All lectures PDE, Engineering Mathematics.pdf

Differential Equation
Definition: An equation involving derivatives or differential
coefficients of one or more dependent variables with respect
to one or more independent variables is called a differential
equation.
Examples:
x
x
dx
dy
dx
y
d
sin
2
2
=
+
0
2
2
2
2
=


+


y
z
x
z
Introduction

Introduction
There are two kinds of differential equations:
1. Ordinary differential equation (O.D.E.)
2. Partial differential equation (P.D.E.)
Ordinary differential equation (O.D.E.)
Definition: A differential equation involving derivatives or
differential coefficients with respect to a single independent
variable is called an ordinary differential equation.
Examples: x
y
dx
dy
dx
y
d
=
+
+ 2
2
2
;
y = f(x)

Introduction
Partial differential equation (P.D.E.)
Definition: A differential equation involving partial
derivatives or differential coefficients with respect to more
than one independent variables is called a partial differential
equation.
Examples: 0
2
5
2 2
2
2
2
2
=


+



+


y
z
y
x
z
x
z
;
z = f(x, y)

Introduction
“A partial differential equation is one which
contains more than one independent variables.”
This type of equation plays a very important role in
sciences and engineering dealing with wave
motions, such as heat, light, electricity, magnetism,
radio, radar, television and weather etc.

Introduction
In general, z is taken as dependent variable whereas x and y
are treated as independent variables so that
z = f(x, y)
The partial derivatives of z with respect to x and y are
denoted by p and q respectively, so that
y
z
q
x
z
p


=


= ,
we will also use
x
p
x
z
r


=


= 2
2
;
y
q
y
z
t
x
q
y
p
y
x
z
s


=


=


=


=



= 2
2
2
,

Introduction
Order and Degree: The order of a PDE is the
order of the highest derivative of the equation.
The degree of the PDE is the power of the
highest derivative of the equation.
 Equation 0
6
5
2
2
=


+


+


y
z
x
z
x
z
or, 0
6
5
2
2
2
=


+








+


y
z
x
z
x
z
is the 2nd order first degree PDE

Introduction
Formation:
The PDE may be derived in the following two ways
i) By eliminating arbitrary constants.
ii) By eliminating arbitrary functions.

Introduction
Formation of PDE by eliminating arbitrary
constants
Let z be a function of two independent variables x
and y connected by
0
)
,
,
,
,
( =
b
a
z
y
x
f (1)
where a and b are arbitrary constants.

Introduction
Formation of PDE by eliminating arbitrary constants
By differentiating (1) partially with respect to x and y we
obtain
0
=


+


=




+




+


z
f
p
x
f
x
z
z
f
x
y
y
f
x
f
(2)
0
=


+


=




+


+




z
f
q
y
f
y
z
z
f
y
f
y
x
x
f
(3)
Now arbitrary constants may be eliminated applying (1), (2),
(3) yielding a partial differential equation of order one given
by
0
)
,
,
,
,
( =
q
p
z
y
x
f

Introduction
Example: Form a PDE by eliminating the arbitrary
constants a and b from
( )( )
2
2
y
b
x
a
z +
+
=
Solution Differentiating
( )( )
2
2
y
b
x
a
z +
+
=
partially with respect to x and y we have
( )
2
2 y
b
x
p
x
z
+
=
=


and ( )
2
2 x
a
y
q
y
z
+
=
=



Introduction
( )( )
2
2
4 y
b
x
a
xy
pq +
+
=
=
xyz
pq 4
=
=
which is the required PDE
.
3
3
ab
by
ax
z +
+
=

Introduction
Solution Differentiating
ab
by
ax
z +
+
= 3
3
partially with respect to x and y we have
2
3ax
p = or 2
3x
p
a =
and
2
3by
q = or 2
3y
q
b =

Introduction
putting the values of a and b in the given equation,
we get
2
2
3
2
3
2
9
3
.
3 y
x
pq
y
y
q
x
x
p
z +
+
=
or, .
3
3
9 3
2
2
3
2
2
pq
y
qx
y
px
z
y
x +
+
=

Introduction
b
y
ax
z +
+
= 2
)
(
2 .
Solution Differentiating partially the given equation
with respect to x and y, we get
a
y
ax
p )
(
2
2 +
= and )
(
2
2 y
ax
q +
=
or, a
y
ax
p )
( +
= or, .
y
ax
q +
=
Now, .
)
( 2
2
q
y
ax
qy
px =
+
=
+

Introduction
Note 1: If the number of arbitrary constants is less than or equal
to the number of independent variables then the differential
equation formed by the elimination of arbitrary constants would
be of the first order.
Note 2: If the number of arbitrary constants are more than the
number of independent variables then the differential equation
will be of the minimum second order.
Note 3: The PDE formed by eliminating arbitrary constants is not
always unique.

Introduction
Formation of PDE by eliminating arbitrary
functions:
Let u = (x, y, z) and v= (x, y, z) be two functions of x, y, z
connected by the relation
0
)
,
( =
v
u
 (1)
z is the dependent variable and x, y independent variable.
Differentiating (1) partially with respect to x, we obtain

Introduction
0
=




+




x
v
v
x
u
u


0
=












+




+






+












+




+






=
x
z
z
v
x
y
y
v
x
x
x
v
v
x
z
z
u
x
y
y
u
x
x
x
u
u


0
=










+




+










+




=
z
v
p
x
v
v
z
u
p
x
u
u


(2)

Introduction
Similarly, differentiating (1) partially with respect to y, we
get
0
=












+




+






+












+




+






=
y
z
z
v
y
y
y
v
y
x
x
v
v
y
z
z
u
y
y
y
u
y
x
x
u
u


0
=










+




+










+




z
v
q
y
v
v
z
u
q
y
u
u


(3)
Eliminating
v
u 


 

,

Introduction
From these, we get 0
=


+




+




+




+


z
v
q
y
v
z
u
q
y
u
z
v
p
x
v
z
u
p
x
u
0
=








+












+


=










+












+


=
z
v
p
x
v
z
u
q
y
u
z
v
q
y
v
z
u
p
x
u
or 











−




+












−




x
v
z
u
z
v
x
u
q
z
v
y
u
y
v
z
u
p
0
=












−




+
x
v
y
u
y
v
x
u

Introduction
which can be rearranged in the form
pP + qQ = R
where,
y
v
z
u
z
v
y
u
p




−




=
z
v
x
u
x
v
z
u
q




−




=
x
v
y
u
y
v
x
u
R




−




=
which is the Linear partial differential equation of 1st order
and 1st degree in p, q and free of the arbitrary function
).
,
( v
u


Introduction
Form a partial differential equation by eliminating
the arbitrary function  from
( ) 0
, 2
2
2
=
−
+
+
+ z
y
x
z
y
x
 .
Solution: Here, ( ) 0
, 2
2
2
=
−
+
+
+ z
y
x
z
y
x
 (1)
Let,
2
2
2
, z
y
x
v
z
y
x
u −
+
=
+
+
= (2)
( ) 0
, =
 v
u
 (3)
1
=


x
u
, x
x
v
2
=


, 1
=


y
u
, y
y
v
=


1
=


z
u
, z
z
v
2
−
=



Introduction
Differentiating (3) partially with respect to x, we obtain
0
=




+




x
v
v
x
u
u


0
=












+




+






+












+




+






=
x
z
z
v
x
y
y
v
x
x
x
v
v
x
z
z
u
x
y
y
u
x
x
x
u
u


0
=










+




+










+




=
z
v
p
x
v
v
z
u
p
x
u
u


( ) ( ) 0
2
2
1 =
−


+
+


= z
p
x
v
p
u

 ( )
( )
0
1
2
=
+
−
−
=




=
p
pz
x
v
u


(4)

