Chapter 1 (maths 3)

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 z = f ( x, y ) .
We will use the following standard notations to denote the partial derivatives.
∂z ∂z ∂2z ∂2z ∂2z
= p, = q, 2 = r , = s, 2 = t
∂x ∂y ∂x ∂x∂y ∂y
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 z = ax + by + a 2 + b 2 .
Solution:

1

Given z = ax + by + a 2 + b 2 ……………... (1)
Here we have two arbitrary constants a & b.
Differentiating equation (1) partially with respect to x and y respectively we get
∂z
=a⇒ p=a ……………… (2)
∂x
∂z
=b⇒q=a ………………. (3)
∂y
Substitute (2) and (3) in (1) we get
z = px + qy + p 2 + q 2 , which is the required partial differential equation.

(2) Form the partial differential equation by eliminating the arbitrary constants

x2 y2 z 2
a, b, c from + + = 1.
a2 b2 c2
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.
x2 y2 z 2
+ + =1 .................. (1)
a2 b2 c2
Differentiating (1) partially with respect to x and then with respect to y, we get

2x 2 y
+ p=0 .....................(2)
a2 c2
2 y 2z
+ q=0 ......................(3)
b2 c2
Differentiating (2) partially with respect to x,
1 1
2
+ 2 ( zr + p 2 ) ……………..(4)
a c
∂2z
Where r = 2 .
∂x
c 2 zp
− 2 = ......................(5)
a x
From (2) and (4) , 2
.
c
− 2 = zr + p 2 ......................(6)
a

2

From (5) and (6), we get
2
∂2z  ∂z  ∂z
xz 2 + x  = z , which is the required partial differential
∂x  ∂x  ∂x
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 ( 0, a, b ) , that lies on the yoz -plane
and having its radius equal to c is
x 2 + ( y − a ) 2 + ( z − b) 2 = c 2 ……………. (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
2 x + 2( z − b ) p = 0 …………… (2)
and 2( y − a ) + 2( z − b ) q = 0 …………….(3)
x
From (2), z −b = − …………….(4)
p
qx
Using (4) in (3), y − b = ……………..(5)
p
Using (4) and (5) in (1), we get
q2x2 x2
x + 2 + 2 = c2
2

p p

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

Problems

3

Formation of partial differential equation by elimination of arbitrary
functions:
(1)Form the partial differential equation by eliminating the arbitrary function ‘f’
from
z = e ay f ( x + by )

solution: Given z = e ay f ( x + by )

i.e. e − ay z = f ( x + by ) ……………(1)
Differentiating (1) partially with respect to x and then with respect to y, we
get
e − ay p = f ' ( u ).1 ………….(2)

e − ay q − ae − ay z = f ' ( u ).b …………….(3)
where u = x + by
Eliminating f’(u) from (2) and (3), we get
q − az
=b
p
i.e. q = az + bp

(2) Form the partial differential equation by eliminating the arbitrary function ‘ φ ’
 x
φ  z 2 − xy,  = 0
 z

 x
Solution: Given φ  z 2 − xy,  = 0 ………………(1)
 z
x
Let u = z 2 − xy , v=
z
Then the given equation is of the form φ ( u , v ) = 0 .
The elimination of φ from equation (2), we get,
∂u ∂v
∂x ∂x = 0
∂u ∂v
∂y ∂y

4

z − px
2 zp − y
i.e. z2 = 0
− xq
2 zq − x
z2

i.e ( 2 zp − y )  − xq  −  z − 2px ( 2 zq − x ) = 0
 2   
 z   z 
i.e px 2 − q ( xy − 2 z 2 ) = zx

(3) Form the partial differential equation by eliminating the arbitrary function ‘f’
from
z = f ( 2 x + y ) + g ( 3x − y )

Solution: Given z = f ( 2 x + y ) + g ( 3x − y ) .…………(1)
Differentiating (1) partially with respect to x,
p = f ′( u ).2 + g ′( v ).3 ………….(2)
Where u = 2 x + y and v = 3 x − y
Differentiating (1) partially with respect to y,
q = f ′( u ).1 + g ′( v ) (−1) …………. (3)
Differentiating (2) partially with respect to x and then with respect to y,
r = f ′′( u ).4 + g ′′( v ).9 …………. (4)
and s = f ′′( u ).2 + g ′′( v ).(−3) ………….. (5)
t = f ′′( u ).1 + g ′′( v ).1 ………….. (6)
Eliminating f ′′( u ) and g ′′( v ) from (4), (5) and (6) using determinants, we
have
4 9 r
2 −3 s = 0
1 1 t

i.e. 5r + 5s − 30t = 0
∂2z ∂2z ∂2z
or + −6 2 = 0
∂x 2 ∂x∂y ∂y

5

(4) Form the partial differential equation by eliminating the arbitrary function ‘ φ ’
from
1
z = φ ( y − x ) + φ ′( y − x ) .
x
1
Solution: Given z = φ ( u ) + φ ′( u ) …………...(1)
x
Where u = y − x
Differentiating partially with respect to x and y, we get
1 1
p = φ ′( u ) (−1) − 2 φ ( u ) + φ ′′( u ) (−1) …………(2)
x x
1
q = φ ′( u ).1 + φ ′′( u ).1 ………….(3)
x
1 2 2
r= φ ′′( u ).1 + 2 φ ′( u ) + 3 φ ( u ) + φ ′′′( u ).1 ………..(4)
x x x
1 1
s=− 2
φ ′( u ) − φ ′′( u ) + φ ′′′( u ) (−1) …………(5)
x x
1
t = φ ′′( u ).1 + φ ′′′( u ).1 …………..(6)
x
2 2
r −t = 2
φ ′( u ) + 3 φ ( u )
x x
2 1 
=  φ ( u ) + φ ′( u ) 
x2 x 
2
= z
x2
 ∂ 2 z ∂ 2z 
i.e. x2  2 − 2
 ∂x  = 2z

 ∂y 

Solutions of partial differential equations
Consider the following two equations

z = ax + by ………..(1)

6

 y
and z = xf   ………..(2)
x
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 xp + yq = z . If we eliminate the arbitrary function f from (2) we get
a partial differential equation of the form xp + yq = z .
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 f ( x, y, z, p, q ) = 0 ………....(1)
is a first order partial differential equation whose complete solution is
φ ( x, y , z , a , b ) = 0 ………..(2)
Where a and b are arbitrary constants.
Let b = f ( a ) , where ‘f’is an arbitrary function.
Then (2) becomes
φ ( x, y , z , a , f ( a ) ) = 0 ……….(3)
Differentiating (3) partially with respect to ‘a’, we get

∂φ ∂φ
+ . f ′( a ) = 0 ……….(4)
∂a ∂b

7

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 f ( p, q ) = 0 ………...(1)
i.e the equation contains p and q only.
Suppose that z = ax + by + c …….....(2)
is a solution of the equation

∂z ∂z
= a, = b
∂x ∂y
⇒ p = a, q = b
substitute the above in (1), we get
f ( a, b ) = 0

on solving this we can get b = φ ( a ) , where φ is a known function.
Using this value of b in (2), the complete solution of the given partial differential
equation is
z = ax + φ ( a ) y + c …………(3)
is a complete solution,
To find the singular solution, we have to eliminate ‘a’ and ‘c’ from
z = ax + φ ( a ) y + c
Differentiating the above with respect to ‘a’ and ‘c’, we get
0 = x + φ ′( a ) y ,
and 0=1.
The last equation is absurd. Hence there is no singular solution for the equation of
Type 1.

