MA6351 TRANSFORMS AND PARTIAL DIFFERENTIAL EQUATIONS€¦ · · 2015-06-04UNIT I PARTIAL DIFFERENTIAL EQUATIONS MA6351 TRANSFORMS AND PARTIAL DIFFERENTIAL EQUATIONS by B.Saravanan

UNIT I PARTIAL DIFFERENTIAL EQUATIONS

MA6351 TRANSFORMS AND PARTIAL DIFFERENTIAL EQUATIONS

byB.Saravanan

Assistant ProfessorDepartment of Applied Mathematics

SVCE

1

�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

Syllabus

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Partial Differential Equation

Partial differential equation is one which involves partial derivatives. The order of PDE is the order of highestderivative occurring in it.

.,,,,

:

.var

,var,),(

2

22

2

2

y

zt

yx

zs

x

zr

y

zq

x

zp

Notation

iabledependentisz

iabletindependenareyxwhereyxfzIf

∂∂=

∂∂∂=

∂∂=

∂∂=

∂∂=

=


Formation of PDE by eliminating arbitrary constant

Let us consider the functional relationf(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

)3(00

)2(00

−−−−−−−=∂∂+

∂∂

⇒=∂∂

∂∂+

∂∂

−−−−−−−=∂∂+

∂∂

⇒=∂∂

∂∂+

∂∂

qz

f

y

f

y

z

z

f

y

f

pz

f

x

f

x

z

z

f

x

f

Equation (2) and (3) will contain a and b. If we eliminate a andb from (1), (2) and (3) we get the PDE (involving p and q) of the first order.


Remarks:

�If the number of constants to be eliminated is equal to numberof 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.

Problem 1Form the partial differential equation by eliminating a and b from

Solution:

Differentiating (1) partially w.r. t x and y we get

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))(( 2222 byaxz ++=

)1())(( 2222 −−−−−−++= byaxzGiven

))(2( 22 byxx

zp +=

∂∂=

)2(2

22 −−−−−−−+=⇒ byx

p

6

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Differentiating (1) partially w.r. t ‘y’ we get

)2)(( 22 yaxy

zq +=

∂∂=

)3(2

22 −−−−−−−+=⇒ axy

q

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

x

p

y

qz

2.

2=

pqzxyei =4.).(

7

Problem 2 Form the partial differential equation by eliminating the arbitrary constants a and b from

Solution:

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α2222 cot)()( zbyax =−+−

)1(cot)()( 2222 −−−−−=−+− αzbyaxGiven

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

α2cot20)(2x

zzax

∂∂=+−

)2(cot 2 −−−−−−−=−⇒ αpzax

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

α2cot2)(20y

zzby

∂∂=−+

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)3(cot 2 −−−−−−−=−⇒ αqzby


ααα 222222 cot)cot()cot( zqzpz =+

αα 222242 cot)(cot zqpz =+

1)(cot 222 =+ qpα

α222 tan.).( =+ qpei

9

Problem 3Form the partial differential equation by eliminating the arbitrary constants a and b from

Solution:

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bayxaz ++= 22

)1(22 −−−−−−−−++= bayxazGiven


)2(2 −−−−−−−=∂∂= a

x

zp


)3(2

2 −−−−−−−=⇒=∂∂=

y

qaay

y

zq

10

Problem 4Form the partial differential equation by eliminating the arbitrary constants a and b from

Solution:

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

2

2

=

y

qp

224 qpy =

nn byaxz +=

)1(−−−−−−+= nn byaxzGiven


1−=∂∂= nxna

x

zp

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x

xnap

n

=

)2(−−−−−−−= nxan

xp


1−=∂∂= nynb

y

zq

y

ynbq

n

=

)3(−−−−−−−= nybn

yq

12

Problem 5Find 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

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n

yq

n

xpz +=

yqxpznei +=.).(

)1(1 −−−−−−=++c

z

a

y

a

x

13

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001 =++

c

p

a

)2(1 −−−−−−−−=ac

p

01

0 =++c

q

a

)3(1 −−−−−−−−=ac

q

Divide (2) by (3), we get

1=q

pqpei =.).(

14

Problem 6Find the partial differential equation of all planes passing through the origin

Solution:

The equation of the plane passing through the origin is

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ax + by + cz = 0

ybxazc −−=⇒

yc

bx

c

az −−=⇒

)1(.).( −−−−−−−+= yBxAzei

where A and B are arbitrary constants

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Problem 7Find the PDE of all planes which are at a constant distance ‘k’from the origin.



Ax

zp =

∂∂=

By

zq =

∂∂=


yqxpz +=

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Solution:

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

)1(0222 −−−−−−=++−++ cbakzcybxa


0=+ pca

)2(−−−−−−−−=⇒ pca

0=+ qcb

)3(−−−−−−−−=⇒ qcb



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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

where ‘a’ is the arbitrary constant.

022222 =++−+−− cqcpckzcyqcxpc

0122 =++−+−− qpkzyqxp

1.).( 22 ++++= qpkyqxpzei

)1()( 2222 −−−−−−−=−++ kazyx

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0)(202 =−++ pazx

)2()( −−−−−−−−−= pazx

0)(220 =−++ qazy

)3()( −−−−−−−−−= qazy


q

p

y

x =

..).( xqypei =

19

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Problem 9Find the partial differential equation of the family of spheres having their centres on the line x = y = z.

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

Solution:

(x – a)2 + (y – a)2 + (z – a)2 = r2 ---------------- (1)

where ‘a’ is the arbitrary constants.


0)(2)(2 =−+− pazax

)2()1(222 −−−−−−−+=+ papzx

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0)(2)(2 =−+− qazay

)3()1(222 −−−−−−−+=+ qaqzy


q

p

qzy

pzx

++=

++

1

1

)(2

)(2

)1)(()1)(( pqzyqpzx ++=++

qpzqzpyyqpzpzqxx +++=+++

yxqxzpzyei −=−+− )()(.).(

21


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.

