ma2211 unit 3
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UNIT III PARTIAL DIFFERENTIAL EQUATIONS
PART – A
Problem 1 Form the partial differential equation by eliminating a and b from
2 2 2 2 z x a y b .
Solution:
2 2 2 2 z x a y b - (1)
Differentiating (1) partially w.r. t x and y we get
2 2 2 z
p y b x x
- (2)
2 2 2 zq x a y y
- (3)
Multiplying Eqn. (2) and Eqn (3)
2 2 2 2 4 pq x a y b xy
4 pq xyz [using (1)]
Problem 2 Obtain the partial differential equation by eliminating arbitrary constants a
and b from 2 2 2 1 x a y b z .
Solution:
2 2 2
1 x a y b z - (1)Differentiating (1) partially w.r. t x and y we get
2 2 0 x a zp
x a zp - (2)
2 2 0 y b zq
y b zq - (3)
Substituting (2) & (3) in (1) we get2 2 2 2 2 1 z p z q z
i.e., 2 2 2 1 1 z p q
Problem 3 From the partial differential equation by eliminating f from
2 2 2 , 0 f x y z x y z .
Solution:
We know that if f (u, v) = 0
then u = f (v)
2 2 2 x y z f x y z - (1)
Differentiating (1) partially w.r. t x and y
We get
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'2 2 1 x zp f x y z p - (2)
'2 2 1 y zq f x y z q
Divide (2) & (3)1
1
x zp p
y zq q
x qx zp zpq y py zq zpq
z y p x z q y x
Problem 4 Form the partial differential equation by eliminating f from
2 12 log z x f x
y
.
Solution:
Let 2 12 log z x f x
y
- (1)
Differentiate (1) w.r. t x and y
' 1 12 2 log
z p x f x
x y x
- (2)
'
2
1 12 log
zq f x
y y y
- (3)
Eliminating ' f from (2) & (3)
22 1 p x y
q x
2 22 px x qy 2 22 px qy x
Problem 5 Obtain the partial differential equation by eliminating the arbitrary constants
a & b from2 2
z xy y x a b .
Solution:
2 2 z xy y x a b - (1)
Differentiating (1) partially w.r. t x and y
2 2
12
2
z p y y x
x x a
2 2
yx p y
x a
2 21
p x
y x a
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2 21
p x
y x a
- (2)
2 2 zq x x a y
2 2q x x a - (3)
Multiplying (2) & (3)
2 2
2 21
1
p xq x x a
y x a
pq x x
y
p y q x xy
pq xp yq xy xy
px qy pq
Problem 6 Find the complete integral of p q pq where z z
p and q x y
.
Solution:
p q pq - (1)
This is of the form , 0 f p q
Let z ax by c - (2) be the complete solution of the partial differential
equation.
z p a
x
zq b
y
(1) reduces to a+b=ab
1
1
a b a
ab
a
1
a z ax y ca
Problem 7 Obtain the complete integral of 2 2 z px qy p q .
Solution:2 2 z px qy p q - (1)
This equation is of the form , z px qy f p q (clairaut’s type)
the complete integral is 2 2 z ax by a b .
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Problem 8 Solve 1 p q qz .
Solution:
1 p q qz - (1)
This equation is of the form , , 0 f z p q
z f x ay be the solution
x ay u z f u
dz adz p q
du du
(1) reduces to
1
1
1
1
1
dz dz dza a z
du du du
dza az
du
dza az
du
dz z
du a
dzdu
za
Integrating 1log z u ba
i.e.,1
log z x ay ba
is the complete solution.
Problem 9 Solve the equation tan tan tan p x q y z .
Solution:
Given tan tan tan p x q y z
This equation is of the form p q
P Q R
When P = tan x Q = tan y R = tan z
The subsidiary equations are
. .,tan tan tan
dx dy dz
P Q R
dx dy dzi e
x y z
Considering the first two,
tan tan
dx dy
x y
Sign,
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cot cot
log sin log sin log
sinlog log
sin
sin
sin
sin. .,
sin
xdx ydy
x y a
xa
y
xa
y
xi e a
y
Take,
tan tan
tan tan
dy dz
y z
dy dz
y z
cot cot
log sin log sin log
sinlog log
sin
ydy zdz
y z b
yb
z
sin
sin
yb
z
Hence the general solution is sin sin, 0sin sin
x y y z
Problem 10 Solve 3 2 33 2 0 D DD D Z .