Introduction
Similarly, Differentiating (3) partially with respect to y, we
obtain
0
=












+




+






+












+




+






=
y
z
z
v
y
y
y
v
y
x
x
v
v
y
z
z
u
y
y
y
u
y
x
x
u
u


0
=










+




+










+




=
z
v
q
y
v
v
z
u
q
y
u
u


( ) ( ) 0
2
2
1 =
−


+
+


= z
q
y
v
q
u


( )
( )
0
1
2
=
+
−
−
=




=
q
qz
y
v
u


(5)

Introduction
From (4) and (5) eliminating
v
u 


 

,
( )
( )
( )
( )
q
qz
y
p
pz
x
+
−
−
=
+
−
−
=
1
2
1
2
)
(
)
(
)
( y
x
z
x
q
z
y
p −
=
+
−
+
which is the required partial differential equation.

Introduction
the arbitrary function  from
)
( 2
2
2
z
y
x
f
z
y
x +
+
=
+
+ .
Solution: Differentiating partially
)
( 2
2
2
z
y
x
f
z
y
x +
+
=
+
+ (1)
with respect to x and y, we have
)
2
2
(
)
(
1 2
2
2
zp
x
z
y
x
f
p +
+
+

=
+ (2)
and )
2
2
(
)
(
1 2
2
2
zq
y
z
y
x
f
q +
+
+

=
+ (3)

Introduction
Using (2) and (3), we get
)
2
2
(
1
)
2
2
(
1
zq
y
q
zp
x
p
+
+
=
+
+
)
(
)
(
)
( y
x
z
x
q
z
y
p −
=
+
+
−
which is the required partial differential
equation.

Introduction
the arbitrary functions f and F from
)
(
)
( at
x
F
at
x
f
y +
+
−
= .
Solution: Differentiating partially
)
(
)
( at
x
F
at
x
f
y +
+
−
= (1)
with respect to x and t, we have
)
(
)
( at
x
F
at
x
f
x
y
+

+
−

=


)
(
)
(
2
2
at
x
F
at
x
f
x
y
+


+
−


=


(2)

Introduction
Similarly ( ) a
at
x
F
a
at
x
f
t
y
)
(
)
( +

+
−
−

=


)
(
)
( 2
2
2
2
at
x
F
a
at
x
f
a
t
y
+


+
−


=


(3)
Using (2) and (3), we get
2
2
2
2
2
x
y
a
t
y


=


which is the required partial differential equation.

1
29 March 2021
Linear partial differential equation of first order
Definition: A partial differential equation
involving partial derivatives p and q only and
not higher partial derivatives is called a first
order partial differential equation.
If the degree of p and q is unity, then it is called a
linear partial differential equation of order one.
Thus
2
2
2
z
qy
px =
+
is a linear partial differential equation of first order.

2
29 March 2021
The general solution of the linear partial
differential equation
R
qQ
pP =
+ (1)
(where P, Q, R are functions of x, y, z)
is 0
)
,
( =
v
u
 (2)
where  is an arbitrary function and
u(x, y, z) = c1 and , v(x, y, z) = c2 (3)
form a solution of the

3
29 March 2021
R
dz
Q
dy
P
dx
=
= (4)
Lagrange’s auxiliary equation.
Proof: Differentiating (3) we get
0
=


+


+


dz
z
u
dy
y
u
dx
x
u
0
=


+


+


dz
z
v
dy
y
v
dx
x
v

4
29 March 2021
Solving above equations for dx, dy, dz, we have










−




=










−




z
v
x
u
x
v
z
u
dy
y
v
z
u
z
v
y
u
dx










−




=
x
v
y
u
y
v
x
u
dz
we obtain the equations
R
dz
Q
dy
P
dx
=
=

5
29 March 2021
Working Rule for Solving R
Qq
Pp =
+ by
Langrange’s Method
Step-1: Put the given first order linear partial
differential equation in the standard form
.
R
Qq
Pp =
+ (1)
Step-2: Write down Lagrange’s auxiliary
equation for (1) namely,
R
dz
Q
dy
P
dx
=
= (2)

6
29 March 2021
Step-3: Find the two independent solutions of
(2) in the form
1
)
,
,
( c
z
y
x
u = and 2
)
,
,
( c
z
y
x
v =
Step-4: The general solution or integral of (1)
is then written in one of the following three
equivalent forms:
)
(
),
(
,
0
)
,
( u
v
v
u
v
u 

 =
=
=

7
29 March 2021
Type -1 for solving R
dz
Q
dy
P
dx
=
= .
Suppose that one of the variables is either
absent or cancels out from any two fractions
of given equations. Then one integral can be
obtained by usual method. The same
procedure can be repeated with another set of
two fractions of the given equations.

8
29 March 2021
Example Solve .
0
2
2
2
=
+
+ z
q
y
p
x
Solution: Lagrange’s auxiliary equations
are, 2
2
2
z
dz
y
dy
x
dx
−
=
=
From first two fractions,
2
2
y
dy
x
dx
=
Integrating, 1
1
1
c
y
x
−
−
=
−

9
29 March 2021
1
1
1
c
y
x
=
−

From the last two fractions, 2
2
z
dz
y
dy
−
=
Integrating, 2
1
1
c
z
y
=
+
Hence the required general solution is
0
1
1
,
1
1
=








+
−
z
y
y
x
 ,
where  is an arbitrary function.

10
29 March 2021
Type -2 for solving R
dz
Q
dy
P
dx
=
= .
Suppose one integral is known by using the
above rule (type-1) but another integral
cannot be obtained by the above rule (type-1).
Then one integral known to us is used to find
the another integral as shown below.

11
29 March 2021
Example Solve ( ) .
)
( 4
3
x
qy
px
xyz
z =
−
+
Solution: The given equation can be written
as
4
2
3
2
3
)
(
)
( x
q
z
xy
yz
p
yz
x
xz =
+
−
+
Lagrange’s auxiliary equations are,
4
2
2
)
(
)
( x
dz
xy
z
yz
dy
xy
z
xz
dx
=
+
−
=
+
From the first two cancelling z(z2
+ xy), we get
0
=
+

y
dy
x
dx

12
29 March 2021
Integrating, 1
ln
ln
ln c
y
x =
+

1
c
xy =
 (1)
Taking 1st and 3rd fractions using (1)
4
1
2
)
( x
dz
c
z
xz
dx
=
+
dz
z
c
z
dx
x )
( 1
3
3
+
=

Integrating,
2
4
4
2
1
4
4
z
c
z
x
+
=


13
29 March 2021
2
2
1
4
4
2 c
z
c
z
x =
−
−
 (2)
From (1) and (2), we get
2
2
4
4
2 c
xyz
z
x =
−
−
Hence the required general solution is
0
)
2
,
( 2
4
4
=
−
− xyz
z
x
xy
 ,
where  is an arbitrary function.

