Chapter 1 (maths 3)

1. CHAPTER 1
PARTIAL DIFFERENTIAL EQUATIONS
A partial differential equation is an equation involving a function of two or
more variables and some of its partial derivatives. Therefore a partial differential
equation contains one dependent variable and one independent variable.
Here z will be taken as the dependent variable and x and y the independent
variable so that( )yxfz ,= .
We will use the following standard notations to denote the partial derivatives.
,, q
y
z
p
x
z
=
¶
¶
=
¶
¶
t
y
z
s
yx
z
r
x
z
=
¶
¶
=
¶¶
¶
=
¶
¶
2
22
2
2
,,
The order of partial differential equation is that of the highest order derivative
occurring in it.
Formation of partial differential equation:
There are two methods to form a partial differential equation.
(i) By elimination of arbitrary constants.
(ii) By elimination of arbitrary functions.

Problems
Formation of partial differential equation by elimination of arbitrary
constants:
(1)Form the partial differential equation by eliminating the arbitrary constants
from
22
babyaxz +++= .
Solution:
1

2. Given
22
babyaxz +++= ……………... (1)
Here we have two arbitrary constants a & b.
Differentiating equation (1) partially with respect to x and y respectively we get
apa
x
z
=Þ=
¶
¶
……………… (2)
aqb
y
z
=Þ=
¶
¶
………………. (3)
Substitute (2) and (3) in (1) we get

22
qpqypxz +++= , which is the required partial differential equation.
(2) Form the partial differential equation by eliminating the arbitrary constants
a, b, c from 1
2
2
2
2
2
2
=++
c
z
b
y
a
x
.
Solution:
We note that the number of constants is more than the number of independent
variable. Hence the order of the resulting equation will be more than 1.
1
2
2
2
2
2
2
=++
c
z
b
y
a
x
.................. (1)
Differentiating (1) partially with respect to x and then with respect to y, we get


)3..(....................0
22
)2.(....................0
22
22
22
=+
=+
q
c
z
b
y
p
c
y
a
x

Differentiating (2) partially with respect to x,
)(
11
2
22
pzr
ca
++ ……………..(4)
Where
2
2
x
z
r
¶
¶
= .
From (2) and (4) ,
)6..(....................
)5..(....................
2
2
2
2
2
pzr
a
c
x
zp
a
c
+=-
=-
.
2

3. From (5) and (6), we get

x
z
z
x
z
x
x
z
xz
¶
¶
=÷
ø
ö
ç
è
æ
¶
¶
+
¶
¶
2
2
2
, which is the required partial differential
equation.
(3) Find the differential equation of all spheres of the same radius c having their
center on the yoz-plane. .
Solution:
The equation of a sphere having its centre at( )ba,,0 , that lies on theyoz-plane
and having its radius equal to c is

2222
)()( cbzayx =-+-+ ……………. (1)
If a and b are treated as arbitrary constants, (1) represents the family of spheres
having the given property.
Differentiating (1) partially with respect to x and then with respect to y, we
have
( )022 =-+ pbzx …………… (2)
and ( ) ( )022 =-+- qbzay …………….(3)
From (2),
p
x
bz -=- …………….(4)
Using (4) in (3),
p
qx
by=- ……………..(5)
Using (4) and (5) in (1), we get

2
2
2
2
22
2
c
p
x
p
xq
x =++
. i.e. ( )
22222
1 pcxqp =++ , which is the required partial differential
equation.

Problems
3

4. Formation of partial differential equation by elimination of arbitrary
functions:
(1)Form the partial differential equation by eliminating the arbitrary function ‘f’
from
( )byxfez
ay
+=
solution: Given ( )byxfez
ay
+=
i.e. ( )byxfze
ay
+=
-
……………(1)
Differentiating (1) partially with respect to x and then with respect to y, we
get
()1.'ufpe
ay
=
-
………….(2)
()bufzaeqe
ayay
.'=-
--
…………….(3)
where byxu+=
Eliminating f’(u) from (2) and (3), we get
b
p
azq
=
-
i.e. bpazq +=
(2) Form the partial differential equation by eliminating the arbitrary function ‘f’
0,
2
=÷
ø
ö
ç
è
æ
-
z
x
xyzf
Solution: Given 0,
2
=÷
ø
ö
ç
è
æ
-
z
x
xyzf ………………(1)
Let xyzu -=
2
,
z
x
v=
Then the given equation is of the form ( )0,=vuf .
The elimination of f from equation (2), we get,
0=
¶
¶
¶
¶
¶
¶
¶
¶
y
v
y
u
x
v
x
u
4

5. i.e. 0
2
2
2
2
=
-
-
-
-
z
xq
xzq
z
pxz
yzp
i.e ( ) ( )022
22
=-÷
ø
ö
ç
è
æ-
-÷
ø
ö
ç
è
æ-
- xzq
z
pxz
z
xq
yzp
i.e ( )zxzxyqpx =--
22
2

(3) Form the partial differential equation by eliminating the arbitrary function ‘f’
from
( ) ( )yxgyxfz -++= 32
Solution: Given ( ) ( )yxgyxfz -++= 32 .…………(1)
Differentiating (1) partially with respect to x,
() ()3.2. vgufp ¢+¢= ………….(2)
Where yxu +=2 and yxv -=3
Differentiating (1) partially with respect to y,
() ())1(1. -¢+¢= vgufq …………. (3)
Differentiating (2) partially with respect to x and then with respect to y,
() ()9.4. vgufr ¢¢+¢¢= …………. (4)
and () ())3.(2. -¢¢+¢¢= vgufs ………….. (5)
Differentiating (3) partially with respect to y,
() ()1.1. vguft ¢¢+¢¢= ………….. (6)
Eliminating ()uf¢¢ and ()vg¢¢from (4), (5) and (6) using determinants, we
have

t
s
r
11
32
94
- = 0
i.e. 03055 =-+ tsr
or 06
2
22
2
2
=
¶
¶
-
¶¶
¶
+
¶
¶
y
z
yx
z
x
z

5

6. (4) Form the partial differential equation by eliminating the arbitrary function ‘f’
from
( ) ( )xyxy
x
z -¢+-= ff
1
.
Solution: Given () ()uu
x
z ff ¢+=
1
…………...(1)
Where xyu-=
Differentiating partially with respect to x and y, we get
() () ())1(
1
)1(
1
2
-¢¢+--¢= uu
x
u
x
p fff …………(2)
() ()1.1.
1
uu
x
q ff ¢¢+¢= ………….(3)
() () () ()1.
22
1.
1
32
uu
x
u
x
u
x
r ffff ¢¢¢++¢+¢¢= ………..(4)
() () ())1(
11
2
-¢¢¢+¢¢-¢-= uu
x
u
x
s fff …………(5)
() ()1.1.
1
uu
x
t ff ¢¢¢+¢¢= …………..(6)
From (4) and (6), we get
() ()u
x
u
x
tr ff
32
22
+¢=-
= () ()
þ
ý
ü
î
í
ì
¢+uu
xx
ff
12
2
= z
x
2
2
i.e. z
yx
z
x
z
2
2
2
2
2
2
=
÷
÷
ø
ö
ç
ç
è
æ
¶
¶
-
¶
¶
Solutions of partial differential equations
Consider the following two equations
byaxz += ………..(1)
6

