TRANSFORMS AND PARTIAL DIFFERENTIAL EQUATION UNIT III

1

TRANSFORMS AND PARTIAL DIFFERENTIAL EQUATION

UNIT III – PARTIAL DIFFERENTIAL EQUATION

PART – A

1. Find the solution of 222 zqypx

Sol. A. E. are

222 z

zd

y

yd

x

xd

Take 1st and 2

nd ratio, we have

22 y

yd

x

xd

Integrating, we get

1

1

11

11

cxy

cyx

Take 2nd

and 3rd

ratio, we have

22 z

zd

y

yd

Integrating, we get

2

2

11

11

cyz

czy

Hence the required solution is

011

,11

yzxyF

2. Solve 0)2( 23 zDDD

Sol. A.E. is m3 – 2m

2 = 0 [Put D = m and D′ = 1]

m2(m – 2) = 0

m2 = 0 (or) m – 2 = 0

m = 0, 0, 2

)2()()( 321 xyfyfxyfz

3. Find the particular integral of yxezDDDD )2( 22

Sol. P.I = yxe

DDDD

22 2

1

yx

yx

e

e

4

1

)1()1)(1(2)1(

122

4. Solve the equation 0)( 3 zDD

Sol. A.E. is (m – 1)3 = 0 [Put D = m and D′ = 1]

(m – 1)(m – 1)(m – 1) = 0

m = 1, 1, 1

)()()( 3

2

21 xyfxxyfxxyfz

www.Vidyarthiplus.com


2

5. Find the particular integral of yxzDDDD )23( 22

Sol. P.I = )(23

122

yxDDDD

)(23

11

)(23

1

1

1

2

2

2

2

22

yxD

DDD

D

yx

D

DDDD

26

2

1

0)(1

)(23

11

23

2

2

2

2

2

yxx

yxx

D

yxD

yxD

DDD

D

6. Solve: 0)2( 22 zDDDD

Sol. A.E. is m2 – 2m + 1 = 0 [Put D = m and D′ = 1]

(m – 1)(m – 1) = 0

m = 1, 1

)()( 21 xyfxxyfz

7. Solve: 0)12)(2( zDDDD

Sol. The given equation is non-homogeneous.

0)12)(2( zDDDD

)2()2( 21

0 xyfexyfez xx

)2()2(.).( 21 xyfexyfzei x

8. Solve 0)( 3223 zDDDDDD

Sol. A.E. is m3 + m

2 – 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

)()()( 321 xyfxxyfxyfz

9. Solve: 08423

3

2

3

2

3

3

3

y

z

yx

z

yx

z

x

z

Sol. The given equation can be written as 0)842( 3223 zDDDDDD

A.E. is m3 – 2m

2 – 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

)2()2()2( 321 xyfxyfxxyfz



3

10. Solve: yx

zsin

2

2

Sol. Given yx

zsin

2

2

)()(sin2

.).(

)(sin

2

yyfxyx

zei

yfyxx

z

11. Solve: yxyx

z

2

Sol. Given yxyx

z

2

)()(4

.).(

)(2

22

2

yxFyx

zei

xfyx

x

z

12. Form the partial differential equation by eliminating the constants a and b from

))(( 2222 byaxz

Sol. Given ))(( 2222 byaxz ---------- (1)

Diff. eqn. (1) p.w.r.t. x, we get

)2(

2

))(2(

22

22

byx

p

byxx

zp

Diff. eqn. (1) p.w.r.t. y, we get

)3(

2

)2)((

22

22

axy

q

yaxy

zq

Substitute (2) and (3) in equation (1), we have

pqzxyei

x

p

y

qz

4.).(

2.

2

13. Form the partial differential equation by eliminating the arbitrary constants ‘a’ and ‘b’

from z = ax + by.

Sol. Given byaxz ---------- (1)


)2(

a

x

zp


)3(

b

y

zq

Substitute (2) and (3) in equation (1), we have qypxz



4

14. Form the partial differential equation by eliminating the arbitrary constants a and b

from 2222 cot)()( zbyax

Sol. Given 2222 cot)()( zbyax ------------ (1)


)2(cot

cot20)(2

2

2

pzax

x

zzax


)3(cot

cot2)(20

2

2

qzby

y

zzby


222

222

222242

222222

tan.).(

1)(cot

cot)(cot

cot)cot()cot(

qpei

qp

zqpz

zqzpz


from bayxaz 22

Sol. Given bayxaz 22 ----------- (1)


)2(2

a

x

zp


)3(2

2

y

qaay

y

zq

Substitute (3) in equation (2), we have

22

2

4

2

qpy

y

qp


from 22 )()( byaxz

Sol. Given 22 )()( byaxz ---------- (1)


)2()(2

ax

x

zp


)3()(2

by

y

zq


22

22

4.).(

22

qpzei

qpz



5


from nn byaxz

Sol. Given nn byaxz ------------ (1)


)2(

1

n

n

n

xan

xp

x

xnap

xnax

zp


)3(

1

n

n

n

ybn

yq

y

ynbq

ynby

zq


yqxpznei

n

yq

n

xpz

.).(

18. Find the partial differential equation of all planes cutting equal intercepts from the

x and y axes.

Sol. The equation of the plane is

1c

z

a

y

a

x ------------ (1)


)2(

1

001

ac

p

c

p

a


)3(

1

01

0

ac

q

c

q

a

Divide (2) by (3), we get

..).(

1

qpei

q

p

19. Form the partial differential equation of all spheres whose centre lies on the z-axis.

Sol. 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

2222 )( kazyx -------------- (1)

where ‘a’ is the arbitrary constant.



