Ordinary differential equations

Laplace transforms
Partial differential equations
Fourier series

DIFFERENTIAL EQUATIONS

Dr. Shafiq Ur Rehman

Department of Mathematics,

University of Engineering and Technology,

Lahore-Pakistan.

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 Ordinary differential equations
Laplace transforms
Partial differential equations
Partial differential equations
Fourier series

Partial differential equations

Basic concepts of partial differential equations
Partial differential equations (PDEs) arise in connection with various
physical and geometrical problems when the function involved depend
upon two or more independent variables. These variables may be the
time t or more space coordinates x , y , z. Here, we give some basic
concepts relating to PDEs, the formation and the methods for finding
solutions of PDEs, and the method of separation of variables for the
solutions of PDEs will also be discussed.
Definition:-
A partial differential equation is an equation involving partial deriva-
tives of a function of two or more independent variables. A partial
differential equation in the dependent variable u and independent vari-
ables x , y , z, and t is a relation of the form:
F (x , y , z, t; u, ux , uy , uz , ut , uxx , uxy , uxxx , · · · ) = 0.

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 Ordinary differential equations
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Partial differential equations

For example:
∂u ∂u
3y 2 + x2 = 2u, (6)
∂x ∂y
∂ 2u 2
2 ∂ u
+ x y = 0. (7)
∂x 2 ∂y 2

Where, u is the dependent variable, while x and y are the independent

variables.
Order and degree of a partial differential equation
The order of a PDE is the order of the highest partial derivative that
appears in the given equation and the greatest exponent of the highest
partial derivative is called the degree of the given PDE. For example,
equations (6) and (7) are of first and second-order, respectively, and
both are of degree 1.

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 Ordinary differential equations
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Whereas, the equation
"2 "3
∂3u ∂2u
! !
+ + 8u = 0 (8)
∂x 3 ∂x 2

is of order 3 and degree 2.

Linear and non-linear partial differential equations
A partial differential equation is said to be linear if
1 it is of the first degree in the dependent variable and all of its
partial derivatives;
2 the coefficients (of the dependent variable and its derivatives)
are functions only of the independent variables.
Whereas, a PDE that is not linear is called as non-linear PDE. For
example, equations (6) and (7) are linear, while the equation (8) is
non-linear.
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 Ordinary differential equations
Laplace transforms
Partial differential equations
Partial differential equations
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Partial differential equations

Homogeneous and non-homogeneous partial differential equa-
tions
A linear PDE is said to be homogeneous if each term in the equation
contains either the dependent variable or one of its derivatives. Other-
wise, it is said to be non-homogeneous or inhomogeneous PDE. For
example, equations (6) and (7) are homogeneous, while the equation
∂ 2u ∂ 2u
2
+ x 2 y 2 = 2xy
∂x ∂y
is non-homogeneous.
General and particular solutions
A solution of a PDE is any function which satisfies the equation iden-
tically. The general solution is a solution which contains the number
of arbitrary independent functions equal to the order of the equation.
Whereas, a particular solution is one which can be obtained from the
general solution by particular choice of arbitrary functions.
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 Ordinary differential equations
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Partial differential equations

For example, as can be observed by substitution

1 2
u = x 2y − xy + F (x ) + G(y )
2
is a solution of the PDE
∂ 2u
= 2x − y .
∂x ∂y

Since it contains two arbitrary independent functions F (x ) and G(y ),

so it is called as general solution. If, in-particular, F (x ) = 2 sin x and
G(y ) = 3y 4 − 5 then we obtain the following particular solution

1 2
u = x 2y − xy + 2 sin x + 3y 4 − 5.
2

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 Ordinary differential equations
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Partial differential equations

Auxiliary conditions
Generally , the PDEs that represent physical systems have infinite
number of solutions. For example, the functions:
! "
y
u = x 2 − y 2 , u = ex cos y , u = log(x 2 + y 2 ), u = tan−1 ,···
x

which are entirely different from each other are solutions of the follow-
ing PDE

∂ 2u ∂ 2u
+ = 0.
∂x 2 ∂y 2

In order to obtain a unique solution of the PDE corresponding to a

given physical problem, one must use additional information (i.e., aux-
iliary conditions) arising from the physical situation. They fall in two
categories: (i) Boundary conditions (ii) Initial conditions.
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 Ordinary differential equations
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Partial differential equations

Boundary conditions
These are the conditions that must be satisfied at points on the bound-
ary S of the region R in which the PDE holds. Here, S may be
the bounding curve of a plane region R or the surface of a three-
dimensional region R. There are three types of boundary conditions:

