Non existence of singular integral type 2b equations

Non-existence of Singular Integral : Type 2(b) equations do not have singular integrals.
EXERCISE 10.5.3.
Find the complete integrals of the following equations:
(a) q
2
= y p
4
(b)
√
p +
√
q = y
(c) p
2
q
3
= y
Ans.
(a) z = ax +
2 a
2
y
3 / 2
3
+ b
(b) z = a
2
x +
( y - a )
3
3
+ b
(c) z = a
3
x +
3 y
4 / 3
4 y 2
+ b
(c) For an equation involving p , q and z :
f ( p , q , z ) = 0 . (10.5.10)
Charpit’s auxiliary equations become
d x
f
p
=
d y
f
q
=
d z
pf
p
+ qf
q
=
d p
- pf
z
=
d q
- qf
z
·
From the last two ratios, we see that p = aq , Inserting this in (10.5.10),
f ( aq , q , z ) = 0 or q = φ ( z , a ) . (10.5.11)
Using p = aq and (10.5.11) in (10.4.4),
d z = q ( a d x + d y ) or { φ ( z , a )}
- 1
d z = a d x + d y
which on integration gives the complete integral of (10.5.10):
Φ ( z , a ) = ( ax + y ) + b . (10.5.12)
We may employ the substitution q = ap also, and obtain the complete integral.
Non-existence of Singular Integral : Type 2(c) equations do not have singular integrals.
EXERCISE 10.5.4.
Find the complete integrals of the following equations:
(a) z = p
2
+ q
2
(b) p
3
= qz

Chapter 10. Partial Differential Equations of First Order
61
(c) z = p
2
+ q
2
+ 1
(d) z ( p
2
+ q
2
+ 1 ) = 1
(e) 4 ( z
3
+ 1 ) = 9 z
4
pq
(f) q
2
= p
2
z
2
( 1 - p
2
)
(g) p
2
z
2
+ q
2
= 1
(h) 9 ( p
2
z + q
2
) = 4
Ans.
(a) 4 ( a
2
+ 1 ) z = ( x + a y + b )
2
(b) 2
√
z =
√
a ( x + a y + b )
(c)
√
a 2 + 1 cosh
- 1
z = x + a y + b
(d) ( a
2
+ 1 )( 1 - z
2
) = ( x + a y + b )
2
(e) a
2
( z
3
+ 1 ) = ( x + a
2
y + b )
2
(f) z = a
2
+ ( x + a
2
y + b )
2
(g)
z
2
√
z 2 + a 2 ) + a
2
sinh
- 1
( z / a ) = x + a y + b
(h) ( z + a )
3 / 2
= x + a y + b
TYPE 3: Separable Equations of the form
Consider an equation of the form
f ( p , x ) = g ( q , y ) . (10.5.13)
For this, the Charpit’s auxiliary equations reduce to
d x
f
p
=
d y
- g
q
=
d z
pf
p
- qg
q
=
d p
- f
x
=
d q
- g
y
·
From these, we have an ordinary differential equation:
d p
d x
+
f
x
f
p
= 0 , which can be written as
f
p
d p + f
x
d x = 0 or d f ( p , x ) = 0 .
Integrating this total differential, we get the solution
f ( p , x ) = a .
Using this in (10.5.13),
f ( p , x ) = a , g ( q , y ) = a .
Solving these for p and q ,
p = μ ( x , a ) , q = ν ( y , a ) . (10.5.14)
Using (10.5.14) in (10.4.4),
d z = μ ( x , a ) d x + ν ( y , a ) d y ,
which on integration gives the complete integral of (10.5.13) as
z =
„
μ ( x , a ) d x +
„
ν ( y , a ) d y + b . (10.5.15)

62
10.5. Special forms of Nonlinear First Order Partial Differential Equations
Non-existence of Singular Integral : Type 3 equations also do not have singular integrals.
EXERCISE 10.5.5.
Find the complete integrals of the following equations:
(a) px
2
= q y
2
(b) pq + qx = y
(c)
p
2
x 2
-
q
2
y 2
= 1
(d) p
2
- q
2
= x - y
(e)
√
p +
√
q = x + y
(f) ( p + q ) x + pq = 0
(g) p
2
= q / x y
(h) px - y
2
q
2
= 1
Ans.
(a) x y z + a ( x + y ) = bx y
(b) z = ax - x
2
+
y
2
2 a
+ b
(c) z =
x
3
a
2
3
+ 2
√
a 2 - 1 y + b
(d) 3 z = 2 ( x + a )
3 / 2
+ 2 ( y + a )
3 / 2
+ κ
(e) 3 z = ( x + a )
3
+ ( y - a )
3
+ κ
(f) 2 a ( a + 1 ) z = -( a + 1 ) x
2
+ a
y
2 + b
(g) 2 z = ( 4
√
x + y
2
) a
2
+ b
(h) z = ( a
2
+ 1 ) log x + a log y + b
EXERCISE 10.5.6.
Find the general and complete integrals of the following equations:
(a) px - q y = y
2
- x
2
(b) p + q = sin x + sin y
(c) px
2
- 2 y
3
q = 1
(d) 2 p - 3 q = z
(e) p + q = 1
Ans.
(a) Auxiliary equations are
d x
x
=
d y
- y
=
d z
y 2 - x 2
. Grouping the first two ratios, we get x y = a . Choosing
( x , y , 1 ) as multipliers, the second solution is x
2
+ y
2
+ 2 z = b . Therefore, the general integral is
f
(
x y , x
2
+ y
2
+ 2 z
)
= 0 . Also, the complete integral is x
2
+ y
2
+ 2 z = 2 a log x y + b .