)2(0 −−−−−−−=

∂∂+

∂∂

∂∂+

∂∂+

∂∂

∂∂

pz

v

x

v

v

fp

z

u

x

u

u

f

),

,,,(

yxoffunctiona

turniniszandzyxoffunctionsarevanduSince

)3(0 −−−−−−−=

∂∂+

∂∂

∂∂+

∂∂+

∂∂

∂∂

qz

v

y

v

v

fq

z

u

y

u

u

f

Differentiating (1) partially with respect to x and y we get


)3()2(lim,lim andfromv

fand

u

finateeusletfinatingeofInstead

∂∂

∂∂

zyxoffunctionsareRQPwhere

RQqPp

formtheofequationangetwe

,,,,

)4(−−−−−−=+

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 1Form the partial differential equation by eliminating an arbitrary function from

Solution:

)( 22 yxfz +=

)1()( 22 −−−−−−+= yxfzGiven



)2()2()( 22 −−−−−−+′= xyxfp

)3()2()( 22 −−−−−−+′= yyxfq


y

x

q

p = xqypei =.).(

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Problem 2Form the partial differential equation by eliminating the arbitrary functions from

Solution:

).()( 21 yfxfz =

)1()()( 21 −−−−−−−= yfxfzGiven


)2()()( 21 −−−−−−′= yfxfp


)3()()( 21 −−−−−−′= yfxfqDiff. eqn. (2) p.w.r.t. x, we get

)4()()( 21 −−−−−−′′= yfxfr


)5()()( 21 −−−−−−′′= yfxfs

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Problem 3Form the partial differential equation by eliminating an arbitrary function from

Solution:


)6()()( 21 −−−−−−′′= yfxft

From (2) and (3) we have

)()()()( 2121 yfxfyfxfqp ′′=

szqpei =.).(

)( 22 yxfxyz ++=

)1()( 22 −−−−−−++= yxfxyzGiven


)2()( 22 xyxfyp +′+=26

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)2()2()( 22 −−−−−−+′=− xyxfyp


)2()( 22 yyxfxq +′+=

)3()2()( 22 −−−−−−+′=− yyxfxq


y

x

xq

yp =−−

22 xxqyyp −=−22.).( xyxqypei −=−

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Problem 4Eliminate the arbitrary function ‘f ’ from the relation

Solution:

++= yx

fyz log1

22

)1(log1

22 −−−−−

++= yx

fyzGiven

)2(1

log1

202

−−−−−−−

−

+′+=x

yx

fp

+′+=y

yx

fyq1

log1

22



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)3(1

log1

22 −−−−−−−

+′=−y

yx

fyq

Dividing (2) by (3), we have

+′

−

+′=

−y

yx

f

xy

xf

yq

p

1log

12

1log

12

2

2

y

x

yq

p

/1

/1

2

2−=−

=>

22 x

y

yq

p −=−

=>

)2(2 yqypx −−==>22 2.).( yqypxei =+

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Problem 5Form the partial differential equation by eliminating the

arbitrary function from

Solution:

0,2 =

−z

xxyzφ

The given equation can be written as

)1(2 −−−−−−−

=−z

xfxyz


)2(.1.

22

−−−−−−

−

′=−z

pxz

z

xfypz

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)3(22

−−−−−−

−

′=−z

qx

z

xfxqz


xq

xpz

xqz

ypz

−−=

−−

2

2

))(2())(2( xpzxqzxqypz −−=−−

22 222 xpxzxqpzqzqyxxqpz +−−=+−

xzqyxzpxei =−+ )2(.).( 22

31

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Problem 6Eliminate the arbitrary function ‘f ’ from the relation

Solution:

0),( 222 =++++ zyxzyxf


)1()(222 −−−−−−−++=++ zyxzyx φDiff. eqn. (1) p.w.r.t. x, we get

)01()(202 pzyxpzx ++++′=++ φ

)2()1()(22 −−−−−−−+++′=+ pzyxpzx φDiff. eqn. (1) p.w.r.t. y, we get

)10()(220 qzyxqzy ++++′=++ φ

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)3()1()(22 −−−−−−−+++′=+ qzyxqzy φ

Dividing (2) by (3), we have

)1()(

)1()(

22

22

qzyx

pzyx

qzy

pzx

+++′+++′

=++

φφ

)1(

)1(

q

p

qzy

pzx

++=

++

)1)(()1)(( pzqyqzpx ++=++

qpzqzpyyqpzpzqxx +++=+++

yxqxzpzyei −=−+− )()(.).(

33


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

R

dz

Q

dy

P

dx ==

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 1Find the solution of

Solution:

This is Lagrange’s linear PDE of the form Pp + Qq =R

R

zd

Q

yd

P

xdareEA ==..

222 z

zd

y

yd

x

xd ==

222 zqypx =+

Take 1st and 2nd ratio, we have

22 y

yd

x

xd =

Integrating, we get 1

11c

yx+−=−

1

11c

xy=−=>

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Take 2nd and 3rd ratio, we have

22 z

zd

y

yd =

Integrating, we get

2

11c

zy+−=−

2

11c

yz=−=>

Hence the required solution is

011

,11 =

−−

yzxyF

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Problem 2

Using multiplier 1/x, 1/y, 1/z and then add, each ratio is

Solution:

Solve: )()()( yxzqxzypzyx −=−+−


R

zd

Q

yd

P

xdareEA ==..

)()()( yxz

zd

xzy

yd

zyx

xd

−=

−=

−

yxxzzy

z

zd

y

yd

x

xd

−+−+−

++0=++⇒

z

zd

y

yd

x

xd

37

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Using multiplier 1, 1, 1 and then add, each ratio is

getwegIntegratin

1loglogloglog czyx =++

1log)log( czyx =⇒

1czyx =⇒

zyzxyxyzxzxy

dzdydx

−+−+−++=

0=++⇒ dzdydx

getwegIntegratin 2czyx =++

Hence the required solution is 0),( =++ zyxzxyF

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Problem 3

Solution:

Solve: 22)( xyyqpxz −=−


R

zd

Q

yd

P

xdareEA ==..

22 xy

zd

zy

yd

zx

xd

−=

−=

Take 1st and 2nd ratio, we have

zy

yd

zx

xd

−=

y

yd

x

xd

−=

Integrating, we get

1logloglog cyx +−=

1logloglog cyx =+

1.).( cyxei =39

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Using multiplier x,y,z and then add, each ratio is

zxzyzyzx

dzzdyydxx2222 −+−

++=

0=++⇒ dzzdyydxx

Integrating, we get

2

222

222c

zyx =++

2222.).( czyxei =++


0),( 222 =++ zyxyxF

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Problem 4

Solution:

Using multiplier 1/x, 1/y, 1/z and then add, each ratio is

Solve: )()()( 222222 yxzqxzypzyx −=−+−

)()()( 222222 yxz

zd

xzy

yd

zyx

xd

−=

−=

−

222222 yxxzzy

z

zd

y

yd

x

xd

−+−+−

++=


R

zd

Q

yd

P

xdareEA ==..