1 Dirichlet condition
u=g on S
2 Neumann (or flux) condition
∂u
=g on S
∂n
3 Mixed (or Robin or radiation) condition
∂u
αu + β =g on S
∂n
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 Ordinary differential equations
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where n denotes the (typically exterior) normal to the boundary and
g, α, β are functions prescribed on S.
Note it that, the boundary must be sufficiently smooth in such a way
that the normal derivative can exist, since, for example, at corner
points on the boundary the normal vector is not well defined.
Initial conditions
These are the conditions that must be satisfied throughout the region
R at the instant when consideration of the physical system begins.
When time t is one of the independent variable and we specify a con-
dition at t = 0, we refer to it as initial condition. A typical initial condi-
tion is said to be of Cauchy type if the values of both u and ∂u ∂t on the
∂u
boundary at t = 0 (i.e., the initial values of u and ∂t ) are given.

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 Ordinary differential equations
Laplace transforms
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Partial differential equations

Formation of PDEs
Partial differential equations may be derived either by the elimination
of arbitrary constants from the given relation between the variables or
by the elimination of arbitrary functions involving the variables.
constants are removed by substituting
Elimination of arbitrary constants thier values in the given general
expression (only)
Case (1): If the number of arbitrary constants is equal to the number
of independent variables, then we proceed as follows:
Let u be a function of two independent variables x and y defined by

f (x , y , u, a, b) = 0, (9)

where a and b are two arbitrary constants. Differentiating equation (9)

partially w.r.t. x and y , we get

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 Ordinary differential equations
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Partial differential equations

∂f ∂f ∂u
+ = 0, (10)
∂x ∂u ∂x
∂f ∂f ∂u
+ = 0. (11)
∂y ∂u ∂y
Generally, the arbitrary constants may be eliminated from equations
(9), (10), (11) yielding a PDE of order one
! "
∂u ∂u
g x , y , u, , = 0.
∂x ∂y
Example:-
Form the PDEs by eliminating the arbitrary constants from the follow-
ing relations:

i) u = ax + by , ii) u = (x 2 + a)(y 2 + b).

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 Ordinary differential equations
Laplace transforms
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Partial differential equations

Solution:- Take

u = ax + by . (12)

Differentiating equation (12) partially w.r.t. x and y , we get

∂u
= a, (13)
∂x
∂u
= b. (14)
∂y

Substituting the values of a and b from equations (13) and (14) in

equation (12), we get

∂u ∂u
u=x +y
∂x ∂y

as the required partial differential equation.

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 Ordinary differential equations
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Case (2): If u is a function of x and y defined by the relation involving
only one arbitrary constant, then it is usually possible to obtain two
distinct PDEs of order one by eliminating the constant.
Example:- Form the PDEs by eliminating the arbitrary constant a from
the following relation
u = a(x + y ).
Solution:- Take
u = a(x + y ). (15)
Differentiating equation (15) partially w.r.t. x , we get
∂u
= a,
∂x
so that the following PDE is obtained
∂u
u = (x + y ) .
∂x
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 Ordinary differential equations
Laplace transforms
Partial differential equations
Partial differential equations
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Partial differential equations

Similarly, differentiating equation (15) partially w.r.t. y , we get
∂u
= a,
∂y
so that the PDE
∂u
u = (x + y )
∂y
is obtained.
Case (3): If the number of arbitrary constants to be eliminated exceeds
the number of independent variables, the resulting PDE is usually of
higher order than the first. However, this PDE may not be unique.
Example:-
Form the PDEs by eliminating the arbitrary constants a, b, c from the
following relation:
u = ax + by + cxy .
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 Ordinary differential equations
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Partial differential equations

Solution:- Take

u = ax + by + cxy . (16)

Differentiating equation (16) partially w.r.t. x and y , we get

∂u
= a + cy , (17)
∂x
∂u
= b + cx . (18)
∂y
These together with the given relation are not sufficient to eliminate 3
arbitrary constants. Differentiating equation (17) partially w.r.t. x , we
have
∂ 2u
=0
∂x 2
a PDE order order 2.
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 Ordinary differential equations
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Differentiating equation (18) partially w.r.t. y , we have
∂ 2u
=0
∂y 2
a PDE order order 2.
Differentiating equation (17) partially w.r.t. y or equation (18) partially
w.r.t. x , we have
∂ 2u
=c
∂x ∂y
From equation(17), we have
∂u ∂ 2u
=a+y ,
∂x ∂x ∂y
∂u ∂ 2u
or a = −y .
∂x ∂x ∂y
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 Ordinary differential equations
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Partial differential equations

