M1 complete notes

TSEDUHUB
www.tseduhub.blogspot.com

JNTUWORLD
MATHEMATICS-I
MEAN VALUE THEOREMS
FUNCTIONS OF SINGLE
&
SEVERAL VARIABLES
I YEAR B.TECH
By
Y. Prabhaker Reddy
Asst
. Professor of Mathematics
Guru Nanak Engineering College
Ibrahimpatnam, Hyderabad.
TSEDUHUB
www.tseduhub.blogspot.com

JNTUWORLD
S
YLLABUS OF MATHEMATICS-I (AS
PER JNTU HYD)
Name of the Unit Name of the Topic
Unit-I
Sequences and Series
B
C
R
C
Int
C
R
A
Unit-II
Functions of single variable
R
L
C
G
F
F
M
Unit-III
Application of single variables
R
E
C
C
Cur
Unit-IV
Integration and its
applications
R
Int
Int
Int
Sur
M
C
C
Unit-V
Differential equations of first
order and their applications
Ove
E
L
B
N
L
Or
Unit-VI
Higher order Linear D.E and
their applications
L
R exp(ax)
R sin ax and cos ax
R
R
M
A
Unit-VII
Laplace Transformations
L
Inve –first shifting property
Tr
Unit
C
Diff
A
Unit-VIII
Vector Calculus
G
L
L
G
G
Sto applications
TSEDUHUB
www.tseduhub.blogspot.com

JNTUWORLD
CONTE
NTS

U
NIT-2
Fu
nctions of Single & Several Variables
Rolle’s Theoremwithout Proof
La
grange’s Mean Value Theoremwithout Proof
Cauchy’s Mean Value Theoremwithout Proof
Generalized Mean Value Theorem (without Proof)
Functions of Several Variables-Functional dependence
Jacobian
Maxima and Minima of functions of two variables
TSEDUHUB
www.tseduhub.blogspot.com

JNTUWORLD
In
troduction
Real valued function:
Any function
is called a Real valued function.
Li
mit of a function: Let
is a real valued function and . Then, a real number is
sai
d to be limit of
at if for each whenever
. It is denoted by .
C
ontinuity: Let
is a real valued function and . Then, is said to be continuous at
if for each whenever . It is
d
enoted by
.
Not
e: 1) The function
is continuous every where
2) E
very
function is continuous every where.
3) E
very
function is not continuous, but in the function is continuous.
4) E
very polynomial is continuous every where.
5) Every exponential function is continuous every where.
6) Every log function is continuous every where.
Differentiability: Let
be a function, then is said to be differentiable or derivable at
a
point
, if exists.
Functions
Algebraic Trignonometric Inverse Exponential Logarithmic Hyperbolic
TSEDUHUB
www.tseduhub.blogspot.com

JNTUWORLD
ME
AN VALUE THEOREMS
Let be a given function.
RO
LLE’S THEOREM
Let such that
(i) is continuous in
(ii)

is differentiable (or) derivable in
(iii
)
then

atleast one point in such that
G
eometrical )nterpretation of Rolle’s Theorem:
From the diagram, it can be observed that
(i) there i
s no gap for the curve
from . Therefore, the function is
conti
nuous
(ii) There exists unique tangent for every intermediate point between
and
(ii) A
lso the ordinates of
andare same, then by Rolle’s theorem, there exists atleast one point
in between and such that tangent at is parallel to axis.
LAG
RANGE’S MEAN VALUE THEOREM
Let such that
(i) is continuous in
(ii)

is differentiable (or) derivable in
then

atleast one point such that
TSEDUHUB
www.tseduhub.blogspot.com

JNTUWORLD
Geomet
rical )nterpretation of Rolle’s Theorem:
From the diagram, it can be observed that
(i) there i
s no gap for the curve
from . Therefore, the function is
conti
nuous
(ii) There exists unique tangent for every intermediate point between
and
T
hen by Lagrange’s mean value theorem, there exists atleast one point
in between and
such that tangent at is parallel to a straight line joining the points and
CA
UCHY’S MEAN VALUE THEOREM
Let are such that
(i) are continuous in
(ii)are differentiable (or) derivable in
(iii
)
then

atleast one point such that
G
eneralised Mean Value Theorems
Taylor’s Theorem: I
f
such that
(i) are continuous on
(ii) are derivable or differentiable on
(iii
)
, then such that
w
here,
is called as Schlomilch –
Roche’s form of remainder and is given by
TSEDUHUB
www.tseduhub.blogspot.com

