Dr. P.

Njori-2024 KyU-SPM 2421-PDE I BMC, BFE, BAST, BED (SC)

WEEK ELEVEN & TWELVE

i). Compatible Systems Of First Order Partial Differential
Equations (Charpit’s Method)
In this section we shall study compatible systems of first order PDE and the Charpit’s
method for solving the non-linear PDEs.

Definition.
Compatible systems of first order PDEs
A system of two first order PDEs
F ( x, y, z , p, q )  0........................................(1)
and
G ( x, y, z , p, q )  0........................................(2)
are said to be compatible if they have a common solution.

Theorem 1. The equations

F ( x, y, z, p, q)  0 and G( x, y, z, p, q)  0
are compatible on a domain D if

Fp Fq
i) 0
Gp Gq
p and q can be solved explicitly from the system of equations (1) and (2) as
p   ( x, y)dx and q   ( x, y
such that the equation
dz   ( x, y)dx   ( x, y )dy
is integrable.

Theorem 2.
The necessary and sufficient condition for the integrability of the equation
dz   ( x, y)dx   ( x, y )dy

and hence, compatible is

F , G   ( F , G)  ( F , G)  p ( F , G)  q ( F , G)  0 …………………….(3)
 ( x, p )  ( y, q ) ( z, p) ( z, q)

In general, equations (1) and (2) are compatible if equation (3) holds.

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Dr. P. Njori-2024 KyU-SPM 2421-PDE I BMC, BFE, BAST, BED (SC)

Example
1. Show that the following equations are compatible and hence solve them.
xp  yq  0, z ( xp  yq)  2 xy

Solution.

Take F  xp  yq  0  Fx  p, Fy   q, Fz  0, F p  x, Fq   y
G  zxp  zyq  2 xy  0  G x  zp  2 y, G y  zq  2 x, G Z  xp  yq, G p  zx, G q  zy
Compatibibilty condition
 F,G   F,G   F,G   F,G 
If  p q 0
 ( x, p )  ( y , q ) ( z, p) ( z, q)
Test
Fx F p Fy Fq Fz F p Fz Fq
  p q
Gx G p G y Gq Gz G p G z Gq
p x q  y 0 x 0 y
  p  q
zp  2 y zx zq  2 x zy xp  yq zx xp  yq zy
  zxp  zxp  2 xy   ( zyq  zyq  2 xy )  p ( x 2 p  xyq )  q ( xyp  y 2 q )
 ( x 2 p 2  y 2 q 2 )
 ( xp  yq ) xp  yq 
0 since, xp  yq  0
Hence compatible.

Start by determining p and q from

xp  yq  0, .........(i) and z ( xp  yq )  2 xy...........(ii)

xp  yq  0  xp  yq from (i)
putting (i) into (ii) we get;
z ( xp  yq )  2 xy
y x
 2 zxp  2 xy  p   2 zyq  2 xy  q 
z z
 dz  pdx  qdy
 zdz  ydx  xdy   zdz   ydx   xdy  z 2  2 xy  c

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Dr. P. Njori-2024 KyU-SPM 2421-PDE I BMC, BFE, BAST, BED (SC)

FIRST SEMESTER EXAMINATIONS 2018/2019

FOURTH YEAR EXAMINATION FOR THE DEGREE OF BACHELOR OF SCIENCE MATHEMATICS
AND COMPUTER SCIENCE (CAT 2 MARKING SCHEME)
SPM 2420: PARTIAL DIFFERENTIAL EQUATIONS I
DATE: 16-11- 2018 TIME: 10.00AM -11.00AM

INSTRUCTIONS:
Answer both questions.

1. Solve the nonlinear differential equation

p 2  x 2  y 2  q (15 marks)

SOLUTION

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Dr. P. Njori-2024 KyU-SPM 2421-PDE I BMC, BFE, BAST, BED (SC)

p 2  x 2  y 2  q  p 2  x 2  y 2  q 2  p 2  x 2  a  y 2  q 2  p  x2  a, q  a  y2
Therefore,
dz  pdx  qdy  dz     a  y dy  z   
x 2  a dx  2

x 2  a dx    a  y dy
2

Take,

 
x 2  a dx and   a  y dy and evaluate separately then put the results together
2

Hence,

 x 2  a dx 
Let x  a tan  dx  a sec 2d
  
x 2  a dx  a  sec 3 φd   sec 3 φd   sec  sec 2d , use,  udv  uv-  vdu
let, u  sec  du  sectand
let, dv  sec 2d  v  tan 

 sec φd   sec  sec 2d , use,  udv  uv-  vdu

3

 sectand   sec  tan 2d

 sectand   sec  (sec 2   1)d

 sectand   sec 3 d   sec d

2  sec 3 φd  sectan   sec d

2  sec 3 φd  sectan  ln[sec   tan  ]  a

 sec
3 1
φd sectan  ln[sec  tan  ]  b
2
a  x 2 dx  a  sec 3 φd  sectan  ln[sec   tan  ]  c
a
 2
a
 
 a  x dx  2 tan   1 tan  ln[ tan   1  tan  ]  c
2 2 2

ax
 
2  a

x 2  a  ln[ 
x 2  a  x 
a
] c



 
ax
2a
  1 
x 2  a  ln[ x 2  a  x]  ln x   c....................................(1)
2 