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0=++⇒z

zd

y

yd

x

xd

getwegIntegratin

1loglogloglog czyx =++

1log)log( czyx =⇒

1czyx =⇒

222222222222 yzxzxyzyzxyx

dzzdyydxx

−+−+−++=

0=++⇒ dzzdyydxx

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Problem 5

Solution:

Integrating, we get

2

222

222c

zyx =++

2222.).( czyxei =++


0),( 222 =++ zyxxyzF

Solve: mxlyqlznxpnymz −=−+− )()(


43

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Using multiplier l,m,n and then add, each ratio is

R

zd

Q

yd

P

xdareEA ==..

mxly

zd

lznx

yd

nymz

xd

−=

−=

−

nmxnlylmzmnxnlylmz

ndzmdyldx

−+−+−++=

0=++⇒ ndzmdyldx

getwegIntegratin

1cnzmylx =++

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xzmlyzlyzxynxynxzm

dzzdyydxx

−+−+−++=

0=++⇒ dzzdyydxx

Integrating, we get

2

222

222c

zyx =++

2222.).( czyxei =++


0),( 222 =++++ zyxnzmylxF

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Problem 6

Solution:

Using multiplier 1/x, (-1/y), 1/z and then add, each ratio is

Solve: )()()( 2222 yxzqzxypzyx −=+++

)()()( 2222 yxz

zd

zxy

yd

zyx

xd

−=

+=

+

)()()( 2222 yxzxzy

z

zd

y

yd

x

xd

−++−+

+−=


R

zd

Q

yd

P

xdareEA ==..

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Using multiplier x,(-y),-1 and then add, each ratio is

0=+−⇒z

zd

y

yd

x

xd

getwegIntegratin

1loglogloglog czyx =+−

1loglog)log( cyzx =−⇒

1.).( cy

zxei =

zyzxzyyxzxyx

dzdyydxx22222222 +−−−+

−−=

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0=−−⇒ dzdyydxx

Integrating, we get

2

22

22cz

yx =−−

222 2.).( czyxei =−−


02, 22 =

−− zyx

y

zxF

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Problem 7

Solution:

Solve: xzqxypzyx 22)( 222 =+−−

xz

zd

xy

yd

zyx

xd

22222==

−−


R

zd

Q

yd

P

xdareEA ==..

Take 2nd and 3rd ratio, we have

xz

zd

yx

yd

22=

z

zd

y

yd ==>

49

6/4/2015B.Saravanan


Integrating, we get

1logloglog czy +=

1logloglog czy =−

1.).( cz

yei =

xzxyzyxx

dzzdyydxx22222 22)( ++−−

++=

xzxyx

dzzdyydxx223 ++

++=)( 222 zyxx

dzzdyydxx

++++=

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Equate this to 2nd ratio, we have

xy

dy

zyxx

dzzdyydxx

2)( 222=

++++

y

dy

zyx

dzzdyydxx

2222=

++++

⇒

getwegIntegratin

2222 loglog

2

1)log(

2

1cyzyx +=++

2222 loglog)log( cyzyx +=++

2222 loglog)log( cyzyx =−++

2

222

.).( cy

zyxei =++

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Problem 8

Solution:


0,222

=

++y

zyx

z

yF

Solve: yxzqxzypzyx −=−+− 222 )()(


R

zd

Q

yd

P

xdareEA ==..

yxz

zd

xzy

yd

zyx

xd

−=

−=

− 222

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(Subtracting 1st and 2nd ratio)

(Subtracting 2nd and 3rd ratio)

-----------(A)

)()( 22 xzyzyx

dydxratioEach

−−−−=

)()(

)(22 zyxzyx

yxd

−+−−=

)())((

)(

yxzyxyx

yxd

−++−−=

))((

)(

zyxyx

yxd

++−−=

)()( 22 yxzxzy

dzdyratioEach

−−−−=

)()(

)(22 xzyxzy

zyd

−+−−=

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---------(B)

From (A) and (B) we have

)())((

)(

zyxzyzy

zyd

−++−−=

))((

)(

zyxzy

zyd

++−−=

))((

)(

))((

)(

zyxzy

zyd

zyxyx

yxd

++−−=

++−−

)(

)(

)(

)(

zy

zyd

yx

yxd

−−=

−−

⇒

Integrating we get

1log)log()log( czyyx +−=−1.).( c

zy

yxei =

−−

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Using multiplier 1,1,1 and then add, each ratio is


-------(C)

---------(D)

xzzyyxzyx

dzdydxratioEach

−−−++++= 222

xzzyyxzyx

zyxd

−−−++++=

222

)(

zyxzyx

dzzdyydxxratioeach

3333 −++++=

))(( 222 xzzyyxzyxzyx

dzzdyydxx

−−−++++++=

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From (C) and (D)

xzzyyxzyx

zyxd

−−−++++

222

)(

))(( 222 xzzyyxzyxzyx

dzzdyydxx

−−−++++++=

dzzdyydxxzyxdzyx ++=++++ )()(

Integrating we get

2

2222

2222

)(c

zyxzyx +++=++

22222)( czyxzyx +++=++

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2222222 )(2 czyxxzzyyxzyx +++=+++++

2)(2 cxzzyyx =++

2.).( cxzzyyxei =++


0, =

++

−−

xzzyyxzy

yxF

57


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


)3(0

)2(0

0,)1(.

)1(0),,,,(

0),,,,(.3

−−−−−=∂∂

−−−−−=∂∂

−−−−−==

b

a

getwetoequateandbatorespectwithpartiallyDiff

bazyx

issolutioncompletewhosePDEabeqpzyxfLet

φ

φ

φ

Eliminate a and b from (1),(2) and (3), the resulting one is called singular integral


Standard types of first order PDE

f (p ,q)=0 ( i.e. equations containing p and q only)

Type I

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

tconsarbitraryarecandawhere

cyaaxz

egralcompletegetweeqninsubstituteandabei

tan

)3()()1(

int)1.()(..

−−−−−−−++=⇒

=φ

φ


absurdiswhichc

zand

a

z

getwetoequateand

catorespectwithpartiallyDiffegralgularfindTo

0100

0

,)3(.intsin

=⇒=∂∂=

∂∂

Hence there is no singular integral

)()()(0

)(.

