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in
Module – I Engineering Mathematics - IV 2018

## Partial Differential Equations (PDE)

Syllabus:- Formulation of Partial differential equations by elimination
arbitrary constants /functions, solution of non-homogeneous Partial
differential equations by direct integration, solution of homogeneous
Partial differential equations involving derivative with respect to one
independent variable only. Derivation of one dimensional heat and wave
equations and their solutions by variable separable method.

INTRODUCTION:
Many problems in vibration of strings, heat conduction, and electrostatics

in
involve two or more variables. Analysis of these problems lead to partial
derivatives and equations involving them. In this unit we first discuss the

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formation of PDE analogous to that of formation of ODE. Later we discuss
some methods of solving PDE also, discuss wave and heat equations.
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Definitions: Partial Differential Equations (PDE)
x
An equation involving one or more partial derivatives of a function of two or
more variables is called partial differential equations
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## Example:- If z is a function of two independent variable in x and y

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Order of PDE:- The order is the highest derivative present in the equation
called order of PDE.

Degree of PDE:- The positive integral power (or degree) of the highest order
derivative in the equation called PDE.

## Homogeneous and Non- Homogeneous PDE:-In a PDE each term contains

either the dependent variable or one of its partial derivatives, the PDE is said to
be homogeneous. Otherwise it is non homogeneous PDE

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Module – I Engineering Mathematics - IV 2018

Examples:-

## Formation of PDE by eliminatory arbitrary constants

in
and arbitrary functions

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Eliminatory arbitrary constants:-Consider the relation of the form
f(x,y, z, a,b) = 0, where z is a function two independent variable x and y and
arbitrary constants a, b. Differentiate z partially with respect x and y and
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eliminate arbitrary constant a, b to form PDE.

## Note: - 1. If number of arbitrary constant < number of independent variables

x
then PDE required First order derivative
sE

## 2. If number of arbitrary constant > number of independent variables then

PDE required second and higher order derivative
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## Notations: The following notations when z is a function of two independent

variable x, y will be used

z z
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p  zx  q  zy 
x y
2z 2z
r  z xx  s  z xy 
x 2 xy
 2u
r  z yy 
y 2

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Module – I Engineering Mathematics - IV 2018
Examples: - Form a PDE by eliminating the arbitrary constants

1. z  ( x  a )( y  b)
Solution : Given, z  ( x  a )( y  b)  (1)
Diff; partially w.r.t x and y
z
 p  ( y  b)  (2)
x
z
 q  ( x  a)  (3) , Substitute (2) and (3) in (1)
y
 z  z 
we get z      pq is the requrid PDE
 x  y 

in
rt.
2. z  ( x  a ) 2  ( y  b) 2

## Diff; partially w.r.t x and y

x pe
Solution : Given, z  ( x  a ) 2  ( y  b) 2  (1)

z
 p  2( x  a )  (2)
sE

x
z
 q  2( y  b)  (3) , Substitute (2) and (3) in (1)
y
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## we get 4 z  p 2  q 2 is the requrid PDE

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3. z  a log( x 2  y 2 )  b
Solution : Given, z  a log( x 2  y 2 )  b  (1)
Diff; partially w.r.t x and y
z 2ax
 p 2  (2)
x x  y2
z 2ay
q 2  (3) , now , (2) / (3)
y x  y2
we get py-qx  0 is the requrid PDE

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Module – I Engineering Mathematics - IV 2018
Form a PDE by eliminating the arbitrary function

1. z  f ( x 2  y 2 )
Solution : Given, z  f ( x 2  y 2 )  (1)
Differentiating (1); partially w.r.t x and y
z
 p  f ( x 2  y 2 )(2 x)    (2)
x
z
 q  f ( x 2  y 2 )(2 y )    (3) , Now,
2
y 3
we get py  qx  0 is the requrid PDE

in
rt.
1 
2. z  y 2  2 f   log y 
x 
1 
pe
Solution : Given, z  y 2  2 f   log y   (1)
x 
x
Differentiating (1); partially w.r.t x and y
sE

z 1  1 
 p  2 f   log y   2     ( 2)
x x  x 
z 1  1 
 q  2 y  2 f   log y      (3) ,
am

y x  y 

1  1 
q  2 y  2 f   log y      (3) now ,
2
x  y  3
Ex

px 2
we get  1
q  2 y  y
px 2  q  2 y  y
px 2 q  2 y  y  0
px 2  qy  2 y 2 is the requried PDE