Solution:
Substituting D = m, & D1
= 1
The Auxiliary equation is m3
- 3m + 2 = 0
m = 1, 1, -2
Complimentary function is 1 2 32 y x x y x y x
i.e., 1 2 31 2 Z y x y x y x
Problem 11 Find the general solution of
2 2 2
2 24 12 9 0
z z z
x x y y
.
Solution:
2 1 124 12 9 0 D DD D Z
The auxiliary equation is 24 12 9 0m m 4m2 – 6m – 6m +9 = 0
2m (2m – 3) – 3 (2m – 3) = 0
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(2m – 3)2
= 0
m =3
2
(twice)
C.F. =1 2
3 3
2 2 y x x y x
the General solution is Z = C.F. + P.I
1 2
3 3
2 2 z y x x y x
Problem 12 Solve 3 2 2 3 0 D DD D D D z .
Solution:
The Auxiliary equation is
3 2
2
2
1 0
1 1 0
1 1 0
. . 1, ,
m m m
m m m
m m
i e m i i
The general solution is 1 2 3 Z y x y ix y ix
Problem 13 Find the particular integral of 3 2 2 3 cos2 x D D D DD D z e y .
Solution:
P.I.3 2 2 3
1cos2 x
e y D D D DD D
3 2 2 3
2
3 2 2 3
2
cos2
1 1 1
R.P. of 1 1 1
R.P. of 1 2 4 8
1 2R.P. of cos 2 sin 25 1 2 1 2
1 1. cos 2 2sin 2
5 5
x
iy x
iy x
x
x
ye
D D D D D D
ee
D D D D D D
ee
i i
e i y i yi i
e y y
P.I. cos 2 2 sin 225
xe
y y
Problem 14 Solve 2 1 0 D DD D z .
Solution:
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21 0
1 ' 1 0
D DD D
D D D
This is of the form
1 1
1 1 2 2
1 1 2 2
0
0, 1, 1, 1
D m D D m D
m m
C.F. is 1 2
1 1 2 2
x x Z e f y m x e f y m x
1 2
x x Z e f y e f y x
Problem 15 Solve 21 2 x y D D D D z e .
Solution:
This is of the form 1 1 2 2 .... 0
n n D m D C D m D C D m D C Z
Hence1 1 2 2
1 1 1 1m c m c 2
2c
Hence the C.F. is 2
1 2
x x z e y x e y x
P.I.
2
1 2
x ye
D D D D
2
2
2 1 1 2 1 2
1
2
x y
x y
e
e
Hence, the complete solution is 2 2
1 2
1
2
x x x y Z e y x e y x e
PART – B
Problem 16
a. From the partial differential equation by eliminating f and
from z f y x y z .
Solution:
z f y x y z - (1)
Differentiating partially with respect to x and y, we get
1P x y z p - (2)
' ' 1q f y x y z q - (3)
2
' . " 1r x y z r x y z p - (4)
' . " 1 1s x y z s x y z p q - (5)
2
" ' '' 1t f y x y z t x y z q - (6)
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From (4) 2
1 ' 1 "r x y z p x y z - (7)
From (5) 1 ' 1 1 "s x y z p q x y z - (8)
Dividing (7) & (8) we get
1
1
1 1
pr
s q
q r p s
b. Solve 2 2 32 x D DD z e x y .
Solution:
Auxiliary Equation is given by 2 2 0m m i.e., m (m – 2) =0
m = 0, m = 2
C.F. = 1 2 2 f y f y x
2
1 2
2
1
2
(Replace D by 2 and D by 0)4
x
x
PI e D DD
e
3
2 2
13
2
3
2
33
2
5 6
5 6
1
2
1 21
1 21 ......
1 2
2.20 4.5.6
20 60
PI x y D DD
D x y
D D
D x y
D D
x x y
D D
x x y
x x y
The complete solution is
1 2
2 5 6
1 2
. .
24 20 20
x
Z C F PI PI
e x y x Z f y f y x
Problem 17
a. Find the complete integral of p q x y .