)()()(

int)(

Bagyax

getwe

atorespectwithpartiallyADiffand

Aagyaaxz

getweegralcompleteinagcputsolutiongeneralfindTo

−−−−−−−′+′+=

−−−−−++==

φ

φ

Eliminate ‘a’ from (A) and (B) we get general integral or general solution.

6/4/2015B.Saravanan

Problem 1

Find the complete integral of 1=+ qp

Solution:

)1(1 −−−−−=+ qpGiven

The complete solution of equation (1) is

cybxaz ++=

where 1=+ ba ab −=⇒ 1 ( )21 ab −=⇒

Hence the complete integral is

( ) cyaxaz +−+=2

1

This is of the form f(p,q)=0

62

6/4/2015B.Saravanan

Problem 2

Find the complete integral of p-q=0

Solution:

The complete solution of equation (1) is

Given p – q = 0 ------------ (1)

cybxaz ++=

Where a-b=0 => b=a


cyaxaz ++=


63

6/4/2015B.Saravanan

Problem 3

Find the complete integral of 0422 =−+ pqqp

Solution:

)1(0422 −−−−−−−=−+ pqqpGiven

The complete solution of equation (1) is cybxaz ++=

where 0422 =−+ baba

04 22 =+−⇒ abab

1.2

.1.4164 22 aaab

−±=


64

6/4/2015B.Saravanan

2

124 2aa ±=

2

324 aa ±=

)32( ±= a

cyaxaz +±+= )32(


65


f (p, q, z)=0 ( i.e. equations containing p ,q and z only)

Type II

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

6/4/2015B.Saravanan

Problem 1

Solve: qzqp =+ )1(

Solution:

This is of the form f(z , p, q) = 0

)1()1( −−−−=+ qzqpGiven

Let q = ap

Then equation (1) becomes

p(1 + ap) = ap z

1 + ap = aza

zap

1−=⇒

paqNow =,

−=a

zaa

1

1−= za

67

6/4/2015B.Saravanan

Substitute p and q in the relation

dz = p dx + q dy

ydzaxda

zadz )1(

1 −+−=

yda

xd

za

zd +=−1

getwegIntegratin ,

bya

x

a

za ++=− )1log(

)2()1log(.).( −−−−−−++=− byaxzaei

which is the complete integral

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6/4/2015B.Saravanan

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

yza

a =−1

10 =and

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

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


)3()()1log()2( −−−−−++=−=> afyaxza

Diff. eqn. (3) p.w.r.t. ‘a’, we get

)4()(1

−−−−−−−−′+=−

afyza

a

The eliminant of ‘a’ between equations (3) and (4) gives the general integral.

69


(i.e. equations containing p , q , x and y)

Type III

f1 (x, p) = f2 (y, q)

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


On integrating we get

∫∫ ++= bdyygdxxfz )()(

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

6/4/2015B.Saravanan

Problem 1

Solve: 2222 yxqp +=+

Solution:

)1(2222 −−−−−+=+ yxqpGiven

This is of the form f(x , p) = g( y , q)

22222 aqyxp =−=−⇒

22222 axpaxp +±=⇒=−

22222 ayqaqy −±=⇒=−

2222 qyxp −=−⇒

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6/4/2015B.Saravanan


dz = p dx + q dy

dyaydxaxdz 2222 −±+±=getwegIntegratin

)2(cosh22

sinh22

12

22

12

22

−−−−+

−−±

++±=

−

−

ba

yaay

y

a

xaax

xz

which is the complete integral

73

6/4/2015B.Saravanan


10

).(cosh1)/(

1

22

)2(

2

).(sinh)/(1

1

22

2

20

122

2

22

122

2

22

=

−

+

−−

−−

−±

+

−+

++

±=

−

−

and

aa

y

a

y

ax

a

ay

ay

aa

x

a

x

ax

a

ax

ax

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


74

6/4/2015B.Saravanan


)3()(cosh22

sinh22

12

22

12

22

−−−−−−+

−−±

++±=

−

−

afa

yaay

y

a

xaax

xz


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

1

22

)2(

2

).(sinh)/(1

1

22

2

20

122

2

22

122

2

22

−−−′+

−

+

−−

−−

−±

+

−+

++

±=

−

−

afaa

y

a

y

ax

a

ay

ay

aa

x

a

x

ax

a

ax

ax


75

6/4/2015B.Saravanan

Problem 2

Find the complete integral of xpq =

Solution:

This is of the form f(x , p) = g( y , q)

)1(−−−= xpqGiven

Let q = a


xap =a

xp =⇒


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6/4/2015B.Saravanan

dyqdxpdz +=

dyadxa

xdz +=

getwegIntegratin ,

baya

xz ++=

2

2

which is the complete integral.

77


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

)3(0

)2(0

−−−−−−−∂∂+=

−−−−−−−∂∂+=

b

fy

a

fx

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

6/4/2015B.Saravanan

Problem 1

Find the complete integral of pqp

y

q

x

pq

z ++=

Solution:

pqp

y

q

x

pq

zGiven ++=

)1(−−−−−−++= pqqpyqxpz

The complete integral of equation (1) is

babaybxaz ++= (replacing p by a and q by b)

This is in Clairaut’s form

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6/4/2015B.Saravanan

Problem 2

Find the singular integral of z=p x +q y +p q

Solution: Given z=p x +q y +p q


The complete integral of equation is

z= a x + b y + a b -------(1)

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 => z= -x y

(replacing p by a and q by b)

80

6/4/2015B.Saravanan

Problem 3

Find the singular solution of 22 qpqpqypxz ++++=

Solution:

)1(22 −−−−−++++= qpqpqypxzGiven



)2(22 −−−−−−++++= bababyaxz(replacing p by a and q by b)

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

bax ++= 20 )3(2 −−−−−−−=+⇒ xba

81

6/4/2015B.Saravanan

bayand 20 ++= )4(2 −−−−−−−=+⇒ yba

Solving (3) and (4) we get

3

223

xyaxya

−=⇒−=3

223

yxbyxband

−=⇒−=

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

2

2

3

2

3

2

3

2

3

2

3

2

3

2

−+

−

−+

−+

−+

−=

yxyxxy

xyyxy

xyxz

22)2( bababyaxz ++++==>

82

6/4/2015B.Saravanan

2

2

)2()2)(2(

)2()2(3)2(39

yxyxxy

xyyxyxyxz

−+−−+−+−+−=

22 3339 yxxyz −−=223.).( yxxyzei −−=

Problem 4

Find the singular integral of the partial differential equation

22 qpqypxz −++=

Solution:

)1(22 −−−−−−++= qpqypxzGiven


83

6/4/2015B.Saravanan


)2(22 −−−−−−−++= babyaxz(replacing p by a and q by b)


ax 20 += )3(2

−−−−−−−=⇒x

a

by 20 −= )4(2

−−−−−−=⇒y

b

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

22

2222)2(

−

−+

+

−==> yxyy

xxz

84

6/4/2015B.Saravanan

2222 224 yxyxz −++−=224.).( xyzei −=

Problem 5

Find the singular integral of pqqypxz 2++=

Solution:


)1(2 −−−−−−++= pqqypxzGiven


)2(2 −−−−−−−−++= baybxaz


85

6/4/2015B.Saravanan


)(2

20 b

abx += )3(−−−−−−=−⇒

a

bx

)(2

20 a

abyand += )4(−−−−−−=−⇒

b

ay

Multiplying (3) and (4) we get

x y=1, which is the singular integral.

86

6/4/2015B.Saravanan

Problem 6

Solve: 221 qpqypxz ++++=

Solution:

)1(1 22 −−−−−++++= qpqypxzGiven


)2(1 22 −−−−−−−++++= babyaxz




87

6/4/2015B.Saravanan

)2(12

10

22a

bax

+++=

)3(1 22

−−−−−−++

−=⇒ba

ax

)2(12

10

22b

bayand

+++=

)4(1 22

−−−−−−++

−=⇒ba

by

Substitute (3) and (4) in equation (2), we get

22

22

2

22

2

111

baba

b

ba

az +++

++−

++−=

221)2( babyaxz ++++==>

88

6/4/2015B.Saravanan

22

2222

1

1

ba

baba

+++++−−=

221

1

ba ++=

)5(1

1.).(

222 −−−−−−

++=

bazei

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

22

2

22

222

11 ba

b

ba

ayx

+++

++=+

22

22

1

1)1(

ba

ba

++−++=

221

11

ba ++−=

89

6/4/2015B.Saravanan

which is the singular integral.

])5(sin[1 222 guzyx −=+

1.).( 222 =++ zyxei



)6()}({1)( 22 −−−−−−−−++++= afayafxaz


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

1)(0

22−−−′+

+++′+= afafa

afayafx


90

6/4/2015B.Saravanan

Problem 7

Find the complete and singular solutions of

−++= p

p

qqypxz

Solution:

)1(−−−−−

−++= p

p

qqypxzGiven



)2(−−−−−−−

−++= aa

bbyaxz


91

6/4/2015B.Saravanan


102

−−=a

bx )3(1

2−−−−−−=−⇒

a

bx

ayand

10 += )4(

11 −−−−−−−=⇒−=⇒y

aa

y

Substitute (4) in (3) , we get

21

1

−

=−

y

bx

21 ybx =−⇒ )5(1

2−−−−−−−=⇒

y

xb

92

6/4/2015B.Saravanan


+−−+

−+−=yy

x

y

xy

y

xz

1)1(12

y

xxxz

1)1(1 +−−−+−=

xzyei −= 1.).(

which is the singular integral

−++==> aa

bbyaxz)2(

93

6/4/2015B.Saravanan

Problem 1

Find the complete solution of 2zpqxy =

Solution:

)1()()( 2 −−−−= zqypxGiven

yYxXPut log,log ==

x

X

X

z

x

zp

∂∂

∂∂=

∂∂= .

xX

z 1

∂∂=

X

zpx

∂∂=⇒

X

zPwherePpxei

∂∂==.).(

94

6/4/2015B.Saravanan

y

Y

Y

z

y

zq

∂∂

∂∂=

∂∂= .

yY

z 1

∂∂=

Y

zqy

∂∂=⇒

Y

zQwhereQqyei

∂∂==.).(

Equation (1) becomes

)2(2 −−−−−= zQP

2)()()1( zqypx ==>

Let Q = aP Then equation (2) becomes

2.)2( zaPP ==>a

zP ==>

95

6/4/2015B.Saravanan

Substitute P and Q in the relation

dz = P dX + Q dY

YdzaXda

zdz +=

PaQNow =,

=a

za za=

YdaXdz

zda +=

getwegIntegratin ,

baYXza ++=log

byaxzaei ++= logloglog.).(

which is the complete solution.

96

6/4/2015B.Saravanan

Problem 2

yxqpz +=+ )( 222

Solution:

Find the general solution of

)1()()( 22 −−−−−+=+ yxqzpzGiven

211 zzZPut == +

x

zz

x

Z

∂∂=

∂∂

2x

ZPwherepz

P

∂∂==⇒

2

y

zz

y

Z

∂∂=

∂∂

2y

ZQwhereqz

Q

∂∂==⇒

2

Equation (1) becomes yxQP +=

+

22

22

97

6/4/2015B.Saravanan

)(4.).( 22 yxQPei +=+

aQyxP =−=−⇒ 22 44

axPaxPLet +±=⇒=− 442

ayQaQyAlso −±=⇒=− 44 2


dz = p dx + q dy

dyaydxaxdz −±+±= 44

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6/4/2015B.Saravanan

getwegIntegratin

bayax

z +−±+±=)2/3(4

)4(

)2/3(4

)4( 2/32/3

which is the complete integral.


)2(6

)4(

6

)4( 2/32/3

−−−−−−+−±+±= bayax

z

10)4(4

1)4(

4

10 2/12/1 =−±+±= andayax

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

99

6/4/2015B.Saravanan



)3()(6

)4(

6

)4( 2/32/3

−−−−−+−±+±= afayax

z


)4()()4(4

1)4(

4

10 2/12/1 −−−−−′+−±+±= afayax


100


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:

08423

3

2

3

2

3

3

3

=∂∂+

∂∂∂−

∂∂∂−

∂∂

y

z

yx

z

yx

z

x

z

0)2( 23 =′− zDDD


Standard form:

The standard form of homogeneous PDE of nth order with constant coefficients is of the form

DandDinreenthofpolinomialogeneousisDDfwhere

yxFzDDf

yDand

xDwhere

yxFzDaDDaDDaDa nn

nnn

′′−−−−−=′∂∂=′

∂∂=

=′++′+′+ −−

deghom),(

)1(),(),(

),()............( 222

110

The general solution of equation (1) is z=C.F+P.I


Procedure to find C.F:

C.F is the solution of equation

0............

