MA6351
TRANSFORMS AND PARTIAL
DIFFERENTIAL EQUATIONS
UNIT I
PARTIAL DIFFERENTIAL EQUATIONS
by
B.Saravanan
Assistant Professor
Department of Applied Mathematics
SVCE
1
Syllabus
Formation of partial differential equations
Lagrange’s linear equation
Solutions of standard types of first order partial
differential equations
Linear partial differential equations of second and
higher order with constant coefficients
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Partial Differential Equation
Partial differential equation is one which involves
partial derivatives. The order of PDE is the order of highest
derivative occurring in it.
If z = f ( x, y ) where x, y are independen t var iable ,
z is dependent var iable .
Notation :
∂z
∂z
∂2z
∂2z
∂2z
p=
, q=
, r= 2 , s=
, t= 2 .
∂x
∂y
∂x ∂y
∂ x
∂ y
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Formation of PDE by eliminating arbitrary constant
Let us consider the functional relation
f(x, y, z, a, b) = 0 -------- (1)
Where a and b are arbitrary constant to be eliminated
Differentiating (1) partially with respect to x and y, we get
∂f ∂f
+
∂x ∂z
∂f ∂f
+
∂y ∂z
∂z
∂f ∂f
=0 ⇒
+
p = 0 − − − − − − − ( 2)
∂x
∂x ∂z
∂z
∂f ∂f
=0 ⇒
+ q = 0 − − − − − − − (3)
∂y
∂y ∂z
Equation (2) and (3) will contain a and b. If we eliminate a and
b from (1), (2) and (3) we get the PDE (involving p and q)
of the first order.
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Remarks:
If the number of constants to be eliminated is equal to number
of independent variables, the PDE got after elimination
will be of first order.
If the number of constants to be eliminated is more than the
number independent variables, the resulting PDE will be
of second or higher order.
Answer is not unique.
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Problem 1
Form the partial differential equation by eliminating a and b
from z = ( x 2 + a 2 )( y 2 + b 2 )
Solution:
Given z = ( x 2 + a 2 )( y 2 + b 2 ) − − − − − −(1)
Differentiating (1) partially w.r. t x and y we get
∂z
p=
= (2 x)( y 2 + b 2 )
∂x
p
⇒
= y 2 + b 2 − − − − − − − ( 2)
2x
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Differentiating (1) partially w.r. t ‘y’ we get
∂z
q=
= ( x 2 + a 2 )(2 y )
∂y
q
⇒
= x 2 + a 2 − − − − − − − (3)
2y
Substitute (2) and (3) in equation (1), we have
q p
z=
.
2 y 2x
(i.e.) 4 xy z = pq
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Problem 2
Form the partial differential equation by eliminating the
arbitrary constants a and b from ( x − a ) 2 + ( y − b ) 2 = z 2 cot 2 α
Solution:
Given ( x − a) 2 + ( y − b) 2 = z 2 cot 2 α − − − − − (1)
Diff. eqn. (1) p.w.r.t. x, we get
∂z
2( x − a) + 0 = 2 z
cot 2 α
∂x
⇒ x − a = z p cot 2 α − − − − − − − (2)
Diff. eqn. (1) p.w.r.t. y, we get
∂z
0 + 2( y − b ) = 2 z
cot 2 α
∂y
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⇒ y − b = z q cot 2 α − − − − − − − (3)
Substitute (2) and (3) in equation (1), we have
( z p cot 2 α ) 2 + ( z q cot 2 α ) 2 = z 2 cot 2 α
z 2 cot 4 α ( p 2 + q 2 ) = z 2 cot 2 α
cot 2 α ( p 2 + q 2 ) = 1
(i.e.) p 2 + q 2 = tan 2 α
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Problem 3
Form the partial differential equation by eliminating the
2
2
arbitrary constants a and b from z = a x + ay + b
Solution:
Given z = a 2 x + ay 2 + b − − − − − − − −(1)
Diff. eqn. (1) p.w.r.t. x, we get
∂z
p=
= a 2 − − − − − − − (2)
∂x
Diff. eqn. (1) p.w.r.t. y, we get
∂z
q
q=
= 2ay ⇒ a =
− − − − − − − (3)
∂y
2y
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Substitute (3) in equation (2), we have
 q 
p =  
 2y 
2
4y2 p = q2
Problem 4
Form the partial differential equation by eliminating the
n
n
arbitrary constants a and b from z = ax + by
Solution:
Given z = ax n + by n − − − − − −(1)
Diff. eqn. (1) p.w.r.t. x, we get
∂z
p=
= a n x n −1
∂x
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a n xn
p=
x
px
= a x n − − − − − − − ( 2)
n
Diff. eqn. (1) p.w.r.t. y, we get
∂z
q=
= b n y n −1
∂y
bn yn
q=
y
qy
= b y n − − − − − − − (3)
n
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Substitute (2) and (3) in equation (1), we have
px qy
z=
+
n
n
(i.e.) n z = p x + q y
Problem 5
Find the partial differential equation of all planes cutting equal
intercepts from the x and y axes.
Solution:
The equation of the plane cutting equal intercept from x and y
axes is
x y z
+ + = 1 − − − − − −(1)
a a c
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Diff. eqn. (1) p.w.r.t. x, we get
1
p
+0+ =0
a
c
p
1
= − − − − − − − − ( 2)
c
a
Diff. eqn. (1) p.w.r.t. y, we get
1 q
0+ + = 0
a c
q
1
= − − − − − − − − (3)
c
a
Divide (2) by (3), we get
p
=1
q
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(i.e.) p = q
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Problem 6
Find the partial differential equation of all planes passing
through the origin
Solution:
The equation of the plane passing through the origin is
ax + by + cz = 0
⇒ c z = −a x − b y
a
b
⇒ z =− x− y
c
c
(i.e.) z = A x + B y − − − − − − − (1)
where A and B are arbitrary constants
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Diff. eqn. (1) p.w.r.t. x, we get
∂z
p=
=A
∂x
Diff. eqn. (1) p.w.r.t. y, we get
∂z
q=
=B
∂y
Substitute (2) and (3) in equation (1), we have
z = px+qy
Problem 7
Find the PDE of all planes which are at a constant distance ‘k’
from the origin.
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Solution:
The equation of the plane having constant distance ‘k’ from
the origin is
a x + b y + c z − k a + b + c = 0 − − − − − −(1)
2
2
2
Diff. eqn. (1) p.w.r.t. x, we get
a+c p =0
⇒ a = − c p − − − − − − − (2)
Diff. eqn. (1) p.w.r.t. y, we get
b + cq = 0
⇒ b = − c q − − − − − − − (3)
Substitute (2) and (3) in equation (1), we have
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−c p x −c q y + c z − k c p + c q + c = 0
2
2
2
2
2
− p x − q y + z − k p2 + q2 +1 = 0
(i.e.) z = p x + q y + k p 2 + q 2 + 1
Problem 8
Form the partial differential equation of all spheres whose
centre lies on the z-axis.
Solution:
Any point on the z-axis is of the form (0, 0, a)
Then the equation of the sphere with centre (0, 0, a) and
radius k (say) is
x 2 + y 2 + ( z − a) 2 = k 2 − − − − − − − (1)
where ‘a’ is the arbitrary constant.
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Diff. eqn. (1) p.w.r.t. x, we get
2 x + 0 + 2( z − a ) p = 0
x = −( z − a ) p − − − − − − − (2)
Diff. eqn. (1) p.w.r.t. y, we get
0 + 2 y + 2( z − a ) q = 0
y = −( z − a ) q − − − − − − − (3)
Divide (2) by (3), we get
x p
=
y q
(i.e.) p y = q x.
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Problem 9
Find the partial differential equation of the family of spheres
having their centres on the line x = y = z.
Solution:
Since the centre (a, b, c) lies on the line x = y = z,
we have a = b = c
Hence the equation of the sphere is
(x – a)2 + (y – a)2 + (z – a)2 = r2 ---------------- (1)
where ‘a’ is the arbitrary constants.
Diff. eqn. (1) p.w.r.t. x, we get
2( x − a) + 2( z − a ) p = 0
2 x + 2 z p = 2 a (1 + p) − − − − − − − (2)
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Diff. eqn. (1) p.w.r.t. y, we get
2( y − a ) + 2( z − a ) q = 0
2 y + 2 z q = 2 a (1 + q) − − − − − − − (3)
Divide (2) by (3), we get
2 ( x + z p) 1 + p
=
2 ( y + z q) 1 + q
( x + z p)(1 + q) = ( y + z q )(1 + p )
x + xq + z p + z pq = y + y p + zq + z pq
(i.e.) ( y − z ) p + ( z − x) q = x − y
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Formation of PDE by eliminating arbitrary functions
Let us consider the relation f (u, v) =0 --------(1)
where u and v are functions of x, y ,z and f is an arbitrary
function to be eliminated.
Differentiating (1) partially with respect to x and y we get
∂f  ∂u ∂u  ∂f  ∂v ∂v
+
p +  +

∂u  ∂x ∂z  ∂v  ∂x ∂z

p  = 0 − − − − − − − (2)