Solution:
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The given equation does not contain z explicitly and is variable separable.
That is the equation can be rewritten as p – x = y – q =a, say - (!)
and p a x q y a
Now dz = pdx + qdy
dz a x dx y a dy (2)
Integrating both sides with respect to he concerned variables, we get
2 2
2 2
a x y a z b
- (3)
when a and b are arbitrary constants.
b. Solve 2 2 y p xyq x z y .
Solution:
This is a Lagrange’s linear equation of the form Pp Qq R
The subsidiary equations are
2 2
dx dy dz
P Q R
dx dy dz
y xy x z y
Taking I & II ratios
2
dx dy
y xy
xdx ydy
Integrating,2 2
2 2
1
2 2
2
x yC
x y c c
Taking II & III ratios 2
dy dz
xy x z y
2
2
2
Z y dy ydz
ydz zdy ydy
d yz ydy
Integrating,2
2
2
2
yz y c
yz y c
the solution is 2 2 2, 0 x y yz y
Problem 18
a. Find the singular integral of the partial differential equation2 2 z px qy p q .
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Solution:
The given equation 2 2 z px qy p q is a clairaut’s type equation.
Hence the complete solution is2 2
z ax by a b -(1)To get singular Solution :Differentiate (1) w.r.t. a and b
0 = x + 2a - (2)
0 = y – 2b - (3)
and2 2
x ya b
Substituting in (1),2 2 2 2
2 2 2 2
2 2
2 2 4
2 24
4 is the singular solution.
x y x y z
x y x y z
z y x
b. Solve 2 2 24 5 3 sin 2 x y D DD D z e x y .
Solution:
The Auxiliary equation,
2
2
4 5 0
5 5 0
5 5 0
1 5 0
1, 5
m m
m m m
m m m
m m
m
C.F. = 5 f y x g y x
2
1 2 2
2
2
2
13
4 5
3
4 8 5
3
9
1
3
x y
x y
x y
x y
PI e D DD D
e
e
e
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2 2 2
1sin 2
4 5
1 sin 21 4 2 5 4
1sin 2
1 8 20
1sin 2
27
PI x y D DD D
x y
x y
x y
the complete solution is1 2 Z CF PI PI
i.e. 21 15 sin 2
3 27
x y Z f y x g y x e x y
Problem 19 a. Find the general solution of 3 4 4 2 2 3 z y p x z q y x .
Solution:
This is a Lagrange’s linear equation of the form Pp Qq R .
The subsidiary equations aredx dy dz
P Q R
3 4 4 2 2 3
dx dy dz
z y x z y x
- (1)
Use Lagrangian multipliers x, y, z we get each ratio is
3 4 4 2 2 3 xdx ydy zdz
xz xy xy yz yz xz
2 2 21
20
d x y z
Hence 2 2 2 0d x y z
Integrating both the sides,2 2 2
1 x y z C - (2)
using multipliers 2, 3, 4 each of equation (1) is
2 3 4
6 8 12 6 8 12
2 3 4
0
dx dy dz
z y x z y x
dx dy dz
2 3 4 0dx dy dz
Hence22 3 4 x y z C - (3)
the General solution is 2 2 2 , 2 3 4 0 x y z x y z
b. Solve 2
2 2 3 22 3 3 2 2 x y D DD D D D z e e .
Solution:
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2 2
6 3 2 4
1 1 2 2
2 3 3 2 ' 2 ' 1
' 2 ' 1 4 41, 2, 1, 1
x x y y
D DD D D D z D D D D z
D D D D z e e em m
2C.F. x xe f y x e f y x
6
1
6
6
1
2 1
1
6 0 2 6 0 1
1
20
x
x
x
PI e D D D D
e
e
3 2
2
3 2
3 2
3 2
14
2 1
4
3 2 2 3 2 1
4
3 4
1
3
x y
x y
x y
x y
PI e D D D D
e
e
e
4
3
4
4
4
14
2 1
4
0 4 2 0 4 1
4
2 3
2
3
y
y
y
y
PI e D D D D
e
e
e
the general solution
2 6 3 2 41 1 2
20 3 3
x x x x y y Z e f y x e f y x e e e
Problem 20
a. Form the differential equation by eliminating the arbitrary function f and
g in z f x y g x y Solution:
z f x y g x y
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Let u x y v x y
. Z f u g v - (1)
Differentiating partially with respect to x and y, we get . ' ' . p f u g v f u g v - (2)
' 1 'q f u g v f u g v - (3)
" 2 ' ' " .r f u g v f u g v f u g v - (4)
" 1 " .s f u g v f u g v - (5)
" 2 ' ' "t f u g v f u g v f u g v - (6)
Subtracting (4) from (6) , we get
4 .r t f u g v - (7)
From (2) & (3), we get 2 2 4 . . ' . ' p q f u g v f u g v
= Z (r – t) from (1) & (7)
i.e.,
222 2
2 2
z z z z z
x y x y
b. Solve the equation pq p q z px qy pq .