Type 3: Method of multipliers
14
29 March 2021
nR
mQ
lP
ndz
mdy
ldx
R
dz
Q
dy
P
dx
+
+
+
+
=
=
=
l, m, n are so chosen that lP + mQ + nR = 0
where l, m, n may be constants or functions of
x, y, z
Thus ldx + mdy + ndz = 0

15
29 March 2021
Solving this we get u(x, y, z) = c1
This method may be repeated to get
v(x, y, z) = c2
Sometimes one can be obtained by
multipliers and other by usual way.

29 March 2021 16
Solve: ( ) ( ) ( )
y
x
z
q
x
z
y
p
z
y
x −
=
−
+
−
Solution: The Lagrange's auxiliary equations for (1)
are
( ) ( ) ( )
y
x
z
dz
x
z
y
dy
z
y
x
dx
−
=
−
=
−
Choosing 1, 1, 1 as multipliers each fraction of (2)
zy
xz
xy
yz
xz
xy
dz
dy
dx
−
+
−
+
−
+
+
=
(1)
(2)

29 March 2021 17
0
dz
dy
dx +
+
=
i.c. dx + dy + dz = 0
Integrating, x + y + z = c1 (3)
Choosing 1/x, 1/y, 1/z as multipliers, each
fraction of (2)

29 March 2021 18
( ) ( ) ( )
y
x
x
z
z
y
dz
z
dy
y
dx
x
−
+
−
+
−
+
+
=
1
1
1
0
1
1
1
dz
z
dy
y
dx
x
+
+
=
i.c. 0
1
1
1
=
+
+ dz
z
dy
y
dx
x

29 March 2021 19

So that, x y z = c2 --------------(4)
Hence the general solution is
Where is an arbitrary function.
(x + y + z , x y z ) = 0

Integrating, log x + log y + log z = log c2

20
29 March 2021
( ) xz
xyq
p
z
y
x 2
2
2
2
2
=
+
−
−
( ) xz
dz
xy
dy
z
y
x
dx
2
2
2
2
2
=
=
−
−
From the last two fractions we get,
xz
dz
xy
dy
2
2
=
Solution: The Lagrange's auxiliary equations for (1) are
Solve: (1)
(2)

21
29 March 2021
cancelling 2x
z
dz
y
dy
=
Integrating, log y = log z + log c1
1
c
z
y
=
Choosing x, y, z as multipliers, each fraction
of (2)
( ) 2
2
2
2
2
2
2 xz
xy
z
y
x
x
zdz
ydy
xdx
+
+
−
−
+
+
=
(3)

29 March 2021 22
( )
2
2
2
z
y
x
x
zdz
ydy
xdx
+
+
+
+
=
Combining the third fraction of (2) with fraction (4) ,
we get
( ) xz
dz
z
y
x
x
zdz
ydy
xdx
2
2
2
2
=
+
+
+
+
( )
( ) z
dz
z
y
x
z
y
x
d
2
2 2
2
2
2
2
2
=
+
+
+
+
(4)

29 March 2021 23
Integrating, log (x2 + y2 + z2) = log z + log c2
( )
2
2
2
2
c
z
z
y
x
=
+
+
Hence the general solution is
So that,

0
,
2
2
2
=







 +
+
z
z
y
x
z
y

Where is an arbitrary function.
(5)

24
29 March 2021
Find the integral surface of the partial
differential equation
y
x
q
x
z
p
z
y −
=
−
+
− )
(
)
(
through .
2
,
0 x
y
z =
=
Solution: Lagrange’s auxiliary equations are,
y
x
dz
x
z
dy
z
y
dx
−
=
−
=
−
(1)
Taking 1, 1, 1 as multipliers, each fraction of (1)
0
dz
dy
dx +
+
=

25
29 March 2021
So that 0
=
+
+ dz
dy
dx
Integrating, 1
c
z
y
x =
+
+ (2)
Taking x, y, z as multipliers, each fraction of (1)
)
(
)
(
)
( yz
zx
xy
yz
zx
xy
zdz
ydy
xdx
−
+
−
+
−
+
+
=
0
zdz
ydy
xdx +
+
=
or, 0
=
+
+ zdz
ydy
xdx

26
29 March 2021
Integrating, 2
2
2
2
c
z
y
x =
+
+ (3)
Given that x
y
z 2
,
0 =
=
Putting z = 0 in (2) and (3), we get
1
c
y
x =
+ and 2
2
2
c
y
x =
+ (4)
From (4) using x
y 2
= becomes
1
3 c
x = and 5x2
= c2
=> x = c1/3 or, 2
2
1
3
5 c
c
=







27
29 March 2021
=> ( ) 2
2
1 9
5 c
c =
Putting the values of c1 and c2 from (2)
and (3) we get the required surface as
( ) )
(
9
5 2
2
2
2
z
y
x
z
y
x +
+
=
+
+
of the given partial differential equation.

1
Non-Linear partial differential equation
of first order:
The equations which involve p and q other than in the
first degree and products of the derivatives occur are
called non-linear partial differential equations of the
first order. For such equations, the complete solution
consists of only two arbitrary constants (i.e. equal to
the number of independent variables involved).
Examples: px + qy = pq
1
2
2

 q
p

2
Solutions or integrals: Solutions of partial
differential equations appear in almost four
forms, namely
1. Complete solution or integral;
2. Particular solution or integral;
3. Singular solution or integral and
4. General solution or integral
Non-Linear partial differential equation of first order

3
Complete integral: A solution which
contains as many arbitrary constants as the
number of independent variables of a
differential equation is called the complete
integral.

4
ab
by
ax
z 

 3
3
z
y
x
pq
x
qy
y
px 2
2
2
3
2
3
9
2
3 


The differential equation formed by eliminating
arbitrary constants a, b from
Equation (1) contains two arbitrary constants equal to the
number of independent variables of differential equation (2)
and is called the complete integral.
is
(1)
(2)

5
Particular integral: Assigning particular
values for arbitrary constants, particular
integrals of the differential equation are
obtained.
1
3
3


 y
x
z
is a particular integral of the PDE (2) which is
obtained by letting a = 1 and b = 1 in the
complete integral (1).

6
  0
,
,
,
, 
b
a
z
y
x

Singular integral: Singular solution is a solution of
differential equation which is independent of arbitrary
constants and which is obtained by following a definite
rule.
Let
be the complete integral of
F(x, y, z, p, q) = 0
(3)
(4)

7
0



a

0



b

The singular integral of PDE (4) is
obtained by eliminating a and b between
CI (3) and
The relation between x, y and z so obtained is called the
singular integral.
and

8
  0
(
,
,
,
, 
a
F
a
z
y
x

General integral: General solution does not
contain any arbitrary constant but is different from
singular solution.
Assume that in (3), one of the constants is a
function of the other, say b = F(a), then (3)
becomes
Hence the general integral is obtained by
eliminating a between
.
  0
(
,
,
,
, 
a
F
a
z
y
x
 0



a

and

9
Charpit’s method
We will now give a general method for obtaining the
complete solution of a nonlinear partial differential
equation. This method is due to Charpit and is
applicable to all nonlinear partial differential
equations of first order but any degree.