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Module – I Engineering Mathematics - IV 2018

3. lx  my  nz   ( x 2  y 2  z 2 )
Solution : Given, lx  my  nz   ( x 2  y 2  z 2 )    (1)
Differentiating (1); partially w.r.t x and y
l  np   ( x 2  y 2  z 2 )(2 x  2 zp)    (2)

## m  nq   ( x 2  y 2  z 2 )(2 y  2 zq)    (3) , Now,

2
3
l  np x  zp
 on simplification
m  nq y  zq
mz  ny  p  nx  lz q  ly  mx is the requrid PDE

in
4.  ( x  y  z , x 2  y 2  z 2 )  0

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Solution : Given,  ( x  y  z , x 2  y 2  z 2 )    (1)
Let us assume , u  x  y  z v x2 y2z 2
Differentiating (u , v); partially w.r.t x and y
ux  1  p v x  2( x  zp)
x pe
uy  1 q v y  2( y  zq)
sE

u x vx 1  p 2( x  zp)
Now,  ,  on simplifica tion
u y vy 1  q 2( y  zq)
 y  z  p  x  z q  x  y is the requrid PDE
am

5.  ( x 2 2 yz , y 2 2 zx)  0
Ex

## Solution : Given,  ( x 2 2 yz , y 2 2 zx)  0    (1)

Let us assume , u  x 2 2 yz v  y 2 2 zx
Differentiating (u , v); partially w.r.t x and y
u x  2 x  2 yp v x  2( z  xp )
u y  2( yq  z ) v y  (2 y  2 xq )
ux v x  yp z  xp
Now,  x,  on simplification
u y vy yq  z y  xq
y 2
    
 zx p  x 2  yz q  z 2  xy is the requrid PDE

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Module – I Engineering Mathematics - IV 2018
Form a PDE by eliminating the arbitrary functions

1. z  f ( x  at )  g ( x  at )    (1)
Differntiate z partially w.r.t x and y
z x  p  f ( x  at )  g ( x  at )
z xx  r  f ( x  at )  g ( x  at )  (2)
z y  q  f ( x  at )(a )  g ( x  at )( a )
z yy  q  f ( x  at )(a ) 2  g ( x  at )( a ) 2
z yy  t  a 2  f ( x  at )  g ( x  at )  (3)
substitute(2) in (3) we get

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z yy  a 2 z xx or r  a 2t is the requried PDE

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2. z  f ( y  x )  g ( y  2 x )
Sol : z  f ( y  x)  g ( y  2 x)
x pe
Differentiate z partially w.r.t x and y
z x  p  f ( y  x)  2 g ( y  2 x)
sE

z y  p  f ( y  x)  g ( y  2 x)
z xx  r  f ( y  x)  4 g ( y  2 x)    (1)
am

## z xy  t  f ( y  x)  2 g ( y  2 x)    (2)

z yy  t  f ( y  x)  g ( y  2 x)    (3)
(1)  (2) will gives, r-s  2 g ( y  2 x)  (4)
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## (2)  (3) will gives, s-t  g ( y  2 x)    (5)

4 will gives, r-s  2
5 s-t 1
r-s  2( s  t ) or r  3s  2t  0
z xx  3 z xy  2 z yy  0

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3. z  f ( y  2 x )  g ( 2 y  x )
Sol : z  f ( y  2 x)  g (2 y  x)
Differentiate z partially w.r.t x and y
z x  p  2 f ( y  2 x)  g (2 y  x)
z y  q  f ( y  2 x)  2 g (2 y  x)
z xx  r  4 f ( y  2 x)  g (2 y  x)    (1)
z xy  t  2 f ( y  2 x)  2 g (2 y  x)    (2)
z yy  t  f ( y  2 x)  4 g (2 y  x)    (3)
2  (1)  (2) will gives, 2r  s  6 f ( y  2 x)  (4)

in
2  (2)  (3) will gives, 2 s  t  3 f ( y  2 x)    (5)
4 will gives, 2r  s   2

rt.
5 2s  t 1
2r  s  2(2 s  t ) or 2r  5s  2t  0 pe
2 z xx  5 z xy  2 z yy  0 is the requried PDE.
x
sE