Problems:

(1) Solve p2 + q2 = 1.

8

Solution:
Given: p 2 + q 2 = 1 ………….(1)
Equation (1) is of the form f ( p, q ) = 0 .
Assume z = ax + by + c ………….(2)
be the solution of equation (1).
From (2) we get p = a, q = b .
(1) ⇒ a 2 + b 2 = 1

⇒ b = ± 1− a2 ……….(3)
Substitute (3) in (2) we get

z = ax ± 1 − a 2 y + c ……....(4)

This is a complete solution.

To find the general solution:
We put c = f ( a ) in (4),where ‘f’ is an arbitrary function.

i.e. z = ax ± 1 − a 2 y + f ( a ) …………(5)


a
x y + f ′( a ) = 0 ……………(6)
1− a2
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
a
0= x± ,
1− a2
0=1.(which is absurd)
so there is no singular solution.

(2) Solve p + q = pq

9

Solution:
Given: p + q = pq ………..(1)
Equation (1) is of the form f ( p, q ) = 0
Assume z = ax + by + c ………(2)
be the solution of equation (1).
From (2) we get p = a, q = b
(1) ⇒ a + b = ab
a
⇒b= …….....(3)
a −1
Substituting (3) in (2), we get
a
z = ax + y+c ………(4)
a −1
This is a complete solution.

We put c = f ( a ) in (4), we get
a
z = ax + y + f ( a) ……..(5)
a −1

1
x− y + f ′( a ) ………..(6)
( a − 1) 2
Eliminating ‘a’ between equations (5) and (6), we get the required general solution

Differentiating (4) with respect to ‘a’ and ‘c’.
1
0= x− y,
( a − 1) 2
and 0=1 (which is absurd).
So there is no singular solution.

10

Type 2: (Clairaut’s type)
The equation of the form
z = px + qy + f ( p, q ) ………(1)
is known as Clairaut’s equation.

Assume z = ax + by + c …………(2)
be a solution of (1).
∂z ∂z
= a, = b
∂x ∂y
⇒ p = a, q = b
Substitute the above in (1), we get
z = ax + by + f (a, b) ………..(3)
which is the complete solution.

Problem:

(1) Solve z = px + qy + pq

Solution:
Given: z = px + qy + pq ……….(1)
Equation (1) is a Clairaut’s equation
Let z = ax + by + c ………..(2)
be the solution of (1).
Put p = a, q = b in (1), we get

z = ax + by + ab ……….(3)
which is a complete solution.

We put b = f ( a ) in (3), we get

z = ax + f ( a ) y + af ( a ) ………(4)
Differentiate (4) partially with respect to ‘a’, we get

11

1
0 = x + f ′( a ) y + [ af ′( a ) + f ( a ) ] ………..(5)
2 af ( a )

Eliminating ‘a’ between equations (4) and (5), we get the required general solution

To find singular solution,
1
0= x+ .b
2 ab

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

a
⇒ y=− ………..(7)
2 b
Multiplying equation (6) and (7),we get
 b  a 
xy =  −  
 2 a  − 2 b 
  
1
xy =
4
4 xy = 1
q
(2) Solve z = px + qy + −p
p
Solution:
q
Given: z = px + qy + −p ……….(1)
p
Equation (1) is a Clairaut’s equation
Let z = ax + by + c ………...(2)
be the solution of (1).
Put p = a, q = b in (1), we get
b
z = ax + by + −a ……….(3)
a

12

which is the complete solution.

We put b = f ( a ) in (3), we get
f ( a)
z = ax + f ( a ) y + −a ……..(4)
a
af ′( a ) − f ( a )
0 = x + f ′( a ) y + −1 ……..(5)
a
Eliminating ‘a’ between equations (4) and (5), we get the required general
solution

Differentiate (3) partially with respect to ‘a’,
b
0= x− −1
a2
b
⇒x= +1
a2
⇒ a2x = b +1
⇒ b = a2 x −1 .............. (4)
Differentiate (3) partially with respect to ‘b’,
1
0= y+
a
1
⇒a=− ………….(5)
y
Substituting equation (4) and (5) in equation (3), we get

 −1
z =   x + ( a 2 x − 1) y +
( a 2 x − 1) −  − 1 
 
 y  −1  y 
   
y
−x 1
z= + a 2 yx − y − a 2 xy + y +
y y

13

−x 1
z= +
y y
yz = 1 − x

Type 3:
Equations not containing x and y explicitly, i.e. equations of the form
f ( z , p, q ) = 0 ……….(1)
For equations of this type ,it is known that a solution will be of the form
z = φ ( x + ay ) ……….(2)
Where ‘a’ is the arbitrary constant and φ is a specific function to be found out.
Putting x + ay = u , (2) becomes z = φ ( u ) or z ( u )
dz ∂u dz
∴ p= . =
du ∂x du
dz ∂u dz
and q= . =a
du ∂y du
If (2) is to be a solution of (1), the values of p and q obtained should satisfy (1).
 dz dz 
i.e. f  z, , a  = 0 ……..(3)
 du du 
From (3), we get
dz
= ψ ( z, a ) ……….(4)
du
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 g ( z, a ) = u + b or g ( z, a ) x + ay + b , is the
complete solution of (1).
The general and singular solution of (1) can be found out by usual method.

