PARTIAL DIFFERENTIAL EQUATIONS

Weightage for university exam: 18Marks

 Def n :
A diff. equation which involves partial derivatives is called partial differential
z z
equation. for e.g. x  y  1 .
 Formation of a differential equation:

To form a differential equation from a given relationship of variables we will

eliminate arbitrary constants or arbitrary functions of these variables using
differentiation.

 Solution of a partial differential Equation:

A solution or integral of a partial differential equation is a relation between the

variables, which is free from derivatives and satisfies the partial differential
equation.
 2u  2u
For e.g. u = e x siny is a solution of   0.
x 2 y 2
 Complete Integral:

The solution which contains a number of arbitrary constants equal to the order of the
differential equation is called a complete integral.

 Methods To Solve Partial Differential Equations:

 Method of Direct Integration:

This method is applicable to those problems, where direct Integration is

possible. This solution depends on definition of P.D.E.

 Lagrange’s equation :

The partial differential equation Pp + Qq =R,

Where P, Q, R are functions of x, y, z.
is known as Lagrange’s (linear) equation.

The general solution of the partial differential equation Pp + Qq =R, is

 (u(x,y,z), v(x,y,z)) = 0,

Where  is an arbitrary function and u(x,y,z) = c1and v(x,y,z) = c2 are two

dx dy dz
independent solutions of P  Q  R (subsidiary auxiliary equations).

 Solutions of Nonlinear Partial Differential Equations:

Lecture notes by Mr. Zalak Patel Page: - 1 -

 Standard forms:

(1) Equations involving only P and Q:

Form : f ( p, q ) = 0
Method to get solution : Replace p by a and q by b in f ( p, q ) =0
Get form f ( a, b) = 0 and find relation b = F(a).
Solution : z = ax + F (a) + c.

(2) Equations not involving the independent variables:

Form : f ( z , p, q ) = 0 ….(1)
Method to get solution : assume q = ap.
Then the (1) becomes f ( z , p, ap) = 0
Get form p =  ( z , a )
Use dz =  ( z , a ) dx + a  ( z , a ) dy
dz
Solution :   ( z , a) = x + ay + b.

(3) Separable Equations:

Form : f 1 ( x, p )  f 2  y , q  ….(1)
Method to get solution : assume f1 ( x, p ) = a = f 2  y, q 
Solving f1 ( x, p ) = a, get p = 1 ( x, a )
Solving f 2 ( y, q ) = a, get p =  2 ( y , a )
Use dz = 1 ( x, a ) dx +  2 ( y, a ) dy
Solution : z =  1 ( x, a ) dx +   2 ( y, a ) dy + b.

(4) Clairaut’s Form:

Form : z = px +qy + f ( p, q ) ….(1)

Method to get solution : Replace p by a and q by b in (1)
Solution : z = ax + by + f ( a, b) .

(5) Equations Reducible to standard Forms:

This category includes those p.d.e. which do not fall directly under any
of the above four forms
We can reduce our given equations in one of the above four forms

 Homogeneous Linear Partial Differential Equations With Constant Coefficient:

A diff. equation
nz nz nz nz
 a1  a 2  ..............................  a n  F ( x, y )
x n x n 1 y x n  2 y 2 y n
Where a1 , a 2 ,............., a n are constants
Lecture notes by Mr. Zalak Patel Page: - 2 -
 is called Linear partial differential equation of nth order first degree.

Above equation can be written by

( D n  a1 D n 1 D' a 2 D n  2 D' 2 ..........................  a n D' n ) z  F ( x, y ) ……(1)
d d
Where D = and D’ = dy
dx
(1) can be written as f (D,D’) z= F(x,y)

 General Solution of Differential Equation:

General solution = Complementary function + Particular integral

G.S. = C.F. + P.I.

 Complementary function (C.F.)of Differential Equation

The solution which contains a number of arbitrary constants equal to the order
of the differential equation is called the complementary function (C.F.) of a
Differential equation.

 Auxiliary Equation:

An equation f (D,D’) = 0 is Auxiliary Equation .

 Method to find C.F. :

Solve equation ( D n  a1 D n 1 D' a2 D n  2 D'2 ..........................  an D'n ) z = o

for values of D & D’.
Say factorization is  D  m1D' D  m2 D' D  m3 D'............. D  mn D'  0 .
Then we can classify roots and we can find C.F. by following way

Sr.
Classification of roots C.F.
No.
If roots D = m1, m2 ,m3 ,………,mn y  f 1 ( y  m1 x)  f 2 ( y  m 2 x)  ..........
1
of Auxiliary equation are real and distinct
Two roots are equal (real roots) y  f 1 ( y  m1 x)  xf 2 ( y  m1 x)
2
i.e. If roots are m1= m2 , m3 ……,mn  f 3 ( y  m3 x).........

 Method to find P.I. :

Consider f (D, D’) z = F(x, y)

1
Particular integral can be given by P.I. = f ( D, D ' ) F ( x, y )
 Short cut methods to find P.I.:

Let given L.P.D.E. is. f (D, D’) z = F(x, y)

We can use following short cut methods to find P.I.

1 1
(1) e ax  by  e ax by if f ( a, b)  0 .
f ( D, D ' ) f ( a , b)

Lecture notes by Mr. Zalak Patel Page: - 3 -

 1 1
(2) sin  ax  by   sin  ax  by 
2 2
f ( D , DD' , D ' ) f (a ,ab,b 2 )
2

if f (  a 2 , ab,b 2 )  0
and
1 1
cos ax  by   cos ax  by 
2 2
f ( D , DD' , D' ) f ( a ,ab,b 2 )
2

if f (  a 2 , ab,b 2 )  0

1 1
xm yn  xm yn
(3) f ( D, D ' ) D
1   
 D' 

1
Use 1  t 1  t  t t ...................
2 3

1 1
Or 1  t 1  t  t t ................... On 1   ( D ) and use
2 3

1 1
F ( x, y )   F ( x, y ) dx and F ( x, y ) dy to get P.I.
D D'

1 1
(4) e ax  by  ( x, y )  e ax  by  ( x, y )
f ( D, D ' ) f ( D  a, D'b)

(5) General method to find P.I.:

1
P.I. = f ( D, D ' ) F ( x, y )
1
= F ( x, y )
 D  m1 D' D  m 2 D' D  m3 D'............. D  m n D'
 1  1
  F ( x, y )
  D  m1  D  m 2  D  m3 .............  D  m1
 1 
   F ( x, c  m1 x) dx
  D  m1  D  m 2  D  m3 ............. 
Where c1 is replaced by y + m1x.
Continue above process for all factors of f (D) . And get P.I. for y.

Lecture notes by Mr. Zalak Patel Page: - 4 -