4. z  f ( y  x )  g ( y  2 x )
Sol : z  f ( y  x)  g ( y  2 x)
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## Differentiate z partially w.r.t x and y

z x  p  f ( y  x)  2 g ( y  2 x)
z y  q  f ( y  x)  g ( y  2 x)
Ex

## z xx  r  f ( y  x)  4 g ( y  2 x)    (1)

z xy  s  f ( y  x)  2 g ( y  2 x)    ( 2)
z yy  t  f ( y  x)  g ( y  2 x)    (3)
(1)  ( 2) will gives, r-s  2 g ( y  2 x)  (4)
( 2)  (3) will gives, s-t  g ( y  2 x)    (5)
4 will gives, r-s  2
5 s-t 1
r-s  2( s  t ) or r  3s  2t  0
z xx  3 z xy  2 z yy  0

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Module – I Engineering Mathematics - IV 2018
5. z  f ( y  2 x )  g ( 2 y  x )
Sol : z  f ( y  2 x)  g (2 y  x)
Differentiate u partially w.r.t x and y
z x  p  2 f ( y  2 x)  g (2 y  x)
z y  q  f ( y  2 x)  2 g ( 2 y  x)
z xx  r  4 f ( y  2 x)  g ( 2 y  x)    (1)
z xy  s  2 f ( y  2 x)  2 g ( 2 y  x)    (2)
z yy  t  f ( y  2 x)  4 g (2 y  x)    (3)
2  (1)  ( 2) will gives, 2r  s  6 f ( y  2 x)  (4)

in
2  ( 2)  (3) will gives, 2 s  t  3 f ( y  2 x)    (5)
4 will gives, 2r  s   2

rt.
5 2s  t 1
2r  s  2(2 s  t ) or 2r  5s  2t  0 pe
2 z xx  5 z xy  2 z yy  0 is the requried PDE.
x
Partial Differential Equations (PDE)
sE

## Equation solvable by Direct Integration

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The PDE which can be solved by direct integration, In the place of usual
constants of integration, we must, however, use arbitrary functions of the
variable held is fixed.
Ex

1. Solve :  z2  x  y
2

x
Solution :  z2  x  y
2

x
Integratin g twice with respect to x (keeping y fixed)
z  x  yx  f ( y)
2

x 2
x 3 yx 2
z   xf ( y)  g ( y) is the required solution
6 2

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Module – I Engineering Mathematics - IV 2018

2.Solve :  z   a
2
x
xy y
Solution : Integratin g w.r.t x (keeping y fixed)
z   1  x  ax  f ( y)
2

y  y  2
Integratin g w.r.t y (keeping x fixed)
 2
x 
z   log y  axy   f ( y)dy  g ( x)
 2 
 

3. Solve :  z  18 xy 2  sin( 2 x  y)  0
3

x y
2

## Solution : Integratin g twice w.r.t x (keeping y fixed)

in
2 z
 9 x 2 y 2  cos(2 x  y )  f ( y)
xy 2

rt.
z sin( 2 x  y )
 3x 3 y 2   xf ( y)  g ( y)
y 4
Integratin g w.r.t y (keeping x fixed)
pe
z  x 3 y 3  cos(2 x  y )  x  f ( y)dy   g ( y)dy  w( x)
4
cos(2 x  y )
zx y   xF ( y)  G ( y)  w( x)
x
3 3

4
sE

2

xy y
am

## and z  0 if y is odd multiples of  2.

Solution : Integratin g w.r.t x (keeping y fixed)
z
  sin y cos x  f ( y)    (1)
y
Ex

z
Given,  2 sin y when x  0
y
 2 sin y   sin y cos 0  f ( y)
 2 sin y  sin y  f ( y)
- sin y  f ( y)
(1) becomes
z
  sin y cos x  sin y    (2)
y
integrate w.r.t y
z  cos y cos x  cos y  g(x)
given, z  0 if y  (2n  1) 2 in eq n (2)
we get g(x)  0
z  cos y cos x  cos y  cos y(cos x  1)

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## 5. Solve :  z2  xy for which z  log( 1  y) when x  1,

