Chapter 2 PARTIAL DIFFERENTIAL EQUATIONS OF SECOND ORDER

Draft PDE Lecture Notes

Khanday M.A.

Chapter 2

PARTIAL DIFFERENTIAL EQUATIONS OF SECOND ORDER

INTRODUCTION: An equation is said to be of order two, if it involves at least one of the differential coefficients r = (2z / 2x), s = (2z / x y), t =(2z / 2y), but now of higher order; the quantities p and q may also enter into the equation. Thus the general form of a second order Partial differential equation is

( , , , , , , , ) = 0

...(1)

The most general linear partial differential equation of order two in two independent variables x and y with variable coefficients is of the form

+ + + + + =

. . . (2)

where , , , , , , are functions of and only and not all , , are zero.

Ex.1: Solve = 6.

Sol.

The

given

equation

can

be

written

as

2 2

= 6

Integrating (1) w. r. t.

=

32

+

1()

where 1() is an arbitrary function of .

Integrating (2) w. r. t. we get

= 3 + 1() + 2()

where 2(y) is an arbitrary function of y.

Ex.2. =

Sol:

Given

equation

can

be

written

as

2 2

=

1

Integrating (1) w. r. t., , we get

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

...(1)

=

2 2

+

1(y)

where 1(y) is an arbitrary function of y

...(2)

Integrating (2) w. r. t., x,

z =

3 6

+

x

1(y)

+

2(y)

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Draft PDE Lecture Notes

or

z

=

2

+

x

1(y)

+

2(y)

where 2(y) is an arbitrary function of y.

Ex.3: Solve r = 2y2

Sol: Try yourself.

Ex. 4. Solve = sin()

Sol.

Given

equation

can

be

written

as

2 2

= sin()...(1)

Integrating (1) w. r. t., y

= -

1

cos() +

1()

Integrating (2) w. r. t., y

= -

1 2

sin

+ 1 + 2

which is the required solution, 1, 2 being arbitrary functions.

Exercises: = 1

Sol:

We

know

that

=

2

Therefore

2 = 1

or

2 = 1

Integrating w.r.t., y we have

1 = log + Again integrating w.r.t., x we get

= log log + +

0r

= log log + +

Khanday M.A. . . . (2)

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Draft PDE Lecture Notes

Exercises:2 + 2 =

Sol: The given equation can be written as

2 = 2 + 2

Integrating w.r.t., , we have

= 2 + 2 +

Integrating w.r.t., , we have

= 2 + 2 + +

= 2 + 2 + +

Exercises: + = 923

Sol: The given equation can be written as

2 2

+

=

9 2 3

+

=

9 2 3

+ = 93

which is linear first order differential equation in

I. F. is log =

Multiplying (1) by we get

+

= 923

= 9 23

33 = 9 3 + = 333 +

Khanday M.A. ... (1)

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Draft PDE Lecture Notes

Khanday M.A.

333 +

=

=

3 2 3

+

Integrating with respect to we get

= 33 + log +

Exercises: - =

Sol: Please try yourself.

Exercises: - = 2

Sol: Please try yourself.

Exercises: = 22

Sol: The given equation can be written as

2 2

=

22

=

22

Integrating with respect to we get

= 22 +

=

22 +

Integrating we get

= 22 + +

= 22 + +

Exercises: = sin

Sol: Please try yourself.

Exercises:log = +

Sol: The given equation can be written as

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Draft PDE Lecture Notes

Khanday M.A.

Integrating w.r.t. we get Integrating w.r.t., , we get

log = +

=

+

=

= +

=

+

= + +

or

= + +

Exercises:

-

=

2

Sol: Please try yourself.

Exercises: + + = 0

Sol: The given equation can be written as

+ + = 0

Integrating with respect to , we get

+ + =

+ = -

It is of the form + =

Its auxiliary system is

= =

1 1 -

...(1)

From first two fractions of (1) we get

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