Problems:

(1)Solve z = p 2 + q 2 .
Solution:

14

Given: z = p 2 + q 2 …………(1)

Equation (1) is of the form f ( z , p, q ) = 0
Assume z = φ ( u ) where, u = x + ay be a solution of (1).
∂z dz ∂u dz dz
p= = . = ⇒ p= …….(2)
∂x du ∂x du du
∂z dz ∂u dz dz
q= = . =a ⇒q=a ……(3)
∂y du ∂y du du
Substituting equation (2) & (3) in (1), we get

2 2
 dz   dz 
z =   + a2 
 du   du 
2

z =   (1 + a 2 )
 dz 
 du 
2
 dz  z
  =
 du  1+ a2
1
dz z2
=
du (1 + a 2 )
By variable separable method,
dz du
=
(1 + a )
1 1
2 2
z 2

By integrating, we get
dz 1
∫ = ∫ du + c
(1 + a )
1 1
2 3
z 2

2 z=
( x + ay ) +c
(1 + a )
1
2 2

z=
( x + ay ) +
c

2(1 + a )
1
2 2 2

z=
( x + ay ) +k ……….(4)
2(1 + a )
1
2 2

15

This is the complete solution.

We put k = f ( a ) in (4), we get

z=
( x + ay ) + f ( a) ……..(5)
2(1 + a )
1
2 2

y − xa
0= + f ′( a ) ………..(6)
2(1 + a )
3
2 2

solution.

Differentiate (4) partially with respect to ‘a’ and ‘k’, we get
y − xa
0= ………..(7)
2(1 + a )
3
2 2

and 0 = 1 (which is absurd)

(2)Solve 9( p 2 z + q 2 ) = 4 .
Solution:
Given: 9( p 2 z + q 2 ) = 4 ………(1)

Equation (1) is of the form f ( z , p, q ) = 0
Assume z = φ ( u ) where , u = x + ay be a solution of (1).
∂z dz ∂u dz dz
p= = . = ⇒ p= …….(2)
∂x du ∂x du du

∂z dz ∂u dz dz
q= = . =a ⇒q=a ……(3)
∂y du ∂y du du

16

Substituting equation (2) & (3) in (1), we get
  dz  2 2
2  dz  

9   z + a   = 4
  du   du  
 
2
 dz 
[
9  z + a 2 = 4 ]
 du 
2
 dz  4
  =
 du  9 z + a2 [ ]
dz 2 1
=± .
du 3 z + a2

2
z + a 2 dz = ± du
3
Integrating the above, we get

∫ (z + a )
1
2
3∫
2 2
dz = ± du + c

(z + a2 ) 2 = ± 2 u + c
3
2
3 3

( z + a 2 ) 2 = ± 2 ( x + ay ) + c
3
2
3 3

⇒ ( z + a 2 ) 2 = ±( x + ay ) + k
3
………..(4)


We put k = f ( a ) in (4), we get

(z + a )
3
2 2
= ±( x + ay ) + f ( a ) ……..(5)


− 3a ( z − a 2 ) 2 = ± y + f ′( a )
1
……..(6)

solution.
Differentiate (4) partially with respect to ‘a’ and ‘k’, we get

17

3a ( z + a 2 ) 2 = ± y
1
………..(7)

and 0 = 1 (which is absurd)

Type 4:
Equations of the form
f ( x, p ) = g ( y , q ) ………..(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 f ( x, p ) = g ( y , q ) = a .where ‘a’ is an arbitrary constant.
Solving f ( x, p ) = a ,we can get p = φ ( x, a ) and solving g ( y , q ) = a ,we can get

q = ψ ( y, a ) .
Now
∂z ∂z
dz = dx + dy or pdx + qdy
∂x ∂y

i.e. dz = φ ( x, a ) dx + ψ ( y, a ) dy
Integrating with respect to the concerned variables, we get
z = ∫ φ ( x, a )dx + ∫ψ ( y, a ) dy + b ……….(2)

The complete solution of (1) is given by (2), which contains two arbitrary constants ‘a’
and ‘b’.

Problems:
(1)Solve pq = xy .
Solution:
Given: pq = xy
p y
⇒ = ………..(1)
x q

Equation (1) is of the form f ( x, p ) = g ( y , q )

18

p y
Let = = a (say)
x q
p
=a ⇒ p = ax ………….(2)
x
y y
Similarly, =a ⇒q= …………(3)
q a
Assume dz = pdx + qdy be a solution of (1)
Substitute equation (2) and (3) to the above, we get
y
dz = axdx + dy
a
Integrating the above we get,
1
∫ dz = a ∫ xdx + a ∫ ydy + c
ax 2 y 2
z= + +c
2 2a
y2
2 z = ax +2
+k ………..(4)
a

(2) Solve p + q = x + y .
Solution:
Given: p + q = x + y
⇒p−x =q− y ………….. (1)
Equation (1) is of the form f ( x, p ) = g ( y , q )
Let p − x = y − q = a (say)
p−x =a ⇒ p = x+a …………. (2)
Similarly, y − q = a ⇒ q = y−a ……………(3)
Assume dz = pdx + qdy be a solution of (1)
Substitute equation (2) and (3) to the above, we get
dz = ( x + a ) dx + ( y − a ) dy
Integrating the above we get,

19

∫ dz = ∫ ( x + a ) dx + ∫ ( y − a ) dy + c
z=
(x 2
+ y2 )
+ a( x − y ) + c
2
2 z = x 2 + y 2 + 2a ( x − y ) + c ……………(4)

Equations reducible to standard types-transformations:
Type A:
Equations of the form f ( x m p, y n q ) = 0 or f ( x m p, y n q, z ) = 0 .
Where m and n are constants, each not equal to 1.
We make the transformations x 1− m = X and y 1− n = Y .
∂z ∂z ∂X ∂z
Then p= = . = (1 − m ) x − m P, where P ≡ and
∂x ∂X ∂x ∂X

∂z ∂z ∂Y ∂z
q= = . = (1 − n ) y −n Q, where Q ≡
∂y ∂Y ∂y ∂Y

Therefore the equation f ( x m p, y n q ) = 0 reduces to f { (1 − m ) P, (1 − n ) Q} = 0, .which is a
type 1 equation.
The equation f ( x m p, y n q, z ) = 0 reduces to f { (1 − m ) P, (1 − n ) Q, z} = 0, .which is a type 3
equation.

Problem:
(1)Solve p 2 x 4 + y 2 zq = 2z 2 .
Solution:
Given: p 2 x 4 + y 2 zq = 2z 2
This can be written as

( px ) + ( qy ) z = 2z .
2 2 2 2

Which is of the form f ( x p, y q, z ) = 0 , where m=2,n=2.
m n

20

1− m 1 1
Put X = x = ; Y = y 1− n =
x y
∂z ∂z ∂x
P= = . = p ( − x 2 ) = − px 2
∂X ∂x ∂X
∂z ∂z ∂y
Q= = . = q ( − y 2 ) = −qy 2
∂Y ∂y ∂Y
Substituting in the given equation,
P 2 − qz = 2z 2 .