6


)2()(

0)(202

pazx

pazx


)3()(

0)(220

qazy

qazy


..).( xqypei

q

p

y

x

20. Find the partial differential equation of all planes passing through the origin.

Sol. The equation of the plane passing through the origin is

ax + by + cz = 0

)1(.).(

yBxAzei

yc

bx

c

az

ybxazc

where A and B are arbitrary constants.


Ax

zp

------------ (2)


By

zq

------------ (3)

Substitute (2) and (3) in equation (1), we have yqxpz

21. Find the partial differential equation of the family of spheres having their centres on the

line x = y = z.

Sol. 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 = r

2 ---------------- (1)

where ‘a’ is the arbitrary constants.


)2()1(222

0)(2)(2

papzx

pazax


)3()1(222

0)(2)(2

qaqzy

qazay


yxqxzpzyei

qpzqzpyyqpzpzqxx

pqzyqpzx

q

p

qzy

pzx

)()(.).(

)1)(()1)((

1

1

)(2

)(2



7

22. Form the partial differential equation by eliminating the arbitrary function from

0,2

z

xxyz

Sol. The given equation can be written as

z

xfxyz 2

-------------- (1)


)2(.1.

22

z

pxz

z

xfypz


)3(22

z

qx

z

xfxqz


xzqyxzpxei

xpxzxqpzqzqyxxqpz

xpzxqzxqypz

xq

xpz

xqz

ypz

)2(.).(

222

))(2())(2(

2

2

22

22

23. Form the partial differential equation by eliminating an arbitrary function from

)( 22 yxfz

Sol. Given )( 22 yxfz -------------- (1)


)2()2()( 22 xyxfp


)3()2()( 22 yyxfq


xqypei

y

x

q

p

.).(


)( 22 yxfxyz

Sol. Given )( 22 yxfxyz -------------- (1)


)2()2()(

)2()(

22

22

xyxfyp

xyxfyp


)3()2()(

)2()(

22

22

yyxfxq

yyxfxq


22

22

.).( xyxqypei

xxqyyp

y

x

xq

yp



8


yxyxfz )( 22

Sol. Given yxyxfz )( 22 -------------- (1)


)2()2()(1

1)2()(

22

22

xyxfp

xyxfp


)3()2()(1

1)2()(

22

22

yyxfq

yyxfq


xyxqypei

xxqyyp

y

x

q

p

.).(

1

1

26. Form the partial differential equation by eliminating the arbitrary functions from

).()( 21 yfxfz

Sol. Given )()( 21 yfxfz ----------- (1)


)2()()( 21 yfxfp


)3()()( 21 yfxfq


)4()()( 21 yfxfr


)5()()( 21 yfxfs


)6()()( 21 yfxft

From (2) and (3) we have

szqpei

yfxfyfxfqp

.).(

)()()()( 2121

27. Find the complete integral of 1 qp

Sol. Given 1 qp --------------- (1)

The solution of equation (1) is

cybxaz

where 1 ba

21

1

ab

ab

Hence the complete integral is

cyaxaz 2

1

28. Find the complete integral of p – q = 0

Sol. Given p – q = 0 ------------ (1)

The solution of equation (1) is cybxaz

where a – b = 0 b = a

Hence the complete integral is cyaxaz



9

29. Find the complete solution of the partial differential equation 0422 pqqp

Sol. Given 0422 pqqp

The solution of equation (1) is

cybxaz

where 0422 baba

)32(2

324

2

124

1.2

.1.4164

04

2

22

22

aaa

aa

aaab

abab

Hence the complete integral is

cyaxaz )32(

30. Solve the partial differential equation xpq

Sol. Given xpq ------------ (1)

Let q = a

Then equation (1) becomes

a

xp

xap

Substitute p and q in the relation

baya

xz

getwegIntegratin

dyadxa

xdz

dyqdxpdz

2

,

2

which is the complete integral.

31. Find the complete integral of pqp

y

q

x

pq

z

Sol. Given pqp

y

q

x

pq

z

pqqpyqxpz ------------- (1)

The complete integral of equation (1) is

babaybxaz

32. Find the complete integral of the partial differential equation (1 – x)p + (2 – y)q = 3 – z.

Sol. Given (1 – x)p + (2 – y)q = 3 – z

(i.e.) z = px + qy + (3 – p – 2q) ----------- (1)


z = ax + by + (3 – a – 2b)



10

33. Find the singular solution of 22 qpqpqypxz

Sol. Given 22 qpqpqypxz ------------- (1)


22 bababyaxz -------------- (2)

To find singular integral, Diff. eqn. (2) p.w.r.t. ‘a’ and ‘b’, in turn, we get

)3(2

20

xba

bax

)4(2

20

yba

bayand

Solving (3) and (4) we get

3

223

xyaxya

3

223

yxbyxb

Substitute the values of a and b in equation (2) we have

22

22

22

22

3.).(

3339

)2()2)(2()2()2(3)2(39

3

2

3

2

3

2

3

2

3

2

3

2

yxxyzei

yxxyz

yxyxxyxyyxyxyxz

yxyxxyxyyxy

xyxz

34. Find the singular integral of the partial differential equation 22 qpqypxz

Sol. Given 22 qpqypxz ------------- (1)