Similarly, from equation(18), we have
∂u ∂ 2u
=b+x ,
∂y ∂x ∂y
∂u ∂ 2u
or b = −x .
∂y ∂x ∂y
Substituting the values of a, b and c in the given relation, we get
∂ 2u ∂ 2u ∂ 2u
! " ! "
∂u ∂u
u= −y x+ −x y + xy ,
∂x ∂x ∂y ∂y ∂x ∂y ∂x ∂y
∂u ∂u ∂2u
=x +y − xy ,
∂x ∂y ∂x ∂y
which is a partial differential equation of order 2. Thus, we have three
PDEs of the same order associated with the given relation:
∂ 2u ∂2u ∂u ∂u ∂ 2u
2
= 0, 2
= 0, u = x +y − xy .
∂x ∂y ∂x ∂y ∂x ∂y
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 Ordinary differential equations
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Partial differential equations

Example:- Obtain the PDEs by eliminating the arbitrary functions from
the following relations:
! "
y
i) u = f ii) u = F (x − ct) + G(x + ct)
x
Solution:- i) Take
! "
y
u=f (19)
x
Differentiating equation (19) partially w.r.t. x and y , we get
! "
∂u y ′ y
= − 2f (20)
∂x x x
! "
∂u 1 ′ y
= f (21)
∂y x x
Multiplying equation (20) by x and equation (21) by y and then adding,
we get
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∂u ∂u
x +y =0 (22)
∂x ∂y
as the required partial differential equation of order one.
ii) Take
u = F (x − ct) + G(x + ct) (23)
Differentiating twice equation (23) partially w.r.t. x , we get
∂u ′ ′
= F (x − ct) + G (x + ct),
∂x
∂ 2u ′′ ′′
= F (x − ct) + G (x + ct). (24)
∂x 2
Differentiating twice equation (23) partially w.r.t. t, we get
∂u ′ ′
= −cF (x − ct) + cG (x + ct)
∂t
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∂ 2u ′′ ′′

2
= c 2 F (x − ct) + c 2 G (x + ct) (25)
∂t
From equations (24) and (25), we have

∂2u 1 ∂ 2u
2
= 2 2
∂x c ∂t
a partial differential equation of order two.
Remarks:- The above equation is called as one-dimensional wave
equation describing the transverse vibrations of a stretched string.
The most generalised solution of the above equation is the equation
(23). From these examples (as illustrated above) it is clear that a PDE
can result both from the elimination of arbitrary constants and from the
elimination of arbitrary functions.

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 Ordinary differential equations
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Solutions of partial differential equations
If a partial differential equation involves derivatives w.r.t. one of the
independent variables only, we can solve it like an ordinary differential
equation treating the other independent variables as parameters. We
explain the method with the following examples.
Example:- Find the solution of the following PDEs:
i) uxx = 0 ii) uy = 2xyu.
Solution:- i) Take
integrating pde results in add of
uxx = 0. arb function while integrating (26)
arbitrary function results in add
Integrating equation (26) w.r.t. x , we get of arb const

ux = f (y )
where f (y ) is an arbitrary function. Integrating the above equation
again w.r.t. x , we get
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u(x , y ) = xf (y ) + g(y )
where g(y ) is an arbitrary function.
Solution:- ii) Take
uy = 2xyu. (27)
The above equation can be written as
uy
= 2xy .
u
Integrating the above equation w.r.t. y , we get
uy
# #
dy = 2xy dy
u
ln u = xy 2 + ln f (x )

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 Ordinary differential equations
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where ln f (x ) is an arbitrary function. Taking antilog of both sides of
the above equation, we get
2
u(x , y ) = f (x )exy .

Example:-
Solve the following partial differential equation

uxx + 3ux − 4u = 12.