Solution:
Rewriting the given equation aspq
Z px qy
pq p q
- (1)
we identify it as a clairaut’s type equation. Hence its complete solution is
ab Z ax by
ab a b
- (2)
The general solution of (1) is found out as usual from (2).
Let us now find the singular solution of (1).
Differentiating (2) partially with respect to a and then b, we get
2
10
ab a b b ab b x
ab a b
i.e.,
2
20
b
x ab a b
- (3)
and similarly
2
20
a y
ab a b
- (4)
From (3) & (4), we get2
2
a y
b x or
a bk
y x , say
a k y and b k x
Using there values in (3) we get
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22 2
2 2
0k x k xy k y k x x
k x k xy k y k x x
i.e., 1k xy x y
1 x yk
xy
Hence1 1
and x y x y
a b x y
Also
1 1
1 11 1
1 1
1
ab
ab a by xb a x y x y
x y
Using these values in (2), the singular solution of (1) is
2
1 1 1
1
z x x y y x y x y
z x y
Problem 21
a. Form the partial differential equation by eliminating
f from 2 2 2 2, 2 0 f x y z z xy .
Solution:
2 2 2 2, 2 0 f x y z z xy
Let 2 2 2 2and v 2u x y z z xy then
the given equation is , 0 f u v - (1)
Differentiating (1) partially w.r.t. x and y respectively
we get . 0 f u f v
u x v x
and . . 0
f u f v
u y v y
2 2 2 2 0 f f x zp zp yu v
- (4)
2 2 + 2zq-2x 0 f f
y zqu v
-(5)
from (4) and (5)
we get 0 x zp zp y
y zq zq x
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2 2 2 2
2 2
x zp zq x y zq zp y
zqx x z pq xzp zpy y z pq zqy
zp x y zq x y x y
zp x y zq x y x y x y
zp zq x y
zp zq y x
z p q y x
b. Solve the equation 21 1 p q q z .
Solution:
The given equation is of the from , , 0 f z p q
21P q q p z - (1)
Let z f x ay be the solution of (1)
If
If then
and
x ay u z f u
dz adz p q
du du
(1) reduces as
2
21 1 z dz adz
a z
u du du
i.e.,
2
21 0dz dz
a a azdu du
As z is not a constant, 0dz
du
2
21 0dz
a a azdu
i.e.,
2
21
dza az a
du
1dz
a az adu
- (3)
solving (3), we get
2
1
2 1
2 1
4 1
dza du
az a
az a u b
az a x ay b
az a x ay b
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which is the complete solution of (1)
Problem 22
a. Solve the equation 2 2 2 2 2 p q z x y .
Solution:
The given equation dues not belong to any of the standard types.
It can be rewritten as 2 2
1 1 2 2 z p z q x y - (1)
As the Equation (1) contains 1 1and z p z q we make the substitution Z = log z-1 -1z p=P and z q Q
Using there values in (1), it becomes2 2 2 2P Q x y (2)
As Eq. (2) dues not contain Z explicitly, we rewrite it as2 2 2 2 2P x y Q a , say (3)
From (3)2 2 2 2 2 2
2 2 2 2
2 2 2 2
and
and
P a x Q y a
P a x Q y a
dz Pdx Qdy
dz a x dx y a dy
Integrating, we get2 2
2 2 1 2 2 1
sin cos2 2 2 2
x a x y a y
Z x a h y a h ba a
the complete solution of (1) is
2 22 2 1 2 2 1log sin cos
2 2 2
x a x y a y z x a h y a h b
a a a
where a and b are arbitrary constants.