( Since u and v are functions of x, y, z and z is in turn,
a function of x, y )
∂f  ∂u ∂u  ∂f  ∂v ∂v 

+
q  +  + q  = 0 − − − − − − − (3)
∂u  ∂y ∂z  ∂v  ∂y ∂z 
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∂f
∂f
Instead of e lim inating f , let us e lim inate
and
from (2) and (3)
∂u
∂v
we get an equation of the form
Pp + Qq = R − − − − − −(4)
where P, Q, R are functions of x, y, z
Remarks:
Equation (4) is called Lagrange’s linear PDE whose
solution will be discussed later.
The order of PDE formed depends only on the number
of arbitrary functions eliminated
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Problem 1
Form the partial differential equation by eliminating an
arbitrary function from z = f ( x 2 + y 2 )
Solution:
Given z = f ( x 2 + y 2 ) − − − − − −(1)
Diff. eqn. (1) p.w.r.t. x, we get
p = f ′ ( x 2 + y 2 ) (2 x) − − − − − − (2)
Diff. eqn. (1) p.w.r.t. y, we get
q = f ′ ( x 2 + y 2 ) (2 y) − − − − − − (3)
Divide (2) by (3), we get
p x
=
q y
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(i.e.) p y = q x
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Problem 2
Form the partial differential equation by eliminating the
arbitrary functions from z = f 1 ( x ) f 2 ( y ).
Solution:
Given z = f1 ( x) f 2 ( y) − − − − − − − (1)
Diff. eqn. (1) p.w.r.t. x, we get
p = f1′( x) f 2 ( y) − − − − − − (2)
Diff. eqn. (1) p.w.r.t. y, we get
q = f1 ( x) f 2′( y) − − − − − − (3)
Diff. eqn. (2) p.w.r.t. x, we get
r = f1′′( x) f 2 ( y) − − − − − − (4)
Diff. eqn. (2) p.w.r.t. y, we get
s = f1′( x) f 2′( y) − − − − − − (5)
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Diff. eqn. (3) p.w.r.t. y, we get
t = f1 ( x) f 2′′( y) − − − − − − (6)
From (2) and (3) we have
p q = f1 ( x) f 2 ( y ) f1′( x) f 2′( y )
(i.e.) p q = z s
Problem 3
Form the partial differential equation by eliminating an
2
2
arbitrary function from z = xy + f ( x + y )
Solution:
Given z = xy + f ( x 2 + y 2 ) − − − − − −(1)
Diff. eqn. (1) p.w.r.t. x, we get
p = y + f ′ ( x 2 + y 2 ) (2 x)
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p − y = f ′ ( x 2 + y 2 ) ( 2 x ) − − − − − − ( 2)
Diff. eqn. (1) p.w.r.t. y, we get
q = x + f ′ ( x 2 + y 2 ) (2 y )
q − x = f ′ ( x 2 + y 2 ) (2 y ) − − − − − − (3)
Divide (2) by (3), we get
p− y x
=
q−x y
p y − y2 = q x − x2
(i.e.) p y − q x = y 2 − x 2
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Problem 4
Eliminate the arbitrary function ‘f ’ from the relation
1

z = y + 2 f  + log y 

x
2
1

Solution: Given z = y + 2 f  + log y  − − − − − (1)
x

2
Diff. eqn. (1) p.w.r.t. x, we get
1


p = 0 + 2 f ′  + log y 
x

 −1
 2  − − − − − − − (2)
x 
Diff. eqn. (1) p.w.r.t. y, we get
1
1


q = 2 y + 2 f ′  + log y   
x
 y 
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1
1


q − 2 y = 2 f ′  + log y    − − − − − − − (3)
x
 y 
Dividing (2) by (3), we have
1
−1



2 f ′  + log y   2 
p
x
x 



=
q − 2y
1
1


2 f ′  + log y   
x
 y 
=>
p
− 1/ x 2
=>
=
q − 2y
1/ y
p
−y
= 2
q − 2y x
=> x 2 p = − y (q − 2 y )
(i.e.)
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x2 p + y q = 2 y2
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Problem 5
Form the partial differential equation by eliminating the
x
 2
arbitrary function from φ  z − xy,  = 0
z

Solution:
The given equation can be written as
 x
z − xy = f   − − − − − − − (1)
z
2
Diff. eqn. (1) p.w.r.t. x, we get
x   z.1 − x. p 

2 z p − y = f ′  
− − − − − − (2)
2


z z
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Diff. eqn. (1) p.w.r.t. y, we get
 x  − xq
′
2 z q − x = f    2  − − − − − − (3)
z  z 
Divide (2) by (3), we get
2z p− y z − px
=
2zq− x
−qx
(2 z p − y )(−q x) = (2 z q − x)( z − p x)
− 2 z p q x + x y q = 2 z 2q − 2 z p q x − z x + p x2
(i.e.) x 2 p + (2 z 2 − x y ) q = z x
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Problem 6
Eliminate the arbitrary function ‘f ’ from the relation
f ( x 2 + y 2 + z 2 , x + y + z) = 0
Solution:
The given equation can be written as
x 2 + y 2 + z 2 = φ ( x + y + z ) − − − − − − − (1)
Diff. eqn. (1) p.w.r.t. x, we get
2 x + 0 + 2 z p = φ ′( x + y + z ) (1 + 0 + p )
2 x + 2 z p = φ ′( x + y + z ) (1 + p) − − − − − − − (2)
Diff. eqn. (1) p.w.r.t. y, we get
0 + 2 y + 2 z q = φ ′( x + y + z ) (0 + 1 + q)
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2 y + 2 z q = φ ′( x + y + z ) (1 + q) − − − − − − − (3)
Dividing (2) by (3), we have
2 x + 2 z p φ ′( x + y + z ) (1 + p)
=
2 y + 2 z q φ ′( x + y + z ) (1 + q)
x + z p (1 + p)
=
y + z q (1 + q )
( x + zp)(1 + q) = ( y + zq)(1 + p)
x + xq+ z p + z pq = y + y p + zq + z pq
(i.e.) ( y − z ) p + ( z − x) q = x − y
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Lagrange’s linear PDE:(Linear first order PDE)
The linear PDE of first order is known as Lagrange’s linear
equation is of the form
Pp + Qq = R
where P,Q, R are functions of x, y, z
This is got by eliminating arbitrary function f (u, v)=0 or u=F(v)
To solve Pp + Qq = R
1. Form the auxiliary equation of the form
dx dy dz
=
=
P
Q
R
2. Solve these auxiliary simultaneous equation, giving two
independent solution u=C1 and v= C2
3. The general solution is f (u, v)=0 or u=F(v)
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Problem 1
2
2
2
px
+
qy
=
z
Find the solution of
Solution:
This is Lagrange’s linear PDE of the form Pp + Qq =R
dx dy dz
A . E. are
=
=
P
Q
R
dx dy dz
= 2 = 2
2
x
y
z
Take 1st and 2nd ratio, we have
dx dy
= 2
2
x
y
Integrating, we get
B.Saravanan
−1 −1
=
+ c1
x
y
35
1 1
=>
− = c1
y x
6/4/2015
Take 2nd and 3rd ratio, we have
dy dz
= 2
2
y
z
Integrating, we get
−1 −1
=
+ c2
y
z
1 1
=>
− = c2
z y
Hence the required solution is
1 1 1 1
F  − , −  = 0
 y x z y
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Problem 2
Solve: x( y − z ) p + y ( z − x) q = z ( x − y )
Solution:
This is Lagrange’s linear PDE of the form Pp + Qq =R
dx dy dz
A . E. are
=
=
P
Q
R
dx
dy
dz
=
=
x ( y − z ) y ( z − x) z ( x − y )
Using multiplier 1/x, 1/y, 1/z and then add, each ratio is
dx dy dz
+
+
x
y
z
y−z+z−x+x− y
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dx dy dz
⇒
+
+
=0
x
y
z
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Integrating we get
log x + log y + log z = log c1
⇒ log( x y z ) = log c1
⇒ x y z = c1
Using multiplier 1, 1, 1 and then add, each ratio is
dx + dy + dz
=
xy − xz + yz − yx + zx − zy
⇒ dx + dy + dz = 0
Integrating we get x + y + z = c2
Hence the required solution is F ( xy z , x + y + z ) = 0
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Problem 3
2
2
z
(
x
p
−
yq
)
=
y
−
x
Solve:
Solution:
This is Lagrange’s linear PDE of the form Pp + Qq =R
dx dy dz
A . E. are
=
=
P
Q
R
dx
dy
dz
=
= 2
x z − y z y − x2
Take 1st and 2nd ratio, we have
log x = − log y + log c1
dx
dy
=
xz − yz
dx dy
=
x
−y
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Integrating, we get
log x + log y = log c1
(i.e.) x y = c1
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Using multiplier x,y,z and then add, each ratio is
x dx + y dy + z dz
= 2
x z − y2 z + y2 z − x2 z
⇒ x dx + y dy + z dz = 0
Integrating, we get
x2 y2 z 2
+
+
= c2
2
2
2
(i.e.) x 2 + y 2 + z 2 = c2
Hence the required solution is
F ( x y, x 2 + y 2 + z 2 ) = 0
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Problem 4
Solve: x ( y 2 − z 2 ) p + y ( z 2 − x 2 ) q = z ( x 2 − y 2 )
Solution:
This is Lagrange’s linear PDE of the form Pp + Qq =R
dx dy dz
A . E. are
=
=
P
Q
R
dx
dy
dz
=
=
2
2
2
2
x ( y − z ) y (z − x ) z (x2 − y2 )
Using multiplier 1/x, 1/y, 1/z and then add, each ratio is
dx dy dz
+
+
x
y
z
= 2
y − z 2 + z2 − x2 + x2 − y2
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dx dy dz
⇒
+
+
=0
x
y
z
Integratin g we get
log x + log y + log z = log c1
⇒ log( x y z ) = log c1
⇒ x y z = c1
Using multiplier x,y,z and then add, each ratio is
x dx + y dy + z dz
= 2 2
x y − x2 z 2 + y2 z2 − y2 x2 + z2 x2 − z 2 y2
⇒ x dx + y dy + z dz = 0
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Integrating, we get
x2 y2 z 2
+
+
= c2
2
2
2
(i.e.) x 2 + y 2 + z 2 = c2
Hence the required solution is
F ( xyz, x 2 + y 2 + z 2 ) = 0
Problem 5
Solve: (mz − ny ) p + (nx − lz )q = ly − mx
Solution:
This is Lagrange’s linear PDE of the form Pp + Qq =R
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dx dy dz
A . E. are
=
=
P
Q
R
dx
dy
dz
=
=
mz − ny nx − lz ly − mx
Using multiplier l,m,n and then add, each ratio is
ldx + mdy + ndz
=
lmz − nly + mnx − lmz + nly − nmx
⇒ ldx + mdy + ndz = 0
Integratin g we get
lx + my + nz = c1
B.Saravanan
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Using multiplier x,y,z and then add, each ratio is
x dx + y dy + z dz
=
m z x − n xy + n xy − lyz + lyz − m z x
⇒ x dx + y dy + z dz = 0
Integrating, we get
x2 y2 z 2
+
+
= c2
2
2
2
(i.e.) x 2 + y 2 + z 2 = c2
Hence the required solution is
F (lx + my + nz , x 2 + y 2 + z 2 ) = 0
B.Saravanan
45
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Problem 6
Solve:
x( y 2 + z ) p + y ( x 2 + z )q = z ( x 2 − y 2 )
Solution:
This is Lagrange’s linear PDE of the form Pp + Qq =R
dx dy dz
A . E. are
=
=
P
Q
R
dx
dy
dz
=
=
2
2
x ( y + z) y ( x + z) z ( x 2 − y 2 )
Using multiplier 1/x, (-1/y), 1/z and then add, each ratio is
dx dy dz
−
+
x
y
z
= 2
( y + z) − ( x 2 + z) + ( x 2 − y 2 )
B.Saravanan
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dx dy dz
⇒
−
+
=0
x
y
z
Integratin g we get
log x − log y + log z = log c1
⇒ log( x z ) − log y = log c1
xz
(i.e.)
= c1
y
Using multiplier x,(-y),-1 and then add, each ratio is
x dx − y dy − dz
= 2 2
2
2 2
2
2
2
x y +x z−x y −y z−x z+ y z
B.Saravanan
47
6/4/2015
⇒ x dx − y dy − dz = 0
Integrating, we get
x2 y2
−
− z = c2
2
2
(i.e.) x 2 − y 2 − 2 z = c2
Hence the required solution is
xz 2