22 babyaxz -------------- (2)


)3(2

20

xa

ax

)4(2

20

yb

by

Substitute the values of a and b in equation (2) we have

22

2222

22

4.).(

224

2222

xyzei

yxyxz

yxyy

xxz



11

PART -B

1. Solve: )()()( yxzqxzypzyx

Sol. A. E. are

)()()( yxz

zd

xzy

yd

zyx

xd

1

1

1

log)log(

loglogloglog

0

czyx

czyx

czyx

getwegIntegratin

z

zd

y

yd

x

xd

yxxzzy

z

zd

y

yd

x

xd

ratioEach

2

0

czyx

getwegIntegratin

dzdydx

zyzxyxyzxzxy

dzdydxratioEach


0),( zyxzxyF

2. Solve: 22)( xyyqpxz

Sol. A. E. are

22 xy

zd

zy

yd

zx

xd

Take 1st and 2

nd ratio, we have

zy

yd

zx

xd

y

yd

x

xd

Integrating, we get

1

1

1

.).(

logloglog

logloglog

cyxei

cyx

cyx

0

2222

dzzdyydxx

zxzyzyzx

dzzdyydxxratioEach

Integrating, we get

2

222

2

222

.).(

222

czyxei

czyx


0),( 222 zyxyxF



12

3. Solve: )()()( 222222 yxzqxzypzyx

Sol. A. E. are

)()()( 222222 yxz

zd

xzy

yd

zyx

xd

1

1

1

222222

log)log(

loglogloglog

0

czyx

czyx

czyx

getwegIntegratin

z

zd

y

yd

x

xd

yxxzzy

z

zd

y

yd

x

xd

ratioEach

0

222222222222

dzzdyydxx

yzxzxyzyzxyx

dzzdyydxxratioEach

Integrating, we get

2

222

2

222

.).(

222

czyxei

czyx


0),( 222 zyxxyzF

4. Solve: mxlyqlznxpnymz )()(

Sol. A. E. are

mxly

zd

lznx

yd

nymz

xd

1

0

cnzmylx

getwegIntegratin

ndzmdyldx

nmxnlylmzmnxnlylmz

ndzmdyldxratioEach

0

dzzdyydxx

xzmlyzlyzxynxynxzm

dzzdyydxxratioEach

Integrating, we get

2

222

2

222

.).(

222

czyxei

czyx


0),( 222 zyxnzmylxF



13

5. Solve: 22)()( yxqyxzpyxz

Sol. A. E. are

22)()( yx

zd

yxz

yd

yxz

xd

0

2222

dzzdyydxx

zyzxzyxyzxyzzx

dzzdyydxxratioEach

Integrating, we get

1

222

1

222

.).(

222

czyxei

czyx

)(

)(2222 yxz

yxd

xyzzxzyxyz

dyxdxyratioEach

Equate this to 3rd

ratio, we have

2

2

2

2

2222

2

2

)(

)(

)(

czxy

cz

xy

getwegIntegratin

dzzxyd

yx

dz

yxz

yxd


0)2,( 2222 zxyzyxF

6. Solve: )()()( 2222 yxzqzxypzyx

Sol. A. E. are

)()()( 2222 yxz

zd

zxy

yd

zyx

xd

1

1

1

2222

.).(

loglog)log(

loglogloglog

0

)()()(

cy

zxei

cyzx

czyx

getwegIntegratin

z

zd

y

yd

x

xd

yxzxzy

z

zd

y

yd

x

xd

ratioEach

0

22222222

dzdyydxx

zyzxzyyxzxyx

dzdyydxxratioEach



14

Integrating, we get

2

22

2

22

2.).(

22

czyxei

czyx


02, 22

zyx

y

zxF

7. Solve: )()()( 2222 yxzqzxyxpzyyx

Sol. A. E. are

)(2222 yxz

zd

zxyx

yd

zyyx

xd

0

2222

dzdydx

zyzxzxyxzyyx

dzdydxratioEach

Integrating, we get

1czyx

))((

)()()()(

22

22222222

yxyx

dyydxx

yxyyxx

dyydxx

zxyxyzyyxx

dyydxxratioEach

Equate this to 3rd

ratio, we have

22

22

2

222

2

22

2

22

22

22

.).(

loglog)log(

loglog2)log(

loglog)log(2

1

)())((

cz

yxei

czyx

czyx

czyx

getwegIntegratin

z

dz

yx

dyydxx

yxz

dz

yxyx

dyydxx


0,2

22

z

yxzyxF

8. Solve: xzqxypzyx 22)( 222

Sol. A. E. are

xz

zd

xy

yd

zyx

xd

22222

Take 2nd

and 3rd

ratio, we have

xz

zd

yx

yd

22

)log(2

1log

2

1

2

12/

)(2

22

22

22