10
Let the given equation be
0
)
,
,
,
,
( 
q
p
z
y
x
F (1)
The corresponding Charpit’s auxiliary equations of
(1) (by Lagrange’s method) are
q
F
q
p
F
p
dz
z
F
q
y
F
dq
z
F
p
x
F
dp


















0
dF
q
F
dy
p
F
dx









(2)

11
Working rule while using Charpit’s method
Step 1: Transfer all terms of the given equation
to L.H.S. and denote the entire expression by F.
Step 2: Write down the Charpit’s auxiliary
equation (2).
Step 3: Using the value of F in step 1 write
down the values of y
F
x
F




, etc. occurring
in step 2 and put these in Charpit’s equations

12
Step 4: After simplifying the step 3, select two
proper fractions so that the resulting integral
may come out to be the simplest relation
involving at least one of p and q.
Step 5: The simplest relation of step 4 is solved
along with the given equation to determine p
and q. Put these values of p and q in equation
qdy
pdx
dz 
 which on integration gives the
complete solution of the given equation since

13
the solution will contain as many arbitrary
constants i.e. two which is equal to the number
of independent variables.
The Singular and General integral may be
obtained in the usual manner.

14
Find complete integral, singular integral and
general integral of
pq
zx
qxy
px 

 2
2
2
Solution The given equation can be written
as Here F = 0
2
2 2



 pq
qxy
px
zx
so that p
xy
q
F
q
x
p
F










2
,
2

15
The Charpit’s auxiliary equations are
q
F
q
p
F
p
dz
z
F
q
y
F
dq
z
F
p
x
F
dp


















q
F
dy
p
F
dx








p
xy
dy
q
x
dx
pq
xyq
px
dz
qx
qx
dq
px
qy
px
z
dp














2
2
2
2
2
2
2
2
2
2
2
Taking the two simplest ratios of Charpit’s
auxiliary equations, we have
qy
px
z
x
F
x
z
F
2
2
2
,
2 







.
2qx
y
F




and

16
qx
qx
dq
px
qy
px
z
dp
2
2
2
2
2
2 






or .
0
2
2
dq
qy
z
dp


Integrating, we get q = a an arbitrary constant.
Substituting in the given equation, we get
a
x
ay
z
x
a
x
axy
xz
p





 2
2
)
(
2
2
2
 ady
a
x
dx
ay
z
x
qdy
pdx
dz 




 2
)
(
2

17
or a
x
xdx
ay
z
ady
dz




2
2
Integrating, b
a
x
ay
z log
)
log(
)
log( 2




or )
( 2
a
x
b
ay
z 


i. e. ),
( 2
a
x
b
ay
z 

 ---(1)
where b is another arbitrary constant. This is
the complete integral.

18
Singular integral: Differentiating the complete
integral partially w.r.t. a and b, we have
y
b
b
y 



0 ---(2)
2
2
0 x
a
a
x 


 ---(3)
Putting the value of a and b in (1) , we get
y
x
z 2
 which is the Singular integral.

19
General integral : Writing b = (a) we have
),
)(
( 2
a
x
a
ay
z 

  (4)
Differentiating (3) partially w.r.t. a we have
),
(
)
)(
(
0 2
/
a
a
x
a
y 
 


 (5)
General integral is obtained by eliminating a from
(4) and (5).

20
Examples: Solve px + qy = pq.
Solution: Here pq
qy
px
F 

 = 0 (1)
.
;
;
;
; p
y
q
F
q
x
p
F
o
z
F
q
y
F
p
x
F

















The Charpit’s auxiliary equations are
)
(
)
( p
y
q
q
x
p
dz
q
dq
p
dp






=
𝑑𝑥
−(𝑥−𝑞)
=
𝑑𝑦
−(𝑦−𝑝)

21
Taking the first two fractions, we have
.
q
dq
p
dp

Integrating, a
q
p log
log
log 

or p = aq
where a is an arbitrary constant. Putting aq for p in the
given equation, we get
a
ax
y
q
aq
qy
aqx




 2
 ax
y
aq
p 



22
Putting the values of p and q in ,
qdy
pdx
dz 

we get
dy
a
ax
y
dx
ax
y
dz
)
(
)
(




or )
)(
( adx
dy
ax
y
adz 


Integrating, b
ax
y
az 

 2
)
(
2
1
(2)
which is the complete integral, as b is another
arbitrary constant

23
Singular integral:
Differentiating the complete integral partially
w.r.t. a and b, we have
)
( ax
y
x
z 
 and 0 = 1 (absurd)
Hence there is no singular integral.

24
General integral : Writing b = (a) we have
)
(
)
(
2
1 2
a
ax
y
az 


 (3)
Differentiating (3) partially w.r.t. a we have
)
(
)
( a
ax
y
x
z 


 (4)
General integral is obtained by eliminating a from
(3) and (4).

1
Linear PD Equations of higher order with constant
coefficients
An equation of the form
)
,
(
1
0
1
1
1
2
1
1
1
1
0
1
1
0
y
x
f
Lz
y
z
K
x
z
K
y
z
B
y
x
z
B
x
z
B
y
z
A
y
x
z
A
x
z
A
n
n
n
n
n
n
n
n
n
n
n
n
n
n
=
+


+


+
+


+
+



+




+
+



+


−
−
−
−
−
−
−
−






is called a linear partial differential equation of order n with
variable coefficients.
L
k
K
B
B
B
A
A
A n
n ,
,
,
,
,
,
,
,
,
, 1
0
1
1
0
1
0 −




If

2
The given equation can be written as
are all constants, is called a linear partial differential
equation of order n with constant coefficients.
For convenience we will use
y
D
x
D y
x


=


= ,
( )
( )
)
,
(
]
..
..........
......
..........
[
1
0
1
1
2
1
1
0
1
1
0
y
x
f
z
L
D
K
D
K
D
B
D
D
B
D
B
D
A
D
D
A
D
A
y
x
n
y
n
y
n
x
n
x
n
y
n
y
n
x
n
x
=
+
+
+
+
+
+
+
+
+
+
+
−
−
−
−
−


coefficients

3
are constants, is called a homogeneous linear
partial differential equation of order n with
constant coefficients.
)
,
(
1
1
0 y
x
f
y
z
A
y
x
z
A
x
z
A n
n
n
n
n
n
n
=


+
+



+


−


where n
A
A
A ,
,
, 1
0 

Homogeneous Linear PD Equations of higher order
with constant coefficients

4
.
, n
n
n
y
n
n
n
x
y
D
x
D


=


=
( ) )
,
(
......
..........
1
1
0 y
x
f
z
D
A
D
D
A
D
A n
y
n
y
n
x
n
x =
+
+
+ −
)
,
(
)
,
( y
x
f
z
D
D
F y
x =
i.e.
The given equation can be written in symbolic
form as
where

5
The particular integral is the particular solution of
Complete solution = complementary function +
particular integral
= C. F. + P. I.
0
)
,
( =
z
D
D
F y
x
)
,
(
)
,
( y
x
f
z
D
D
F y
x =
The complementary function is the general
solution of
.
coefficients

6
Rules for finding the complementary function
(C.F.):
0
)
(
)
)(
(
)
,
( 2
1 =
−
−
−
= z
D
m
D
D
m
D
D
m
D
z
D
D
F y
n
x
y
x
y
x
y
x 

0
)
,
( =
z
D
D
F y
x
Let,
i. e.
Where, are roots,
n
m
m
m ,
,
, 2
1 

.
0
)
( =
− z
D
m
D y
r
x
consider

7
0
1
dz
m
dy
dx
r
=
−
=
1
c
x
m
y r =
+ and z = c2
Using Lagrange's auxiliary equations we have
)
( x
m
y
z r
r +
=
C. F.
0
=
− q
m
p r
or,