2
F 
, x − y − 2 z  = 0
 y

B.Saravanan
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6/4/2015
Problem 7
Solve: ( x 2 − y 2 − z 2 ) p + 2 xy q = 2 z x
Solution:
This is Lagrange’s linear PDE of the form Pp + Qq =R
dx dy dz
A . E. are
=
=
P
Q
R
dx
dy
dz
=
=
2
2
2
x − y −z
2 xy 2 z x
Take 2nd and 3rd ratio, we have
dy
dz
dy dz
=
=>
=
2x y 2 z x
y
z
B.Saravanan
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Integrating, we get
log y = log z + log c1
log y − log z = log c1
y
(i.e.)
= c1
z
Using multiplier x,y,z and then add, each ratio is
x dx + y dy + z dz
=
x( x 2 − y 2 − z 2 ) + 2 y 2 x + 2 z 2 x
x dx + y dy + z dz
x dx + y dy + z dz
= 3
=
2
2
x + y x+z x
x( x 2 + y 2 + z 2 )
B.Saravanan
50
6/4/2015
Equate this to 2nd ratio, we have
x dx + y dy + z dz
dy
x dx + y dy + z dz dy
=
⇒
=
2
2
2
2
2
2
x( x + y + z )
2 xy
x +y +z
2y
Integrating we get
1
1
2
2
2
log( x + y + z ) = log y + log c 2
2
2
log( x 2 + y 2 + z 2 ) = log y + log c2
log( x 2 + y 2 + z 2 ) − log y = log c2
x2 + y2 + z 2
(i.e.)
= c2
y
B.Saravanan
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Hence the required solution is
 y x2 + y2 + z2 
 = 0
F  ,
y
z

Problem 8
Solve: ( x 2 − y z ) p + ( y 2 − z x) q = z 2 − x y
Solution:
This is Lagrange’s linear PDE of the form Pp + Qq =R
dx dy dz
A . E. are
=
=
P
Q
R
dx
dy
dz
= 2
= 2
2
x −yz y −zx z −xy
B.Saravanan
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6/4/2015
dx − dy
Each ratio =
( x 2 − y z ) − ( y 2 − z x)
(Subtracting 1st and 2nd ratio)
d ( x − y)
= 2
( x − y 2 ) + ( z x − y z)
d ( x − y)
=
( x − y )( x + y ) + z ( x − y )
d ( x − y)
=
( x − y )( x + y + z )
-----------(A)
dy − dz
Each ratio =
( y 2 − z x) − ( z 2 − x y )
(Subtracting 2nd and 3rd ratio)
d ( y − z)
= 2
( y − z 2 ) + ( x y − z x)
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d ( y − z)
=
( y − z )( y + z ) + x( y − z )
d ( y − z)
=
( y − z )( x + y + z )
---------(B)
From (A) and (B) we have
d ( x − y)
d ( y − z)
=
( x − y )( x + y + z ) ( y − z )( x + y + z )
d ( x − y) d ( y − z)
⇒
=
( x − y)
( y − z)
Integrating we get
log( x − y ) = log( y − z ) + log c1
B.Saravanan
54
x− y
(i.e.)
= c1
y−z
6/4/2015
Using multiplier 1,1,1 and then add, each ratio is
dx + dy + dz
Each ratio = 2
x + y2 + z2 − x y − y z − z x
d ( x + y + z)
= 2
x + y2 + z2 − x y − y z − z x
-------(C)
Using multiplier x,y,z and then add, each ratio is
x dx + y dy + z dz
each ratio = 3
x + y 3 + z 3 − 3x y z
x dx + y dy + z dz
=
( x + y + z )( x 2 + y 2 + z 2 − x y − y z − z x)
---------(D)
B.Saravanan
55
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From (C) and (D)
d ( x + y + z)
x2 + y2 + z2 − x y − y z − z x
x dx + y dy + z dz
=
( x + y + z )( x 2 + y 2 + z 2 − x y − y z − z x)
( x + y + z ) d ( x + y + z ) = x dx + y dy + z dz
Integrating we get
( x + y + z)2 x 2 y 2 z 2
=
+
+
+ c2
2
2
2
2
( x + y + z) 2 = x 2 + y 2 + z 2 + c2
B.Saravanan
56
6/4/2015
x 2 + y 2 + z 2 + 2( x y + y z + z x ) = x 2 + y 2 + z 2 + c 2
2( x y + y z + z x ) = c 2
(i.e.) x y + y z + z x = c2
Hence the required solution is
x− y