7. and ÷
ø
ö
ç
è
æ
=
x
y
xfz ………..(2)
Equation (1) contains arbitrary constants a and b, but equation (2) contains only one
arbitrary function f.
If we eliminate the arbitrary constants a and b from (1) we get a partial differential
equation of the form zyqxp =+ . If we eliminate the arbitrary function f from (2) we get
a partial differential equation of the form zyqxp =+ .
Therefore for a given partial differential equation we may have more than one type of
solutions.
Types of solutions:
(a) A solution in which the number of arbitrary constants is equal to the number of
independent variables is called Complete Integral (or) Complete solution.
(b) In complete integral if we give particular values to the arbitrary constants we get
Particular Integral.
(c) The equation which does not have any arbitrary constants is known as Singular
Integral.
To find the general integral:
Suppose that ( )0,,,, =qpzyxf ………....(1)
is a first order partial differential equation whose complete solution is
( )0,,,, =bazyxf ………..(2)
Where a and b are arbitrary constants.
Let ()afb= , where ‘f’is an arbitrary function.
Then (2) becomes
()( )0,,,, =afazyxf ……….(3)
Differentiating (3) partially with respect to ‘a’, we get

()0. =¢
¶
¶
+
¶
¶
af
ba
ff
……….(4)
7

8. The eliminant of ‘a’ between the two equations (3) and (4), when it exists, is called the
general integral of (1).
Methods to solve the first order partial differential equation:
Type 1:
Equation of the form 0),( =qpf ………...(1)
i.e the equation contains p and q only.
Suppose that cbyaxz ++= …….....(2)
is a solution of the equation
b
y
z
a
x
z
=
¶
¶
=
¶
¶
,
bqap ==Þ ,
substitute the above in (1), we get
0),(=baf
on solving this we can get ()abf= , where f is a known function.
Using this value of b in (2), the complete solution of the given partial differential
equation is
()cyaaxz ++=f …………(3)
is a complete solution,
To find the singular solution, we have to eliminate ‘a’ and ‘c’ from
()cyaaxz ++=f
Differentiating the above with respect to ‘a’ and ‘c’, we get
()yaxf¢+=0 ,
and 0=1.
The last equation is absurd. Hence there is no singular solution for the equation of
Type 1.
Problems:
(1) Solve 1
22
=+qp .
8

9. Solution:
Given: 1
22
=+qp ………….(1)
Equation (1) is of the form 0),( =qpf .
Assume cbyaxz ++= ………….(2)
be the solution of equation (1).
From (2) we get bqap ==, .
(1) 1
22
=+Þ ba

2
1ab -±=Þ ……….(3)
Substitute (3) in (2) we get
cyaaxz +-±=
2
1 ……....(4)
This is a complete solution.
To find the general solution:
We put ()afc= in (4),where ‘f’ is an arbitrary function.
i.e. ()afyaaxz +-±=
2
1 …………(5)
Differentiating (5) partially with respect to ‘a’, we get
()0
1
2
=¢+
-
afy
a
a
x ……………(6)
Eliminating ‘a’ between equations (5) and (6), we get the required general solution.
To find the singular solution:
Differentiate (4) partially with respect to ‘a’ and ‘c’, we get

2
1
0
a
a
x
-
±= ,
0=1.(which is absurd)
so there is no singular solution.
(2) Solve pqqp=+
9

10. Solution:
Given: pqqp=+ ………..(1)
Equation (1) is of the form 0),( =qpf
Assume cbyaxz ++= ………(2)
be the solution of equation (1).
From (2) we get bqap ==,
(1) abba=+Þ

1-
=Þ
a
a
b …….....(3)
Substituting (3) in (2), we get
cy
a
a
axz +
-
+=
1
………(4)
This is a complete solution.
To find the general solution:
We put ()afc= in (4), we get
()afy
a
a
axz +
-
+=
1
……..(5)
Differentiating (5) partially with respect to ‘a’, we get

( )
()afy
a
x ¢+
-
-
2
1
1
………..(6)
Eliminating ‘a’ between equations (5) and (6), we get the required general solution
To find the singular solution:
Differentiating (4) with respect to ‘a’ and ‘c’.

( )
,
1
1
0
2
y
a
x
-
-=
and 0=1 (which is absurd).
So there is no singular solution.
10

11. Type 2: (Clairaut’s type)
The equation of the form
),(qpfqypxz ++= ………(1)
is known as Clairaut’s equation.
Assume cbyaxz ++= …………(2)
be a solution of (1).
b
y
z
a
x
z
=
¶
¶
=
¶
¶
,
bqap ==Þ ,
Substitute the above in (1), we get
),(bafbyaxz ++= ………..(3)
which is the complete solution.
Problem:
(1) Solve pqqypxz ++=
Solution:
Given: pqqypxz ++= ……….(1)
Equation (1) is a Clairaut’s equation
Let cbyaxz ++= ………..(2)
be the solution of (1).
Put bqap ==, in (1), we get
abbyaxz ++= ……….(3)
which is a complete solution.
To find the general solution:
We put ()afb= in (3), we get
() ()aafyafaxz ++= ………(4)
Differentiate (4) partially with respect to ‘a’, we get
11

12. ()
()
() ()[ ]afafa
aaf
yafx +¢+¢+=
2
1
0
………..(5)
Eliminating ‘a’ between equations (4) and (5), we get the required general solution
To find singular solution,
Differentiate (3) partially with respect to ‘a’, we get
b
ab
x .
2
1
0+=

a
b
x
2
-=Þ ………..(6)
Differentiate (3) partially with respect to ‘b’, we get
a
ab
y .
2
1
0+=

b
a
y
2
-=Þ ………..(7)
Multiplying equation (6) and (7),we get

÷
÷
ø
ö
ç
ç
è
æ
-
÷
÷
ø
ö
ç
ç
è
æ
-=
b
a
a
b
xy
22

4
1
=xy
14=xy
(2) Solve p
p
q
qypxz -++=
Solution:
Given: p
p
q
qypxz -++= ……….(1)
Equation (1) is a Clairaut’s equation
Let cbyaxz ++= ………...(2)
be the solution of (1).
Put bqap ==, in (1), we get
a
a
b
byaxz -++= ……….(3)
12