8
The auxiliary equation is simply obtained by
replacing by m and by 1 in
z
Dx
z
Dy
0
)
,
( =
z
D
D
F y
x
Auxiliary equation:
The equation
is called an auxiliary equation.
0
1
1
0 =
+
+
+ −
n
n
n
A
m
A
m
A 


9
Case 1
If are real and distinct
roots, then the complementary function is
n
m
m
m ,
,
, 2
1 

)
(
)
(
)
( 2
2
1
1 x
m
y
x
m
y
x
m
y
z n
n +
+
+
+
+
+
= 

 


10
Example Solve 0
6
5 2
2
2
2
2
=


+



+


y
z
y
x
z
x
z
Solution The given equation can be written as
( ) .
0
6
5 2
2
=
+
+ z
D
D
D
D y
y
x
x
Replacing 𝐷𝑥 𝑧 by m and 𝐷𝑦 𝑧 by 1, we get
the auxiliary equation 0
6
5
2
=
+
+ m
m
=> .
3
,
2 −
−
=
m
Hence the complete solution is
( ) ( ).
3
2 2
1 x
y
x
y
z −
+
−
= 


11
Case2
If the roots of the auxiliary equation are
equal say m is repeated r times then the
complementary function is
)
(
)
(
)
(
)
(
1
3
2
2
1
mx
y
x
mx
y
x
mx
y
x
mx
y
z
r
r
+
+
+
+
+
+
+
+
=
−







12
Example Solve 0
8
12
6 3
3
2
3
2
3
3
3
=


+



+



+


y
z
y
x
z
y
x
z
x
z
( ) .
0
8
12
6 3
2
2
3
=
+
+
+ z
D
D
D
D
D
D y
y
x
y
x
x
Replacing by m and by 1, we get
the auxiliary equation 0
8
12
6 2
3
=
+
+
+ m
m
m
=> .
2
,
2
,
2 −
−
−
=
m
Hence the complete solution is
).
2
(
)
2
(
)
2
( 3
2
2
1 x
y
x
x
y
x
x
y
z −
+
−
+
−
= 



13
Method 1: When 0
)
,
( =
z
D
D
f y
x can be expressed
as product of factors linear in Dx and Dy
Case 1
If 0
)
( =
−
− z
c
bD
aD y
x non- repeated factor
and 0

a then the complementary function becomes
)
( bx
ay
e
z a
cx
+
=  ; where a, b, c are arbitrary constants
Non - Homogeneous Linear PD Equations of
higher order with constant coefficients

14
Case 2
If 0
)
( =
−
− z
c
bD
aD y
x repeated ' r ' times then
the complementary function becomes






+
+
−
−
−
−
−
−
−
+
+
+
+
+
+
= −
)
(
)
(
)
(
)
(
1
3
2
2
1
bx
ay
x
bx
ay
x
bx
ay
x
bx
ay
e
z
r
r
a
cx





15
Example Solve .
0
)
( 2
2
=
−
+
− z
D
D
D
D y
x
y
x
0
)
1
)(
( =
+
+
− z
D
D
D
D y
x
y
x
Hence the complementary function is
C. F. )
(
)
( 2
1 x
y
e
x
y x
−
+
+
= −



16
Example Solve .
0
)
1
2
2
2
( 2
2
=
+
+
+
+
+ z
D
D
D
D
D
D y
x
y
y
x
x
  0
1
)
(
2
)
( 2
=
+
+
+
+ z
D
D
D
D y
x
y
x
=>  0
)
1
( 2
=
+
+ z
D
D y
x
C. F.  
)
(
)
( 2
1 x
y
x
x
y
e x
−
+
−
= −



17
Method 2
When z
D
D
F y
x )
,
( can not be factorized into linear factors:
Let, ky
hx
Ae
z +
= be a trial solution ; where A, h, k are constants
Substituting this in 0
)
,
( =
z
D
D
f y
x we get,
0
)
,
( =
+ky
hx
e
k
h
Af if 0
)
,
( =
k
h
f ;
0

+ky
hx
e
then the complementary function becomes


=
+
=
1
r
y
k
x
h
r
r
r
e
A
z

18
Example Solve .
0
)
( 2
=
− z
D
D y
x
Solution The given equation )
( 2
y
x D
D − is not linear in
Dx and Dy
Let,
ky
hx
Ae
z +
= be the trial solution of the given equation ;
where A, h, k are arbitrary constants
ky
hx
x Ahe
z
D +
=
ky
hx
y Ake
z
D +
=
ky
hx
y e
Ak
z
D +
= 2
2

19
Solution Substituting these in the given equation, we get
( ) 0
2
=
− +ky
hx
e
k
h
A
=> ( ) 0
2
=
− k
h ; 0

+ky
hx
e
=> 2
k
h =
C. F =  +
= ky
x
k
Ae
z
2

20
Case (i) If Dx is non-repeated:
Therefore C. F. )
(y

=
Case (ii) If Dx is repeated ' n ' times then the
complementary function becomes
C. F. )
(
)
(
)
(
)
( 1
3
2
2
1 y
x
y
x
y
x
y n
n



 −
+
−
−
−
−
−
−
+
+
+
=
coefficients

21
Case (iii) If Dy is non-repeated:
Therefore C. F. )
(x

=
Case (iv) If Dy is repeated ' n ' times then the
complementary function becomes
C. F. )
(
)
(
)
(
)
( 1
3
2
2
1 x
y
x
y
x
y
x n
n



 −
+
−
−
−
−
−
−
+
+
+
=
coefficients

22
Example Solve .
0
)
( 3
2
2
3
=
+ z
D
D
D
D y
x
y
x
Solution .
0
)
(
2
2
=
+ z
D
D
D
D y
x
y
x
.
0
)
( =
+ z
D
D y
x
Replacing by m and by 1, we get
the auxiliary equation m + 1 = 0
=> m = −1
Therefore
C. F )
(
)
(
)
(
)
(
)
( 5
4
3
2
1 x
y
x
y
x
y
x
y −
+
+
+
+
= 




coefficients

1
Rules for finding the particular integral (P.I.):
  )
,
(
, y
x
f
z
D
D
F y
x 
i. e. Particular integral
  )
,
(
,
1
y
x
f
D
D
F y
x

coefficients

2
Case-1
When
by
ax
e
y
x
f 

)
,
( ; where a and b are arbitrary constants
by
ax
by
ax
y
x
y
x
e
b
a
F
e
D
D
F
y
x
f
D
D
F
I
P





)
,
(
1
)
,
(
1
)
,
(
)
,
(
1
.
.
provided .
0
)
,
( 
b
a
F
coefficients

3
Example Solve   .
3
2
2
2 y
x
y
x
y
x e
z
D
D
D
D 




0
)
1
)(
( 


 z
D
D
D
D y
x
y
x
 )
(
)
(
.
. 2
1 x
y
e
x
y
F
C x



 


y
x
y
x
y
x
e
D
D
D
D
I
P 3
2
2
2
1
.
. 




y
x
y
x
e
e 3
2
3
2
2
2
6
1
3
2
3
2
1 
 





Hence the complete solution is z = C. F. + P. I.
y
x
x
e
x
y
e
x
y
z 3
2
2
1
6
1
)
(
)
( 





 
 Ans.
coefficients

4
coefficients

5
y
x
y
y
x
x
e
D
D
D
D
I
P 2
3
2
3
4
3
1
.
. 