22

yxt

t

dt

t

dtI

dtydyxdx

dtydyxdx

tyxput

yx

ydyxdxI



15

z

zd

y

yd

Integrating, we get

1

1

1

.).(

logloglog

logloglog

cz

yei

czy

czy

)(

22)(

222

223

22222

zyxx

dzzdyydxx

xzxyx

dzzdyydxx

xzxyzyxx

dzzdyydxxratioEach

Equate this to 2nd

ratio, we have

y

dy

zyx

dzzdyydxx

xy

dy

zyxx

dzzdyydxx

2

2)(

222

222

2

222

2

222

2

222

2

222

.).(

loglog)log(

loglog)log(

loglog2

1)log(

2

1

cy

zyxei

cyzyx

cyzyx

cyzyx

getwegIntegratin


0,222

y

zyx

z

yF

9. Solve: xyqzxpyz 32)24()43(

Sol. A. E. are

xy

zd

zx

yd

yz

xd

322443

1432

0432

12861286

432

czyx

getwegIntegratin

dzdydx

xyzxyz

dzdydxratioEach

0

322443

dzzdyydxx

xzyzyzxyxyxz

dzzdyydxxratioEach

)log(2

1log

2

1

2

12/

)(2

222

222

222

222

zyxt

t

dt

t

dtI

dtzdzydyxdx

dtzdzydyxdx

tzyxput

zyx

zdzydyxdxI



16

Integrating, we get

2

222

2

222

.).(

222

czyxei

czyx


0),432( 222 zyxzyxF

10. Solve: yxzqxzypzyx 222 )()(

Sol. A. E. are

yxz

zd

xzy

yd

zyx

xd

222

))((

)(

)())((

)(

)()(

)(

)()( 2222

zyxyx

yxd

yxzyxyx

yxd

zyxzyx

yxd

xzyzyx

dydxratioEach

))((

)(

)())((

)(

)()(

)(

)()( 2222

zyxzy

zyd

zyxzyzy

zyd

xzyxzy

zyd

yxzxzy

dzdyratioEach

))((

)(

))((

)(

zyxzy

zyd

zyxyx

yxd

)(

)(

)(

)(

zy

zyd

yx

yxd

Integrating we get

1

1

.).(

log)log()log(

czy

yxei

czyyx

xzzyyxzyx

zyxd

xzzyyxzyx

dzdydxratioEach

222222

)(

))((

3

222

333

xzzyyxzyxzyx

dzzdyydxx

zyxzyx

dzzdyydxxratioeachAlso

))((

)(222222 xzzyyxzyxzyx

dzzdyydxx

xzzyyxzyx

zyxd

dzzdyydxxzyxdzyx )()(

Integrating we get

2

2222

2222

)(c

zyxzyx



17

2

2

2

222222

2

2222

.).(

)(2

)(2

)(

cxzzyyxei

cxzzyyx

czyxxzzyyxzyx

czyxzyx


0,

xzzyyx

zy

yxF

11. Solve: xezDDDD yx sin)44( 2322

Sol. A.E. is 4m2 – 4m + 1 = 0 [Put D = m and D′ = 1]

4m2 – 2m – 2m + 1 = 0

2m(2m – 1) – 1(2m – 1) = 0

(2m – 1)(2m – 1) = 0

m = 2

1,

2

1

C.F =

xyfxxyf

2

1

2

121

P.I1 = yxe

DDDD

23

22 44

1

yx

yx

e

e

23

23

22

64

1

)2()2)(3(4)3(4

1

P.I2 = )0sin(44

122

yxDDDD

x

yx

sin4

1

)0sin(00)1(4

1

z = C.F + P.I1 + P.I2

xexyfxxyfzei yx sin4

1

64

1

2

1

2

1.).( 23

21

12. Solve: yxeyxzDDDD 222 )2(

Sol. A.E. is m2 + 2m + 1 = 0 [Put D = m and D′ = 1]

(m + 1)(m + 1) = 0

m = –1, –1

C.F = )()( 21 xyfxxyf

P.I1 = yxe

DDDD

22 2

1

yx

yx

yx

ex

eDD

x

e

2

22

)1()1)(1(2)1(

1

2

22

Since the denominator = 0, we have to

multiply x on Nr. and Diff. Dr. w.r.t.‘D’



18

P.I2 = yxDDDD

2

22 2

1

3012

12

2

3

1

3

21

)(21

)(2

)(1

21

1

21

1

21

1

21

1

54

43

32

2

22

2

22

2

2

2

2

2

2

2

2

1

2

2

2

2

2

22

xyx

xyx

D

xyx

D

xD

yxD

yxD

Dyx

D

yxD

D

D

yxD

DDD

D

yxD

DDD

D

yx

D

DDDD

z = C.F + P.I1 + P.I2

30122

)()(.).(542

21

xyxe

xxyfxxyfzei yx

13. Solve: )cos()( 23223 yxezDDDDDD yx

Sol. A.E. is m3 + m

2 – 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 = )()()( 321 xyfxxyfxyf