Singular solution does not exist and the General solution is found out as usual.
b. Solve the equation 2 2 2sin 2 sin 3 2 sin D D z x y x y .
Solution:
The auxiliary equation is m2
+ 1 = 0i.e., m i
C.F. = f y ix g y ix
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1 2 2
2 2
1sin 2 sin3
'
1 1. cos 2 3 cos 2 3' 2
1 1 1cos 2 3 cos 2 3
2 4 9 4 9
1 1. cos 2 3 cos 2 3
13 2
1sin 2 sin3
13
PI x y D D
x y x y D D
x y x y
x y x y
x y
2
2 2 2
2 2
2 2 2 2
0
2 2
2
12sin
'
11 cos 2 2
'
1 11 cos 2 2
' '
1 1cos 2 2
' 4 4
1cos 2 2
2 8
x
PI x y D D
x y D D
x y D D D D
e x y D D
x x y
The complete solution is1 2
. . Z C F PI PI
i.e., 21 1sin 2 sin 3 cos 2 2
13 2 8
x Z f y i x g y i x x y x y
Problem 23
a. Solve 1 p q .
Solution:
This is of the form , 0 f p q .
The complete integral is given by z ax by c where
2
1
1
1
a b
b a
b a
The complete solution is 2
1 z ax a y c - (1)
Differentiating partially w.r.t. c we get 0 = 1 (absurd)
There is no singular integral
Taking c f a where f is arbitrary,
2
1 z ax a y f a - (2)
Differentiating partially w.r.t a, we get
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1
0 2 1 '2
x a y f aa
- (3)
Eliminating ‘a’ between (2) & (3) we get the general solution.
b. Solve 2 2 2 x yz p y zx q z xy .
Solution:
This is a Lagrange’s linear equation of the form Pp Qq R
The subsidiary equations are2 2 2
dx dy dz
x yz y zx z xy
- (1)
2 2 2 22 2
dx dy dy dz dx dz
y zx z xy x yz z xy x yz y zx
i.e.,
2 2 2 22 2 1
d x y d y z d x z
y z x y z x z y z x y z x y
i.e.,
d x y d y z d x z
x y x y z y z x y z x z x y z
- (2)
i.e.,
d x y d y z d x z
x y y z x z
- (3)
Taking the first two ratios, and integrating log log log x y y z a
x ya
y z
- (4)
Similarly taking the last two ratios of (3) we get,
y zb
x z
- (5)
Butx y
y z
and
y z
x z
are not independent solutions for 1
x y
y z
gives
x z
y z
which is the reciprocal of the second solution.Therefore solution given by (4) and (5) are not independent. Hence we have to
search for another independent solution.
Using multipliers x, y, z in equation (1) each ratio is3 3 3
3
xdx ydy zdz
x y z xyz
Using multipliers 1, 1, 1 each ratio is
2 2 2
dx dy dz
x y z xy yz zx
3 3 3 2 2 23
xdx ydy zdz dx dy dz
x y z xyz x y z xy yz zx
2 2 2
2 2 22 2 2
1
2d x y z d x y z
x y z xy yz zx x y z x y z xy yz zx
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Hence 2 2 21
2d x y z x y z d x y z
Integrating 2
2 2 212 2
x y z x y z k
22 2 2
2 2 2 2 2 2
2
2 2 2 2
x y z x y z k
x y z x y z xy yz zx k
i.e., xy yz zx b
the general solution is , 0 x y
xy yz zx y z
Problem 24a. Solve 2 log yp xy q .
Solution:
2 log
2 log
2 log
log2 ( )
yp xy q
yp xy q
p x y q
q p x a say
y
12 and log p x a q a
y
- (1)
2 , log
ay
p x a q ay
q e
Now dz pdx qdy
2 aydz x a dx e dy - (2)
Integrating (2), we get
2 1 ay z x ax e ba
- (3)
Where a and b are arbitrary constant.
Equation (3) is the complete solution of the given equation.
Differentiating (3) partially w.r.t b we get 0 = 1 (absurd)
There is no singular integral.