This is of the form f ( p, q, z ) = 0 .
Let Z = f ( X + aY ) , where u = X + aY
dz dz
P= ,Q=a
du du
Equation becomes, .
2
 dz  dz
  − az − 2z 2 = 0
 du  du
dz dz az ± a 2 z 2 + 8 z 2
Solving for , =
du du 2
dz a ± a 2 + 8
= du
z 2

a ± a2 + 8
log z = ( X + aY ) + b
2

a ± a2 + 8  1 a 
log z =  +  + b is a complete solution.
x y
2  
The general and singular solution can be found out by usual method.

Type B:
Equations of the form f ( z k p, z k q ) = 0 or f ( z k p , z k q, x, y ) = 0 .
Where k is a constant, which is not equal to -1.
We make the transformations Z = z k +1 .
∂Z
Then P= = ( k + 1) z k p and
∂x

21

∂Z
Q= = ( k + 1) z k p
∂y

 P Q 
Therefore the equation f ( z k p, z k q ) = 0 reduces to f  ,  = 0, which is a type 1
 k +1 k + 1
equation.
 P 
f ( z k p, z k q, x, y ) = 0 reduces to f 
Q
The equation , , x, y  = 0, which is a
 k +1 k +1 
type 4 equation.

Problems:
(1)Solve: z 4 q 2 − z 2 p = 1.
Solution:
Given: z 4 q 2 − z 2 p = 1.

The equation can be rewritten as ( z 2 q ) − ( z 2 p ) = 1
2
………(1)

Which contains z 2 p and z 2 q .
Hence we make the transformation Z = z 3
∂Z
P= = 3z 2 p
∂x
P
⇒ z2 p =
3
Q
Similarly z q =
2

3
Using these values in (1), we get
Q 2 − 3P = 9 ………..(2)
As (2) is an equation containing P and Q only, a solution of (2) will be of the form
Z = ax + by + c ………….(3)
Now P = a and Q = b, obtained from (3) satisfy equation (2)
b 2 − 3a = 9
i.e. b = ± 3a + 9

Therefore the complete solution of (2) is Z = ax ± 3a + 9 y + c

i.e complete solution of (1) is z 3 = ax ± 3a + 9 y + c

22

Singular solution does not exist. General solution is found out as usual.
Type C:
Equations of the form f ( x m z k p, y n z k q ) = 0 , where m, n ≠ 1; k ≠ −1

We make the transformations
X = x 1− m , Y = y 1−n and Z = z k +1
∂Z dZ ∂z dx
Then P= = . .
∂X dz ∂x dX
xm
= ( k + 1) z p.
k

1− m
yn
and Q = ( k + 1) z q.
k

1− n
Therefore the given equation reduces to
 1 − m   1 − n  
f   P,  Q  = 0
 k + 1   k + 1  
This is of type 1 equation.

Problem:
 p2 q2 
z2 2 + 2
(1)Solve   =1

x y 
Solution:

2 p q2 
2

Given: z  2 + 2  = 1
x
 y 

It can be rewritten as (x −1
zp ) + ( y −1 zq ) = 1
2 2
………..(1)

which is of the form ( x m z k p ) + ( y n z k q ) = 1
2 2

we make the transformations
X = x 1− m , Y = y 1−n and Z = z k +1

i.e. X = x 2 , Y = y 2 and Z = z 2
∂z dz ∂Z dX 1
Then p= = . . = .P.2 x
∂x dZ ∂X dx 2 z
⇒ P = x −1 zp ,

23

Similarly, Q = y −1 zq ,
Using these in (1),it becomes
P2 + Q2 = 1 …………(2)
As (2) contains only P and Q explicitly, a solution of the equation will be of the
form
Z = aX + bY + c ………….(3)
Therefore P = a and Q = b, obtained from (3) satisfy equation (2)
a 2 + b 2 = 1,
i.e.
⇒b = ± 1− a2
Therefore the complete solution of (2) is

Z = aX ± 1 − a 2 Y + c
Therefore the complete solution of (1) is

z 2 = ax 2 ± 1 − a 2 y 2 + c
Singular solution does not exist. General solution is found out as usual.

Type D:
 px qy 
Equation of the form f  ,  = 0
 z z 
By putting X = log x, Y = log y and Z = log z the equation reduces to

f ( P, Q ) = 0,
∂Z ∂Z
where P = and Q = .
∂X ∂Y

Problems:
(1)Solve pqxy = z 2 .
Solution:
Given: pqxy = z 2 . …………….(1)
Rewriting (1),
 px  qy 
   = 1 . …………….(2)
 z  z 

24

 px   qy 
As (2) contains   and   , we make the substitutions
 z   z 
X = log x, Y = log y and Z = log z
∂z dz ∂Z dX 1
Then P= = . . = z.P.
∂x dZ ∂X dx x
px
i.e. =P
z
qx
Similarly, =Q
z
Using these in (2), it becomes
PQ = 1 …………..(3)
which contains only P and Q explicitly. A solution of (3) is of the form
Z = aX + bY + c …………(4)
Therefore P = a and Q = b, obtained from (4) satisfy equation (3)
1
i.e. ab = 1 or b =
a
1
Therefore the complete solution of (3) is Z = aX + Y + c
a
1
Therefore the complete solution of (1) is log z = a log x + log y + c ……..(5)
a
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

∂ 2u ∂u
(1)Solve the equation = e −t cos x, if u = 0 when t = 0 and = 0 when x = 0.
∂x∂t ∂t
Also show that u → sin x, when t → ∞ .
Solution:
∂ 2u
Given: = e −t cos x, …………….(1)
∂x∂t
Integrating (1) partially with respect to x,
∂u
= e −t sin x + f ( t ) …………….(2)
∂t
∂u
When t = 0 and = 0 in (2), we get f ( t ) = 0 .
∂t
∂u
Equation (2) becomes = e −t sin x …………….(3)
∂t
Integrating (3) partially with respect to t, we get
u = −e −t sin x + g ( x ) ……………(4)
Using the given condition, namely u = 0 when t = 0 , we get
0 = − sin x + g ( x ) or g ( x ) = sin x
Using this value in (4), the required particular solution of (1) is
u = sin x (1 − e − t )

Now t →∞
{
lim( u ) = sin x lim(1 − e −t )
t →∞
}
= sin x
i.e. when t → ∞, u → sin x .