2

x x
and z  0 when x  0
Solution : Integratin g twice w.r.t x (keeping y fixed)
z yx
2

  f ( y)    (1)
x 2
z
Given,  log( 1  y) when x  1
x
y
log( 1  y)   f ( y)
2
y
log( 1  y) -  f ( y)
2

in
(1) becomes
z yx
2
y
  log( 1  y) -    (2)

rt.
x 2 2
integrate w.r.t x
z
yx 3 
6 
y
  log( 1  y) -  x  g ( y)
given, z  0 when x  0
2
x pe
we get g(y)  0
sE

yx 3  y
z   log( 1  y) -  x
6  2
am

## Partial Differential Equations (PDE)

Solution of homogeneous PDE involving derivatives with
Ex

## respect to one indecent variable only

Suppose that the dependent variable has been differentiated partially w.r.t.
one independent variable say x only. Then PDE can be treated as an ordinary
differential equation (ODE). The arbitrary constants in the solution are then
replaced by arbitrary function of the other variable (y) giving a solution of the
PDE

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Module – I Engineering Mathematics - IV 2018

2z z
1. Solve :  z given that z  0 and  sin x, when y  0,
y 2 y
Solution :
Here z is function of y only.The PDE can be assume
in the form ODE
d 2z
 z  0 or  D  1 z  0 where
2 d
D
dy 2   dy
Auxillary Equation is
D 2  1  0 or D 2  1

in
D   1   1 (real and different roots )
complementary function

rt.
z  c1e y  c 2 e  y (for ODE)
The solution for PDE, replace c1 & c 2 by function y
z  f(x)e y  g(x)e  y - - - - - (1)
pe
Now find the value of f(x) & g(x) using the given data
x
When y  0, z  0 hence eq (1)becomes
sE

0  f(x)e 0  g(x)e 0
0  f(x)  g(x)    (2)
z
am

## Also by data, when y  0 ,  sin x

y
differentiate Eq.(1) w.r.t. y partially we get
z
 f(x)e y  g(x)e  y
Ex

y
sin x  f(x)e 0  g(x)e 0
sin x
Eq.(2)  Eq.(3) f ( y) 
2
 sin x
Eq.(2)  Eq.(3) g ( x) 
2
Eq.1 becomes
 sin x   y   sin x 
z  ey e  
 2   2 
 e y  e y 
z    sin x
 2 
z
Dr. sinh y  sinMIT,
A.H.Srinivasa, x is Mysore
the required solution Page 11
sin x  f(x)  g(x)    (3)
solve Eq.(2) and Eq.(3)For More Notes Visit ExamsExpert.in
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Module – I Engineering Mathematics - IV 2018

2z z
2. Solve : 3  4 z  0 subject to conditions that
x 2 x
z
z  1 and  y when x  0,
x
Solution :
Here z is function of x only . The PDE can be assume
in the form ODE
d 2z dz  4 z  0 or  D 2  3D  4  z  0 where d  D
3  
dx  
2
dx dx

in
Auxillary Equation is
D 2  3D  4  0

rt.
D  1, D  4, ( real and different roots )
comp lement ary function
z  c1e x  c 2 e  4 x (for ODE) pe
The solution for PDE, rep lace c1 & c 2 by function y
z  f(y)e x  g(y)e  4 x - - - - - (1)
x
Now find the value of f(y) & g(y) using the given data
sE

## When x  0, z  1 hence Eq.(1) becomes

z
1  f(y)  g(y)    ( 2) Also by data,  y , when x  0 ,
x
am

## differentiate Eq.(1) w.r.t. x p artially we get

z
 f(y)e x  4 g(y)e  4 x
x
Ex

y  f(ye 0  4 g(y)e 0
y  f(y )  4 g(y)    (3)
solve Eq.(2) and Eq.(3)
solveing Eq.(2) & Eq.(3)
4 y 1 y
f(y)  & g(y) 
5 5
 4  y  x  1  y  4 x
z  e   e is the required solution
 5   5 

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## 3. Solve : 2  z given that when y  0, z  e and z  e

2z x x

y y
Solution :
Here z is function of y only. The PDE can be assume
in the form ODE
d 2z
 z  0 or  D 2  1 z  0 where
d
D
 
2
dy dx
Auxillary Equation is
D 2  1  0 or D 2  1
D   1   1 (real and different roots )

in
complementary function

rt.
z  c1e y  c 2 e  y (for ODE)
The solution for PDE, replace c1 & c 2 by function y
z  f(x)e y  g(x)e  y - - - - - (1) pe
Now find the value of f(x) & g(x) using the given data
When y  0, z  e x hence Eq.(1)becomes
x
e x  f(x)e 0  g(x)e 0
sE