y
x
e 2
32
6
1
1 



y
x
e 2
27
1 

y
x
e
x
y
x
x
y
x
y
z 2
3
2
1
27
1
)
2
(
)
2
(
)
( 






 


Ans.
coefficients

6
coefficients
Case 2: When )
,
( y
x
f be of the form
)
cos(
or
)
sin( by
ax
by
ax 

Here
  )
cos(
)
sin(
,
,
1
.
. 2
2
by
ax
or
by
ax
D
D
D
D
F
I
P
y
y
x
x



  )
cos(
)
sin(
,
,
1
2
2
by
ax
or
by
ax
b
ab
a
F






provided   0
,
, 2
2



 b
ab
a
F

7
Example Solve  
y
x
z
D
D
D
D y
y
x
x 



 3
sin
)
1
4
( 2
Solution The given equation is not factorized into linear
factors in Dx and Dy
Let,
ky
hx
Ae
z 
 be the trial solution of
0
)
1
4
( 2



 z
D
D
D
D y
y
x
x (1)
ky
hx
x Ahe
z
D 
 ;
ky
hx
x e
Ah
z
D 
 2
2
ky
hx
y Ake
z
D 

ky
hx
y
x Ahke
z
D
D 


8
Solution Substituting these into the equation (1), we get
  0
1
4
2



 ky
hx
e
k
hk
h
A
=>   0
1
4
2



 k
hk
h ; 0

ky
hx
e
=>
h
h
k
4
1
1 2



C. F. = 










 y
h
h
hx
Ae
4
1
1 2

9
 
y
x
D
D
D
D
I
P
y
y
x
x




 3
sin
1
4
1
.
. 2
 
y
x
Dy






 3
sin
1
))
1
.
3
(
(
4
3
1
2
 
y
x
Dy


 3
sin
2
1

10
   
y
x
D
D
y
y



 3
sin
4
2
2
   
y
x
Dy




 3
sin
4
1
2
   
 
y
x
y
x 



 3
sin
2
3
cos
5
1
   
 
y
x
y
x
Ae
z
y
h
h
hx




 











3
sin
2
3
cos
5
1
4
1
1 2

11
Case 3: When n
m
y
x
y
x
f 
)
,
( ; where
m and n are constants and non-negative.
Here
 
 
  n
m
y
x
n
m
y
x
y
x
D
D
F
y
x
D
D
F
I
P
1
,
,
1
.
.



and to evaluate it expand  
 1
,

y
x D
D
F in powers of
x
y
D
D or
y
x
D
D or Dx or Dy and then operate on
xm
yn
term by term.
coefficients

12
y
x
z
D
D
D x
y
x 


 2
2
2
)
2
(
Let,
ky
hx
Ae
z 
0
)
2
( 2
2


 z
D
D
D x
y
x (1)
ky
hx
x Ahe
z
D 

;
ky
hx
x e
Ah
z
D 
 2
2
ky
hx
y e
Ak
z
D 
 2
2

13
Solution Substituting these into the equation (1), we get
  0
2 2
2


 ky
hx
e
h
k
h
A
=>   0
2 2
2


 h
k
h ; 0

ky
hx
e
=> 2
2h
h
k 

C. F.  

 y
h
h
hx
Ae
2
2

14
 
y
x
D
D
D
I
P
x
y
x



 2
2
2
2
1
.
.
 
y
x
D
D
D
D
x
y
x
x











 2
2
2
1
1
 
y
x
D
D
D
D x
y
x
x













2
1
2
2
1
1

15
 
y
x
D
D
D
D
D
D
D x
y
x
x
y
x
x

































 2
2
2
2
2
2
1
1
 
  
y
x
D
D
D
x
x
x



 2
2
2
2
1
1
 
2
.
4
2
.
2
1 2



 x
y
x
Dx

16
x
x
xy
x
8
2
3
2
3




x
x
xy
x
Ae
z y
h
h
hx
8
2
3
2
3
2 2




  


17

18
 
y
x
D
D
I
P
y
x


 2
2
2
1
.
.
 
y
x
D
D
D
x
y
x










 2
2
2
2
1
1
 
y
x
D
D
D x
y
x












2
1
2
2
2
1
1

19
 
y
x
D
D
D x
y
x













 2
2
2
2
1
1
 
0
1 2
2


 y
x
Dx







 xy
x
Dx 3
1 3

20








2
12
2
4
y
x
x












2
12
)
(
)
(
2
4
2
1
y
x
x
x
y
x
y
z 


21
Alternative:
 
y
x
D
D
I
P
y
x


 2
2
2
1
.
.
 
y
x
D
D
D
y
x
y











 2
2
2
2
1
1
 
y
x
D
D
D y
x
y













2
1
2
2
2
1
1

22
 
y
x
D
D
D y
x
y
















 2
2
2
2
1
1











 2
1
1
2
2
2
y
y D
y
x
D
 
2
2
2
1
y
y
x
Dy





23










3
2
1 3
2
2 y
y
y
x
Dy










12
6
2
4
3
2
2
y
y
y
x













12
6
2
)
(
)
(
4
3
2
2
2
1
y
y
y
x
x
y
x
y
z 


1
coefficients
Case 4: If V
e
y
x
f by
ax

)
,
( then
V
b
D
a
D
F
e
y
y
x
by
ax
p
)
,
(
1


 
(i)
n
m
by
ax
y
x
e
y
x
f 

)
,
(
(ii)    
dy
cx
e
or
dy
cx
e
y
x
f by
ax
by
ax


 

cos
sin
)
,
(
(iii)    
by
ax
y
x
or
by
ax
y
x
y
x
f n
m
n
m


 cos
sin
)
,
(

2
Example Solve
y
x
y
x xe
z
D
D 

 2
2
)
(
Let,
ky
hx
Ae
z 
0
)
( 2

 z
D
D y
x (1)
ky
hx
x Ahe
z
D 
 ;
ky
hx
x e
Ah
z
D 
 2
2
ky
hx
y Ake
z
D 


3
Substituting these into the equation (1), we get
  0
2

 ky
hx
e
k
h
A
=>   0
2

 k
h ; 0

ky
hx
e
=> 2
h
k 
C. F.  
 y
h
hx
Ae
2

4
y
x
y
x
xe
D
D
I
P 

 2
2
1
.
.
   
x
D
D
e
y
x
y
x
1
2
1
2
2



 
x
D
D
D
e
y
x
x
y
x
3
4
1
2
2



 

5
x
D
D
D
e x
y
x
y
x
1
2
2
3
4
1
3
1









 



x
D
D
D
D
D
D
e
x
y
x
x
y
x
y
x

























 









 