∂z ∂z
(2) Solve the equation = 3 x − y and = − x + cos y simultaneously.
∂x ∂y
Solution: Given
∂z
= 3x − y ................(1)
∂x
∂z
= − x + cos y ..................(2)
∂y
Integrating (1) partially with respect to x,

26

3x 2
z= − yx + f ( y ) …………..(3)
2
∂z
= − x + f ′( y ) …………...(4)
∂y
Comparing (2) and (4), we get
f ′( y ) = cos y
f ( y ) = sin y + c ...................(5)
Therefore the required solution is
3x 2
z= − yx + sin y + c , where c is an arbitrary constant.
2

Lagrange’s linear equation of the first order:
A linear partial differential equation of the first order , which is of the form
Pp + Qq = R

where P, Q, R are functions of x, y , z is called Lagrange’s linear equation.
working rule to solve Pp + Qq = R
(1)To solve Pp + Qq = R , we form the corresponding subsidiary simultaneous equations
dx dy dz
= = .
P Q R
(2)Solving these equations, we get two independent solutions u = a and v = b .
(3)Then the required general solution is f ( u , v ) = 0 or u = φ ( v ) or v = ψ ( u ) .
dx dy dz
Solution of the simultaneous equations = = .
P Q R
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

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 lP + mQ + nR = 0 , then the solution of the
simultaneous equation is found out as follows.
dx dy dz ldx + mdy + ndz
= = =
P Q R lP + mQ + nR
Since lP + mQ + nR = 0, ldx + mdy + ndz = 0. If ldx + mdy + ndz = 0 is an exact

differential of some function u ( x, y , z ) , then we get du = 0. Integrating this, we get
dx dy dz
u = a , which is a solution of = = .
P Q R
Similarly, if we can find another set of independent multipliers l ′, m′, n ′, we can get
another independent solution v = b .

Problems:
(1)Solve xp + yq = x .
Solution:
Given: xp + yq = x .
This is of Lagrange’s type of PDE where P = x, Q = y, R = x .
dx dy dz
The subsidiary equations are = = .
x y x
dx dy
Taking first two members =
x y

Integrating we get log x = log y + log c1
x
= c1
y
i.e. ………….(1)
x
u=
y
dx dz
Taking first and last members = .
x x
i.e. dx = dz .
x − z = c2
Integrating we get ………......(2)
v = x−z

28

x 
Therefore the solution of the given PDE is φ ( u , v ) = 0 i.e. φ  , x − z  = 0 .
y 
 
(2)Solve the equation ( x − 2 z ) p + ( 2 z − y ) q = y − x.
Solution:
Given: ( x − 2 z ) p + ( 2 z − y ) q = y − x.
This is of Lagrange’s type of PDE where P = x − 2 z , Q = 2 z − y, R = y − x .
dx dy dz
The subsidiary equations are = = . ……….(1)
x − 2z 2z − y y − x
dx + dy + dz
Using the multipliers 1,1,1, each ratio in (1)= .
0
dx + dy + dz = 0 .
Integrating, we get x+ y+z =a ……………(2)
ydx + xdy + 2zdz
Using the multipliers y,x,2z, each ratio in (1)= .
0
d ( xy ) + 2 zdz = 0 .

Integrating, we get xy + z 2 = b ……………(3)

Therefore the general solution of the given equation is f ( x + y + z, xy + z 2 ) = 0 .

(3)Show that the integral surface of the PDE x( y 2 + z ) p − y ( x 2 + z ) q = ( x 2 − y 2 ) z .

Which contains the straight line x + y = 0, z = 1 is x 2 + y 2 + 2 xyz − 2 z + 2 = 0 .
Solution:
The subsidiary equations of the given Lagrange ‘s equation are
dx dy dz
= = 2 .
x( y + z ) − y ( x + z ) ( x − y 2 ) z
2 2 ……………(1)

1 1 1
1 1 1 dx + dy + dz
Using the multipliers , , , . each ratio in (1)= x y z .
x y z
0
1 1 1
dx + dy + dz = 0 .
x y z
Integrating, we get xyz = a ……………(2)

29

xdx + ydy − dz
Using the multipliers y,x,-1, each ratio in (1)= .
0
xdx + ydy − dz = 0 .

Integrating, we get x 2 + y 2 − 2 z = b ……………(3)
The required surface has to pass through
x+ y =0
……………(4)
z =1
Using (4) in (2) and (3), we have
− x2 = a
……………(5)
2x 2 − 2 = b
Eliminating x in (5) we get,
2a + b + 2 = 0 …………….(6)
Substituting for a and b from (2) and (3) in (6), we get
x 2 + y 2 + 2 xyz − 2 z + 2 = 0 , 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
n th order with constant coefficients is
∂nz ∂nz ∂nz ∂n z
a 0 n + a1 n −1 + a 2 n −2 2 + ... + a n n = R ( x, y ) …………….(1)
∂x ∂x ∂y ∂x ∂y ∂y
where a’s are constants.
∂ ∂
If we use the operators D ≡ and D′ ≡ , we can symbolically write equation (1)
∂x ∂y
as
(a D
0
n
+ a1 D n −1 D ′ + ... + a n D ′ n ) z = R( x, y ) …………….(2)

f ( D, D ′) z = R ( x, y ) …………….(3)
where f ( D, D ′) is a homogeneous polynomial of the n th degree in D and D ′ .
The method of solving (3) is similar to that of solving ordinary linear differential
equations with constant coefficients.

30

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

1
f ( D, D ′) z = 0 and the particular integral is given symbolically by R ( x, y ) .
f ( D, D ′)

Complementary function of f ( D, D ′) z = R( x, y ) :
C.F of the solution of f ( D, D ′) z = R( x, y ) is the R.H.S of the solution of
f ( D, D ′) z = 0 . …………(1)
In this equation, we put D = m, D ′ = 1 ,then we get an equation which is called the
auxiliary equation.
Hence the auxiliary equation of (1) is
a 0 m n + a1 m n −1 + ... + a n = 0 ………….(2)

Let the roots of this equation be m1 , m2 ,...mn .