Put b a in (3)
2aye
z x ax aa
- (4)
Differentiating partially w.r.t a we get
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2
2
1 10 0 '
10 ' -(5)
ay ay
ay ay
x e y e aa a
y x e e aa a
Eliminating ‘a’ between (4) & (5) we obtain the general solution.
b. Solve4 2 22 3 0 x p yzq z .
Solution:
Rewriting this equation, we get
22
2 3 0 x p yq
z z
This is an equation of the form , 0m k n k f x z p y z q may be transformed into
the standard type , 0 f P Q by putting 1 1,m n X x Y y and1, 1k Z z whereK
(or) log log log X x Y y Z z if 1, 1m n and 1k
Here 2 1m n
Hence 1 log and log X x Y y Z z
2
21
1.
Z Z z x pxP p x
X z x X z z
Z Z z y yQ q y q
Y z y Y Z Z
the equation becomes, 22 3 0P Q
This is of the form , 0 f P Q
Hence the complete integral is Z aX bY c Where
2
2
2 3 0
2 3
a b
b a
the complete solution is Z aX bY c
i.e., 2log 2 3 loga
z a y c x
- (1)
Differentiating equation (1) partially w.r.t to cWe get 0 = 1 (absurd)
Singular integral does not exist.
Taking C = f (a) where f is arbitrary
2log 2 3 loga
z a y f a x
- (2)
Differentiating partially w.r.t a we get
1
0 4 log 'a y f a x
- (3)
Eliminating ‘a’ between (2) & (3) we get the general solution.
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Problem 25
a. Find the complete solution of 2 2
1 x pz y qz .Solution:
2 2
1 x pz y qz - (1)
This equation is of the form , 0k k f z p z q
If 1k and 1k Z z
Hence put Z = z
2
. 2 Z Z z
P zp x z x
2
pPz
. 2 Z Z z
Q zq y z y
2
Qqz
Substituting there results in (1)2 2
12 2
P Q x y
This is a separable equation
2 2
2
12 2
P Q x y a
22 2 1
dZ Pdx Qdy
dZ a x dx a y dy
Integrating,
22 2
22 2 2
2 12
2 1
y Z a x a y dy
Z a x a y y b
is the complete solution.
b. Solve 2 22 z yz y p xy zx q xy zx .
Solution:
This is a Lagrange’s linear equation of the form Pp Qq R
The subsidiary equations are2 22
dx dy dz
z yz y xy zx xy zx
Taking x, y, z as multipliers, we have each fraction0
xdx ydy zdz
0 xdx ydy zdz
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22
Integrating2 2 2
2 2 2
x y zc
i.e.,2 2 2
1 x y z C - (1)
Again, taking the last two members, we have
dy dz
x y z x y z
i.e.,dy dz
y z y z
0
0
y z dy y z dz
ydy zdy ydz zdz
ydy zdy ydz zdz
ydy d yz zdz z
Integrating we get 2 2
22 y yz z C - (2)
From (1) & (2) the general solution is
2 2 2 2 2, 2 0 x y z y yz z
Problem 26
a. Solve the equation 4 39 4 1 pqz z .
Solution:
4 39 4 1 pqz z - (1)
The given equation is of the form , , 0 f z p q Let z f x ay be the solution of (1)
If x ay u then z f u
dzP
du and
dzq a
du
Substituting the values of p & q in (1), we get
2
4 3
2 3
2
3
9 4 1
3 2 1
3
2 1
dza z z
du
dz
az zdu
a z dzdu
z
Integrating we get2
3
3
2 1
a zdz du
z
Put 1 + z
3= t
2
23 2 z dz tdt
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2
2
a tdt du
t
a dt du
a t u b
i.e, 31a z x ay b
231a z x ay b - (2)
The singular solution is found as usual
Put b a
231a z x ay a - (3)
Differential w.r.t. a
31 2 z x ay a y a - (4)
The elimination of ‘a’ between (3) & (4) gives the General solution.
b. Solve 2 2 23 2 2 4 x y D DD D z x e .
Solution:
The Auxiliary equation is
2 3 2 0
1 2 0
1, 2
m m
m m
m
C.F. 1 22 y x y x
2
2 2
2
2
2
2
2 4. .