2
2
2
2
3
4
3
4
1
3
x
D
e x
y
x









3
4
1
3
2

6









3
4
3
2
x
e y
x











 3
4
3
2
2
x
e
Ae
z
y
x
y
h
hx

7
y
x
e
z
D
D y
x
y
x 

 
cos
)
( 3
2
2
Let,
ky
hx
Ae
z 
0
)
( 2

 z
D
D y
x (1)
ky
hx
x Ahe
z
D 

;
ky
hx
x e
Ah
z
D 
 2
2
ky
hx
y Ake
z
D 


8
  0
2

 ky
hx
e
k
h
A
=>   0
2

 k
h ; 0

ky
hx
e
=> 2
h
k 
C. F.  
 y
h
hx
Ae
2

9
 
y
x
e
D
D
I
P y
x
y
x


 
cos
1
.
. 3
2
2
   
 
y
x
D
D
e
y
x
y
x




 
cos
3
2
1
2
3
2
 
y
x
D
D
D
e
y
x
x
y
x




 
cos
1
4
1
2
3
2

10
 
y
x
D
D
e
y
x
y
x





 
cos
1
4
1
1
3
2
 
y
x
D
D
e
y
x
y
x


 
cos
4
1
3
2
 
y
x
D
D
D
D
e
y
x
x
x
y
x


 
cos
4 2
3
2

11
   
 
y
x
D
e x
y
x




 
cos
1
1
4
3
2
 
y
x
e y
x



sin
3
3
2
 
y
x
e
Ae
z
y
x
y
h
hx





 sin
3
3
2
2

12
Example Solve   x
y
x
z
D
D y
x cos
)
( 2



Let,
ky
hx
Ae
z 
0
)
( 2

 z
D
D y
x (1)
ky
hx
x Ahe
z
D 
 ;
ky
hx
x e
Ah
z
D 
 2
2
ky
hx
y Ake
z
D 


13
  0
2

 ky
hx
e
k
h
A
=>   0
2

 k
h ; 0

ky
hx
e
=> 2
h
k 
C. F.  
 y
h
hx
Ae
2

14
  x
y
x
D
D
I
P
y
x
cos
1
.
. 2


 We know
= Real Part of   ix
y
x
e
y
x
D
D


2
1
= Real Part of
   
 
y
x
D
i
D
e
y
x
ix



 0
1
2
 
x
i
x
eix
sin
cos 


15
= Real Part of  
y
x
D
iD
D
e
y
x
x
ix



 1
2
1
2
= Real Part of     
y
x
D
iD
D
e y
x
x
ix





1
2
2
1
1
= Real Part of    
  
y
x
D
iD
D
e y
x
x
ix










 2
1 2

16
= Real Part of   
1
2
0 



 i
y
x
eix
= Real Part of   
1
2
0
sin
cos 




 i
y
x
x
i
x
 
 
x
x
y
x sin
2
cos
1 




 
 
x
x
y
x
Ae
z y
h
hx
sin
2
cos
1
2




  

1
Case-5 General method of finding P. I of HL PDE
when )
,
( y
x D
D
F is homogeneous of y
x D
D and , we
use the following
(a) )
,
(
1
. y
x
f
mD
D
I
P
y
x 

dx
mx
c
x
f
 
 )
,
( , where c
mx
y 

(b) )
,
(
1
. y
x
f
mD
D
I
P
y
x 

dx
mx
c
x
f
 
 )
,
( 1 , where 1
c
mx
y 


2
After performing integration , 1
and c
c must be
replaced mx
y  and mx
y  respectively .
To find P. I , we factories )
,
( y
x D
D
f into linear
factors
)
,
(
)
(
)
)(
(
1
.
2
1
y
x
F
D
m
D
D
m
D
D
m
D
I
P
y
n
x
y
x
y
x 








3
Example Solve
x
y
z
D
D
D
D y
y
x
x
2
1
)
6
5
( 2
2




Solution Replace x
D by m and y
D by 1,
the auxiliary equation becomes
3
,
2
0
6
5
2






 m
m
m
)
3
(
)
2
(
.
. 2
1 x
y
x
y
F
C 


 


4
P. I. =
x
y
D
D
D
D y
y
x
x 2
1
)
6
5
(
1
2
2



x
y
D
D
D
D y
x
y
x 2
1
)
3
)(
2
(
1




dx
x
x
c
D
D y
x
 



2
3
1
)
2
(
1
put c
x
y 
3
[as corresponding to c
x
y
z
D
D y
x 



 3
0
)
3
( ]
dx
x
c
D
D y
x
 


1
)
2
(
1

5
 
x
c
D
D y
x


 log
)
2
(
1
 
x
x
y
D
D y
x



 3
log
)
2
(
1
, put c
x
y 
 3
 
x
y
D
D y
x
2
log
)
2
(
1




6
  dx
x
x
c
 

 2
2
log 1 put 1
2 c
x
y 

[as corresponding to 1
2
0
)
2
( c
x
y
z
D
D y
x 



 ]
  dx
c

 1
log
x
c1
log
  x
x
y 2
log 

Hence complete solution is z = C. F. + P. I.
 x
x
y
x
y
x
y
z 2
log
)
3
(
)
2
( 2
1 




 

which is the required solution of the given equation.

7
Alternative
P. I. =
x
y
D
D
D
D y
y
x
x 2
1
)
6
5
(
1
2
2



x
y
D
D
D
D y
x
y
x 2
1
)
3
)(
2
(
1




dx
x
x
c
D
D y
x
 



2
2
1
)
3
(
1
put c
x
y 
 2
x
y
z
D
D y
x 



 2
0
)
2
( ]

8
c
x
D
D y
x )
3
(
1


x
y
x
D
D y
x 2
)
3
(
1


 , put c
x
y 
 2
dx
x
x
c
x
 


2
3
, put c
x
y 
 3
x
y
z
D
D y
x 



 3
0
)
3
( ]

9
dx
x
c
x
 

dx
x
c
c
x
c
 


  
x
c
c
x 

 log
   
x
y
x
y
x 2
log
3 


 put c
x
y 
 3
   
x
y
x
y
x
x
y
x
y
z 2
log
3
)
3
(
)
2
( 2
1 





 

which is the required solution of the given equation.

10
Short method of finding P. I. of
homogeneous LPDE:
When  
by
ax
y
x
f 
 
)
,
(
  dv
dv
dv
v
b
a
F
I
P 







 
 
)
,
(
1
.
.
where v = ax + by
)
,
( y
x D
D
F is a homogeneous function of
degree n and 0
)
,
( 
b
a
F

11
Example Solve
   
y
x
or
y
x
or
e
z
D
D
D
D y
x
y
y
x
x 3
2
tan
3
2
)
2
( 3
2
2
2




 
Solution Replace x
D by m and y
D by 1,
1
,
1
0
1
2
2






 m
m
m
C. F. = )
(
)
( 2
1 x
y
x
x
y 

 


12
P. I. =  
y
x
D
D
D
D y
y
x
x
3
2
2
1
2
2



dv
v
d
v



 2
2
3
3
.
2
.
2
2
1
6
25
1 3
v
  3
3
2
150
1
y
x 

 3
2
1 3
2
150
1
)
(
)
( y
x
x
y
x
x
y
z 




 

which is the required solution of the given equation

13
Solve  
y
x
z
D
D
D
D y
y
x
x 2
log
)
4
4
( 2
2




Solution Replace x
D by m and y
D by 1,
0
1
4
4 2


 m
m
 
2
1
,
2
1
0
1
2
2



 m
m
C. F. = )
2
1
(
)
2
1
( 2
1 x
y
x
x
y 

 


14
 
y
x
D
D
D
D
I
P
y
y
x
x
2
log
4
4
1
.
. 2
2




dv
v
d
v



 log
2
2
.
1
.
4
1
.
4
1
2
2
The denominator vanishes when 1

x
D and
2

y
D .
So differentiating the denominator w. r. to
x
D and multiplying the numerator by x .

15
dv
v
D
D
x
y
x


 log
4
8
dv
v
x


 log
2
.
4
1
.
8
Again case of failure.
So differentiating the denominator w. r. to
x
D and multiplying the numerator by x .