Case 1:
The roots of (2) are real and distinct.
The general solution is given by
z = φ1 ( y + m1 x ) + φ 2 ( y + m2 x ) + ... + φ n ( y + mn x )

Case 2:
Two of the roots of (2) are equal and others are distinct.
z = φ1 ( y + m1 x ) + φ 2 ( y + m1 x ) + ... + φ n ( y + mn x )

Case 3:
‘r’ of the roots of (2) are equal and others distinct.
z = φ1 ( y + m1 x ) + xφ 2 ( y + m1 x ) + ... + x r −1φ r ( y + mr x )
To find particular integral:
Rule (1): If the R.H.S of a given PDE is f ( x, y ) = e ax +by , then
1
P.I = e ax +by
f ( D, D ′ )
Put D = a, D ′ = b

31

1
P.I = e ax +by if f ( a, b ) ≠ 0
f ( a, b )

If f ( a, b ) = 0, refer to Rule (4).
Rule (2): If the R.H.S of a given PDE is f ( x, y ) = sin ( ax + by ) or cos( ax + by ) , then
1
P.I = sin ( ax + by ) or cos( ax + by )
f ( D, D ′ )

Replace D 2 = −a 2 , D ′ 2 = −b 2 and DD ′ = −ab in f ( D, D ′) 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 f ( x, y ) = x m y n , then
1
P.I = xm yn
f ( D, D ′ )
= { f ( D, D ′ ) }
−1
(x m
, yn )

Expand { f ( D, D ′)} −1 by using Binomial Theorem and then operate on x m y n .

Rule (4): If the R.H.S of a given PDE f ( x, y ) is any other function [other than Rule(1),

(2) and(3)] resolve f ( D, D ′) into linear factors say ( D − m1 D ′)( D − m2 D ′)
etc. then the
1
P.I = f ( x, y )
( D − m1 D ′)( D − m2 D ′)
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 sin ( ax + by ) or cos( ax + by ) or e ax +by

1 xn
Then P.I = . f ( ax + by ) = n f ( ax + by )
( bD − aD ′) b n!
This rule can be applied only for equal roots.

Problems:
(1) Solve ( D 3 − 3DD ′ 2 + 2 D ′ 3 ) z = e 2 x − y + e x + y
Solution:
Given: ( D 3 − 3DD ′ 2 + 2 D ′ 3 ) z = e 2 x − y + e x + y

32

The auxiliary equation is m 3 − 3m + 2 = 0
⇒ m = 1,1,−2
C.F = xf 1 ( y + x ) + f 2 ( y + x ) + f 3 ( y − 2 x )

P.I =
1
(e 2 x− y + e x+ y )
D − 3D D ′ + 2 D ′
3 22 3

1 1
= e 2 x− y + e x+ y
( D + 2 D ′)( D − D ′) 2
( D − D ′) ( D + 2 D ′ )
2

1 1 1 1
= . e 2 x− y + . e x+ y
9 D + 2D ′ 9 ( D − D ′) 2

1  2 x− y x 2 x+ y 
=  xe + e 
9 2 
The general solution of the given equation is
x 2 x− y x 2 x+ y
z = xf1 ( y + x ) + f 2 ( y + x ) + f 3 ( y − 2 x ) + e + e
9 18

(2)Solve ( D 2 + 4 DD ′ − 5 D ′ 2 ) z = xy + sin ( 2 x + 3 y )
Solution:
Given: ( D 2 + 4 DD ′ − 5 D ′ 2 ) z = xy + sin ( 2 x + 3 y )
The auxiliary equation is m 2 + 4m − 5 = 0
⇒ m = 1,−5
C.F = φ1 ( y − 5 x ) + φ 2 ( y + x )
1
( P.I ) 1 = ( xy )
D + 4 DD ′ − 5 D ′ 2
2

1
= ( xy )
2 D′ 
D 1 + 2 ( 4 D − 5 D ′) 
 D 
−1
1  D′ 
= 2 1 + 2 ( 4 D − 5 D ′)  ( xy )
D  D 
1  D′ 
= 2 1 − 2 ( 4 D − 5D ′) + ...( xy )
D  D 

33

1
= 2
( xy ) − 14 .4 D ′( xy )
D D
3
x y 1
= − 4 ( 4x)
6 D
1 3 1 5
= x y− x
6 30
1
( P.I ) 2 = sin ( 2 x + 3 y )
D + 4 DD ′ − 5 D ′ 2
2

1
= sin ( 2 x + 3 y )
( − 2) + 4.( − 2.3) − 5( − 3) 2
2

1
= sin ( 2 x + 3 y )
17
Therefore the general solution is

z = φ1 ( y − 5 x ) + φ 2 ( y + x ) +
6
( x y ) − 30 x 5 + 17 sin ( 2 x + 3 y )
1 3 1 1

(3)Solve ( D 2 − 2 DD ′ + D ′ 2 ) z = x 2 y 2 e x + y
Solution:
Given: ( D 2 − 2 DD ′ + D ′ 2 ) z = x 2 y 2 e x + y
The auxiliary equation is m 2 − 2m + 1 = 0
⇒ m = 1,1
C.F = xf 1 ( y + x ) + f 2 ( y + x )

e x+ y ( x 2 y 2 )
1
P.I =
( D − D ′) 2

1
= e x+ y x2 y2
{ ( D + 1) − ( D ′ + 1)} 2

1
= e x+ y x2 y2
( D − D ′) 2

−2
1  D′ 
1 −  ( x y )
x+ y
=e 2 2

D2  D
1  2 D′ D′ 2 
= e x + y 2 1 + + 3 2 ( x 2 y 2 )
D  D D  

34

x y + ( 2 x y ) + 2 ( 2 x )
1  2 2 2 3 2 
= e x+ y 2

D2  D D 

= e x + y  y 2 2 ( x 2 ) + 4 y. 3 ( x 2 ) + 6. 4 ( x 2 ) 
 1 1 1 
 D D D 
1 1 1 6  x+ y
=  x4 y 2 + x5 y + x e
 12 15 60 
Therefore the general solution is
1 1 1 2  4 x+ y
z = xf1 ( y + x ) + f 2 ( y + x ) +  y 2 + xy + x x e
 12 15 60 

35

UNIT 1
Part A
(1)Form a partial differential equation by eliminating arbitrary constants a and b from
z = ( x + a ) + ( y + b)
2 2