## e x  f(x)  g(x)    (2)

z
Also by data, when y  0 ,  ex
y
am

## differentiate Eq.(1) w.r.t. y partially we get

z
 f(x)e y  g(x)e  y
y
Ex

e  x  f(x)e 0  g(x)e 0
e  x  f(x)  g(x)    (3)
Solve Eq.(2) and Eq.(3)
Eq. (2)  Eq. (3)
2 f ( x)  e x  e  x
e x  ex
f ( x)   cosh x
2
eq (2)  eq (3)
2 f ( x)  e x  e  x
e x  ex
g ( x)   sinh x
2
z  e y cosh x  e  y sinh x is the required solution
Dr. A.H.Srinivasa, MIT, Mysore Page 13

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Module – I Engineering Mathematics - IV 2018

## Application of Partial Differential equation

In physical problems, we always seek a solution of the differential equation
which satisfies some specified conditions known as the boundary conditions.
The differential equation together with these boundary conditions, constitute a
boundary value problems.In problems involving ordinary differential equations,
we may first find general solution and then determine the arbitrary constants
from the initial values. But the same process is not applicable to problems
involving PDE’s for the general solution of a PDE contains arbitrary functions

in
which are difficult to adjust so as to satisfy the given boundary conditions.

## ONE DIMENSTIONAL WAVE EQUATIONS

rt.
Consider a flexible string tightly stretched between two fixed points at a
pe
distance l. Let  be the mass per unit length of the string.

## We shall assume the following.

x
i). The tension T of the string is small through out.
sE

am

Ex

## Since there is no motion in the horizontal direction, the horizontal components

T1 and T2 must cancel each other.

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Module – I Engineering Mathematics - IV 2018
Where  &  are the angles made by T1 and T2 with horizontal. Vertical
components of tension are –T1 sin and T2 cos, where negative sign indicate
T1 directed to downwards. Hence the resultant force acting vertically upwards
is T2 sin–T1 sin

## T2 sin–T1 sin = (x)utt

[x is the mass of the element portion AB and utt represents acc^n]

in
Divide throughout by T we have

  2u

rt.
T2 T
sin   1 sin   x 2
T T T t
T 1 T2 1
Using Eq : 1 1  
T cos 
sin  sin    2 u
  x
;
T cos  pe
cos  cos  T t 2
x
  2u
sE

## tan   tan   x 2      (2)

T t
But tan  and tan  represents the slope at B( x  x) and A(x) respectively
 u   u 
am

## tan     and tan   

 x  x x  x  x
 u   u    2u
      x 2 divide by x and apply limit x  0
 x  x x  x  x T t
Ex

 u   u 
   
 x  x x  x  x   u
2
Lim 
x 0 x T t 2
 2u   2u T
  c 2 we get
x 2
T t 2

 2u 2  u
2
 c or u tt  c 2 u xx
t 2
x 2

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Module – I Engineering Mathematics - IV 2018
VARIOUS POSSIBLE SOLUTIONS OF ONE DIMENSTIONAL WAVE
EQUATIONS

## Let the one dimension wave equation

 2u 2  u
2
Consider  c
t 2 x 2
Let u  XT - - - - - - - - - - - - - - - (1)
where X ( x), T (t ) be the solution
of the PDE
1 d 2T 1 d 2 X

c 2T dt 2 X dx 2

in
equating to common constant
1 d 2T 1 d2X

rt.
2 2
 k and k
c T dt X dx 2
2
d T d2X
 c 2 kT  0 and  kX  0
dt

2

D 2  c 2 k T  0 and 
dx 2

D2  k X  0
pe
2 2
x
d d
where, D 2  2
 2
dx dt
sE
am
Ex

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Module – I Engineering Mathematics - IV 2018
Case 1 :
Let k  0
D 2 X  0 and D 2T  0
Solution are given
X  c1  c 2 x e 0 x and T  c 3  c 4 t e 0t
X  c1  c 2 x  and T  c 3  c 4 t 
Hence the solution of PDE
U  XT  c1  c 2 x c 3  c 4 t 
Case 2 : Let k be negative say, k  p 2
D 2
 
 p 2 X  0 and D 2  c 2 p 2 T  0

in
A.E is
D 2  p 2  0 and D 2  c 2 p 2  0

rt.
D  p & D  cp
Solution are given
  
X  c 5 e px  c 6 e  px and T  c 7 e pct  c 8 e  pct
Hence the solution of PDE
 pe
Case 3 : Let k be negative say, k   p 2
x
D 2
 
 p 2 X  0 and D 2  c 2 p 2 T  0
sE

A.E is
D 2  p 2  0 and D 2  c 2 p 2  0
am

## D  ip & D  icp

Solution are given
X  c 9 cos px  c10 sin px  and T  c11 cos pct  c12 sin pct 
Ex