13. which is the complete solution.
To find the general solution:
We put ()afb= in (3), we get
()
()
a
a
af
yafaxz -++= ……..(4)
Differentiate (4) partially with respect to ‘a’, we get
()
() ()
10 -
-¢
+¢+=
a
afafa
yafx ……..(5)
Eliminating ‘a’ between equations (4) and (5), we get the required general
solution
To find the singular solution:
Differentiate (3) partially with respect to ‘a’,
10
2
--=
a
b
x
1
2
+=Þ
a
b
x
1
2
+=Þ bxa
1
2
-=Þ xab .............. (4)
Differentiate (3) partially with respect to ‘b’,

a
y
1
0+=

y
a
1
-=Þ ………….(5)
Substituting equation (4) and (5) in equation (3), we get


( )
( )
÷
÷
ø
ö
ç
ç
è
æ-
-
-
-
+-+
÷
÷
ø
ö
ç
ç
è
æ-
=
y
y
xa
yxax
y
z
1
1
1
1
1
2
2

y
yxyayyxa
y
x
z
1
22
++--+
-
=
13

14. yy
x
z
1
+
-
=
xyz-=1
Type 3:
Equations not containing x and y explicitly, i.e. equations of the form
( )0,,=qpzf ……….(1)
For equations of this type ,it is known that a solution will be of the form
( )ayxz +=f ……….(2)
Where ‘a’ is the arbitrary constant and f is a specific function to be found out.
Putting uayx =+ , (2) becomes () ()uzoruzf=
\
du
dz
x
u
du
dz
p =
¶
¶
=.
and
du
dz
a
y
u
du
dz
q =
¶
¶
=.
If (2) is to be a solution of (1), the values of p and q obtained should satisfy (1).
i.e. 0,, =÷
ø
ö
ç
è
æ
du
dz
a
du
dz
zf ……..(3)
From (3), we get
( )az
du
dz
,y= ……….(4)
Now (4) is a ordinary differential equation, which can be solved by variable separable
method.
The solution of (4), which will be of the form( ) ( ) bayxazgorbuazg +++= ,, , is the
complete solution of (1).
The general and singular solution of (1) can be found out by usual method.
Problems:
(1)Solve
22
qpz += .
Solution:
14

15. Given:
22
qpz += …………(1)
Equation (1) is of the form ( )0,,=qpzf
Assume ()uzf= where, ayxu+= be a solution of (1).

du
dz
p
du
dz
x
u
du
dz
x
z
p =Þ=
¶
¶
=
¶
¶
= . …….(2)

du
dz
aq
du
dz
a
y
u
du
dz
y
z
q =Þ=
¶
¶
=
¶
¶
= . ……(3)
Substituting equation (2) & (3) in (1), we get


2
2
2
÷
ø
ö
ç
è
æ
+÷
ø
ö
ç
è
æ
=
du
dz
a
du
dz
z
( )
2
2
1a
du
dz
z +÷
ø
ö
ç
è
æ
=

2
2
1a
z
du
dz
+
=÷
ø
ö
ç
è
æ

( )
2
2
1
1a
z
du
dz
+
=
By variable separable method,

( )
2
1
2
2
1
1a
du
z
dz
+
=
By integrating, we get

( )
òò
+
+
= cdu
az
dz
3
1
2
2
1
1
1

( )
( )
c
a
ayx
z +
+
+
=
2
1
2
1
2

( )
( )
2
12
2
1
2
c
a
ayx
z +
+
+
=

( )
( )
k
a
ayx
z +
+
+
=
2
1
2
12
……….(4)
15

16. This is the complete solution.
To find the general solution:
We put ()afk= in (4), we get

( )
( )
()af
a
ayx
z +
+
+
=
2
1
2
12
……..(5)
Differentiate (5) partially with respect to ‘a’, we get

( )
()af
a
xay
¢+
+
-
=
2
3
2
12
0
………..(6)
Eliminating ‘a’ between equations (4) and (5), we get the required general
solution.

To find the singular solution:
Differentiate (4) partially with respect to ‘a’ and ‘k’, we get

( )
2
3
2
12
0
a
xay
+
-
=
………..(7)
and 10= (which is absurd)
So there is no singular solution.
(2)Solve( )49
22
=+qzp .
Solution:
Given: ( )49
22
=+qzp ………(1)
Equation (1) is of the form ( )0,,=qpzf
Assume ()uzf= where , ayxu+= be a solution of (1).

du
dz
p
du
dz
x
u
du
dz
x
z
p =Þ=
¶
¶
=
¶
¶
= . …….(2)

du
dz
aq
du
dz
a
y
u
du
dz
y
z
q =Þ=
¶
¶
=
¶
¶
= . ……(3)
16

17. Substituting equation (2) & (3) in (1), we get
49
2
2
2
=
÷
÷
ø
ö
ç
ç
è
æ
÷
ø
ö
ç
è
æ
+÷
ø
ö
ç
è
æ
du
dz
az
du
dz
[ ]49
2
2
=+÷
ø
ö
ç
è
æ
az
du
dz

[ ]
2
2
9
4
azdu
dz
+
=÷
ø
ö
ç
è
æ

2
1
.
3
2
azdu
dz
+
±=
dudzaz
3
2
2
±=+
Integrating the above, we get
( ) cdudzaz +±=+ òò
3
2
2
1
2

( ) cuaz +±=+
3
2
3
2
2
3
2
( ) ( )cayxaz ++±=+
3
2
3
2
2
3
2
( ) ( )kayxaz ++±=+Þ
2
3
2 ………..(4)
This is the complete solution.
To find the general solution:
We put ()afk= in (4), we get
( ) ( ) ()afayxaz ++±=+
2
3
2 ……..(5)
Differentiate (5) partially with respect to ‘a’, we get
( ) ()afyaza ¢+±=--
2
1
2
3
……..(6)
Eliminating ‘a’ between equations (4) and (5), we get the required general
solution.
To find the singular solution:
Differentiate (4) partially with respect to ‘a’ and ‘k’, we get
17

18. ( ) yaza ±=+
2
1
2
3
………..(7)
and 10= (which is absurd)
So there is no singular solution.
Type 4:
Equations of the form
( ) ( )qygpxf ,,= ………..(1)
i.e. equation which do not contain z explicitly and in which terms containing p and x can
be separated from those containing q and y.
To find the complete solution of (1),
We assume that ( ) ( )aqygpxf ==,, .where ‘a’ is an arbitrary constant.
Solving( )apxf =, ,we can get ( )axp ,f= and solving ( )aqyg =, ,we can get
( )ayq ,y= .
Now
qdypdxordy
y
z
dx
x
z
dz +
¶
¶
+
¶
¶
=
i.e. ( ) ( )dyaydxaxdz ,,yf +=
Integrating with respect to the concerned variables, we get
( ) ( )òò
++= bdyaydxaxz ,, yf ……….(2)
The complete solution of (1) is given by (2), which contains two arbitrary constants ‘a’
and ‘b’.
The general and singular solution of (1) can be found out by usual method.
Problems:
(1)Solvexypq=.
Solution:
Given:xypq=

q
y
x
p
=Þ ………..(1)
Equation (1) is of the form ( ) ( )qygpxf ,,=
18

19. Let a
q
y
x
p
== (say)
a
x
p
= axp=Þ ………….(2)
Similarly, a
q
y
=
a
y
q=Þ …………(3)
Assume qdypdxdz += be a solution of (1)
Substitute equation (2) and (3) to the above, we get
dy
a
y
axdxdz +=
Integrating the above we get,
cydy
a
xdxadz ++= òòò
1
c
a
yax
z ++=
22
22
k
a
y
axz ++=
2
2
2 ………..(4)
This is the complete solution.
The general and singular solution of (1) can be found out by usual method.