Ans:
Given z = ( x + a ) 2 + ( y + b ) 2 ……. (1)
∂z
p= = 2( x + a )  ( 2)
∂x
∂z
q= = 2( y + b )  ( 3)
∂y
p2 q2
Substituting (2) & (3) in (1), we get z = +
4 4
(2) Solve: ( D 2 − 2 DD ′ + D 2 ) z = 0
Ans:
Auxiliary equation
m 2 − 2m + 1 = 0
( m − 1) 2 = 0
m = 1,1
z = f 1 ( y + x ) + xf 2 ( y + x )
(3)Form a partial differential equation by eliminating the arbitrary constants a and b from
the equation ( x − a ) 2 + ( y − b ) 2 = z 2 cot 2 α .
Ans:
Given: ( x − a ) 2 + ( y − b ) 2 = z 2 cot 2 α ….. (1)
Partially differentiating with respect to ‘x’ and ‘y’ we get
2( x − a ) = 2 zp cot 2 α ….. (2)
2( y − b ) = 2 zq cot α 2
…... (3)
(2) ⇒ x − a = zp cot 2 α …… (4)
(3) ⇒ y − b = zq cot 2 α ….. (5)
Substituting (4) and (5) in (1) we get
z 2 p 2 cot 4 α + z 2 q 2 cot 4 α = z 2 cot 2 α .
⇒ p 2 + q 2 = tan 2 α .
(4)Find the complete solution of the partial differential equation p 2 + q 2 − 4 pq = 0.
Ans:
Given:
p 2 + q 2 − 4 pq = 0. …….. (1)
Let us assume that
z = ax + by + c ……… (2)

36

be the solution of (1)
∂z
p= =a
∂x …….. (3)
∂z
q= =b
∂y
Substituting (3) in (1) we get
a 2 + b 2 − 4ab = 0
From the above equation we get,
4b ± 16b 2 − 4a 2 b 2
a=
2
a = b ± b 4 − a2 ………. (4)
( )
z = b 1 ± 4 − a 2 x + by + c
(5)Find the PDE of all planes having equal intercepts on the x and y axis.
Ans:
The equation of such plane is
x y z
+ + =1 ………. (1)
a a b
Partially differentiating (1) with respect to ‘x’ and ‘y’ we get
1 p b
+ =0⇒ p =− ……….. (2)
a b a
1 q b
+ =0⇒q =− ……….. (3)
a b a
p=q
(6)Find the solution of px 2 + qy 2 = z 2 .
Ans:
The S.E is
dx dy dz
= =
x2 y2 z 2
Taking first two members, we get
dx dy
=
x2 y2
Integrating we get
−1 −1
= + c1
x y
1 1
i.e u  −  = c1
 y x
 
Taking last two members, we get

37

dy dz
=
y2 z2
Integrating we get
−1 −1
= + c2
y z
1 1
i.e v −  = c 2
z y
 
The complete solution is
1 1 1 1
φ − , −  = 0
 y x z y
 
(7)Find the singular integral of the partial differential equation z = px + qy + p 2 − q 2 .
Ans:
The complete integral is
z = ax + by + a 2 − b 2 .
∂z x
= x + 2a = 0 ⇒ a = −
∂a 2
∂z y
= y − 2b = 0 ⇒ b =
∂b 2

Therefore
x2 y2 x2 y2 x2 y2
z=− + + − =− +
2 2 4 4 4 4
⇒ y − x = 4z
2 2

(8)Solve: 2
p + q2 = m2.
Ans:
Given p2 + q2 = m2 …….. (1)
Let us assume that
z = ax + by + c …… (2)
be the solution of (1)
∂z
p= =a
∂x …….. (3)
∂z
q= =b
∂y
a2 + b2 = m2
This is the required solution.

38

(9)Form a partial differential equation by eliminating the arbitrary constants a and b from
z = ax n + by n .
Ans:
Given z = ax n + by n . ……… (1)
∂z p
p= = a.nx n −1 ⇒ a = n −1
∂x nx …….. (2)
∂z n −1 q
q= = b.ny ⇒ b = n −1
∂y ny
p q
z = n −1 x n + n −1 y n
nx ny
1
z = ( px + qy )
n
This is the required PDE.
(10)Solve:
(D 3
+ D 2 D ′ + DD ′ 2 + D ′ 3 ) z = 0.
Ans:
Auxiliary equation
m3 + m2 + m + 1 = 0
( m + 1) 3 = 0
m = −1,−1,−1
z = f 1 ( y − x ) + xf 2 ( y − x ) + x 2 f 3 ( y − x )
(11)Form a partial differential equation by eliminating the arbitrary constants a and b
from
z = ( x 2 + a 2 )( y 2 + b 2 ).
Ans: Given z = ( x 2 + a 2 )( y 2 + b 2 ) ……. (1)
∂z
= 2 x( y 2 + b 2 ) ⇒ y 2 + b 2 =
p
p=  ( 2 )
∂x 2x
∂z
= 2 y( x 2 + a 2 ) ⇒ x 2 + a 2 =
q
q=  ( 3)
∂y 2y
q p
Substituting (2) & (3) in (1), we get z = .
2 y 2x
pq = 4 xyz
(12)Solve:
( D 2 − DD′ + D′ − 1) z = 0
Ans:
The given equation can be written as

39

( D − 1)( D − D ′ + 1) z = 0
We know that the C.F corresponding to the factors
( D − m1 D ′ − α 1 )( D − m2 D′ − α 2 ) z = 0 is
z = e α1x f 1 ( y + m1 x ) + e α 2 x f 2 ( y + m2 x )
In our problem
α 1 = 1, α 2 = −1, m1 = 0, m2 = 1
C.F = e x f 1 ( y + x ) + e − x f 2 ( y + x )
z = e x f1 ( y + x ) + e − x f 2 ( y + x )
(13)Form a partial differential equation by eliminate the arbitrary function f from
 xy 
z = f  .
 z 

Ans:
 xy 
Given : z = f  .
 z 
 xy  zy − xy. p
p = f ′ .. ............(1)
 z  z2
 xy  zx − xy.q
q = f ′ .. ............(2)
 z  z2

From (1), we get
 xy  pz 2
f ′  = .............(3)
 z  zy − xyp

Substituting (3) in(2), we get
pz 2 zy − xy. p
p= .
zy − xyp z2
(14)Solve:
(D 3
+ 2 D 2 D ′ − DD ′ 2 − 2 D ′ 3 ) z = 0.
Ans: Auxiliary equation
m 3 + 2m 2 − m − 2 = 0
m = 1,−1,−2
Solution is z = f 1 ( y + x ) + f 2 ( y − x ) + f 3 ( y − 2 x )
(15)Obtain partial differential equation by eliminating arbitrary constants a and b from
( x + a ) 2 + ( y + b) 2 + z 2 = 1.
Ans:
Given ( x − a ) 2 + ( y − b ) 2 + z 2 = 1 ……. (1)

