ??? ?=???? [?? (?? )+?? '
(?? )]+???? (?? )+???? (?? )
?=????
?
2
?? ??? ??? +?? ???? ??? ??? +?? ??? ??? -????
?
2
?? ??? ??? -?? ?=0
is the required partial differential equation.
1.6 Find the surface which intersects the surface of the system
?? (?? +?? )=?? (?? ?? +?? )(?? being a constant )
Orthogonally and which passes through the circle ?? ?? +?? ?? =?? ,?? =?? .
(2013: 15 marks)
Solution:
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Oct 31, 2024 Last updated |
Document Description: Formulation of P.D.E. for UPSC 2024 is part of Mathematics Optional Notes for UPSC preparation.
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The content of Formulation of P.D.E. is prepared as per the latest UPSC syllabus.
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