P.I1 = yxe

DDDDDD

2

3223

1

yx

yx

e

e

2

2

3223

9

1

)1()1)(2()1()2()2(

1

P.I2 = )cos(1

3223yx

DDDDDD

)cos(1

yxDDDD

D3 = D

2D D′

3 = D′

2D′

= (–1)D = (–1)D′

= – D = – D′



19

)cos(4

)cos()1()1(2)1(3

)cos(23 22

yxx

yxx

yxDDDD

x

z = C.F + P.I1 + P.I2

)cos(49

1)()()(.).( 2

321 yxx

exyfxxyfxyfzei yx

14. Solve: 22322 )2()2332( yx eezDDDDDD

Sol. The given equation is non-homogeneous and it can be written as

yxyx eeezDDDD 2346 44)2)(1(

C.F = )()( 2

2

1 xyfexyfe xx

P.I1 = yxe

DDDD

06

)2)(1(

1

x

yx

e

e

6

06

20

1

)206)(106(

1

P.I2 = yxe

DDDD

404)2)(1(

1

y

yx

e

e

4

40

3

2

)240)(140(

14

P.I3 = yxe

DDDD

234)2)(1(

1

yx

yx

e

e

23

23

3

1

)223)(123(

14

z = C.F + P.I1 + P.I2 + P.I3

yxyxxx eeexyfexyfezei 2346

2

2

13

1

3

2

20

1)()(.).(

15. Solve: )2sin()67( 2323 yxezDDDD yx

Sol. 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

C.F = )3()2()( 321 xyfxyfxyf



–1 1 0 –7 – 6

0 –1 1 6

1 –1 – 6 0



20

P.I1 = yxe

DDDD

2

323 67

1

yx

yx

e

e

2

2

323

12

1

)1(6)1)(2(7)2(

1

P.I2 = )2sin(67

1323

yxDDDD

)2sin()4(6)4(7

1yx

DDD

)2cos(75

1

)]2cos(7[525

1

)]2cos(16)2cos(9[525

1

525

)]2[sin(8)]2[sin(9

)2sin()]4(64)1(81[3

89

)2sin()6481(3

89

)2sin()89)(89(3

89

)2sin()89(3

1

)2sin(2427

1

22

yx

yx

yxyx

yxDyxD

yxDD

yxDD

DD

yxDDDD

DD

yxDD

yxDD

z = C.F + P.I1 + P.I2

)2cos(75

1

12

1)3()2()(.).( 2

321 yxexyfxyfxyfzei yx

16. Solve: xyyxy

z

yx

z

x

z

)sinh(2

2

22

2

2

(or) xyyxtsr )sinh(2

Sol. The given equation can be written as xyyxzDDDD )sinh()2( 22

A.E. is m2 + m – 2 = 0 [Put D = m and D′ = 1]

(m + 2)(m – 1) = 0

m = –2, 1

C.F = )()2( 21 xyfxyf

P.I1 = )sinh(2

122

yxDDDD

22

1 )(

22

yxyx ee

DDDD

yxyx e

DDDDe

DDDD 2222 2

1

2

1

2

1

D3 = D

2D D′

3 = D′

2D′

= (–1)D = (–4)D′

= – D = – 4D′

2sinh

xx eex



21

yxyx

yxyx

yxyx

yxyx

ex

ex

ex

ex

eDD

xe

DD

x

ee

66

12122

1

222

1

)1(2)1)(1()1(

1

)1(2)1)(1()1(

1

2

12222

P.I2 = xyDDDD 22 2

1

246

62

1

2

1

)(11

)()(1

11

21

1

21

1

21

1

43

32

2

2

2

2

2

2

2

2

1

2

2

2

2

22

xyx

xyx

D

xxy

D

xD

xyD

xyD

Dxy

D

xyD

D

D

xyD

DDD

D

xyD

DDD

D

xy

D

DDDD

z = C.F + P.I1 + P.I2

24666

)()2(.).(43

21

xyxe

xe

xxyfxyfzei yxyx

17. Solve: xyy

z

yx

z

x

zsin65

2

22

2

2

Sol. The given equation can be written as xyzDDDD sin)65( 22

A.E. is m2 – 5m + 6 = 0 [Put D = m and D′ = 1]

(m – 2)(m – 3) = 0

m = 2, 3

C.F = )3()2( 21 xyfxyf





22

P.I = xyDDDD

sin65

122

xyx

xxxy

xxxxc

dxxxxc

xxyDD

xxxcDD

dxxxcDD

xyDDDD

xyDDDD

sincos5

cos3cos2sin

)cos(3)]cos)(2())(sin2([

]sin3cos)2([

]sin3cos[2

1

)sin)(3()cos)(3(2

1

sin)3(2

1

sin3

1

2

1

sin)3()2(

1

z = C.F + P.I

xyxxyfxyfzei sincos5)3()2(.).( 21

18. Solve: 7)33( 22 xyzDDDD


7)3)(( xyzDDDD

C.F = )()( 2

3

1

0 xyfexyfe xx

)()( 2

3

1 xyfexyf x

P.I = )7()3)((

1

xy

DDDD

9

65

3313

1

9

2

337

)(3

1

)7(9

2)7(

3)7(

3)7(

)(3

1

)7(9

2

331

)(3

1

)7(33

1)(3

1

)7(3

1)(3

1

)7(

313

1

)(

1

2

1

xyxy

D

DD

xyxy

DD

xyDD

xyD

xyD

xyDD

xyDDDD

DD

xyDDDD

DD

xyDD

DD

xyDDDD

where y = c – 3x

where y = c – 2x



23

9

65

33623

1

669

65

6323

1

329

65

333

1

3

11

9

65

333

1

9

65

339

65

333

1

9

65

331

3

1

9

65

331

3

1

232

2322

2

1

xxxyxyx

xxxxxyyx

xxxyxy

D

xD

xyxy

D

xyxy

D

Dxyxy

D

xyxy

D

D

D

xyxy

D

D

D

z = C.F + P.I

9

65

33623

1)()(.).(

232

2

3

1

xxxyxyxxyfexyfzei x

19. Solve: yxezDDDDDD 222 )1222(


yxezDDDD 2)1)(1(

C.F = )()( 21 xyfxexyfe xx

P.I = yxe

DDDDDD

2

22 1222

1

yx

yx

e

e

2

2

22

16

1

1)1(2)2(2)1)(2(2)1()2(

1

z = C.F + P.I

yxxx exyfxexyfezei 2

2116

1)()(.).(

20. Solve: yxzDDDD sin)43( 22

Sol. A.E. is m2 + 3m – 4 = 0 [Put D = m and D′ = 1]