40

2( x − a ) + 2 zp = 0 ⇒ ( x − a ) = − zp .........(2)
2( y − b ) + 2 zq = 0 ⇒ ( y − b ) = − zq  ( 3)
Substituting (2) & (3) in (1), we get
z 2 p2 + z 2q2 + z 2 = 1
1
p2 + q2 +1 =
z2
(16)Find the general solution of
∂2 z ∂2 z ∂2 z
4 2 − 12 + 9 2 = 0.
∂y ∂x∂y ∂ y
Ans:
Auxiliary equation is
4m 2 − 12m + 9 = 0
12 ± 144 − 144
m=
8
3 3
m= ,
2 2
General solution is
 3   3 
z = f 1  y + x  + xf 2  y + x 
 2   2 
(17)Find the complete integral of
∂z ∂z
p + q = pq, where p= ,q = .
∂x ∂y
Ans: Let us assume that
z = ax + by + c ……… (1)
be the solution of the given equation.
∂z
p= =a
∂x …….. (2)
∂z
q= =b
∂y
a
a + b = ab ⇒ b =
a −1
Substituting the above in (1) we get
 a 
z = ax +  y + c
 a −1
This gives the complete integral.
(18)Solve:

41

(D 3
− 3 DD ′ 2 + 2 D ′ 3 ) z = 0
Ans:
Auxiliary equation
m 3 − 3m + 2 = 0
m = 1,1,−2
Solution is z = f 1 ( y + x ) + xf 2 ( y + x ) + f 3 ( y − 2 x )
(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
( x − a) 2 + ( y − a) 2 + ( z − a) 2 = r 2
2( x − a ) + 2( z − a ) p = 0 ...........(1)
2( y − a ) + 2( z − a ) q = 0 ...........(2)
From (1),
x + zp
a= ...........(3)
1+ p
From (2),
y + zq
a= ...........(4)
1+ q
x + zp y + zq
=
1+ p 1+ q
This is the required PDE.
(20)Solve:
∂3 z ∂2z ∂3z ∂3 z
−2 2 −4 + 8 3 = 0.
∂x 3 ∂x ∂y ∂x∂y 2 ∂y
Ans:
Auxiliary equation
m 3 − 2m 2 − 4 m + 8 = 0
m = 2,2,−2
Solution is z = f 1 ( y + 2 x ) + xf 2 ( y + 2 x ) + f 3 ( y − 2 x )

Part B
(1)(i) Form a partial differential equation by eliminating arbitrary functions from
z = xf ( 2 x + y ) + g ( 2 x + y )
(ii) Solve: p 2 y (1 + x 2 ) = qx 2

(2)(i) Solve: x( z 2 − y 2 ) p + y ( x 2 − z 2 ) q = z ( y 2 − x 2 )

42

∂3z ∂3 z
(ii) Solve: 3 − 2 2 = e x + 2 y + 4 sin ( x + y )
∂x ∂x ∂y

(3)(i) Solve: ( x 2 − y 2 − z 2 ) p + 2 xyq − 2 xz = 0
(ii) Solve: ( D 2 − DD ′ − 2 D ′ 2 ) z = 2 x + 3 y + e 3 x + 4 y

(4)(i) Solve: z 2 ( p 2 + q 2 ) = x 2 + y 2
(ii) Solve: ( D 2 + 3DD ′ − 4 D ′ 2 ) z = sin y

(5)(i) Solve: ( x 2 + y 2 + yz ) p + ( x 2 + y 2 − xz ) q = z ( x + y )
(ii) Solve: ( D 2 − DD ′ − 20 D ′ 2 ) z = e 5 x + y + sin ( 4 x − y )

(6)(i) Solve: z = p 2 + q 2
(ii) Solve: ( D 2 + DD ′ − 6 D ′ 2 ) z = x 2 y + e 3 x + y

(7)(i) Solve: ( y 2 + z 2 ) p − xyq + xz = 0
(ii) Solve: ( D 2 − 6 DD ′ + 5D ′ 2 ) z = xy + e x sinh y

(8)(i) Solve: p (1 − q 2 ) = q (1 − z )
(ii) Solve: ( D 2 − 4 DD ′ + 4 D ′ 2 ) z = xy + e 2 x + y

(9)(i) Solve: ( x 2 − yz ) p + ( y 2 − zx ) q = z 2 − xy
(ii) Solve: x( y − z ) p + y ( z − x ) q = z ( x − y )

(10)(i) Solve: p 2 + x 2 y 2 q 2 = x 2 z 2
(ii)Solve: ( D 2 − D ′ 2 − 3D + 3D ′) z = xy

(11)(i) Form the partial differential equation by eliminating f and φ from
z = f ( y) + φ( x + y + z)
(ii) Solve: ( D ′ 2 − 2 DD ′) z = x 3 y + e 2 x

(12)(i) Find the complete integral of p + q = x + y
(ii) Solve: y 2 p − xyq = x( z − 2 y )

(13)(i) Solve: ( 3 z − 4 y ) p + ( 4 x − 2 z ) q = 2 y − 3 x
(ii) Solve: ( D 2 − 2 DD ′ − D ′ 2 + 3D + 3D ′ + 2 ) z = ( e 3 x + 2e −2 y )
2

(iii) Solve: ( D 2 + 4 DD ′ − 5 D ′ 2 ) z = sin ( x − 2 y ) + e 2 x − y

(14)(i) Solve z 2 = 1 + p 2 + q 2
(ii) Solve: ( y − z ) p − ( 2 x + y ) q = 2 x + z

43

(15)(i) Form a partial differential equation by eliminating arbitrary functions f and g in
z = x 2 f ( y ) + y 2 g ( x)
(ii) Solve: ( D 2 − DD ′ − 20 D ′ 2 ) z = sin ( 4 x − y ) + e 5 x + y

(16)(i) Form a partial differential equation by eliminating arbitrary functions f and g in
z = f ( x3 + 2 y) + g( x3 − 2 y)
(ii) Solve: ( y − xy ) p + ( yz − x ) q = ( x + y )( x − y )

(17)(i) Find the singular solution of z = px + qy + p2 + q2 +1
(ii) Solve: ( D 2 − DD ′ − 30 D ′ 2 ) z = xy + e 6 x + y

(18) (i) Solve: ( D 3 − 7 DD ′ 2 − 6 D ′ 3 ) z = sin ( x + 2 y ) + e 2 x + y
(ii) Find the singular integral of a partial differential equation z = px + qy + p 2 − q 2

(19)(i) Solve: ( 4 D 2 − 4 DD ′ + D ′ 2 ) z = 16 log( x + 2 y )
(ii) Form a partial differential equation by eliminating arbitrary functions ‘f’ from
f ( z − xy, x 2 + y 2 ) = 0

(20)(i) Solve: ( D 2 + D ′ 2 ) z = sin 2 x sin 3 y + 2 sin 2 ( x + y )
(ii) Solve: p 2 + x 2 y 2 q 2 = x 2 z 2

44

Chapter 1 (maths 3)

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