F 
, x y + y z + z x  = 0
 y−z

B.Saravanan
57
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Non-Linear first order PDE (Standard types)
Those equations in which p and q occur other than the first
degree and product of p, q terms are called non linear first
order PDE
Ex:
p2 +q2+pq = 4
Types of Solutions:
1. A solution in which the number of arbitrary constants is
equal to number of independent variable is called complete
integral or complete solution.
2. In the complete integral if we give particular value to
arbitrary constant we get particular integral
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3. Let f ( x, y, z , p, q) = 0 be a PDE whose complete solution is
φ ( x, y, z, a, b) = 0 − − − − − (1)
Diff . (1) partially with respect to a , b and equate to 0 we get
∂φ
= 0 − − − − − (2)
∂a
∂φ
= 0 − − − − − (3)
∂b
Eliminate a and b from (1),(2) and (3), the resulting one
is called singular integral
B.Saravanan
59
6/4/2015
Standard types of first order PDE
Type I
f (p ,q)=0 ( i.e. equations containing p and q only)
Its complete integral is given by z = a x + b y + c -------- (1)
where a and b are connected by f (a, b)=0----------(2)
From (2) express b as function of a
i.e. b = φ (a) and substitute in eqn.(1) we get complete int egral
(1) ⇒ z = ax + φ (a) y + c − − − − − − − (3)
where a and c are arbitrary cons tan t
B.Saravanan
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To find sin gular int egral Diff . (3) partially with respect to a , c
and equate to 0 we get
∂z
∂z
= 0 ⇒ 1 = 0 which is absurd
= 0 and
∂a
∂c
Hence there is no singular integral
To find general solution put c = g ( a) in complete int egral we get
z = ax + φ (a) y + g (a) − − − − − ( A)
and Diff . ( A) partially with respect to a
we get
0 = x + φ ′(a) y + g ′(a) − − − − − − − ( B )
Eliminate ‘a’ from (A) and (B) we get general integral
or general solution.
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Problem 1
Find the complete integral of
p + q =1
Solution:
Given
p + q = 1 − − − − − (1)
This is of the form f(p,q)=0
The complete solution of equation (1) is
z = ax+b y+c
where
(
a + b = 1 ⇒ b = 1− a ⇒ b = 1− a
)
2
Hence the complete integral is
(
z = a x + 1− a
B.Saravanan
)
2
62
y+c
6/4/2015
Problem 2
Find the complete integral of p-q=0
Solution:
Given p – q = 0 ------------ (1)
This is of the form f(p,q)=0
The complete solution of equation (1) is
z = ax+b y+c
Where a-b=0 => b=a
Hence the complete integral is
z = ax+ay+c
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63
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Problem 3
2
2
p
+
q
− 4 pq = 0
Find the complete integral of
Solution:
Given p 2 + q 2 − 4 pq = 0 − − − − − − − (1)
This is of the form f(p,q)=0
The complete solution of equation (1) is z = a x + b y + c
where a 2 + b 2 − 4 a b = 0
⇒ b2 − 4 a b + a2 = 0
4a ± 16a 2 − 4.1.a 2
b=
2.1
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4a ± 12 a 2
=
2
4a ± 2 a 3
=
2
= a (2 ± 3 )
Hence the complete integral is
z = a x + a (2 ± 3 ) y + c
B.Saravanan
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Type II
f (p, q, z)=0 ( i.e. equations containing p ,q and z only)
The given PDE is f (p, q, z)=0--------(1)
Put q=ap in equation (1) and find p and q as function of z
Substitute p and q in dz = p dx + q dy
(keep z terms in LHS and remaining in RHS)
On integrating we get the complete integral
Procedure for obtaining Singular integral and general
solution are same as explained in type I
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Problem 1
Solve: p (1 + q ) = qz
Solution:
This is of the form f(z , p, q) = 0
Given p(1 + q) = qz − − − −(1)
Let q = ap
Now, q = a p
Then equation (1) becomes
p(1 + ap) = ap z
a z −1
1 + ap = az ⇒ p =
a
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67
 a z −1
= a

 a 
= a z −1
6/4/2015
Substitute p and q in the relation
dz = p dx + q dy
a z −1
dz =
d x + (a z − 1) d y
a
dz
dx
=
+d y
a z −1 a
Integratin g , we get
log(a z − 1) x
= + y+b
a
a
(i.e.) log(a z − 1) = x + a y + b − − − − − − (2)
which is the complete integral
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To find singular integral, Diff. eqn. (2) p.w.r.t. ‘a’ and ‘b’, in
turn, we get
a
= y and 0 = 1
a z −1
The last equation is absurd and shows that there is no singular
integral.
To find general integral, assume b = f(a)
Then equation (2) becomes
(2) => log(a z − 1) = x + a y + f (a) − − − − − (3)
Diff. eqn. (3) p.w.r.t. ‘a’, we get
a
= y + f ′(a) − − − − − − − − (4)
a z −1
The eliminant of ‘a’ between equations (3) and (4) gives the
general integral.
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Type III
f1 (x, p) = f2 (y, q)
(i.e. equations containing p , q , x and y)
Let f1 (x, p) = f2 (y, q) = a
f1 (x, p) = a and f2 (y, q) = a
solve for p and q (write p as function of x and q as function of y)
p=f(x) and q=g(y)
Substitute p and q in dz = pdx + qdy
=> dz= f(x) dx +g(y )dy
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On integrating we get
z = ∫ f ( x)dx + ∫ g ( y )dy + b
Which is the complete integral contains two arbitrary
constant a and b
Procedure for obtaining Singular integral and general solution
are same as explained in type I
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71
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Problem 1
2
2
2
2
p
+
q
=
x
+
y
Solve:
Solution:
This is of the form f(x , p) = g( y , q)
p 2 + q 2 = x 2 + y 2 − − − − − (1)
Given
⇒ p2 − x2 = y2 − q2
⇒ p2 − x2 = y2 − q2 = a2
p −x =a
2
2
2
⇒ p=±
y2 − q2 = a2 ⇒ q = ±
B.Saravanan
x +a
2
2
y2 − a2
72
6/4/2015
Substitute p and q in the relation
dz = p dx + q dy
dz = ± x 2 + a 2 dx ±
Integrating we get
y 2 − a 2 dy
2
x 2
a
2
−1  x  
z=± 
x +a +
sinh  
2
 a 
2
y
±
2
2
a
2
2
−1  y  
y −a −
cosh   + b − − − − (2)
2
 a 
which is the complete integral
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To find singular integral, Diff. eqn. (2) p.w.r.t. ‘a’ and ‘b’,
in turn, we get
x

2a
a2
1
−x
−1  x 
0=± 
+
 2  + sinh  .(a)
2
2
2
2 1 + ( x / a)  a 
 a  
 2 2 x + a
 y ( −2 a )
a2
−
±
2
2
2
 2 2 y − a
and 0 = 1

− y
−1  y 
 2  + cosh  .(− a )
2
a

( x / a) − 1  a 
1
The last equation is absurd and shows that there is no
singular integral
To find general integral, assume b = f(a)
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Then equation (2) becomes
2
x 2
a
 x 
z=± 
x + a 2 + sinh −1  
2
 a 
2
2
y
a
2
2
−1  y 
±
y − a − cosh   + f (a) − − − − − −(3)
2
 a 
2
Diff. eqn. (3) p.w.r.t. ‘a’, we get
x

2a
a2
1
−x
−1  x 
0=± 
+
 2  + sinh  .(a)
2
2
2
2 1+ (x / a)  a 
 a  
 2 2 x + a
 y (−2a)

a2
1
− y
−1  y 
±
−
+ cosh  .(−a) + f ′ (a) − − − (4)


2
a
 2 2 y 2 − a2 2 (x / a)2 −1  a 

The eliminant of ‘a’ between equations (3) and (4) gives
the general integral.
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Problem 2
Find the complete integral of
pq = x
Solution:
This is of the form f(x , p) = g( y , q)
Given pq = x − − − (1)
Let q = a
Then equation (1) becomes
pa = x ⇒ p =
x
a
Substitute p and q in the relation
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dz = p dx + q dy
x
dz = dx + a dy
a
Integratin g , we get
x2
z=
+ ay + b
2a
which is the complete integral.
B.Saravanan
77
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Type IV (Clairaut’s form)
An equation of the form z = p x + q y + f (p, q) is known
as Clairaut’s equation
Its complete integral is z = a x + b y + f (a, b) ------(1)
(by replacing p by a and q by b)
To find singular integral diff. (1) partially with respect to a, b
we get
∂f
0=x+
− − − − − − − ( 2)
∂a
∂f
0= y+
− − − − − − − (3)
∂b
Eliminate a and b from (1), (2) and (3) we get singular integral.
Procedure for obtaining general solution are same as
explained in type I
B.Saravanan
78
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Problem 1
Find the complete integral of
z
x y
= + +
pq q p
pq
Solution:
z
x y
Given
= + +
pq q p
pq
z = p x + q y + p q pq − − − − − −(1)
This is in Clairaut’s form
The complete integral of equation (1) is
z = a x + b y + ab ab
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79
(replacing p by a and q by b)
6/4/2015
Problem 2
Find the singular integral of z=p x +q y +p q
Solution:
Given z=p x +q y +p q
This is in Clairaut’s form
The complete integral of equation is
z= a x + b y + a b -------(1)
(replacing p by a and q by b)
To find the singular integral, diff. (1) partially w.r.to a and b
0 = x + b => b = -x
0 = y + a => a = -y
(1)=> z= -x y – x y + x y
B.Saravanan
=> z= -x y
80
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Problem 3
Find the singular solution of z = px + qy + p 2 + pq + q 2
Solution:
Given z = px + qy + p 2 + pq + q 2 − − − − − (1)
This is in Clairaut’s form
The complete integral of equation (1) is
z = ax + by + a 2 + ab + b 2 − − − − − −(2)
(replacing p by a and q by b)
To find singular integral, Diff. eqn. (2) p.w.r.t. ‘a’ and ‘b’,
we get
0 = x + 2a + b ⇒ 2a + b = − x − − − − − − (3)
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and 0 = y + a + 2b ⇒ a + 2b = − y − − − − − − (4)
Solving (3) and (4) we get
y − 2x
x − 2y
3a = y − 2 x ⇒ a =
and 3b = x − 2 y ⇒ b =
3
3
Substitute the values of a and b in equation (2) we have
( 2) => z = ax + by + a 2 + ab + b 2
 y − 2x 
z = x
+
 3 
 x − 2 y   y − 2x 
y
+