16
v
x
log
8
2

 
y
x
x
2
log
8
2


 
y
x
x
x
y
x
x
y
z 2
log
8
)
2
1
(
)
2
1
(
2
2
1 




 

which is the required solution of the given
equation

17
Example ).
2
cos(
)
6
( 2
2
y
x
z
D
D
D
D y
y
x
x 



Solution Replace x
D by m and y
D by 1,
0
6
2


 m
m giving m = 2, -3
 C. F. = )
3
(
)
2
( 2
1 x
y
x
y 

 

)
2
cos(
)
6
(
1
.
. 2
2
y
x
D
D
D
D
I
P
y
y
x
x




The denominator becomes zero when
2
,
1
,
2 2
2
2





 y
x
y
x D
D
D
D . It is case of failure

18
Differentiating denominator w.r.to Dx and
multiplying numerator by x, we get
)
2
cos(
2
1
.
. y
x
D
D
x
I
P
y
x



)
2
cos(
2 2
y
x
D
D
D
D
x
y
x
x
x


 Multiplying by Dx
)
2
cos(
2
8
y
x
D
x x



 )
2
sin(
5
y
x
x


)
2
sin(
5
)
3
(
)
2
( 2
1 y
x
x
x
y
x
x
y
z 





 


)
,
(
1
0
2
2
2
2
2
2
2
1
1
1
1
0
y
x
f
Kz
y
z
y
L
x
z
x
L
y
z
y
B
x
z
x
B
y
z
y
A
y
x
z
y
x
A
x
z
x
A n
n
n
n
n
n
n
n
n
n
=
+


+


+


+


+
+


+
+



+


−
−






where K
L
A
A ,
,
,
, 1
1
0  are constant, is called a linear
partial differential equation of order n with variable
coefficients.
Also called homogeneous linear equation of order n.
Equations of the second order with variable coefficients

A partial differential equation of the form
)
,
(
)
,
( y
x
f
yD
xD
F y
x = having variable coefficients
can be reduced to linear partial differential
equation of order n with constant coefficients by
suitable transformations.
Let us change the independent variables x, y to u
and v by the substitutions

v
u
e
y
e
x =
= ,
or, y
v
x
u log
,
log =
=
u
z
x
x
u
u
z
x
z
z
Dx


=




=


=
1
or,
u
z
x
z
x


=


i. e. z
D
z
xD u
x =











=










=


u
z
x
x
x
z
x
x
z 1
2
2












−


+




= 2
2
2
1
1
x
u
z
x
u
u
z
x










−


=
u
z
x
u
z
x 2
2
2
2
1
1










−


=
u
z
u
z
x 2
2
2
1

or, z
D
D
u
z
u
z
x
z
x u
u )
( 2
2
2
2
2
2
−
=


−


=


i. e. )
1
(
2
2
2
2
2
−
=
=


u
u
x D
D
D
x
x
x
Similarly, )
2
)(
1
(
3
3
3
−
−
=


u
u
u D
D
D
x
x

Proceeding in a similar way we obtain
v
z
y
z
y


=


v
D
y
y =


)
1
(
2
2
2
−
=


v
v D
D
y
y
)
2
)(
1
(
3
3
3
−
−
=


v
v
v D
D
D
y
y etc.

( )( ) ( )
( ) ( )
( )
v
u
f
z
K
D
L
D
L
D
D
B
D
D
B
n
D
D
D
D
A
v
u
v
v
u
u
u
u
u
u
,
1
1
1
2
1
1
0
2
1
0
=






+
+
+
−
+
−
+
−
−
−
−
−
−
−
−
+
+
−
−
−
−
−
This is the linear partial differential
equation of higher order with constant
coefficients with the independent variables
u and v.

Solve xy
y
z
y
x
z
x =


−


2
2
2
2
2
2
Solution Let
v
u
e
y
e
x =
= ,
or, y
v
x
u log
,
log =
=
u
z
x
x
u
u
z
x
z


=




=

 1
=> z
D
u
z
x
z
x u
=


=



and z
D
D
u
z
u
z
x
z
x u
u )
1
(
2
2
2
2
2
−
=


−


=


Proceeding in a similar way we obtain
z
D
v
z
y
z
y v
=


=


)
1
(
2
2
2
−
=


v
v D
D
y
y

Now the given equation becomes
v
u
v
v
u
u e
z
D
D
D
D +
=
−
−
− )]
1
(
)
1
(
[
=>
v
u
v
u
v
u e
z
D
D
D
D +
=
+
−
− )
( 2
2
This is the non-homogeneous linear partial
differential equation with constant coefficients.
Here dependent variable z and independent
variables u and v.

0
)
( 2
2
=
+
−
− z
D
D
D
D v
u
v
u
or, 0
)
1
)(
( =
−
+
− z
D
D
D
D v
u
v
u
)
(
)
(
.
. 2
1 u
v
e
u
v
F
C u
−
+
+
= 

)
log
(log
)
log
(log 2
1 x
y
x
x
y −
+
+
= 







+
=
x
y
xf
xy
f 2
1 )
(

Particular integral
P. I. v
u
v
u
v
u
e
D
D
D
D
+
+
−
−
= 2
2
1
v
u
u
ue
D
+
−
=
1
2
1
Case of failure.
The denominator vanishes when 1
=
u
D and
1
=
v
D . So differentiating the denominator w. r.
to u
D and multiplying by u.

v
u
e
u +
=
= (xy) logx
Retuning to the original independent variable x
and y by putting eu
= x and ev
= y we get the
general solution y = C. F. + P. I.= yc + yp is
x
xy
x
y
xf
xy
f
I
P
F
C
z log
)
(
.
.
.
. 2
1 +






+
=
+
=

Solve x
y
z
y
x
z
x
y
z
y
x
z
x log
2
2
2
2
2
2
=


−


+


−


Solution Let
v
u
e
y
e
x =
= ,
or, y
v
x
u log
,
log =
=
u
z
x
x
u
u
z
x
z


=




=

 1
=> z
D
u
z
x
z
x u
=


=



Now the given equation becomes
u
z
D
D
D
D
D
D v
u
v
v
u
u =
−
+
−
−
− ]
)
1
(
)
1
(
[
=> u
z
D
D v
u =
− )
( 2
2
This is the homogeneous linear partial
differential equation with constant coefficients.
Here dependent variable z and independent
variables u and v.

Replace u
D by m and v
D by 1,
0
1
2
=
−
m or, 1

=
m
)
(
)
(
.
. 2
1 u
v
u
v
F
C −
+
+
= 

)
log
(log
)
log
(log 2
1 x
y
x
y −
+
+
= 







+
=
x
y
f
xy
f 2
1 )
(

Particular integral
P. I. u
D
D v
u
2
2
1
−
=
du
du
u

−
=
0
1
1
6
3
u
=
6
)
(log 3
x
=

Alternative method:
P. I. u
D
D v
u
2
2
1
−
=
u
D
D
D u
v
u






+
= 2
2
2
1
1
 
0
1
2
+
= u
Du 6
3
u
=
6
)
(log 3
x
=

Retuning to the original independent variable x
and y by putting eu
= x and ev
= y we get the
general solution y = C. F. + P. I. = yc + yp is
6
)
(log
)
(
.
.
.
.
3
2
1
x
x
y
f
xy
f
I
P
F
C
z +






+
=
+
=

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