(2) Solve yxqp +=+ .
Solution:
Given: yxqp +=+
yqxp -=-Þ ………….. (1)
Equation (1) is of the form ( ) ( )qygpxf ,,=
Let aqyxp =-=- (say)
axp=- axp+=Þ …………. (2)
Similarly, aqy=- ayq-=Þ ……………(3)
Assume qdypdxdz += be a solution of (1)
Substitute equation (2) and (3) to the above, we get
( ) ( )dyaydxaxdz -++=
Integrating the above we get,
19

20. ( ) ( )cdyaydxaxdz +-++= òòò

( )
( )cyxa
yx
z +-+
+
=
2
22
( )cyxayxz +-++= 22
22
……………(4)
This is the complete solution.
The general and singular solution of (1) can be found out by usual method.
Equations reducible to standard types-transformations:
Type A:
Equations of the form ( ) ( )0,,0, == zqypxforqypxf
nmnm
.
Where m and n are constants, each not equal to 1.
We make the transformations YyandXx
nm
==
-- 11
.
Then ( )
X
z
PwherePxm
x
X
X
z
x
z
p
m
¶
¶
º-=
¶
¶
¶
¶
=
¶
¶
=
-
,1. and
( )
Y
z
QwhereQyn
y
Y
Y
z
y
z
q
n
¶
¶
º-=
¶
¶
¶
¶
=
¶
¶
=
-
,1.
Therefore the equation ( )0,=qypxf
nm
reduces to ( )( ){ },01,1 =-- QnPmf .which is a
type 1 equation.
The equation ( )0,, =zqypxf
nm
reduces to ( )( ){ },0,1,1 =-- zQnPmf .which is a type 3
equation.
Problem:
(1)Solve
2242
2zzqyxp =+ .
Solution:
Given:
2242
2zzqyxp =+
This can be written as
( ) ( )
22
2
2
2zzqypx =+ .
Which is of the form( )0,, =zqypxf
nm
, where m=2,n=2.
20

21. Put
y
yY
x
xX
nm 1
;
1
11
====
--
( )
22
. pxxp
X
x
x
z
X
z
P -=-=
¶
¶
¶
¶
=
¶
¶
=
( )
22
. qyyq
Y
y
y
z
Y
z
Q -=-=
¶
¶
¶
¶
=
¶
¶
=
Substituting in the given equation,

22
2zqzP =- .
This is of the form( )0,,=zqpf .
Let( )aYXfZ += , where aYXu +=

du
dz
aQ
du
dz
P ==,
Equation becomes, .
Solving for
du
dz
,
du
aa
z
dz
zzaaz
du
dz
z
du
dz
az
du
dz
2
8
2
8
02
2
222
2
2
+±
=
+±
=
=--÷
ø
ö
ç
è
æ
( )baYX
aa
z ++
+±
=
2
8
log
2
b
y
a
x
aa
z +
÷
÷
ø
ö
ç
ç
è
æ
+
+±
=
1
2
8
log
2
is a complete solution.
The general and singular solution can be found out by usual method.
Type B:
Equations of the form( ) ( )0,,,0, == yxqzpzforqzpzf
kkkk
.
Where k is a constant, which is not equal to -1.
We make the transformations
1+
=
k
zZ .
Then ( )pzk
x
Z
P
k
1+=
¶
¶
= and
21

22. ( )pzk
y
Z
Q
k
1+=
¶
¶
=
Therefore the equation ( )0,=qzpzf
kk
reduces to ,0
1
,
1
=÷
ø
ö
ç
è
æ
++k
Q
k
P
f which is a type 1
equation.
The equation ( )0,,, =yxqzpzf
kk
reduces to ,0,,
1
,
1
=÷
ø
ö
ç
è
æ
++
yx
k
Q
k
P
f which is a
type 4 equation.
Problems:
(1)Solve: .1
224
=-pzqz
Solution:
Given: .1
224
=-pzqz
The equation can be rewritten as ( ) ( )1
2
2
2
=-pzqz ………(1)
Which contains qzandpz
22
.
Hence we make the transformation
3
zZ=

3
3
2
2
P
pz
pz
x
Z
P
=Þ
=
¶
¶
=
Similarly
3
2Q
qz=
Using these values in (1), we get
93
2
=-PQ ………..(2)
As (2) is an equation containing P and Q only, a solution of (2) will be of the form
cbyaxZ ++= ………….(3)
Now ,bQandaP == obtained from (3) satisfy equation (2)
93
2
=-ab
i.e. 93+±= ab
Therefore the complete solution of (2) is cyaaxZ ++±= 93
i.e complete solution of (1) is cyaaxz ++±= 93
3
22

23. Singular solution does not exist. General solution is found out as usual.
Type C:
Equations of the form ( )0, =qzypzxf
knkm
, where 1;1, -¹¹knm
We make the transformations

111
,
+--
===
knm
zZandyYxX
Then
dX
dx
x
z
dz
dZ
X
Z
P ..
¶
¶
=
¶
¶
=
( )
m
x
pzk
m
k
-
+=
1
.1
and ( )
n
y
qzkQ
n
k
-
+=
1
.1
Therefore the given equation reduces to
0
1
1
,
1
1
=
þ
ý
ü
î
í
ì
÷
ø
ö
ç
è
æ
+
-
÷
ø
ö
ç
è
æ
+
-
Q
k
n
P
k
m
f
This is of type 1 equation.
Problem:
(1)Solve 1
2
2
2
2
2
=
÷
÷
ø
ö
ç
ç
è
æ
+
y
q
x
p
z
Solution:
Given: 1
2
2
2
2
2
=
÷
÷
ø
ö
ç
ç
è
æ
+
y
q
x
p
z
It can be rewritten as ( ) ( )1
2
1
2
1
=+
--
zqyzpx ………..(1)
which is of the form ( ) ( )1
22
=+ qzypzx
knkm
we make the transformations

111
,
+--
===
knm
zZandyYxX
i.e.
222
, zZandyYxX ===
Then xP
zdx
dX
X
Z
dZ
dz
x
z
p 2..
2
1
.. =
¶
¶
=
¶
¶
=
zpxP
1-
=Þ ,
23

24. Similarly, zqyQ
1-
= ,
Using these in (1),it becomes
1
22
=+QP …………(2)
As (2) contains only P and Q explicitly, a solution of the equation will be of the
form
cbYaXZ ++= ………….(3)
Therefore ,bQandaP == obtained from (3) satisfy equation (2)
i.e.
2
22
1
,1
ab
ba
-±=Þ
=+
Therefore the complete solution of (2) is
cYaaXZ +-±=
2
1
Therefore the complete solution of (1) is
cyaaxz +-±=
2222
1
Singular solution does not exist. General solution is found out as usual.
Type D:
Equation of the form 0,=÷
ø
ö
ç
è
æ
z
qy
z
px
f
By putting zZandyYxX loglog,log === the equation reduces to
( ),0,=QPf
where
Y
Z
Qand
X
Z
P
¶
¶
=
¶
¶
= .
Problems:
(1)Solve
2
zpqxy=.
Solution:
Given:
2
zpqxy=. …………….(1)
Rewriting (1),
1=÷
ø
ö
ç
è
æ
÷
ø
ö
ç
è
æ
z
qy
z
px
. …………….(2)
24