 3   3 
 y − 2x   x − 2 y   x − 2 y 
+

+

 3  3   3 
B.Saravanan
82
2
2
6/4/2015
9 z = 3 x( y − 2 x) + 3 y ( x − 2 y ) + ( y − 2 x) 2
+ ( y − 2 x)( x − 2 y ) + ( x − 2 y ) 2
9 z = 3 xy − 3 x 2 − 3 y 2
(i.e.) 3 z = xy − x − y
2
2
Problem 4
Find the singular integral of the partial differential
2
2
equation z = px + qy + p − q
Solution:
Given z = px + qy + p − q − − − − − (1)
2
2
This is in Clairaut’s form
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83
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The complete integral of equation (1) is
z = ax + by + a 2 − b 2 − − − − − −(2)
(replacing p by a and q by b)
To find singular integral, Diff. eqn. (2) p.w.r.t. ‘a’ and ‘b’,
we get
0 = x + 2a
0 = y − 2b
x
⇒ a=−
− − − − − − (3)
2
y
⇒b=
− − − − − − (4)
2
Substitute the values of a and b in equation (2) we have
2
 x
 y  x  y
(2) => z = x  −  + y   +  −  −  
 2
 2  2  2
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84
2
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4 z = −2 x 2 + 2 y 2 + x 2 − y 2
(i.e.) 4 z = y 2 − x 2
Problem 5
Find the singular integral of
z = px + qy + 2 pq
Solution:
Given z = px + qy + 2 pq − − − − − −(1)
This is in Clairaut’s form
The complete integral of equation (1) is
z = a x + b y + 2 a b − − − − − − − −(2)
(replacing p by a and q by b)
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85
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To find singular integral, Diff. eqn. (2) p.w.r.t. ‘a’ and ‘b’,
we get
2
0= x+
(b)
2 ab
⇒ −x=
b
− − − − − − (3)
a
2
a
and 0 = y +
(a) ⇒ − y =
− − − − − − (4)
2 ab
b
Multiplying (3) and (4) we get
x y=1, which is the singular integral.
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86
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Problem 6
2
2
z
=
px
+
qy
+
1
+
p
+
q
Solve:
Solution:
Given z = px + qy + 1 + p 2 + q 2 − − − − − (1)
This is in Clairaut’s form
The complete integral of equation (1) is
z = ax + by + 1 + a + b − − − − − − − (2)
2
2
(replacing p by a and q by b)
To find singular integral, Diff. eqn. (2) p.w.r.t. ‘a’ and ‘b’,
we get
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0= x+
⇒ x =−
1
2 1+ a + b
a
2
2
( 2a )
− − − − − − (3)
1 + a2 + b2
1
and 0 = y +
( 2b)
2
2
2 1+ a + b
b
⇒ y =−
− − − − − − ( 4)
1 + a2 + b2
Substitute (3) and (4) in equation (2), we get
(2) => z = ax + by + 1 + a + b
a2
b2
z=−
−
+ 1 + a2 + b2
1 + a2 + b2
1 + a2 + b2
2
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2
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=
=
(i.e.) z =
2
− a2 − b2 + 1 + a2 + b2
1 + a2 + b2
1
1 + a2 + b2
1
1+ a + b
2
2
− − − − − − (5)
Squaring and adding (3) and (4), we have
2
2
a
b
x2 + y2 =
+
2
2
1+ a + b
1 + a2 + b2
(1 + a 2 + b 2 ) − 1
1
=
=1−
2
2
1+ a + b
1 + a2 + b2
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x2 + y2 = 1 − z2
[ u sin g (5) ]
(i.e.) x 2 + y 2 + z 2 = 1
which is the singular integral.
To find general integral, assume b = f(a)
Then equation (2) becomes
z = a x + f (a) y + 1 + a 2 + { f (a)}2 − − − − − − − −(6)
Diff. eqn. (6) p.w.r.t. ‘a’, we get
0 = x + f ′(a) y +
1
2 1 + a + { f (a)}
2
2
[2a + 2 f (a). f ′(a).1] − − − (7)
The eliminant of ‘a’ between equations (6) and (7)
gives the general integral.
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Problem 7
Find the complete and singular solutions of
q
z = px + qy +  −
p

p 

Solution:
q

Given z = px + qy +  − p  − − − − − (1)
p

This is in Clairaut’s form
The complete integral of equation (1) is
b

z = ax + by +  − a  − − − − − − − (2)
a

(replacing p by a and q by b)
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To find singular integral, Diff. eqn. (2) p.w.r.t. ‘a’ and ‘b’,
we get
b
0 = x − 2 −1
a
1
and 0 = y +
a
b
⇒ x −1 = 2
a
⇒ y =−
− − − − − − (3)
1
1
⇒ a = − − − − − − − ( 4)
a
y
Substitute (4) in (3) , we get
x −1 =
b
 1
 − 
 y
⇒ x −1 = b y
B.Saravanan
2
2
x −1
⇒ b = 2 − − − − − − (5)
y
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Substitute (4) and (5) in equation (2), we have
b