## Hence the solution of PDE

U  XT  c 9 cos px  c10 sin px c11 cos pct  c12 sin pct 
 
U  XT  c 5 e px  c 6 e  px c 7 e pct  c 8 e  pct 
ONE DIMENSTIONAL HEAT EQUATIONS
We shall assume the following.

## i). Heat flows from higher temperature to a lower temperature.

ii). The amount of heat in a body is proportional to its mass and temperature.

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Module – I Engineering Mathematics - IV 2018
iii) The rate of inflow across an area is proportional to the area and to the
temperature gradient normal to the area where the constant of proportionality
(K) is called the thermal conductivity

Fig

Consider the homogeneous bar of constant cross – section area A. Let  be the
density, S be the specific heat and K is the thermal conductivity of the
material. Let the sides be insulated so that the stream lines of heat flow are
parllel and perpendicular to the area A.

in
Let one end of the bar be taken as a origin O and the direction of the heat flow
be the positive X – axis.

rt.
Let U (x, t) be the temperature of the slab at a distance x from the origin.

pe
Consider an element of bar between the planes PQRS and P`Q`R`S` at a
distance x and x + x from the end O.

x
sE

am

= A  s x u

## Hence the rate of increase of heat in this slab element

Ex

u
R   Asx       (1)
t
If R 1 is the rate in flow of heat and R 0
is the rate of out flow of heat we have
 u   u 
R1   KA  and R o   KA       (2)
 t  x  t  x x
where the negative sign is due to emprical law (1)
Hence we have from (1) and (2)

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Module – I Engineering Mathematics - IV 2018

R  R 1  R0

 Asx  u  KA u   u 
 KA 
t  x  x x  x 
  u   u  
 KA   KA  
u K x  x  x 
    x x
t s  x 
 
 
Apply limit x  0
u K  2u

t s x 2

in
u  2u
 c 2 2 This is called One dimentiona l heat equation

rt.
t x
VARIOUS POSSIBLE SOLUTIONS OF ONE DIMENSTIONAL HEAT
EQUATIONS pe
Let the one dimension wave equation
x
sE
am
Ex

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Module – I Engineering Mathematics - IV 2018

u 2  u
2
Consider c
t x 2
Let u  XT - - - - - - - - - - - - - - - (1)
where X ( x), T (t ) be the solution of the PDE
1 dT 1 d 2 X

c 2T dt X dx 2
equating to common constant
1 dT 1 d2X
2
 k and k
c T dt X dx 2
dT d2X
 c 2 kT  0 and  kX  0
dt dx 2

in
 
D  c 2 k T  0 and  
D2  k X  0

rt.
d d
where, D  D
dx dt
Case 1 :
Let k  0
pe
DT  0 and D 2 X  0
x
Solution are given
sE

X  c1 e 0 x and T  c 3  c 4 x e 0 x
X  c1  and T  c 3  c 4 x 
am

## Hence the solution of PDE

U  XT  c1 c 3  c 4 x 
Ex

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## Case 2 : Let k be negative say, k  p 2

D  
 p 2 X  0 and D  c 2 p 2 T  0
2

A.E is
D 2  p 2  0 and D  c 2 p 2  0
D   p & D  cp
Solution are given
  
X  c 5 e px  c 6 e  px and T  c 7 e pct 
Hence the solution of PDE
Case 3 : Let k be negative say, k  p 2

in
D  
 p 2 X  0 and D  c 2 p 2 T  0
2

A.E is

rt.
D 2  p 2  0 and D  c 2 p 2  0
D  ip & D  cp
Solution are given

X  c 9 cos px  c10 sin px  and T  c11e  pct
x

pe
Hence the solution of PDE

U  XT  c 9 cos px  c10 sin px  c11e  pct 
sE
am
Ex
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