25. As (2) contains ÷
ø
ö
ç
è
æ
÷
ø
ö
ç
è
æ
z
qy
and
z
px
, we make the substitutions
zZandyYxX loglog,log ===
Then
x
Pz
dx
dX
X
Z
dZ
dz
x
z
P
1
.... =
¶
¶
=
¶
¶
=
i.e. P
z
px
=
Similarly, Q
z
qx
=
Using these in (2), it becomes
1=PQ …………..(3)
which contains only P and Q explicitly. A solution of (3) is of the form
cbYaXZ ++= …………(4)
Therefore ,bQandaP == obtained from (4) satisfy equation (3)
i.e.
a
borab
1
1 ==
Therefore the complete solution of (3) is cY
a
aXZ ++=
1
Therefore the complete solution of (1) is cy
a
xaz ++= log
1
loglog ……..(5)
General solution of (1) is obtained as usual.
General solution of partial differential equations:
Partial differential equations, for which the general solution can be obtained
directly, can be divided in to the following three categories.
(1)Equations that can be solved by direct (partial) integration.
(2)Lagrange’s linear equation of the first order.
(3)Linear partial differential equations of higher order with constant coefficients.
Equations that can be solved by direct (partial) integration:
Problems:
25

26. (1)Solve the equation ,cos
2
xe
tx
u
t-
=
¶¶
¶
if 0=u when 00 =
¶
¶
=
t
u
andt when .0=x
Also show that ,sinxu® when¥®t .
Solution:
Given: ,cos
2
xe
tx
u
t-
=
¶¶
¶
…………….(1)
Integrating (1) partially with respect to x,
()tfxe
t
u
t
+=
¶
¶
-
sin …………….(2)
When 00 =
¶
¶
=
t
u
andt in (2), we get ()0=tf .
Equation (2) becomes xe
t
u
t
sin
-
=
¶
¶
…………….(3)
Integrating (3) partially with respect to t, we get
()xgxeu
t
+-=
-
sin ……………(4)
Using the given condition, namely 0=u when ,0=t we get
() () xxgorxgx sinsin0 =+-=
Using this value in (4), the required particular solution of (1) is
( )
t
exu
-
-= 1sin
Now () ( ){ }
t
tt
exu
-
¥®¥®
-= 1limsinlim
xsin=
i.e. when xut sin,®¥® .
(2) Solve the equation yx
y
z
andyx
x
z
cos3 +-=
¶
¶
-=
¶
¶
simultaneously.
Solution: Given
)1......(..........3yx
x
z
-=
¶
¶
)2........(..........cosyx
y
z
+-=
¶
¶
Integrating (1) partially with respect to x,
26

27. ()yfyx
x
z +-=
2
3
2
…………..(3)
Differentiating (1) partially with respect to y,
()yfx
y
z
¢+-=
¶
¶
…………...(4)
Comparing (2) and (4), we get

()
() )5.........(..........sin
cos
cyyf
yyf
+=
=¢

Therefore the required solution is
cyyx
x
z ++-= sin
2
3
2
, where c is an arbitrary constant.
Lagrange’s linear equation of the first order:
A linear partial differential equation of the first order , which is of the form
RQqPp =+
where RQP,, are functions of zyx,, is called Lagrange’s linear equation.
working rule to solve RQqPp =+
(1)To solve RQqPp =+ , we form the corresponding subsidiary simultaneous equations
.
R
dz
Q
dy
P
dx
==
(2)Solving these equations, we get two independent solutions bvandau == .
(3)Then the required general solution is ( ) () ()uvorvuorvuf yf ===0, .
Solution of the simultaneous equations .
R
dz
Q
dy
P
dx
==
Methods of grouping:
By grouping any two of three ratios, it may be possible to get an ordinary
differential equation containing only two variables, eventhough P;Q;R are in
general, functions of x,y,z. By solving this equation, we can get a solution of the
simultaneous equations. By this method, we may be able to get two independent
solutions, by using different groupings.
27

28. Methods of multipliers:
If we can find a set of three quantities l,m,n which may be constants or
functions of the variables x,y,z, such that 0=++ nRmQlP , then the solution of the
simultaneous equation is found out as follows.

nRmQlP
ndzmdyldx
R
dz
Q
dy
P
dx
++
++
===
Since .0,0 =++=++ ndzmdyldxnRmQlP If 0=++ ndzmdyldx is an exact
differential of some function ( )zyxu,,, then we get .0=duIntegrating this, we get
au=, which is a solution of .
R
dz
Q
dy
P
dx
==
Similarly, if we can find another set of independent multipliers ,,,nml ¢¢¢ we can get
another independent solution bv=.
Problems:
(1)Solve xyqxp =+ .
Solution:
Given: xyqxp =+ .
This is of Lagrange’s type of PDE where xRyQxP === ,, .
The subsidiary equations are .
x
dz
y
dy
x
dx
==
Taking first two members
y
dy
x
dx
=
Integrating we get
1
logloglog cyx +=
i.e.
y
x
u
c
y
x
=
=
1
………….(1)
Taking first and last members
x
dz
x
dx
=.
i.e. dzdx=.
Integrating we get
zxv
czx
-=
=-
2
………......(2)
28

29. Therefore the solution of the given PDE is ( ) 0,..0, =
÷
÷
ø
ö
ç
ç
è
æ
-= zx
y
x
eivu ff .
(2)Solve the equation ( ) ( ) .22 xyqyzpzx -=-+-
Solution:
Given: ( ) ( ) .22 xyqyzpzx -=-+-
This is of Lagrange’s type of PDE where xyRyzQzxP -=-=-= ,2,2 .
The subsidiary equations are .
22 xy
dz
yz
dy
zx
dx
-
=
-
=
-
……….(1)
Using the multipliers 1,1,1, each ratio in (1)=
0
dzdydx ++
.
0=++ dzdydx .
Integrating, we get azyx =++ ……………(2)
Using the multipliers y,x,2z, each ratio in (1)=
0
2zdzxdyydx ++
.
() 02=+zdzxyd .
Integrating, we get bzxy =+
2
……………(3)
Therefore the general solution of the given equation is ( )0,
2
=+++ zxyzyxf .
(3)Show that the integral surface of the PDE ( ) ( ) ( )zyxqzxypzyx
2222
-=+-+ .
Which contains the straight line 02221,0
22
=+-++==+ zxyzyxiszyx .
Solution:
The subsidiary equations of the given Lagrange ‘s equation are

( ) ( )( )
.
2222
zyx
dz
zxy
dy
zyx
dx
-
=
+-
=
+
……………(1)
Using the multipliers ,
1
,
1
,
1
zyx
. each ratio in (1)=
0
111
dz
z
dy
y
dx
x
++
.
0
111
=++ dz
z
dy
y
dx
x
.
Integrating, we get axyz= ……………(2)
29

30. Using the multipliers y,x,-1, each ratio in (1)=
0
dzydyxdx -+
.
0=-+ dzydyxdx .
Integrating, we get bzyx =-+ 2
22
……………(3)
The required surface has to pass through