(2) => z = ax + by +  − a 
a

 x − 1   ( x − 1) 1 
x
z = − + y 2  +  −
+ 
y
y
y
 y  
− x + x − 1 − ( x − 1) + 1
z=
y
(i.e.) y z = 1 − x
which is the singular integral
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Problem 1
Find the complete solution of
pqxy = z 2
Solution:
Given ( x p) ( y q) = z 2 − − − −(1)
Put X = log x , Y = log y
∂z 1
∂z ∂z ∂X
=
=
.
p=
∂x ∂X ∂x ∂X x
∂z
⇒ xp=
∂X
∂z
(i.e.) x p = P where P =
∂X
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∂z 1
∂ z ∂ z ∂Y
=
q=
=
.
∂ y ∂Y ∂ y ∂Y y
∂z
⇒ yq =
∂Y
∂z
(i.e.) y q = Q where Q =
∂Y
Equation (1) becomes
(1) => ( x p) ( y q) = z 2
P Q = z 2 − − − − − (2)
Let Q = aP Then equation (2) becomes
(2) => P.aP = z => P =
2
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z
a
95
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 z 
Now, Q = a P = a 
 =
 a
a z
Substitute P and Q in the relation
dz = P dX + Q dY
z
dz =
d X + a zdY
a
dz
a
= d X +a dY
z
Integratin g , we get
a log z = X + aY + b
(i.e.)
a log z = log x + a log y + b
which is the complete solution.
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Problem 2
Find the general solution of z 2 ( p 2 + q 2 ) = x + y
Solution:
Given ( z p) 2 + ( z q) 2 = x + y − − − − − (1)
Put Z = z 1 +1 = z 2
P
∂Z
∂Z
∂z
⇒ = z p where P =
= 2z
2
∂x
∂x
∂x
∂Z
∂z
Q
∂Z
= 2z
⇒ = z q where Q =
∂y
∂y
2
∂y
2
Equation (1) becomes
B.Saravanan
2
 P Q
  +  = x+ y
2 2
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(i.e.) P 2 + Q 2 = 4 ( x + y )
⇒ P2 − 4 x = 4 y − Q2 = a
Let P 2 − 4 x = a ⇒ P = ±
4x + a
Also 4 y − Q 2 = a ⇒ Q = ±
4y − a
Substitute p and q in the relation
dz = p dx + q dy
dz = ± 4 x + a dx ± 4 y − a dy
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Integrating we get
(4 x + a) 3 / 2 (4 y − a) 3 / 2
z=±
±
+b
4(3 / 2)
4(3 / 2)
(4 x + a) 3 / 2 (4 y − a) 3 / 2
z=±
±
+ b − − − − − − (2)
6
6
which is the complete integral.
To find singular integral, Diff. eqn. (2) p.w.r.t. ‘a’ and ‘b’, in
turn, we get
1
1
1/ 2
0 = ± ( 4 x + a ) ± ( 4 y − a )1 / 2
4
4
and 0 = 1
The last equation is absurd and shows that there is no
singular integral.
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To find general integral, assume b = f(a)
Then equation (2) becomes
(4 x + a) 3 / 2 (4 y − a) 3 / 2
z=±
±
+ f (a ) − − − − − (3)
6
6
Diff. eqn. (3) p.w.r.t. ‘a’, we get
1
1
1/ 2
0 = ± (4 x + a ) ± (4 y − a )1 / 2 + f ′(a) − − − − − (4)
4
4
The eliminant of ‘a’ between equations (3) and (4) gives the
general integral.
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Linear PDE of second and higher order
with constant coefficients:
Homogeneous PDE
Non-Homogeneous PDE
Homogeneous linear PDE
An equation in which the partial derivatives occurring are
all of same order and coefficients are constant is called
homogeneous PDE
Ex:
∂3z
∂3z
∂3z
∂3z
−2 2 −4
+8 3 = 0
3
2
∂x
∂x ∂y
∂x∂y
∂y
( D 3 − 2 D 2 D ′) z = 0
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Standard form:
The standard form of homogeneous PDE of nth order with
constant coefficients is of the form
( a 0 D n + a1 D n −1 D ′ + a 2 D n − 2 D ′ 2 + ............ + a n D ′ n ) z = F ( x, y )
∂
∂
′
where D =
and D =
∂x
∂y
f ( D, D ′) z = F ( x, y ) − − − − − (1)
where f ( D , D ′)is hom ogeneous polinomial of nth deg ree in D and D ′
The general solution of equation (1) is z=C.F+P.I
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Procedure to find C.F:
C.F is the solution of equation f ( D, D ′) = 0
a0 D n + a1 D n−1 D′ + a2 D n−2 D′ 2 + .......... .. + an D ′n = 0
put D = m and D′ = 1
⇒ a0 m n + a1m n−1 + a2 m n−2 + .......... .. + an = 0
The above equation is called auxiliary equation and solving
the A.E we get n roots m1, m2, m3, ......... mn,
Case I
(All the roots are distinct)
C.F = f1 ( y + m1 x) + f 2 ( y + m2 x) + ........... + f n ( y + mn x)
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Case II
(All the roots are equal)
C.F = f1 ( y + m1 x) + xf 2 ( y + m1 x) + x 2 f 3 ( y + m1 x)........... + x n−1 f n ( y + m1 x)
Note: roots may be real or complex
Particular Integral:
(1) R.H.S is an exponential function then
1
1
ax +by
P.I =
e
=
e ax+by if f (a, b) ≠ 0
f ( D, D′)
f ( a, b )
If f (a, b) = 0 then multiply x in the Nr. and differentiate
the Dr. w.r.to D and apply the same procedure.
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(2) R.H.S is an Trigonometric function then
1
P.I =
Sin (or )Cos(ax + by )
2
2
f ( D , DD′, D′ )
1
2
2
=
Sin
(
or
)
Cos
(
ax
+
by
)
if
f
(
−
a
,
−
ab
,
−
b
)≠0
2
2
f (−a ,−ab,−b )
if f (− a 2 ,− ab,−b 2 ) = 0 then multiply x in the Nr. and differentiate
the Dr. w.r.to D and apply the same procedure.
(3) R.H.S is an Polynomial function then
1
P.I =
x m y n = [1 + φ ( D, D′)]−1 x m y n
f ( D, D′)
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(4) Exponential shift rule
1
1
ax + by
ax +by
P.I =
e
φ ( x, y ) = e
φ ( x, y )
f ( D, D′)
f ( D + a, D ′ + b )
(5) General rule
If F ( x, y ) is any other function
1
P.I =
F ( x, y )
f ( D, D′)
factorize f ( D, D′) and then operate each factor by
1
F ( x, y ) = ∫ F ( x, a − mx)dx
D − mD′
after int egration a is replaced by y + mx
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Problem 1
3
2
(
D
−
2
D
D ′) z = 0
Solve
Solution:
A.E. is
m3 – 2m2 = 0
[Put D = m and D′ = 1]
m2(m – 2) = 0
m2 = 0 (or) m – 2 = 0
m = 0, 0, 2
∴ z = f1 ( y ) + x f 2 ( y ) + f 3 ( y + 2 x )
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Problem 2
3
′
(
D
−
D
)
z=0
Solve
Solution:
( D − D ′) 3 z = 0
A.E. is
(m – 1)3 = 0
[Put D = m and D′ = 1]
(m – 1)(m – 1)(m – 1) = 0
m = 1, 1, 1
∴ z = f1 ( y + x ) + x f 2 ( y + x ) + x 2 f 3 ( y + x )
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Problem 3
Solve ( D 2 − 2 DD ′ + D ′ 2 ) z = 0
Solution:
A.E. is
m2 – 2m + 1 = 0
[Put D = m and D′ = 1]
(m – 1)(m – 1) = 0
m = 1, 1
∴ z = f1 ( y + x ) + x f 2 ( y + x )
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∴
Problem 4
Solve ( D 3 + D 2 D ′ − DD ′ 2 − D ′ 3 ) z = 0
Solution:
A.E. is
m3 + m2 – m – 1 = 0
[Put D = m and D′ = 1]
m2(m + 1) –1(m + 1) = 0
(m + 1)(m2 – 1) = 0
m = –1, m2 = 1
m=±1
m = 1, –1, –1
∴ z = f1 ( y + x ) + f 2 ( y − x ) + x f 3 ( y − x )
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∴
Problem 5 3
∂ z
∂3z
∂3z
∂3z
−2 2 −4
+8 3 = 0
Solve
3
2
∂x
∂x ∂y
∂x∂y
∂y
Solution: The given equation can be written as
( D 3 − 2D 2 D′ − 4DD′2 + 8D′3 ) z = 0
A.E. is
m3 – 2m2 – 4m + 8 = 0
[Put D = m and D′ = 1]
m2(m – 2) – 4(m – 2) = 0
(m – 2)(m2 – 4) = 0
m = 2, m2 = 4
m=± 2
=> m = 2, 2, –2
∴ z = f1 ( y + 2 x) + x f 2 ( y + 2 x) + f 3 ( y − 2 x)
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Problem 6
Solve ( D 3 − 7 DD′ 2 − 6D′3 ) z = e 2 x + y + sin( x + 2 y)
Solution:
A.E. is m3 – 7m – 6 = 0
[Put D = m and D′ = 1]
m = –1 is a root
The other roots are
m2 – m – 6 = 0
(
m – 3)(m + 2) = 0
m = 3, –2
m = –1, –2, 3
B.Saravanan
C .F = f 1 ( y − x ) + f 2 ( y − 2 x ) + f 3 ( y + 3 x )
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1
2x + y
P. I1 = 3
e
D − 7 DD ′ 2 − 6 D ′ 3
1
2x + y
=
e
(2) 3 − 7(2)(1) 2 − 6(1) 3
1 2x+ y
e
12
1
P. I 2 = 3
sin( x + 2 y )
2
3
D − 7 DD ′ − 6 D ′
=−
=
1
sin( x + 2 y )
− D − 7 D ( −4) − 6( −4 D ′)
1
=
sin( x + 2 y )
27 D + 24 D ′
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=
1
sin( x + 2 y )
3(9 D + 8D ′)
9 D − 8D′
=
sin( x + 2 y )
3(9 D + 8D′)(9 D − 8D ′)
9 D − 8D′
=
sin( x + 2 y )
2
2
3(81D − 64 D ′ )
9 D − 8D′
=
sin( x + 2 y )
3[81(−1) − 64(−4)]
9 D[sin( x + 2 y )] − 8 D ′[sin( x + 2 y )]
=
525
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∴
1
[9 cos( x + 2 y ) − 16 cos( x + 2 y )]
=
525
1
=
[−7 cos( x + 2 y )]
525
1
cos( x + 2 y )
=−
75
z = C.F + P.I1 + P.I2
(i.e.) z = f 1 ( y − x) + f 2 ( y − 2 x) + f 3 ( y + 3 x)
1 2x + y 1
−
e
− cos( x + 2 y )
12
75
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Problem 7
Solve ( D 3 + D 2 D′ − DD′ 2 − D′3 ) z = e 2 x + y + cos(x + y)
Solution:
A.E. is
m3 + m2 – m – 1 = 0
[Put D = m and D′ = 1]
m2(m + 1) –1(m + 1) = 0
(m + 1)(m2 – 1) = 0
m = –1, m2 = 1
m=±1
m = 1, –1, –1
C. F = f1 ( y + x) + f 2 ( y − x) + x f 3 ( y − x)
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1
2x + y
P. I1 = 3
e
D + D 2 D ′ − DD ′ 2 − D ′ 3
1
2x + y
=
e
(2) 3 + (2) 2 (1) − (2)(1) 2 − (1) 3
1 2x+ y
= e
9
1
P.I 2 = 3
cos( x + y )
2
2
3
D + D D ′ − DD ′ − D ′
1
=
cos( x + y )
− D − D′ + D + D′
1
= cos( x + y )
0
B.Saravanan
Since the denominator = 0, we
have to multiply x on Nr. and
Diff. Dr. w.r.t.‘D’
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x
=
cos( x + y )
2
2
3D + 2 DD ′ − D′
x
=
cos( x + y )
3(−1) + 2(−1) − (−1)
x
= − cos( x + y )
4
z = C.F + P.I1 + P.I2
(i.e.) z = f 1 ( y + x) + f 2 ( y − x) + x f 3 ( y − x)
1 2x + y x
+ e
− cos( x + y )
9
4
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Problem 8
Solve (4 D 2 − 4 DD′ + D′ 2 ) z = e 3 x − 2 y + sin x
Solution:
A.E. is
4m2 – 4m + 1 = 0
[Put D = m and D′ = 1]
(2m – 1)2 = 0
1 1
m= ,
2 2
1 
1 