1

0............

22

110

222

110

=++++⇒

=′==′++′+′+

−−

−−

nnnn

nn

nnn

amamama

DandmDput

DaDDaDDaDa

0),( =′DDf

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)

)(...........)()(. 2211 xmyfxmyfxmyfFC nn ++++++=


Case II (All the roots are equal)

)(..).........()()(. 11

132

1211 xmyfxxmyfxxmyxfxmyfFC nn +++++++= −

Note: roots may be real or complex

Particular Integral:

(1) R.H.S is an exponential function then

0),(),(

1

),(

1. ≠=

′= ++ bafife

bafe

DDfIP byaxbyax

....

.0),(

proceduresametheapplyandDtorwDrthe

atedifferentiandNrtheinxmultiplythenbafIf =


(2) R.H.S is an Trigonometric function then

)()(),,(

1.

22byaxCosorSin

DDDDfIP +

′′=

0),,()()(),,(

1 2222 ≠−−−+

−−−= babafifbyaxCosorSin

babaf

....

.0),,( 22

proceduresametheapplyandDtorwDrthe

atedifferentiandNrtheinxmultiplythenbabafif =−−−

(3) R.H.S is an Polynomial function then

nmnm yxDDyxDDf

IP 1)],(1[),(

1. −′+=

′= φ


(4) Exponential shift rule

),(),(

1),(

),(

1. yx

bDaDfeyxe

DDfIP byaxbyax φφ

+′+=

′= ++

(5) General rule

),(),(

1.

),(

yxFDDf

IP

functionotheranyisyxFIf

′=

byfactoreachoperatethenandDDffactorize ),( ′

dxmxaxFyxFDmD ∫ −=

′−),(),(

1

mxybyreplacedisaegrationafter +int

6/4/2015B.Saravanan

Problem 1

Solve 0)2( 23 =′− zDDD

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

)2()()( 321 xyfyfxyfz +++=∴

107

6/4/2015B.Saravanan

Problem 2

Solve 0)( 3 =′− zDD

Solution:

0)( 3 =′− zDD

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

(m – 1)(m – 1)(m – 1) = 0

m = 1, 1, 1

)()()( 32

21 xyfxxyfxxyfz +++++=∴

108

6/4/2015B.Saravanan

Problem 3

0)2( 22 =′+′− zDDDDSolve

Solution:

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

(m – 1)(m – 1) = 0

m = 1, 1

)()( 21 xyfxxyfz +++=∴

109

6/4/2015B.Saravanan

Problem 4

0)( 3223 =′−′−′+ zDDDDDDSolve

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

1±=m

∴

m = 1, –1, –1

)()()( 321 xyfxxyfxyfz −+−++=∴

110

6/4/2015B.Saravanan

Problem 5

Solve 08423

3

2

3

2

3

3

3

=∂∂+

∂∂∂−

∂∂∂−

∂∂

y

z

yx

z

yx

z

x

z

Solution: The given equation can be written as

0)842( 3223 =′+′−′− zDDDDDD

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

2±=m

∴

=> m = 2, 2, –2

)2()2()2( 321 xyfxyfxxyfz −++++=∴

111

6/4/2015B.Saravanan

Problem 6

Solve )2sin()67( 2323 yxezDDDD yx ++=′−′− +

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 )3()2()(. 321 xyfxyfxyfFC ++−+−=

112

6/4/2015B.Saravanan

yxeDDDD

IP +

′−′−= 2

3231 67

1.

yxe +

−−= 2

323 )1(6)1)(2(7)2(

1

yxe +−= 2

12

1

)2sin(67

1.

3232 yxDDDD

IP +′−′−

=

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

1yx

DDD+

′−−−−−=

)2sin(2427

1yx

DD+

′+=

113

6/4/2015B.Saravanan

)2sin()89(3

1yx

DD+

′+=

)2sin()89)(89(3

89yx

DDDD

DD +′−′+

′−=

)2sin()6481(3

8922

yxDD

DD +′−

′−=

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

89yx

DD +−−−

′−=

525

)]2[sin(8)]2[sin(9 yxDyxD +′−+=

114

6/4/2015B.Saravanan

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

1yxyx +−+=

)]2cos(7[525

1yx +−=

)2cos(75

1yx +−=

∴

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

)2cos(75

1

12

1

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

2

321

yxe

xyfxyfxyfzei

yx +−−

++−+−=

+

115

6/4/2015B.Saravanan

Problem 7

)cos()( 23223 yxezDDDDDD yx ++=′−′−′+ +Solve

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

1±=m

m = 1, –1, –1

)()()(. 321 xyfxxyfxyfFC −+−++=116

6/4/2015B.Saravanan

yxeDDDDDD

IP +

′−′−′+= 2

32231

1.

yxe +

−−+= 2

3223 )1()1)(2()1()2()2(

1

yxe += 2

9

1

)cos(1

.32232 yx

DDDDDDIP +

′−′−′+=

)cos(1

yxDDDD

+′++′−−

=

Since the denominator = 0, we have to multiply x on Nr. and Diff. Dr. w.r.t.‘D’

)cos(0

1yx +=

117

6/4/2015B.Saravanan

)cos(23 22

yxDDDD

x +′−′+

=

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

yxx +

−−−+−=

)cos(4

yxx +−=

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

)cos(49

1

)()()(.).(

2

321

yxx

e

xyfxxyfxyfzei

yx +−+

−+−++=

+

118

6/4/2015B.Saravanan

Problem 8

Solve xezDDDD yx sin)44( 2322 +=′+′− −

Solution:

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

(2m – 1)2 = 0

2

1,

2

1=m

++

+= xyfxxyfFC2

1

2

1.. 21

P.I1 = yxe

DDDD23

22 44

1 −

′+′−119

6/4/2015B.Saravanan

yxe 2322 )2()2)(3(4)3(4

1 −

−+−−=

yxe 23

64

1 −=

P.I2 = )0sin(44

122

yxDDDD

+′+′−

)0sin(00)1(4

1yx +

+−−=

xsin4

1−=

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

xexyfxxyfzei yx sin4

1

64

1

2

1

2

1.).( 23

21 −+

++

+= −

120

6/4/2015B.Saravanan

Problem 9

yxeyxzDDDD −+=′+′+ 222 )2(Solve

Solution:

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

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

m = –1, –1

)()(.. 21 xyfxxyfFC −+−=

P.I1 = yxe

DDDD−

′+′+ 22 2

1

yxe −

−+−+=

22 )1()1)(1(2)1(

1 Since the denominator = 0, we have to multiply x on Nr. and Diff. Dr. w.r.t.‘D’

121

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yxeDD

x −

′+=

22

yxex −=2

2

P.I2 = yxDDDD

222 2

1′+′+

yx

D

DDDD

2

2

22 2

1

1

′+′+

=

yxD

DDD

D2

1

2

2

2

21

1−

′+′+= yx

D

DDD

D2

2

2

2

21

1

′+′−=

122

6/4/2015B.Saravanan

yxD

D

D2

2

21

1

′−=

′−= )(

2)(

1 222

yxD

Dyx

D

−= )(21 22

2x

Dyx

D

−=

3

21 32

2

xyx

D

−=

12

2

3

1 43 xyx

D

123

6/4/2015B.Saravanan

3012

54 xyx −=

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

30122)()(.).(

542

21

xyxe

xxyfxxyfzei yx −++−+−= −

Problem 10

Solve yxzDDDD sin)43( 22 +=′−′+

Solution:

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

(m – 1)(m + 4) = 0

m = 1, – 4 )4()(.. 21 xyfxyfFC −++==>124

6/4/2015B.Saravanan

P.I1 = xDDDD 22 43

1′−′+

x

D

DDDD

′−′+

=

2

22 43

1

1

xD

DDD

D

1

2

2

2

431

1−

′−′+=

xD

DDD

D

′−′−= 2

2

2

431

1

[ ]01

2−= x

D

=

2

1 2x

D 6

3x=

125

6/4/2015B.Saravanan

P.I2 = )0sin(43

122

yxDDDD

+′−′+

)0sin()1(400

1yx +

−−+=

ysin4

1=

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

yx

xyfxyfzei sin4

1

6)4()(.).(

3

21 ++−++=

126

6/4/2015B.Saravanan

Problem 11

Solve xyyxy

z

yx

z

x

z ++=∂∂−

∂∂∂+

∂∂

)sinh(22

22

2

2

Solution:


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

)()2(.. 21 xyfxyfFC ++−=

127

6/4/2015B.Saravanan

P.I1 = )sinh(2

122

yxDDDD

+′−′+

−′−′+

=+−+

22

1 )(

22

yxyx ee

DDDD

′−′+

−′−′+

= −−+ yxyx eDDDD

eDDDD 2222 2

1

2

1

2

1

−−−−+−−

−+= −−+ yxyx ee

2222 )1(2)1)(1()1(

1

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

1

2

1

Since the denominator = 0, we have to multiply x on Nr. and Diff. Dr. w.r.t. ‘D’ we get

128

6/4/2015B.Saravanan

yxyx ex

ex −−+ +=

66

−−−

+= −−+ yxyx e

xe

x

12122

1

′+

−′+

= −−+ yxyx eDD

xe

DD

x

222

1

P.I2 = xyDDDD 22 2

1′−′+

xy

D

DDDD

′−′+

=

2

22 2

1

1

129

6/4/2015B.Saravanan

xyD

DDD

D

1

2

2

2

21

1−

′−′+=

xyD

DDD

D

′−′−=

2

2

2

21

1

xyD

D

D

′−= 1

12

′−= )()(

12

xyD

Dxy

D

−= )(11

2x

Dxy

D

130

6/4/2015B.Saravanan

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

24666)()2(.).(

43

21

xyxe

xe

xxyfxyfzei yxyx −+++++−= −−+

246

43 xyx −=

−=

62

1 32 xyx

D

−=

2

1 2

2

xxy

D

131

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Problem 12

Solve xyy

z

yx

z

x

zsin65

2

22

2

2

=∂∂+

∂∂∂−

∂∂

Solution:


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

)3()2(.. 21 xyfxyfFC +++=

132

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P.I = xyDDDD

sin65

122 ′+′−

xyDDDD

sin)3()2(

1′−′−

=

′−′−

= xyDDDD

sin3

1

2

1

∫ −′−

= dxxxcDD

sin)3(2

1

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

1xxxc

DD−−−−−

′−=

where y = c – 3x

133

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z = C.F + P.I

xyxxyfxyfzei sincos5)3()2(.).( 21 −++++=

xyx sincos5 −=

xxxy cos3cos2sin ++−=

)cos(3)]cos)(2())(sin2([ xxxxc −−−−−−−=

∫ −−−= dxxxxc ]sin3cos)2([

]sin3cos[2

1xxy

DD−−

′−=

where y = c – 2x

134

6/4/2015B.Saravanan

Problem 13

Solve )32sin()( 22 yxezDD yx +=′− −

Solution:

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

m2 = 1

1±=m

)()(.. 21 xyfxyfFC −++=

P.I = )32sin(1

22yxe

DDyx +

′−−

)32sin()1()1(

122 yx

DDe yx +

−′−+= −

135

6/4/2015B.Saravanan

)32sin(1212

122

yxDDDD

e yx +−′+′−++

= −

)32sin(22

122

yxDDDD

e yx +′+′−+

= −

)32sin(2)9(24

1yx

DDe yx +

′+−−+−= −

)32sin(5)(2

1yx

DDe yx +

+′+= −

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

]5)(2[yx

DDDD

DDe yx +

−′++′+−′+= −

136

6/4/2015B.Saravanan

)32sin(25)(4

]5)(2[2

yxDD

DDe yx +

−′+−′+= −

)32sin(25)2(4

]5)(2[22

yxDDDD

DDe yx +

−′+′+−′+= −

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

]5)(2[yx

DDe yx +

−−+−+−−′+= −

)32sin(125

]5)(2[yx

DDe yx +

−−′+= −

125

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

−+−+′++= − yxyxDyxD

e yx

137

6/4/2015B.Saravanan

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

yxyxyxe yx

+−+++−=−

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

yxyxe yx

+−+−=−

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

yxyxe yx

+−+=−

z = C.F + P.I

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

)()(.).( 21 yxyxe

xyfxyfzeiyx

+−++−++=−

138


Non-Homogeneous PDE

.hom)1.(

hom),(

)1(),(),(

PDElinearogeneousnoncalledeqnthen

ogeneousnotisDDfpolonomialtheif

yxFzDDfequationtheIn

−′

−−−−−=′

The general solution of equation (1) is z=C.F+P.I

The method to find P.I. is same as those for homogeneousPDE

To find C.F.