1
0
=
=+
z
yx
……………(4)
Using (4) in (2) and (3), we have

bx
ax
=-
=-
22
2
2
……………(5)
Eliminating x in (5) we get,
022 =++ba …………….(6)
Substituting for a and b from (2) and (3) in (6), we get
0222
22
=+-++ zxyzyx , which is the equation of the required
surface.
Linear P.D.E.S of higher order with constant coefficients:
The standard form of a homogeneous linear partial differential equation of the

th
norder with constant coefficients is
( )yxR
y
z
a
yx
z
a
yx
z
a
x
z
a
n
n
nn
n
n
n
n
n
,...
222110
=
¶
¶
++
¶¶
¶
+
¶¶
¶
+
¶
¶
-- …………….(1)
where a’s are constants.
If we use the operators
y
Dand
x
D
¶
¶
º¢
¶
¶
º , we can symbolically write equation (1)
as
( ) ( )yxRzDaDDaDa
n
n
nn
,...
1
10
=¢++¢+
-
…………….(2)
( ) ( )yxRzDDf ,,=¢ …………….(3)
where ( )DDf ¢, is a homogeneous polynomial of the
th
n degree in DandD ¢.
The method of solving (3) is similar to that of solving ordinary linear differential
equations with constant coefficients.
30

31. The general solution of (3) is of the form z = (complementary function)+(particular
integral),where the complementary function is the R.H.S of the general solution of
( )0,=¢zDDf and the particular integral is given symbolically by
( )
( )yxR
DDf
,
,
1
¢
.
Complementary function of ( ) ( )yxRzDDf ,,=¢ :
C.F of the solution of ( ) ( )yxRzDDf ,,=¢ is the R.H.S of the solution of
( )0,=¢zDDf . …………(1)
In this equation, we put1,=¢=DmD ,then we get an equation which is called the
auxiliary equation.
Hence the auxiliary equation of (1) is
0...
1
10
=+++
-
n
nn
amama ………….(2)
Let the roots of this equation be n
mmm ,...,
21 .
Case 1:
The roots of (2) are real and distinct.
The general solution is given by
( ) ( ) ( )xmyxmyxmyz
nn
++++++= fff ...
2211
Case 2:
Two of the roots of (2) are equal and others are distinct.
The general solution is given by
( ) ( ) ( )xmyxmyxmyz
nn
++++++= fff ...
1211
Case 3:
‘r’ of the roots of (2) are equal and others distinct.
The general solution is given by
( ) ( ) ( )xmyxxmyxxmyz
rr
r
++++++=
-
fff
1
1211
...
To find particular integral:
Rule (1): If the R.H.S of a given PDE is ( )
byax
eyxf
+
=, , then

( )
byax
e
DDf
IP
+
¢
=
,
1
.
Put bDaD =¢=,
31

32. ( )
( )0,
,
1
. ¹=
+
bafife
baf
IP
byax
If ( ),0,=baf refer to Rule (4).
Rule (2): If the R.H.S of a given PDE is( ) ( ) ( )byaxorbyaxyxf ++= cossin, , then

( )
( ) ( )byaxorbyax
DDf
IP ++
¢
= cossin
,
1
.
Replace ( )DDfinabDDandbDaD ¢-=¢-=¢-= ,,
2222
provided the
denominator is not equal to zero.
If the denominator is zero, refer to Rule (4).
Rule (3): If the R.H.S of a given PDE is( )
nm
yxyxf =, , then
( )
( ){ }( )
nm
nm
yxDDf
yx
DDf
IP
,,
,
1
.
1-
¢=
¢
=
Expand ( ){ }
1
,
-
¢DDf by using Binomial Theorem and then operate on
nm
yx.
Rule (4): If the R.H.S of a given PDE( )yxf, is any other function [other than Rule(1),
(2) and(3)] resolve ( )DDf ¢, into linear factors say ( )( )DmDDmD ¢-¢-
21
etc. then the

( )( )
( )yxf
DmDDmD
IP ,
1
.
21
¢-¢-
=
Note: If the denominator is zero in Rule (1) and (2) then apply Rule (4).
Working rule to find P.I when denominator is zero in Rule (1) and Rule (2).
If the R.H.S of a given PDE is of the form ( ) ( )
byax
eorbyaxorbyax
+
++ cossin
Then
( )
( ) ( )byaxf
nb
x
byaxf
DabD
IP
n
n
+=+
¢-
=
!
.
1
.
This rule can be applied only for equal roots.

Problems:
(1) Solve ( )
yxyx
eezDDDD
+-
+=¢+¢-
2323
23
Solution:
Given: ( )
yxyx
eezDDDD
+-
+=¢+¢-
2323
23
32

33. The auxiliary equation is 023
3
=+-mm
2,1,1-=Þm
( ) ( ) ( )xyfxyfxyxfFC 2.
321
-++++=

( )
( )( ) ( )( )
yxyx
yxyx
e
DDDD
e
DDDD
ee
DDDD
IP
+-
+-
¢+¢-
+
¢-¢+
=
+
¢+¢-
=
2
1
2
1
23
1
.
2
2
2
2
3223

( )
þ
ý
ü
î
í
ì
+=
¢-
+
¢+
=
+-
+-
yxyx
yxyx
e
x
xe
e
DD
e
DD
29
1
1
.
9
1
2
1
.
9
1
2
2
2
2
The general solution of the given equation is
( ) ( ) ( )
yxyx
e
x
e
x
xyfxyfxyxfz
+-
++-++++=
189
2
2
2
321
(2)Solve ( ) ( )yxxyzDDDD 32sin54
22
++=¢-¢+
Solution:
Given: ( ) ( )yxxyzDDDD 32sin54
22
++=¢-¢+
The auxiliary equation is 054
2
=-+mm
5,1-=Þm
( ) ( )xyxyFC ++-=
21
5. ff

( ) ()
( )
()
( ) ()
( ) ()xyDD
D
D
D
xyDD
D
D
D
xy
DD
D
D
D
xy
DDDD
IP
þ
ý
ü
î
í
ì
+¢-
¢
-=
þ
ý
ü
î
í
ì
¢-
¢
+=
þ
ý
ü
î
í
ì
¢-
¢
+
=
¢-¢+
=
-
...541
1
541
1
541
1
54
1
.
22
1
22
2
2
221
33

34. () ()
()
53
4
3
42
30
1
6
1
4
1
6
4.
11
xyx
x
D
yx
xyD
D
xy
D
-=
-=
¢-=

( ) ( )yx
DDDD
IP 32sin
54
1
.
222
+
¢-¢+
=

( ) ( ) ( )
( )
( )yx
yx
32sin
17
1
32sin
353.2.42
1
22
+=
+
---+-
=
Therefore the general solution is
( ) ( ) ( ) ( )yxxyxxyxyz 32sin
17
1
30
1
6
1
5
53
21
++-+++-= ff
(3)Solve ( )
yx
eyxzDDDD
+
=¢+¢-
2222
2
Solution:
Given:( )
yx
eyxzDDDD
+
=¢+¢-
2222
2
The auxiliary equation is 012
2
=+-mm
1,1=Þm
( ) ( )xyfxyxfFC +++=
21
.