C. F . = f1  y + x  + x f 2  y + x 
2 
2 


1
3x − 2 y
P.I1 =
e
4 D 2 − 4 DD ′ + D ′ 2
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1
3x − 2 y
=
e
4(3) 2 − 4(3)(−2) + (−2) 2
1 3x − 2 y
= e
64
1
sin( x + 0 y )
P.I2 =
2
2
4 D − 4 DD ′ + D ′
1
=
sin( x + 0 y )
4(−1) − 0 + 0
1
= − sin x
4
z = C.F + P.I1 + P.I2
1 
1  1 3x − 2 y 1


(i.e.) z = f1  y + x  + x f 2  y + x  + e
− sin x
2 
2  64
4


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Problem 9
Solve ( D 2 + 2 DD′ + D′ 2 ) z = x 2 y + e x − y
Solution:
A.E. is
m2 + 2m + 1 = 0
[Put D = m and D′ = 1]
(m + 1)(m + 1) = 0
m = –1, –1
C. F . = f 1 ( y − x ) + x f 2 ( y − x )
1
x− y
e
P.I1 = 2
D + 2 DD ′ + D ′ 2
1
x− y
= 2
e
(1) + 2(1)(−1) + (−1) 2
B.Saravanan
121
Since the denominator = 0, we
have to multiply x on Nr. and Diff.
Dr. w.r.t.‘D’
6/4/2015
=
x
ex− y
2 D + 2 D′
x2 x− y
=
e
2
1
2
x
y
P.I2 = 2
2
D + 2 DD ′ + D ′
1
2
=
x
y
2
2 DD′ + D′ 
2
D 1 +

2
D


1
= 2
D
B.Saravanan
  2 DD ′ + D ′
1 + 
2
D
 
2



−1
2
x y
122
1
= 2
D
  2 DD′ + D′ 2  2
 x y
1 − 
2
D

 
6/4/2015
1
= 2
D
 2 D′  2
1 − D  x y
1
= 2
D
2 D′ 2 
 2
( x y ) − D ( x y )
1
= 2
D
2 2 
 2
 x y − D ( x )
1
= 2
D
 2
2 x3 
x y −

3


1  x3 y 2x 4 
= 
−

12 
D 3
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x 4 y x5
=
−
12 30
z = C.F + P.I1 + P.I2
x 2 x − y x4 y x5
(i.e.) z = f1 ( y − x) + x f 2 ( y − x) + e +
−
2
12 30
Problem 10
Solve ( D 2 + 3DD′ − 4D′2 ) z = x + sin y
Solution:
A.E. is
m2 + 3m – 4 = 0
[Put D = m and D′ = 1]
(m – 1)(m + 4) = 0
m = 1, – 4
B.Saravanan
=> C. F . = f1 ( y + x) + f 2 ( y − 4 x)
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P.I1 =
=
1
x
2
2
D + 3 DD ′ − 4 D ′
1
x
 3DD ′ − 4 D ′ 
D 1 +

2
D


2
2
−1
1
= 2
D
  3DD ′ − 4 D ′
1 + 
2
D
 
1
= 2
D
  3DD′ − 4 D′ 2 
 x
1 − 
2
D

 
2



x
3
1
1  x2 
x
= 2 [x − 0] =   =
D
D2
6
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1
P.I2 = 2
sin( 0 x + y )
2
D + 3 DD ′ − 4 D ′
1
=
sin(0 x + y )
0 + 0 − 4(−1)
1
= sin y
4
z = C.F + P.I1 + P.I2
x3 1
(i.e.) z = f1 ( y + x) + f 2 ( y − 4 x) +
+ sin y
6 4
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Problem 11 2
∂ z
∂2 z
∂2 z
Solve
+
− 2 2 = sinh(x + y) + xy
2
∂x
∂ x∂y
∂y
Solution:
The given equation can be written as
( D 2 + DD′ − 2 D′ 2 ) z = sinh(x + y) + xy
A.E. is
m2 + m – 2 = 0
[Put D = m and D′ = 1]
(m + 2)(m – 1) = 0
m = –2, 1
C.F . = f1 ( y − 2 x) + f 2 ( y + x)
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1
P.I1 = 2
sinh( x + y )
2
D + DD ′ − 2 D ′
 e x+ y − e − ( x+ y ) 
1
= 2

2
D + DD ′ − 2 D ′ 2 

1
=
2
1
1

x+ y
−x−y 
−
e
e
 D 2 + DD′ − 2 D′ 2

D 2 + DD ′ − 2 D ′ 2
1
=
2

1
1
x+ y
− x−y 
 (1) 2 + (1)(1) − 2(1) 2 e − (−1) 2 + (−1)(−1) − 2(−1) 2 e



Since the denominator = 0, we have to multiply x on Nr.
and Diff. Dr. w.r.t. ‘D’ we get
B.Saravanan
128
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1 x
x

x+ y
= 
e −
e− x −y 
2  2 D + D′
2 D + D′

1  x x+ y
x
−x−y 
= 
e −
e

− 2 −1
2 2 +1
x x+ y x − x − y
= e + e
6
6
1
P.I2 = 2
xy
2
D + DD ′ − 2 D ′
1
=
xy
2
DD′ − 2 D′ 
2
D 1 +

2
D


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B.Saravanan
−1
1
= 2
D
  DD′ − 2 D′
1 + 
2
D
 
1
= 2
D
  DD′ − 2 D′ 2 
 xy
1 − 
2
D

 
1
= 2
D
 D′ 
1 − D  xy
1
= 2
D
D′


(
xy
)
−
(
xy
)


D
1
= 2
D
1


xy
−
(
x
)

D 
2



130
xy
6/4/2015
1
= 2
D

x2 
 xy − 2 


1  x 2 y x3 
= 
− 
D 2
6
x3 y x 4
=
−
6
24
z = C.F + P.I1 + P.I2
x x + y x − x − y x3 y x 4
(i.e.) z = f1 ( y − 2 x) + f 2 ( y + x) + e + e
+
−
6
6
6
24
B.Saravanan
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Problem 12 2
∂ z
∂2 z
∂2 z
−5
+ 6 2 = y sin x
Solve
2
∂x
∂ x∂y
∂y
Solution:
The given equation can be written as
( D 2 − 5DD′ + 6D′2 ) z = y sin x
A.E. is
m2 – 5m + 6 = 0
[Put D = m and D′ = 1]
(m – 2)(m – 3) = 0
m = 2, 3
C. F . = f 1 ( y + 2 x ) + f 2 ( y + 3 x)
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1
P.I =
y sin x
2
2
D − 5 DD ′ + 6 D ′
1
=
y sin x
( D − 2 D′) ( D − 3D′)
1
 1