3 ' 2 '
2 4
2 ' '
Re 1, ' ' 2
., , ' '
2 4
1 2 ' 2 1 ' 2
2 4
2 ' 3 ' 1
2 4
2 '3 1 1 '
3
x y
x y
x y
x y
x y
e xP I
D DD D
e x
D D D D
placeD D D D
ie D D a D D b
x
e D D D D
xe
D D D D
xe
D D D D
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112
2
2
2
2 2
1 2 '1 1 ' 2 4
3 3
1 2 '1 ... 1 ' ... 2 4
3 3
1 4 5 ' ...1 2 4
3 3
1 12 4 16
3 3
1 22 24 11 6
3 3 9
x y
x y
x y
x y
x y x y
D De D D x
D De D D x
D De x
e x
e x e x
Hence the general solution is
2
1 2
. . . .
22 11 6
9
x y
Z C F P I
Z y x y x e x
Problem 27
a. Solve2 4 2 22 p x y zq z .
Solution:
This can be written as 2
2 2 22Px qy z z - (1)
Which is of the form , , 0m n f x p y q z Where m = 2, n= 2
put 1 11 1;m n
X x y y x y
2 2
2 2
.
.
Z Z xP p x px
X x X
Z Z yQ q y qy
Y y X
Substituting in (1) the given equation reduces to 2 22P Qz z
This is of the form , , 0 f p q z Let Z f X aY where u = X + aY
,dz adz
P Qdu du
Equation becomes,2
22 0dz dz
az zdu du
Solving fordz
du,we get
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2 2 2
2
2
2
2
8
2
8
2
8log
2
8log
2
8 1log
2
dz az a z z
du
dz a adu
z
a a z u b
a a z X ay b
a a a z b
x y
is the complete solution.
The singular and general integrals are found out as usual.
b. Find the general solution of y z p z x q x y .
Solution:
This is a Lagrange’s linear equation of the form Pp Qq R .
The auxiliary equations aredx dy dz
y z z x x y
each is equal to
2
dx dy dz dx dy dy dz
x y z y x z y
- (1)
Taking the first two ratios,
2
d x y z d x y
x y z x y
Integrating,
2
2
1log log log
2
log log log
x y z x y c
x y z x y k
x y z k x y
i.e., 2 x y z x y k - (2)
Taking the last two ratios of equation (1),
d x y d y z
x y y z
Integrating, log log log x y y z b
x yb
y z
- (3)
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Hence the general solution is 2
, 0 x y
f x y z x y y z
Problem 28
a. Find the equation of the cone satisfying px qy z and passing through the
circle2 2 2 4, 2 x y z x y z .
Solution:
The auxiliary equations aredx dy dz
x y z
Taking the first two ratios,log x = log y + log a
x a y
- (1)
Taking the second and third ratios,
log y = log z + log b
yb
z - (2)
the general solution is , 0 x y
y z
- (3)
We have to find that function satisfying (3) and
also x2+ y2 + z2 = 4 - (4)
and x + y + z = 2 - (5)
Hence, we will eliminate x, y, z from (1),(2),(4),(5)
From (2) y = bz
From (1) x = ay
x =abzusing the value of x & y in (4) & (5)
2 2 2 2 2 2 4a b z b z z
2 2 2 2 1 4 z a b b - (6)
Also abz + bz + z = 2
1 2b ab z - (7)
Eliminating z between (6) & (7)
Squaring (7)
2 21 4b ab z - (8)
From (6) & (8),
22 2 2 2 2 2 21 1 1 2 2 2b a b b ab b a b b ab ab
Simplifying b + ab2
+ ab = 0
1 0ab a
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Using , x y
a b y z
in (9)
1 0 xy x yz y
1 0 x x
z y
i.e., 0 yz xy xz is the required surface.
b. Solve 2 26 cos D DD D z y y .
Solution:
The Auxiliary equation is m2
+ m – 6 = 0
m = 2, –3
C.F. = 1 22 3 y x y x
2 2
cos. .
