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)
Department of Mathematics, University of Kashmir, Srinagar-190006
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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)
Department of Mathematics, University of Kashmir, Srinagar-190006
2
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)
Department of Mathematics, University of Kashmir, Srinagar-190006
3
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
Department of Mathematics, University of Kashmir, Srinagar-190006
4
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
Department of Mathematics, University of Kashmir, Srinagar-190006
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