=
y
sin
x

D − 2 D′  D − 3D′
1
=
D − 2 D′
∫ (c − 3x) sin x dx
where y = c – 3x
1
=
[(c − 3x)(− cos x) − (−3)(− sin x)]
D − 2 D′
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1
=
[− y cos x − 3 sin x]
D − 2 D′
= ∫ [−(c − 2 x) cos x − 3 sin x] dx
where y = c – 2x
= − [ (c − 2 x)(sin x) − (−2)(− cos x)] − 3(− cos x)
= − y sin x + 2 cos x + 3 cos x
= 5 cos x − y sin x
z = C.F + P.I
(i.e.) z = f1 ( y + 2 x) + f 2 ( y + 3x) + 5 cos x − y sin x
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Problem 13
2
2
x− y
′
(
D
−
D
)
z
=
e
sin(2 x + 3 y)
Solve
Solution:
A.E. is m2 – 1 = 0
[Put D = m and D′ = 1]
m2 = 1
m=±1
C. F . = f 1 ( y + x ) + f 2 ( y − x )
1
x− y
P.I = 2
e
sin( 2 x + 3 y )
2
D − D′
=e
B.Saravanan
x− y
1
sin(2 x + 3 y )
2
2
( D + 1) − ( D′ − 1)
135
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=e
x− y
1
sin(2 x + 3 y )
2
2
D + 2 D + 1 − D′ + 2 D′ − 1
=e
x− y
1
sin(2 x + 3 y )
2
2
D + 2 D − D′ + 2 D′
=e
x− y
1
sin(2 x + 3 y )
− 4 + 2 D − (−9) + 2 D′
=e
=e
B.Saravanan
x− y
x− y
1
sin(2 x + 3 y )
2( D + D′) + 5
[2( D + D′) − 5]
sin(2 x + 3 y )
[2( D + D′) + 5][2( D + D′) − 5]
136
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=e
x− y
=e
x− y
=e
x− y
=e
x− y
=e
x− y
B.Saravanan
[2( D + D′) − 5]
sin(2 x + 3 y )
2
4( D + D′) − 25
[2( D + D′) − 5]
sin(2 x + 3 y )
2
2
4( D + 2 DD′ + D′ ) − 25
[2( D + D′) − 5]
sin(2 x + 3 y )
4[(−4) + 2(−6) + (−9)] − 25
[2( D + D′) − 5]
sin(2 x + 3 y )
− 125
2 D[sin(2 x + 3 y )] + 2 D′[sin(2 x + 3 y )] − 5 sin(2 x + 3 y )
− 125
137
6/4/2015
ex− y
=−
[4 cos(2 x + 3 y ) + 6 cos(2 x + 3 y ) − 5 sin(2 x + 3 y )]
125
ex− y
=−
[10 cos(2 x + 3 y ) − 5 sin(2 x + 3 y )]
125
ex− y
=
[sin(2 x + 3 y ) − 2 cos(2 x + 3 y )]
25
z = C.F + P.I
ex− y
(i.e.) z = f1 ( y + x) + f 2 ( y − x) +
[sin(2 x + 3 y ) − 2 cos(2 x + 3 y )]
25
B.Saravanan
138
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Non-Homogeneous PDE
f ( D , D ′) z = F ( x , y ) − − − − − (1)
if the polonomial f ( D , D ′) is not hom ogeneous
then eqn .(1) called non − hom ogeneous linear PDE .
In the equation
The general solution of equation (1) is z=C.F+P.I
The method to find P.I. is same as those for homogeneous
PDE
To find C.F.
Factorize f ( D , D ′) in to
( D − m1 D ′ − c1 )( D − m 2 D ′ − c 2 ) .......... .( D − m n D ′ − c n ) z = 0
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C.F = e c1x f1 ( y + m1 x) + e c2 x f 2 ( y + m2 x) + ........ + e cn x f n ( y + mn x)
for repeated factor say ( D − mD ′ − a ) 3 z = 0
then C.F = e ax f1 ( y + mx) + xe ax f 2 ( y + mx) + x 2 e ax f 3 ( y + mx)
B.Saravanan
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Problem 1
Solve ( D − 2 D ′)( D − 2 D ′ + 1) z = 0
Solution:
The given equation is non-homogeneous.
( D − 2D′)(D − 2D′ + 1) z = 0
∴ z = e 0 x f1 ( y + 2 x ) + e − x f 2 ( y + 2 x )
(i.e.) z = f1 ( y + 2 x ) + e − x f 2 ( y + 2 x )
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Problem 2
Solve ( D 2 − 2 DD′ + D′ 2 − 3D + 3D′ + 2) z = (e3 x + 2 e − 2 y ) 2
Solution:
The given equation is non-homogeneous and it can be written as
( D − D′ − 1)(D − D′ − 2) z = e6 x + 4 e − 4 y + 4 e3 x − 2 y
C .F . = e x f 1 ( y + x ) + e 2 x f 2 ( y + x )
1
P.I1 =
e6x + 0 y
( D − D ′ − 1)( D − D ′ − 2)
1
1 6x
6x + 0 y
=
e
= e
(6 − 0 − 1)(6 − 0 − 2)
20
B.Saravanan
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P.I2 =
1
4 e0x − 4 y
( D − D ′ − 1)( D − D ′ − 2)
=4
P.I3 =
1
e0x − 4 y = 2 e− 4 y
(0 + 4 − 1)(0 + 4 − 2)
3
1
4 e3x − 2 y
( D − D ′ − 1)( D − D ′ − 2)
1
1 3x − 2 y
3x − 2 y
e
=4
= e
(3 + 2 − 1)(3 + 2 − 2)
3
z = C.F + P.I1 + P.I2 + P.I3
1 6 x 2 − 4 y 1 3x − 2 y
(i.e.) z = e f1 ( y + x) + e f 2 ( y + x) +
e + e + e
20
3
3
x
B.Saravanan
2x
143
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Problem 3
2
2
2 x+ y
′
′
′
(
D
+
D
+
2
D
D
+
2
D
+
2
D
+
1
)
z
=
e
Solve
Solution:
The given equation is non-homogeneous and it can be written
as
( D + D′ + 1)(D + D′ + 1) z = e 2 x+ y
C . F . = e − x f 1 ( y − x ) + xe − x f 2 ( y − x )
1
2x + y
e
P.I = 2
D + D ′ 2 + 2 DD ′ + 2 D + 2 D ′ + 1
1
2x + y
1 2x+ y
=
e
= e
(2) 2 + (1) 2 + 2(2)(1) + 2(2) + 2(1) + 1
16
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z = C.F + P.I
1 2 x+ y
(i.e.) z = e f1 ( y − x) + xe f 2 ( y − x) + e
16
−x
−x
Problem 4
Solve (2 D 2 − DD′ − D′2 + 6D + 3D′) z = x e y
Solution:
Given (2D 2 − DD′ − D′ 2 + 6 D + 3D′) z = x e y
(2D + D′)(D − D′ + 3) z = x e y
1
Here c1 = 0, m1 = − , c 2 = −3, m2 = 1
2
B.Saravanan
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C. F . = e
0x
1  −3 x

f1  y − x  + e f 2 ( y + x)
2 

1  −3 x

= f1  y − x  + e f 2 ( y + x)
2 

1
y
P.I =
x
e
2 D 2 − DD ′ − D ′ 2 + 6 D + 3 D ′
1
=e
x
2
2
2 D − D( D′ + 1) − ( D′ + 1) + 6 D + 3( D′ + 1)
y
1
=e
x
2
2
2 D − DD′ − D − D′ − 2 D′ − 1 + 6 D + 3D′ + 3
y
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1
=e
x
2
2
2 + 2 D − DD′ − D′ + 5 D + D′
y
1
=e
x
2
2
 2 D − DD′ − D′ + 5 D + D′ 
2 1 +

2


y
B.Saravanan
−1
e
=
2
 2 D − DD′ − D′ + 5D + D′ 
1 +

2


ey
=
2
  2 D 2 − DD′ − D′ 2 + 5D + D′ 
 x
1 − 
2

 
ey
=
2
5D  e y 

x−
 x − 2 ( x) =

2 
y
2
2
147
x
5
2 
6/4/2015
ey
=
(2 x − 5)
4
z = C.F + P.I
y
1
e


(i.e.) z = f1  y − x  + e − 3 x f 2 ( y + x) +
(2 x − 5)
2 
4

Problem 5
Solve ( D 2 − 3DD′ + 2D′2 + 2 D − 2D′) z = x + y + sin(2 x + y)
Solution:
Given ( D 2 − 3DD′ + 2D′ 2 + 2D − 2D′) z = x + y + sin(2 x + y)
( D − D′)(D − 2D′ + 2) z = x + y + sin(2 x + y)
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Here c1 = 0, m1 = 1, c 2 = −2, m1 = 2
C .F . = e 0 x f 1 ( y + x ) + e − 2 x f 2 ( y + 2 x )
= f 1 ( y + x ) + e −2 x f 2 ( y + 2 x )
1
P.I1 = 2
( x + y)
2
D − 3 DD ′ + 2 D ′ + 2 D − 2 D ′
1
=
( x + y)
( D − D′)( D − 2 D′ + 2)
=
B.Saravanan
 D′ 
D 1 − 
D

1
( x + y)
D − 2 D′ 

21 +

2


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1  D′ 
=
1 − 
2D 
D
−1
1 
D′ 
=
1 + 
2D 
D
D − 2 D′ 

1 +

2


−1
( x + y)
  D − 2 D′   D − 2 D′  2 
+
  ( x + y)
1 − 
2
2
 
 
 

1  1 D′   D
D2
=  + 2  1 − + D′ +
− DD′ ( x + y )
2D D  
2
4

1  1 1 D′ D
D′ D′ D′ 
=  − +
+ − D′ + 2 −
+  ( x + y)
2 D 2 D 4
D
2D 4 
1  1 1 D ′ D D ′ 3D ′ 
=  − +
+ + 2 −
( x + y)

2  D 2 2D 4 D
4 
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x + y D ′( x + y ) D( x + y ) 
1
+
+
( x + y) −


1 D
4
2
2D
= 

D ′( x + y ) 3D ′( x + y )
2

+
−


4
D2
1  x2
x y x 1 x2 3 
=  +xy− − + + +
− 
2 2
2 2 2 4 2 4
1 2
y 1
= x + x y − − 
2
2 2
P.I2 =
B.Saravanan
1
sin( 2 x + y )
2
2
D − 3 DD ′ + 2 D ′ + 2 D − 2 D ′
151
6/4/2015
=
1
sin(2 x + y )
− 4 − 3(−2) + 2( −1) + 2 D − 2 D′
1
=
sin(2 x + y )
2 D − 2 D′
2 D + 2 D′
=
sin(2 x + y )
(2 D − 2 D ′)(2 D + 2 D ′)
2 D + 2 D′
=
sin(2 x + y )
2
2
4D − 4D′
2 D + 2 D′
=
sin(2 x + y )
4(−4) − 4(−1)
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2 D[sin( 2 x + y )] + 2 D ′[sin( 2 x + y )]
=
− 12
1
= − [4 cos(2 x + y ) + 2 cos(2 x + y )]
12
1
= − [6 cos(2 x + y )]
12
1
= − cos(2 x + y )
2
z = C.F + P.I1 + P.I2
(i.e.) z = f 1 ( y + x ) + e −2 x f 2 ( y + 2 x )
1 2
y 1 1
+  x + x y − −  − cos( 2 x + y )
2
2 2 2
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