(m – 1)(m + 4) = 0

m = 1, – 4

C.F = )4()( 21 xyfxyf

P.I1 = xDDDD 22 43

1

xD

DDD

D

x

D

DDDD

1

2

2

2

2

22

431

1

431

1



24

6

2

1

01

431

1

3

2

2

2

2

2

x

x

D

xD

xD

DDD

D

P.I2 = )0sin(43

122

yxDDDD

y

yx

sin4

1

)0sin()1(400

1

z = C.F + P.I1 + P.I2

yx

xyfxyfzei sin4

1

6)4()(.).(

3

21

21. Eliminate the arbitrary function ‘f ’ from the relation 0),( 222 zyxzyxf

Sol. The given relation 0),( 222 zyxzyxf can also be written as

)(222 zyxzyx ------------- (1)

Diff. equation (1) p.w.r.to x, we get

)2()1()(22

)01()(202

pzyxpzx

pzyxpzx

Diff. equation (1) p.w.r.to y, we get

)3()1()(22

)10()(220

qzyxqzy

qzyxqzy

Dividing (2) by (3), we have

yxqxzpzyei

qpzqzpyyqpzpzqxx

pzqyqzpx

q

p

qzy

pzx

qzyx

pzyx

qzy

pzx

)()(.).(

)1)(()1)((

)1(

)1(

)1()(

)1()(

22

22

22. Eliminate the arbitrary function ‘f ’ from the relation

y

xfyz log

122

Sol. Given

y

xfyz log

122

------------------ (1)


)2(1

log1

202

xy

xfp



25


)3(

1log

122

1log

122

yy

xfyq

yy

xfyq

Dividing (2) by (3), we have

22

2

2

2

2

2.).(

)2(

2

/1

/1

2

1log

12

1log

12

2

yqypxei

yqypx

x

y

yq

p

y

x

yq

p

yy

xf

xy

xf

yq

p

23. Eliminate the arbitrary function ‘f ’ and ‘ ’ from the relation )()( yxyxfz

Sol. Given )()( yxyxfz ------------ (1)


)2()()()()(

yxyxfyxyxf

x

zp


)3()()()()(

)()()1)(()(

yxyxfyxyxfq

yxyxfyxyxfy

zq


)4()()()()(2)()(

)()(

)()()()()()(2

2

yxyxfyxyxfyxyxfr

yxyxf

yxyxfyxyxfyxyxfx

zr


)5()()()()(

)()(

)()()()()()(2

yxyxfyxyxfs

yxyxf

yxyxfyxyxfyxyxfyx

zs


)6()()()()(2)()(

)()(

)()()()()()(2

2

yxyxfyxyxfyxyxft

yxyxf

yxyxfyxyxfyxyxfy

zt



26

(2) + (3) )()(2 yxyxfqp

(2) – (3) )()(2 yxyxfqp

)()()()(4))(( yxyxfyxyxfqpqp

)()(4.).( 22 yxyxfzqpei -------------- (7)

(4) – (6) )()(4 yxyxftr

(7) )(22 trzqp

24. Find the PDE of all planes which are at a constant distance ‘k’ from the origin.

Sol. The equation of the plane having constant distance ‘k’ from the origin is

0222 cbakzcybxa ------------------ (1)


)2(

0

pca

pca


)3(

0

qcb

qcb


1.).(

01

0

22

22

22222

qpkyqxpzei

qpkzyqxp

cqcpckzcyqcxpc

25. Solve: )32sin()( 22 yxezDD yx

Sol. A.E. is m2 – 1 = 0 [Put D = m and D′ = 1]

