MA2211 Unit 3

8/8/2019 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:


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


b. Solve 2 22 z yz y p xy zx q xy zx .

Solution:


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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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


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:


+ 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


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:


- 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

.

MA2211 Unit 3

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