∂ ∂2
Solution:- Let D = ∂x and D 2 = ∂x 2
, then the above equation be-
comes

(D 2 + 3D − 4)u = 12

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 Ordinary differential equations
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Partial differential equations of order one
A partial differential equation of order one in the dependent variable u
and independent variables x and y is a relation of the form
F (x , y , u, ux , uy ) = 0.
Thus the general linear partial differential equation of order 1 is
aux + buy + cu = d
where a, b, c, and d may be constants or may dependent on x and y
but not on u. For example, the following are linear partial differential
equations of order 1:
i) ux = 2xyu
ii) ux + uy = 0
iii) yux − xuy = 0
iv ) xux + yuy = 2(x + y )u
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Partial differential equations of order two
A partial differential equation of order two in the dependent variable u
and independent variables x and y is a relation of the form
F (x , y , u, ux , uy , uxx , uxy , uyy ) = 0.
Thus the general linear partial differential equation of order 2 is
auxx + 2buxy + cuyy + dux + euy + fu = g
where a, b, c, d, e, f , and g may be constants or may dependent on
x and y but not on u. If g = 0, then the equation is called homo-
geneous, while if g ̸= 0, then it is called non-homogeneous. For
example, the following are linear partial differential equations of order
2:
1
i) uxx = utt One-dimensional wave equation
c2
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1
ii) uxx = ut One-dimensional heat equation
c2
iii) uxx + uyy = 0 Two-dimensional Laplace equation
iv ) uxx + uyy = f (x , y ) Two-dimensional Poison equation
v ) uxx + uyy + uzz = 0 Three-dimensional Laplace equation

Here, c 2 is a positive constant, t is time, and x , y , z are Cartesian

coordinates. The equation (iv) with f (x , y ) ̸= 0 is non-homogeneous,
whereas all the other equations are homogeneous.

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 Ordinary differential equations
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Classification of second order partial differential equations
Consider the following general linear homogeneous partial differential
equation of order two
auxx + 2buxy + cuyy + dux + euy + fu = 0. (28)
We note that the form of equation (28) resembles that of a general
conic section
ax 2 + 2bxy + cy 2 + dx + ey + f = 0. (29)
The above equation represents an ellipse, parabola, or hyperbola when
the discriminant b2 − ac is negative, zero, or positive, respectively. We
use the similar classification for the PDE (28) and say that it is ellip-
tic, parabolic, or hyperbolic when the discriminant b2 − ac is negative,
zero, or positive, respectively. Note it that, the type of equations (28)
and (29) is determined solely by its principal part, i.e., the terms involv-
ing the highest-order derivatives of u and that the type will generally
change with position in the xy -plane unless a, b, c are constants.
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 Ordinary differential equations
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Example:- Classify the following partial differential equations:

1
i) uxx + uyy = 0 ii) uxx = ut
α2
1
iii) uxx = utt iv ) yuxx + xuyy = 0
α2
Solution:- i) The two-dimensional Laplace’s equation uxx + uyy = 0
may be obtained from equation (28) by putting a = 1, b = 0, c = 1,
and d = e = f = 0. Since b2 − ac = −1 < 0, therefore, the given PDE
is of elliptic type.
Note it that, a PDE may be elliptic in one region of the plane, parabolic
in another region, and hyperbolic in yet another region.

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 Ordinary differential equations
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Method of separation of variables
In this method, it is assumed that the solution, i.e., dependent variable
can be expressed as a product of unknown functions each of which
depends upon only one of the independent variables. With this solu-
tion, the partial differential equation can be written in such a way that
one side depends upon only one variable while the other side depends
upon the remaining variables from which it is concluded that each side
must be a constant.
By the repetition of this, the given PDE reduces to two or more ODEs
each one involving one of the independent variable. Solving these
ODEs, the unknown functions can be determined. The principle of su-
perposition can then be used to find the actual solution of the given
PDE. We illustrate this method of separation of variables with the fol-
lowing examples.

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 Ordinary differential equations
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Example:- Using the method of separation of variables, solve the fol-
lowing partial differential equations:
i) ux + uy = 0 ii) uxy = u.
Solution:- i) ux + uy = 0.
Let
u(x , y ) = X (x )Y (y ) = XY . (30)
′ ′
Then by differentiation, we get ux = X Y and uy = XY so that the
given partial differential equation becomes
′ ′ ′ dX ′ dY
X Y + XY = 0, where X = and Y = .
dx dy
Separating variables, we get
′ ′
X Y
=− .
X Y
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The L.H.S. of the above equation depends upon x while the R.H.S.
depends upon y only. Moreover, since they are equal, their common
value cannot be a function of either x or y and must therefore be a
constant, say, c, i.e.,
′ ′
X Y
=− = c.
X Y
This implies two ordinary differential equations:
′ ′
X − cX = 0 and Y + cY = 0,

whose solutions are: X = Aecx and Y = Be−cy , respectively. Where,

A and B are arbitrary constants. Thus from equation (30), the general
solution of the given partial differential equation is

u(x , y ) = X (x )Y (y ) = XY = Aecx Be−cy = Kec(x−y ) , where K = AB.