0).(..........))((

),(

2211 =−′−−′−−′−′

zcDmDcDmDcDmD

toinDDfFactorize

nn


)(........)()(. 221121 xmyfexmyfexmyfeFC nn

xcxcxc n ++++++=

0)( 3 =−′− zaDmDsayfactorrepeatedfor

)()()(. 32

21 mxyfexmxyfxemxyfeFCthen axaxax +++++=

6/4/2015B.Saravanan

Problem 1

Solve 0)12)(2( =+′−′− zDDDD

Solution:

The given equation is non-homogeneous.

0)12)(2( =+′−′− zDDDD

)2()2( 210 xyfexyfez xx +++=∴ −

)2()2(.).( 21 xyfexyfzei x +++= −

141

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Problem 2

Solve 22322 )2()2332( yx eezDDDDDD −+=+′+−′+′−

Solution:

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

yxyx eeezDDDD 2346 44)2)(1( −− ++=−′−−′−

)()(.. 22

1 xyfexyfeFC xx +++=

P.I1 = yxe

DDDD06

)2)(1(

1 +

−′−−′−

yxe 06

)206)(106(

1 +

−−−−= xe6

20

1=

142

6/4/2015B.Saravanan

P.I2 = yxe

DDDD404

)2)(1(

1 −

−′−−′−

yxe 40

)240)(140(

14 −

−+−+= ye 4

3

2 −=

P.I3 = yxe

DDDD234

)2)(1(

1 −

−′−−′−

yxe 23

)223)(123(

14 −

−+−+= yxe 23

3

1 −=

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

yxyxxx eeexyfexyfezei 23462

21 3

1

3

2

20

1)()(.).( −− ++++++=

143

6/4/2015B.Saravanan

Problem 3

SolveyxezDDDDDD +=+′++′+′+ 222 )1222(

Solution:

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

yxezDDDD +=+′++′+ 2)1)(1(

)()(.. 21 xyfxexyfeFC xx −+−= −−

P.I = yxe

DDDDDD+

+′++′+′+2

22 1222

1

yxe +

+++++= 2

22 1)1(2)2(2)1)(2(2)1()2(

1yxe += 2

16

1

144

6/4/2015B.Saravanan

z = C.F + P.I

yxxx exyfxexyfezei +−− +−+−= 221 16

1)()(.).(

Problem 4

Solve yexzDDDDDD =′++′−′− )362( 22

Solution:

yexzDDDDDDGiven =′++′−′− )362( 22

yexzDDDD =+′−′+ )3)(2(

1,3,2

1,0 2211 =−=−== mcmcHere

145

6/4/2015B.Saravanan

)(2

1.. 2

31

0 xyfexyfeFC xx ++

−= −

)(2

12

31 xyfexyf x ++

−= −

P.I = yexDDDDDD ′++′−′− 362

122

xDDDDDD

e y

)1(36)1()1(2

122 +′+++′−+′−

=

xDDDDDDDD

e y

336122

122 +′++−′−′−−′−

=

146

6/4/2015B.Saravanan

xDDDDDD

e y

′++′−′−+=

2

5212

122

xDDDDDDe y 122

2

521

2

−

′++′−′−+=

xDDDDDDe y

′++′−′−−=2

521

2

22

−= )(2

5

2x

Dx

e y

−=2

5

2x

e y

xDDDDDD

e y

′++′−′−+=

522

122

147

6/4/2015B.Saravanan

z = C.F + P.I

)52(4

)(2

1.).( 2

31 −+++

−= − xe

xyfexyfzeiy

x

)52(4

−= xe y

Problem 5

Solve

Solution:

)2sin()2223( 22 yxyxzDDDDDD +++=′−+′+′−

)2sin()2223( 22 yxyxzDDDDDDGiven +++=′−+′+′−

)2sin()22)(( yxyxzDDDD +++=+′−′−148

6/4/2015B.Saravanan

2,2,1,0 1211 =−=== mcmcHere

)2()(.. 22

10 xyfexyfeFC xx +++= −

)2()( 22

1 xyfexyf x +++= −

)(2223

122

yxDDDDDD

+′−+′+′−

P.I1 =

)()22)((

1yx

DDDD+

+′−′−=

)(

2

2121

1yx

DD

D

DD

+

′−+

′−

=

149

6/4/2015B.Saravanan

)(2

211

2

111

yxDD

D

D

D+

′−+

′−=

−−

)(2

2

2

211

2

12

yxDDDD

D

D

D+

′−+

′−−

′+=

)(42

11

2

1 2

2yxDD

DD

D

D

D

D+

′−+′+−

′+=

)(4242

11

2

12

yxD

D

D

D

DD

D

D

D

D+

′+

′−

′+′−+

′+−=

)(4

3

422

11

2

12

yxD

D

DD

D

D

D+

′−

′++

′+−=

150

6/4/2015B.Saravanan

−−+=2

1

22

1 2 yyxx

−+++−−+=

4

3

24

1

22222

1 22 xxyxyx

x

+′−+′

+

+++′++−+

=

4

)(3)(4

)(

2

)(

2)(

1

2

1

2

yxD

D

yxD

yxD

D

yxDyxyx

D

P.I2 = )2sin(2223

122

yxDDDDDD

+′−+′+′−

151

6/4/2015B.Saravanan

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

1yx

DD+

′−+−+−−−=

)2sin(22

1yx

DD+

′−=

)2sin()22)(22(

22yx

DDDD

DD +′+′−

′+=

)2sin(44

2222

yxDD

DD +′−′+=

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

22yx

DD +−−−′+=

152

6/4/2015B.Saravanan

)2cos(2

1yx +−=

)]2cos(6[12

1yx +−=

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

1yxyx +++−=

12

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

−+′++= yxDyxD

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

)2cos(2

1

2

1

22

1

)2()(.).(

2

22

1

yxy

yxx

xyfexyfzei x

+−

−−++

+++= −

153

MA6351 TRANSFORMS AND PARTIAL DIFFERENTIAL EQUATIONS€¦ · · 2015-06-04UNIT I PARTIAL DIFFERENTIAL EQUATIONS MA6351 TRANSFORMS AND PARTIAL DIFFERENTIAL EQUATIONS by B.Saravanan

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