( )
( )
( )( ){ }
22
2
22
2
11
1
1
.
yx
DD
e
yxe
DD
IP
yx
yx
+¢-+
=
¢-
=
+
+

( )
22
2
1
yx
DD
e
yx
¢-
=
+

( )
( )
22
2
2
2
22
2
2
3
2
1
1
1
1
yx
D
D
D
D
D
e
yx
D
D
D
e
yx
yx
÷
÷
ø
ö
ç
ç
è
æ ¢
+
¢
+=
÷
ø
ö
ç
è
æ ¢
-=
+
-
+
34

35. ( ) ( )
() () ()
yx
yx
yx
exyxyx
x
D
x
D
yx
D
ye
x
D
yx
D
yx
D
e
+
+
+
÷
ø
ö
ç
è
æ
++=
þ
ý
ü
î
í
ì
++=
þ
ý
ü
î
í
ì
++=
6524
2
4
2
3
2
2
2
2
2
222
2
60
1
15
1
12
1
1
.6
1
.4
1
2
3
2
21
Therefore the general solution is
( ) ( )
yx
exxxyyxyfxyxfz
+
÷
ø
ö
ç
è
æ
++++++=
422
21
60
1
15
1
12
1

35

36. UNIT 1
Part A
(1)Form a partial differential equation by eliminating arbitrary constants a and b from
( ) ( )
22
byaxz +++=
Ans:
Given ( ) ( )
22
byaxz +++= ……. (1)

( ) ()
( ) ()32
22


by
y
z
q
ax
x
z
p
+=
¶
¶
=
+=
¶
¶
=
Substituting (2) & (3) in (1), we get
44
22
qp
z +=
(2) Solve: ( )02
22
=+¢- zDDDD
Ans:
Auxiliary equation

( )
( ) ( )xyxfxyfz
m
m
mm
+++=
=
=-
=+-
21
2
2
1,1
01
012
(3)Form a partial differential equation by eliminating the arbitrary constants a and b from
the equation( ) ( ) a
2222
cotzbyax =-+- .
Ans:
Given: ( ) ( ) a
2222
cotzbyax =-+- ….. (1)
Partially differentiating with respect to ‘x’ and ‘y’ we get
( ) a
2
cot22 zpax=- ….. (2)
( ) a
2
cot22 zqby=- …... (3)
(2) a
2
cotzpax=-Þ …… (4)
(3) a
2
cotzqby=-Þ ….. (5)
Substituting (4) and (5) in (1) we get
aaa
22422422
cotcotcot zqzpz =+ .
a
222
tan=+Þ qp .
(4)Find the complete solution of the partial differential equation .04
22
=-+ pqqp
Ans:
Given:

.04
22
=-+ pqqp
…….. (1)
Let us assume that

cbyaxz ++=
……… (2)
36

37. be the solution of (1)
Partially differentiating with respect to ‘x’ and ‘y’ we get


b
y
z
q
a
x
z
p
=
¶
¶
=
=
¶
¶
=

…….. (3)
Substituting (3) in (1) we get
04
22
=-+ abba
From the above equation we get,

2
4164
222
babb
a
-±
=

2
4abba -±= ………. (4)
Substituting (5) in (2) we get
( ) cbyxabz ++-±=
2
41
(5)Find the PDE of all planes having equal intercepts on the x and y axis.
Ans:
The equation of such plane is


1=++
b
z
a
y
a
x
………. (1)
Partially differentiating (1) with respect to ‘x’ and ‘y’ we get


a
b
p
b
p
a
-=Þ=+ 0
1
……….. (2)

a
b
q
b
q
a
-=Þ=+ 0
1
……….. (3)
From (2) and (3), we get

qp=

(6)Find the solution of
222
zqypx =+ .
Ans:
The S.E is

222
z
dz
y
dy
x
dx
==

Taking first two members, we get
22
y
dy
x
dx
=
Integrating we get
1
11
c
yx
+
-
=
-
i.e 1
11
c
xy
u =
÷
÷
ø
ö
ç
ç
è
æ
-
Taking last two members, we get
37

38. 22
z
dz
y
dy
=
Integrating we get
2
11
c
zy
+
-
=
-
i.e 2
11
c
yz
v =
÷
÷
ø
ö
ç
ç
è
æ
-
The complete solution is

0
11
,
11
=
÷
÷
ø
ö
ç
ç
è
æ
--
yzxy
f
(7)Find the singular integral of the partial differential equation .
22
qpqypxz -++=
Ans:
The complete integral is

.
22
babyaxz -++=


2
02
2
02
y
bby
b
z
x
aax
a
z
=Þ=-=
¶
¶
-=Þ=+=
¶
¶
Therefore


zxy
yxyxyx
z
4
444422
22
222222
=-Þ
+-=-++-=

(8)Solve:
.
222
mqp =+
Ans:

222
mqpGiven =+
…….. (1)
Let us assume that

cbyaxz ++=
…… (2)
be the solution of (1)
Partially differentiating with respect to ‘x’ and ‘y’ we get


b
y
z
q
a
x
z
p
=
¶
¶
=
=
¶
¶
=

…….. (3)
Substituting (3) in (1) we get

222
mba =+
This is the required solution.
38

39. (9)Form a partial differential equation by eliminating the arbitrary constants a and b from

.
nn
byaxz +=
Ans:

.
nn
byaxzGiven +=
……… (1)
Partially differentiating with respect to ‘x’ and ‘y’ we get


1
1
1
1
.
.
-
-
-
-
=Þ=
¶
¶
=
=Þ=
¶
¶
=
n
n
n
n
ny
q
bnyb
y
z
q
nx
p
anxa
x
z
p

…….. (2)
Substituting (2) in (1) we get

n
n
n
n
y
ny
q
x
nx
p
z
11 --
+=
( )qypx
n
z +=
1
This is the required PDE.
(10)Solve:
( ).0
3223
=¢+¢+¢+ zDDDDDD
Ans:
Auxiliary equation

( )
( ) ( ) ( )xyfxxyxfxyfz
m
m
mmm
-+-+-=
---=
=+
=+++
3
2
21
3
23
1,1,1
01
01
(11)Form a partial differential equation by eliminating the arbitrary constants a and b
from

( )( ).
2222
byaxz ++=
Ans: Given ( )( )
2222
byaxz ++= ……. (1)

( ) ()
( ) ()3
2
2
2
2
2
2222
2222


y
q
axaxy
y
z
q
x
p
bybyx
x
z
p
=+Þ+=
¶
¶
=
=+Þ+=
¶
¶
=
Substituting (2) & (3) in (1), we get
x
p
y
q
z
2
.
2
=

xyzpq4=
(12)Solve:
( )01
2
=-¢+¢- zDDDD
Ans:
The given equation can be written as
39

40. ( )( )011 =+¢-- zDDD
We know that the C.F corresponding to the factors
( )( ) iszDmDDmD 0
2211
=-¢--¢- aa

( ) ( )xmyfexmyfez
xx
2211
21
+++=
aa
In our problem

1,0,1,1
2121
==-== mmaa
( ) ( )xyfexyfeFC
xx
+++=
-
21
.
( ) ( )xyfexyfez
xx
+++=
-
21
(13)Form a partial differential equation by eliminate the arbitrary function f from

.÷
ø
ö
ç
è
æ
=
z
xy
fz
Ans:
.: ÷
ø
ö
ç
è
æ
=
z
xy
fzGiven

)1..(..........
.
..
2
z
pxyzy
z
xy
fp
-
÷
ø
ö
ç
è
æ
¢=


)2..(..........
.
..
2
z
qxyzx
z
xy
fq
-
÷
ø
ö
ç
è
æ
¢=
From (1), we get

)3...(..........
2
xypzy
pz
z
xy
f
-
=÷
ø
ö
ç
è
æ
¢
Substituting (3) in(2), we get