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Solution:- ii) uxy = u.
Let

u(x , y ) = X (x )Y (y ) = XY . (31)
′ ′
Then by differentiation, we get uxy = X Y so that the given partial
differential equation becomes
′ ′ ′ dX ′ dY
X Y = XY , where X = and Y = .
dx dy

Separating variables, we get

′
X Y
= ′.
X Y

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 Ordinary differential equations
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Example:-
Using the method of separation of variables, solve the following bound-
ary value problems (BVPs):
i) ux = 4uy u(0, y ) = 8e−3y
ii) ux = 4uy u(0, y ) = 8e−3y + 4e−5y
Solution:- i) ux = 4uy .
Let
u(x , y ) = X (x )Y (y ) = XY . (32)
′ ′
Then by differentiation, we get ux = X Y and uy = XY so that the
given partial differential equation becomes
′ ′ ′ dX ′ dY
X Y = 4XY , where X = and Y = .
dx dy

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Separating variables, we get
′ ′
X Y
= .
4X Y
The L.H.S. of the above equation depends upon x while the R.H.S.
depends upon y only. Moreover, since they are equal, their common
value cannot be a function of either x or y and must therefore be a
constant, say, c, i.e.,
′ ′
X Y
=− = c.
4X Y
This implies two ordinary differential equations:
′ ′
X − 4cX = 0 and Y − cY = 0,
whose solutions are: X = Ae4cx and Y = Becy , respectively. Where, A
and B are arbitrary constants. From equation (32), the general solution
of the given partial differential equation is
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u(x , y ) = X (x )Y (y ) = XY = Ae4cx Becy = Kec(4x+y ) , where K = AB.

Using the boundary condition u(0, y ) = 8e−3y , we have

8e−3y = Kecy

which is possible if and if K = 8 and c = −3. Thus, from equation

(32), we have

u(x , y ) = 8e−3(4x+y ) ,

is the required solution.

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 Ordinary differential equations
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Partial differential equations

Example:-
Obtain the partial differential equations by eliminating arbitrary con-
stants from the following relations:
x2 y2
i) u = ax + (1 − a)y + b ii) 2u = 2
+ 2
$ a b
by
iii) u = xy + y x 2 − a2 + b iv ) u = ae sin bx

Example:-
Obtain the partial differential equations by eliminating arbitrary func-
tions from the following relations:
! " ! "
2 1 n y
i) u = y + 2f + ln y ii) u = x f
x x
iii) u = yf (x ) + xg(y ) iv ) u = f (x − 3y ) + g(2x + y )

Dr. Shafiq Ur Rehman, UET, Lahore DIFFERENTIAL EQUATIONS_2025_P2

 Ordinary differential equations
Laplace transforms
Partial differential equations
Partial differential equations
Fourier series

Partial differential equations

Example:-
Show that u(x , y ) = xf (2x + y ) is a general solution of
∂u ∂u
x − 2x = u.
∂x ∂y
Then find a particular solution satisfying u(1, y ) = y 2 .
Example:-
Show that u(x , t) = e−8t sin 2x is a solution to the following initial-
boundary value problem:
∂u ∂ 2u
=2 2
∂t ∂x
(B. C’s) u(0, t) = 0, u(π, t) = 0, t > 0
(I. C.) u(x , 0) = sin 2x

Dr. Shafiq Ur Rehman, UET, Lahore DIFFERENTIAL EQUATIONS_2025_P2

 Ordinary differential equations
Laplace transforms
Partial differential equations
Partial differential equations
Fourier series

Partial differential equations

Example:-
Solve the following partial differential equation

∂ 3u
= cos(2x + 3y ).
∂y ∂x 2

Example:-

Classify the following partial differential equations according to the

type:

i) 3uxx + 2uxy + 5uyy + xuy = 0 ii) uxx + 2yuxy + xuyy − ux + u = 0

iii) 2xyuxy + xuy + yux = 0 iv ) yuxx + 2xuxy + yuyy = 0

Dr. Shafiq Ur Rehman, UET, Lahore DIFFERENTIAL EQUATIONS_2025_P2

 Ordinary differential equations
Laplace transforms
Partial differential equations
Partial differential equations
Fourier series

Partial differential equations

Example:-
Using the method of separation of variables, solve the following partial
differential equations:

i) ux + uy = 2(x + y )u ii) x 2 uxy + 3y 2 u = 0

Example:-

Using the method of separation of variables, solve the following bound-

ary value problems (BVPs):

i) 3ux + 2uy = 0, u(x , 0) = 4e−x

ii) ux = 2uy + u, u(x , 0) = 3e−5x + 2e−3x

Dr. Shafiq Ur Rehman, UET, Lahore DIFFERENTIAL EQUATIONS_2025_P2