Also,

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Dr. P. Njori-2024 KyU-SPM 2421-PDE I BMC, BFE, BAST, BED (SC)

 a  y 2 dy 
Let y  a sin   dy  a cos d

 
a  y 2 dy   a  a sin 2  a cos d

  a  y dy  a  cos d

2 2

But,

 cos d   cos  cos d , let u  cos  du  sind , Let dv  cosd  v  sin
2

use  udv  uv   vdu

 cos d  cos  sin    sin d

2 2

 cos  sin    1  cos  d 2


1
2

cos  sin    d 


1
2

1  sin 2     a 
1 1  y 
  a  y 2  sin 1     a
2 a  a 
but

 
a  y 2 dy  a  cos 2 d

a 1  y 
  a  y 2  sin 1     c......................................(2)
2  a  a 

Hence,

z
ax

2a
x 2
  1
2

 a 1
 a  ln x 2  a  x 2  x  ln a   
 2 a
 y 
a  y 2  sin 1     c
 a 

2. Verify the given Pfaffian differential equation is integrable and determine its primitive

( y 2  yz )dx  ( xz  z 2 )dy  ( y 2  xy)dz  0

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Dr. P. Njori-2024 KyU-SPM 2421-PDE I BMC, BFE, BAST, BED (SC)

(15marks) Solution.

(Work out the test for integrability)

Pdx  Qdy  Rdz  0  P  y 2  yz, Px  0 ,Py  2 y  z, Pz  y

Q  xz  z 2 , Qx  z , Q y  0, Qz  x  2 z
R  y 2  xy,Rx   y, R y  2 y  x, Rz  0
( y 2  yz )dx  ( xz  z 2 )dy  ( y 2  xy)dz  0
If P Q z  R y  Q R x  Pz  R Py  Q x  0, Then, Pdx  Qdy  Rdz  0 is integrable.
test
P Q z  R y  Q R x  Pz  R Py  Q x
     
 y 2  yz 2 z  2 y  2 x   xz  z 2 (2 y )  y 2  xy (2 y )
 2 y 2 z  2 y 3  2 xy 2  2 yz 2  2 xyz  2 xyz  2 yz 2  2 y 3  2 xy 2
0 Hence, integrable.

Solve

( y 2  yz )dx  ( xz  z 2 )dy  ( y 2  xy)dz  0

Solution.

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Dr. P. Njori-2024 KyU-SPM 2421-PDE I BMC, BFE, BAST, BED (SC)

Pdx  Qdy  Rdz  0

Work out the integrating factor
Px  Qy  Rz  g(x,y,z)  0
 g(x,y,z)  xy 2  xyz  xyz  yz 2  y 2 z  xyz
 xy 2  xyz  yz 2  y 2 z
 xy ( y  z )  yz ( z  y )
 ( y  z )( xy  yz )
 y ( y  z )( x  z )
divide the original equation by g(x,y,z) to get
1
 y ( y  z )dx  z ( x  z )dy  y ( y  x)dz  0
y ( y  z )( x  z )
dx z yx
 dy  dz  0
x  z y y  z  ( y  z )( x  z )
But ,
z A B
   Let y  0, A  1, Let y   z , B  1
y y  z  y y  z
z 1 1
  
y y  z  y y  z
Also,
yx C D
   Let y   z , C  1, Let x   z , C  1, D  1
( y  z )( x  z ) y  z x  z
yx 1 1
  
( y  z )( x  z ) y  z x  z

Hence,
dx z yx
 dy  dz  0
x  z y y  z  ( y  z )( x  z )
dx dy dy dz dz
     0
xz y yz yz xz
dx dz dy  dy dz 
      0
x  z x  z y  y  z y  z 

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Dr. P. Njori-2024 KyU-SPM 2421-PDE I BMC, BFE, BAST, BED (SC)

 d ln( x  z )   d (ln( y  z )  d (ln y )  d (ln c)

  d ln( x  z )    d (ln( y  z )   d (ln y )   d (ln c)
 ln( x  z )  ln( y  z )  ln y  ln c
y( x  z)
 c
 y  x

1. Show that the following equations are compatible and hence solve them.

( y  z ) p  ( z  x)q  x  y and z  px  qy  0

Others

1. Solve q   px  p
2

Solution
1
q   px  p 2  a  q  a, and  px  p 2  a  p 2  px  a  p 
2

x  x 2  4a 
1

 dz  pdx  qdy  dz  x  x 2  4a dx  ady  dz  
2

1  x 2  x 2  4a 
2  x  x 2  4a 
dx  ady

1
 z  x2 
4

Solve for the complete and singular solutions .
z  px  qy  p 2  1  p 2  q 2

Solution

z  ax  by  a 2  1  a 2  b 2 .......complete solution.
Singular solution
Differentiating w.r.t " a".
a ( x  2a ) 1
0  x  2a    .......(1)
1  a 2  b2 a 1  a2  b2
Differentiating w.r.t " b".

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Dr. P. Njori-2024 KyU-SPM 2421-PDE I BMC, BFE, BAST, BED (SC)

b y 1
0 y  .......................(2)
1 a  b
2 2 b 1 a2  b2
( x  2a )b bx
 1 y   2b
ay a

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