2
2
.
.
z
pxyzy
xypzy
pz
p
-
-
=

(14)Solve:
( ).022
3223
=¢-¢-¢+ zDDDDDD
Ans: Auxiliary equation

( ) ( ) ( )xyfxyfxyfzisSolution
m
mmm
2
2,1,1
022
321
23
-+-++=
--=
=--+
(15)Obtain partial differential equation by eliminating arbitrary constants a and b from
( ) ( ) .1
222
=++++ zbyax
Ans:
Given ( ) ( ) 1
222
=+-+- zbyax ……. (1)
40

41. ( ) ( )
( ) ( ) ()3022
)2.........(022
zqbyzqby
zpaxzpax
-=-Þ=+-
-=-Þ=+-
Substituting (2) & (3) in (1), we get

2
22
22222
1
1
1
z
qp
zqzpz
=++
=++
(16)Find the general solution of

.09124
2
22
2
2
=
¶
¶
+
¶¶
¶
-
¶
¶
y
z
yx
z
y
z
Ans:
Auxiliary equation is
09124
2
=+-mm

8
14414412 -±
=m
2
3
,
2
3
=m
General solution is

÷
ø
ö
ç
è
æ
++÷
ø
ö
ç
è
æ
+= xyxfxyfz
2
3
2
3
21
(17)Find the complete integral of

.,,
y
z
q
x
z
pwherepqqp
¶
¶
=
¶
¶
==+
Ans: Let us assume that

cbyaxz ++=
……… (1)
be the solution of the given equation.
Partially differentiating with respect to ‘x’ and ‘y’ we get


b
y
z
q
a
x
z
p
=
¶
¶
=
=
¶
¶
=

…….. (2)
Substituting (2) in (1) we get

1-
=Þ=+
a
a
babba
Substituting the above in (1) we get

cy
a
a
axz +÷
ø
ö
ç
è
æ
-
+=
1

This gives the complete integral.
(18)Solve:
41

42. ( )023
323
=¢+¢- zDDDD
Ans:
Auxiliary equation

( ) ( ) ( )xyfxyxfxyfzisSolution
m
mm
2
2,1,1
023
321
3
-++++=
-=
=+-
(19)Find the PDE of the family of spheres having their centers on the line x=y=z.
Ans: The equation of such sphere is
( ) ( ) ( )
2222
razayax =-+-+-

Partially differentiating with respect to ‘x’ and ‘y’ we get
( ) ( ) )1.(..........022 =-+- pazax

( ) ( ) )2.(..........022 =-+- qazay

From (1),

)3.(..........
1p
zpx
a
+
+
=

From (2),

)4.(..........
1q
zqy
a
+
+
=

From (3) and (4), we get

q
zqy
p
zpx
+
+
=
+
+
11

This is the required PDE.
(20)Solve:

.0842
3
3
2
3
2
2
3
3
=
¶
¶
+
¶¶
¶
-
¶¶
¶
-
¶
¶
y
z
yx
z
yx
z
x
z
Ans:
Auxiliary equation

( ) ( ) ( )xyfxyxfxyfzisSolution
m
mmm
222
2,2,2
0842
321
23
-++++=
-=
=+--
Part B
(1)(i) Form a partial differential equation by eliminating arbitrary functions from
( ) ( )yxgyxxfz +++= 22
(ii) Solve: ( )
222
1 qxxyp =+
(2)(i) Solve: ( ) ( ) ( )
222222
xyzqzxypyzx -=-+-
42

43. (ii) Solve: ( )yxe
yx
z
x
z
yx
++=
¶¶
¶
-
¶
¶
+
sin42
2
2
3
3
3
(3)(i) Solve: ( ) 022
222
=-+-- xzxyqpzyx
(ii) Solve: ( )
yx
eyxzDDDD
4322
322
+
++=¢-¢-
(4)(i) Solve: ( )
22222
yxqpz +=+
(ii) Solve: ( ) yzDDDD sin43
22
=¢-¢+
(5)(i) Solve: ( ) ( ) ( )yxzqxzyxpyzyx +=-++++
2222
(ii) Solve: ( ) ( )yxezDDDD
yx
-+=¢-¢-
+
4sin20
522
(6)(i) Solve:
22
qpz +=
(ii) Solve: ( )
yx
eyxzDDDD
+
+=¢-¢+
3222
6
(7)(i) Solve: ( ) 0
22
=+-+ xzxyqpzy
(ii) Solve: ( ) yexyzDDDD
x
sinh56
22
+=¢+¢-
(8)(i) Solve: ( ) ( )zqqp -=- 11
2
(ii) Solve: ( )
yx
exyzDDDD
+
+=¢+¢-
222
44
(9)(i) Solve: ( ) ( ) xyzqzxypyzx -=-+-
222
(ii) Solve: ( ) ( ) ( )yxzqxzypzyx -=-+-
(10)(i) Solve:
222222
zxqyxp =+
(ii)Solve: ( )xyzDDDD =¢+-¢- 33
22
(11)(i) Form the partial differential equation by eliminating fandf from
() ( )zyxyfz +++= f
(ii) Solve: ( )
x
eyxzDDD
232
2 +=¢-¢
(12)(i) Find the complete integral of yxqp +=+
(ii) Solve: ( )yzxxyqpy 2
2
-=-
(13)(i) Solve: ( ) ( ) xyqzxpyz 322443 -=-+-
(ii) Solve: ( ) ( )
2
2322
22332
yx
eezDDDDDD
-
+=+¢++¢-¢-
(iii) Solve: ( ) ( )
yx
eyxzDDDD
-
+-=¢-¢+
222
2sin54
(14)(i) Solve
222
1 qpz ++=
(ii) Solve: ( ) ( ) zxqyxpzy +=+-- 22
43

44. (15)(i) Form a partial differential equation by eliminating arbitrary functions f and g in
() ()xgyyfxz
22
+=
(ii) Solve: ( ) ( )
yx
eyxzDDDD
+
+-=¢-¢-
522
4sin20
(16)(i) Form a partial differential equation by eliminating arbitrary functions f and g in
( ) ( )yxgyxfz 22
33
-++=
(ii) Solve: ( ) ( ) ( )( )yxyxqxyzpxyy -+=-+-
(17)(i) Find the singular solution of 1
22
++++= qpqypxz
(ii) Solve: ( )
yx
exyzDDDD
+
+=¢-¢-
622
30
(18) (i) Solve: ( ) ( )
yx
eyxzDDDD
+
++=¢-¢-
2323
2sin67
(ii) Find the singular integral of a partial differential equation
22
qpqypxz -++=
(19)(i) Solve: ( ) ( )yxzDDDD 2log1644
22
+=¢+¢-
(ii) Form a partial differential equation by eliminating arbitrary functions ‘f’ from
( )0,
22
=+- yxxyzf
(20)(i) Solve: ( ) ( )yxyxzDD ++=¢+
222
sin23sin2sin
(ii) Solve:
222222
zxqyxp =+
44