' 6
1 cos
2 ' 3 '
13 cos where 3
2 '
13 sin 3 sin where 3
2 '1
sin 3 cos2 '
2 sin 3cos where 2
2 cos 2 sin 3sin wher
y xP I
D DD D
y x
D D D D
a x xdx a x y D D
a x x xdx a x y
D D
y x x D D
a x x x dx a x y
a x x x x
e 2a x y
1 2
. . cos sin
2 3 sin cos
P I y x x
Z y x y x x y x
Problem 29
a. Solve the equation2 2 2 2sin cos 1 pz x qz y .
Solution:
The given equation contains (z2p) and (z
2q).
Therefore we make the substitution Z = z3
2 23 and Q=3z Z
P z p q x
Using there values in the given equation, it becomes
2 2sin cos 13 3
P Q x y
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Equation 1 does not contain Z explicitly.
Rewriting (1), we have 2 2sin 1 cos
3 3
P Q x y a , say - (2)
From (2), 2 23 cos and 3 1 secP a ec x Q a y
Now dZ = Pdx + Qdy
= 3a cosec2 xdx + 3 (1 – a) sec
2 y dy - (3)
Integrating (3),
3
3 cot 3 1 tan
3 cot 3 1 tan (4)
Z a x a y b
z a x a y b
is the complete solution.
Singular solution does not exist.
put b a in (4)
3 3 cot 3 1 tan z a x a y a - (5)
Differentiating (5) w.r.t. to a we get
0 3cot 3 tan ' x y a - (6)
Eliminating a between (5) & (6) we get the required solution.
b. Solve the equation 2 2 2 22 x y D DD D z x y e
.
Solution:
The Auxiliary equation is m2
- 2m + 1 = 0
(m – 1)2
= 0
m = 1 twice 1 2. .C F y x x y x
2 2
2 2
2 2
2
2 2
2
2 2
2
2
2 2
2
22 2
2 2
1. .
2
1
1
1 1
1
11
1 2 3 '1
x y
x y
x y
x y
x y
x y
P I e x y D DD D
e x y D D
e x y D D
e x y D D
De x y
D D
D D x ye
D D D
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2 2 2 2
2 2
2 2 2 2
2 3 4
1 2 32 2
1 1 14 6
x y
x y
e x y x y x D D D
e y x y x x D D D
4 2 5 61 1 1. .
12 15 60
x yP I x y x y x e
4 2 5 6
1 2
. .
1 1 1
12 15 60
x y
Z C F P I
y x x y x x y x y x e
Problem 30
a. Solve the equation mz ny p nx lz q ly mx . Hence write down the
solution of the equation 2 2 0 z y p x z q x y .
Solution:
The equation mz ny p nx lz q ly mx is a Lagrange’s linear equation of the
form Pp Qq R .
The Subsidiary equations aredx dy dz
mz ny nx lz ly mx
- (1)
Using l, m, n as a set of multipliers, the ratio in (1)0
ldx mdy ndz
i.e., 0ldx mdy ndz
Integrating we get lx my nz a
Choosing x, y, z as another set of multipliers, the ratio in (1)0
xdx ydy zdz
Integrating we get 2 2 2 x y z b
the general solution of the given equation is 2 2 2, 0 f lx my nz x y z
comparing the equation 2 2 ......... 2 z y p x z q x y
With the pervious equation 1, we get 1, 2, 1l m n
Therefore the solution of equation (2) is
2 2 22 , 0 f x y z x y z
b. Solve 2 2 3 3 7 D D D D z xy .
Solution: 2 2 3 3 7 D D D D z xy
3 7 D D D D z xy
Here1 1 2 2
1 0 1 3m c m c
C.F 3 x y x e y x
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P.I
7
3
xy
D D D D
1 1
2
2
2
12
1 7
1 3 13
11 1 7
3 3
11 ...
3
11 ... 7
3 9
21 ... 7
3 3 9 9 3
xy D D D D
D
D D D xy
D D
D D
D D D
D D D D xy
D D D DD DD xy
D
2
2 2 3
2 2 3
1 1 1 2 47
3 3 3 2 3 27
1 677
3 2 3 27 3 3 6 9
1 677
3 2 3 3 3 9 6 27
D D D D DD xy
D D D
x xy x x x y y x
x y xy x x y x x
Hence the general solution is
2 2 3
3 1 677
3 2 3 3 3 9 6 27
x x y xy x x y x z y x e f y x x
.
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