m2 = 1

m = 1

C.F = )()( 21 xyfxyf

P.I = )32sin(1

22yxe

DD

yx

)32sin(25)(4

]5)(2[

)32sin(]5)(2][5)(2[

]5)(2[

)32sin(5)(2

1

)32sin(2)9(24

1

)32sin(22

1

)32sin(1212

1

)32sin()1()1(

1

2

22

22

22

yxDD

DDe

yxDDDD

DDe

yxDD

e

yxDD

e

yxDDDD

e

yxDDDD

e

yxDD

e

yx

yx

yx

yx

yx

yx

yx



27

)]32cos(2)32[sin(25

)]32sin(5)32cos(10[125

)]32sin(5)32cos(6)32cos(4[125

125

)32sin(5)]32[sin(2)]32[sin(2

)32sin(125

]5)(2[

)32sin(25)]9()6(2)4[(4

]5)(2[

)32sin(25)2(4

]5)(2[22

yxyxe

yxyxe

yxyxyxe

yxyxDyxDe

yxDD

e

yxDD

e

yxDDDD

DDe

yx

yx

yx

yx

yx

yx

yx

z = C.F + P.I

)]32cos(2)32[sin(25

)()(.).( 21 yxyxe

xyfxyfzeiyx

26. Solve: yexzDDDDDD )362( 22

Sol. Given yexzDDDDDD )362( 22

yexzDDDD )3)(2(

1,3,2

1,0 1211 mmHere

C.F = )(2

12

3

1

0 xyfexyfe xx

= )(2

12

3

1 xyfexyf x

P.I = yex

DDDDDD 362

122

xDDDDDDe

xDDDDDDe

xDDDDDD

e

xDDDDDD

e

xDDDDDDDD

e

xDDDDDD

e

y

y

y

y

y

y

2

521

2

2

521

2

2

5212

1

522

1

336122

1

)1(36)1()1(2

1

22

122

22

22

22

22



28

)52(4

2

5

2

)(2

5

2

xe

xe

xD

xe

y

y

y

z = C.F + P.I

)52(4

)(2

1.).( 2

3

1

x

exyfexyfzei

yx

27. Solve: )2sin()2223( 22 yxyxzDDDDDD

Sol. Given )2sin()2223( 22 yxyxzDDDDDD

)2sin()22)(( yxyxzDDDD

2,2,1,0 1211 mmHere

C.F = )2()( 2

2

1

0 xyfexyfe xx

= )2()( 2

2

1 xyfexyf x

P.I1 = )(2223

122

yxDDDDDD

2

1

22

1

4

3

24

1

22222

1

4

)(3)(

4

)(

2

)(

2)(

1

2

1

)(4

3

422

11

2

1

)(4242

11

2

1

)(42

11

2

1

)(2

2

2

211

2

1

)(2

211

2

1

)(

2

2121

1

)()22)((

1

2

22

2

2

2

2

2

2

11

yyxx

xxyxyx

x

yxD

D

yxDyxD

D

yxDyxyx

D

yxD

D

DD

D

D

D

yxD

D

D

D

DD

D

D

D

D

yxDDD

DD

D

D

D

yxDDDD

D

D

D

yxDD

D

D

D

yxDD

D

DD

yxDDDD



29

P.I2 = )2sin(2223

122

yxDDDDDD

)2sin(22

1

)2sin(22)1(2)2(34

1

yxDD

yxDD

)2cos(2

1

)]2cos(6[12

1

)]2cos(2)2cos(4[12

1

12

)]2[sin(2)]2[sin(2

)2sin()1(4)4(4

22

)2sin(44

22

)2sin()22)(22(

22

22

yx

yx

yxyx

yxDyxD

yxDD

yxDD

DD

yxDDDD

DD

z = C.F + P.I1 + P.I2

)2cos(2

1

2

1

22

1)2()(.).( 2

2

2

1 yxy

yxxxyfexyfzei x

28. Solve: )2cos()44( 3223 yxzDDDDDD

Sol. A.E. is m3 + m

2 – 4m – 4 = 0 [Put D = m and D′ = 1]

m2 (m + 1) – 4(m + 1) = 0

(m + 1)(m2 – 4) = 0

m = –1, m2 = 4

m = 2

m = –1, –2, 2

C.F = )2()2()( 321 xyfxyfxyf

P.I = )2cos(44

13223

yxDDDDDD

)2cos(12

)2cos()1(4)2(2)4(3

)2cos(423

)2cos(4444

1

22

yxx

yxx

yxDDDD

x

yxDDDD

z = C.F + P.I

)2cos(12

)2()2()(.).( 321 yxx

xyfxyfxyfzei





30

29. Solve: 22 qpqypxz

Sol. Given 22 qpqypxz ------------- (1)


22babyaxz ------------------- (2)


0 = x + 2ab2 x = – 2ab

2 --------- (3)

and 0 = y + 2a2b y = – 2a

2b --------- (4)

Multiplying (3) × a + (4) × b, we get

a x + by + 4a2b

2 = 0

(a x + by + a2b

2) + 3a

2b

2 = 0

(i.e.) z = – 3(ab)2 -------------- (5)

Now, multiplying (3) and (4), we get

x y = 4a3b

3

)6(

4

4)(

3/1

3

yxba

yxba

Substitute (6) in equation (5) we have

02716.).(

427

43

223

2

3

3/2

yxzei

yxz

yxz

To find general integral, assume b = f(a)

Then equation (2) becomes 22 )}({)( afayafxaz -------------- (7)

Diff. eqn. (7) p.w.r.t. ‘a’, we get

)8(2.)}({)()}({2.)(0 22 aafafafayafx

The eliminant of ‘a’ between equations (7) and (8) gives the general integral.

30. Find the singular integral of pqqypxz 2

Sol. Given pqqypxz 2 ------------- (1)


baybxaz 2 ------------- (2)


)3(

)(2

20

a

bx

bab

x

)4(

)(2

20

b

ay

aab

yand

Multiplying (3) and (4) we get

x y=1, which is the singular integral.



31

31. Solve: 221 qpqypxz

Sol. Given 221 qpqypxz ------------------- (1)


221 babyaxz ------------- (2)


)3(

1

)2(12

10

22

22

ba

ax

aba

x

)4(

1

)2(12

10

22

22

ba

by

bba

yand

Substitute (3) and (4) in equation (1), we get

)5(1

1.).(

1

1

1

1

111

22

2

22

22

2222

22

22

2

22

2

bazei

ba

ba

baba

baba

b

ba

az

Squaring and adding (3) and (4), we have

1.).(

])5(sin[1

1

11

1

1)1(

11

222

222

22

22

22

22

2

22

222

zyxei

guzyx

ba

ba

ba

ba

b

ba

ayx

which is the singular integral.


Then equation (2) becomes 22 )}({1)( afayafxaz -------------- (6)


)7(]1).().(22[)}({12

1)(0

22

afafa

afayafx




32

32. For the equation

p

p

qqypxz , find the complete and singular solutions.

Sol. Given

p

p

qqypxz ------------------- (1)


a

a

bbyaxz ------------- (2)


)3(1

10

2

2

a

bx

a

bx

)4(

11

10

ya

ay

ayand

Substitute (4) in (3) , we get

)5(1

1

11

2

2

2

y

xb

ybx

y

bx


xzyei

y

xxxz

yy

x

y

xy

y

xz

1.).(

1)1(1

1)1(12

which is the singular integral.

33. Solve: qzqp )1(

Sol. Given qzqp )1( --------------- (1)

Let q = ap


p(1 + ap) = ap z

1 + ap = az

a

zap

1

1

1

,

za

a

zaa

paqNow



33


dz = p dx + q dy

)2()1log(.).(

)1log(

,

1

)1(1

byaxzaei

bya

x

a

za

getwegIntegratin

yda

xd

za

zd

ydzaxda

zadz



10

1

and

yza

a

The last equation is absurd and shows that there is no singular integral.



)()1log( afyaxza -------------- (3)


)4()(1

afyza

a


34. Solve: 2222 yxqp

Sol. Given p

2 + q

2 = x

2 + y

2

2222 qyxp

Let 22222 axpaxp

Also 22222 ayqaqy


dz = p dx + q dy

)2(cosh22

sinh22

12

2212

22

2222

ba

yaay

y

a

xaax

xz

getwegIntegratin

dyaydxaxdz



10

).(cosh1)/(

1

22

)2(

2

).(sinh)/(1

1

22

2

20

1

22

2

22

1

22

2

22

and

aa

y

a

y

ax

a

ay

ay

aa

x

a

x

ax

a

ax

ax




34



)(cosh22

sinh22

12

2212

22 afa

yaay

y

a

xaax

xz

-------- (3)


)4()().(cosh1)/(

1

22

)2(

2

).(sinh)/(1

1

22

2

20

1

22

2

22

1

22

2

22

afaa

y

a

y

ax

a

ay

ay

aa

x

a

x

ax

a

ax

ax


35. Find the complete solution of 2zpqxy

Sol. Given 2)()( zqypx ---------- (1)

Put yYxX log,log

X

zPwherePpxei

X

zpx

xX

zx

X

X

z

x

zp

.).(

1

.

Y

zQwhereQqyei

Y

zqy

yY

zy

Y

Y

z

y

zq

.).(

1

.

Equation (1) becomes

2zQP ---------- (2)

Let Q = aP


a

zP

zaPP

2.

za

a

za

PaQNow

,

Substitute P and Q in the relation

dz = P dX + Q dY

YdzaXda

zdz



35

byaxzaei

baYXza

getwegIntegratin

YdaXdz

zda

logloglog.).(

log

,

which is the complete solution.

36. Solve: yxqpz )( 222

Sol. Given yxqzpz 22 )()( ----------- (1)

Put 211 zzZ

x

ZPwherepz

P

x

zz

x

Z

2

2

y

ZQwhereqz

Q

y

zz

y

Z

2

2

Equation (1) becomes

22

22

22

44

)(4.).(

22

QyxP

yxQPei

yxQP

Let axPaxP 442

Also ayQaQy 44 2


dz = p dx + q dy

)2(6

)4(

6

)4(

)2/3(4

)4(

)2/3(4

)4(

44

2/32/3

2/32/3

bayax

z

bayax

z

getwegIntegratin

dyaydxaxdz



10

)4(4

1)4(

4

10 2/12/1

and

ayax




)(6

)4(

6

)4( 2/32/3

afayax

z

-------------- (3)


)4()()4(4

1)4(

4

10 2/12/1 afayax




36

Problems for practice

1. Solve: yxeyxzDDDD 222 )sinh()2(

2. Solve: )2sin()222( 22 yxzDDDDDD



5. Solve: 1)24( 22 yxezDDDD

6. Solve: )2sin()67( 2323 yxezDDDD yx

7. Solve: )()()( yxqxzpzy

8. Solve: ))(()()( yxyxqxyzpxzy

9. Solve: )()()( 222 yxzqxzypzyx

10. Find the general solution of 2

2

yxzqpx

zy

11. Eliminate the arbitrary function ‘ф’ from the relation 0),( 222 czbyaxzyx

12. Form the partial differential equation by eliminating arbitrary function f and ф from

)()( ctxctxfz

13. Form the partial differential equation by eliminating the arbitrary function ‘g’ from the

relation 0),( 222 xyzzyxg

14. Form the partial differential equation by eliminating arbitrary functions ‘f’ and ‘g’ from

)2()2( yxgyxxfz

Answers

yxeyxxyfxxyfz 2

219

1)sinh(

4

1)()(.1

)]2sin(3)2cos(2[39

1)()(.2 2

2

1 yxyxxyfexyfz x

)]2sin(3)2cos(4[50

)()(.32

21 yxyxe

xyfxyfzyx

)]2cos(2)2sin([15

)()(.4 21 yxyxe

xyfxyfzyx

2

)22()22(.52

21

xexyfxyfz yx

)2cos(75

1

12

1)3()2()(.6 2

321 yxexyfxyfxyfz yx

0))((,.7 2

yxzyx

zy

yxF

0],[.8 222 zyxzyxF

0,111

.9 222

zyx

zyxF

0],[.10 2233 zxyxF

xbyaqzaxcpyczb )()(.11 rcT 2.12

zyxyqxzxpzy )()()(.13 222222

)(4.14 tsr



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TRANSFORMS AND PARTIAL DIFFERENTIAL EQUATION UNIT III