JNTUWORLD
La
grange’s form of Remainder:
Substituting
in we get Lagrange’s form of Remainder
i.e
.
C
auchy’s form of Remainder:
Substituting
in we get Cauchy’s form of Remainder
i.e
.
M
aclaurin’s Theorem: If
such that
(i) are continuous on
(ii) are derivable or differentiable on
(iii
)
, then such that
w
here,
is called as Schlomilch –
Roche’s form of remainder and is given by
La
grange’s form of Remainder:
Substituting
in we get Lagrange’s form of Remainder
i.e
.
C
auchy’s form of Remainder:
Substituting
in we get Cauchy’s form of Remainder
i.e
.
TSEDUHUB
www.tseduhub.blogspot.com

JNTUWORLD
Ja
cobian
Let are two functions , then the Jacobian of w.r.t is
d
enoted by
and is defined as
P
roperties:

I
f
are functions of and are functions of then
Funct
ional Dependence:
Two functions
are said to be functional dependent on one another if the Jacobian of
is zero.
I
f they are functionally dependent on one another, then it is possible to find the relation between
these two functions.
MAXIMA AND MINIMA
Maxima and Minima for the function of one Variable:
L
et us consider a function
T
o find the Maxima and Minima, the following procedure must be followed:
Step 1: First find the first derivative and equate to zero. i.e.
S
tep 2: Since
is a polynomial is a polynomial equation. By solving this
eq
uation we get roots.
Step 3: Find second derivative i.e.
S
tep 4: Now substitute the obtained roots in
S
tep 5: Depending on the Nature of
at that point we will solve further. The following cases will
b
e there.
TSEDUHUB
www.tseduhub.blogspot.com

JNTUWORLD
C
ase (i): If
at a point say , then has maximum at and the maximum
v
alue is given by
C
ase (ii): If
at a point say , then has minimum at and the minimum
v
alue is given by
C
ase (iii): If
at a point say , then has neither minimum nor maximum. i.e.
sta
tionary.
Maxima and Minima for the function of Two Variable
Let us consider a function
T
o find the Maxima and Minima for the given function, the following procedure must be followed:
Step 1: First find the first derivatives and equate to zero. i.e.
(
Here, since we have two variables, we go for partial derivatives, but not ordinary
derivatives)
Step 2: By solving
, we get the different values of and .
Wr
ite these values as set of ordered pairs. i.e.
S
tep 3: Now, find second order partial derivatives.
i.e.,
S
tep 4: Let us consider
,
S
tep 5: Now, we have to see for what values of
, the given function is maximum/minimum/
does not have extreme values/ fails to have maximum or minimum.
If at a point, say
, then has ma
at this point and the
maximum value will be obtained by substituting
in the given function.
I
f at a point, say
, then has mi
at this point and the
minimum value will be obtained by substituting
in the given function.
I
f at a point, say
, then has n
and
such points are called as saddle points.
If at a point, say
, then fails to have maximum or minimum and case
ne
eds further investigation to decide maxima/minima. i.e. No conclusion
TSEDUHUB
www.tseduhub.blogspot.com

JNTUWORLD
Pr
oblem
1)Ex

S
ol: Given
T
he first order partial derivatives of
are given by
and
Now
, equating first order partial derivatives to zero, we get
S
olving
we get
S
ubstituting
in
S
ubstituting
in
All possible set of values are
Now
, the second order partial derivatives are given by
Now
,
A
t a point
&
has maximum at and the maximum value will be obtained by substituting in the
f
unction
i.e.
is the maximum value.
TSEDUHUB
www.tseduhub.blogspot.com

JNTUWORLD
A
lso, at a point
&
has minimum at and the minimum value is .
Also,
at a point
has neither minimum nor maximum at this point.
A
gain, at a point
has neither minimum nor maximum at this point.
La
grange’s Method of Undetermined Multipliers
This method is useful to find the extreme values (i.e., maximum and minimum) for the given
f
unction, whenever some condition is given involving the variables.
To find the Maxima and Minima for the given function using Lagrange’s Method , the following
procedure must be followed:
Step 1: Let us consider given function to be
subject to the condition
S
tep 2: Let us define a Lagrangean function
, where is called the Lagrange
mu
ltiplier.
Step 3: Find first order partial derivatives and equate to zero
i.e.
L
et the given condition be
S
tep 4: Solve
, eliminate to get the values of
S
tep 5: The values so obtained will give the stationary point of
S
tep 6: The minimum/maximum value will be obtained by substituting the values of
in the
g
iven function.
Problem
1) Find the minimum value of subj

S
ol: Let us consider given function to be
and
L
et us define Lagrangean function
, where is called the Lagrange multiplier.
TSEDUHUB
www.tseduhub.blogspot.com

JNTUWORLD
Now
,
S
olving
Now
, consider
A
gain, consider
A
gain solving
Giv
en
S
imilarly, we get
H
ence, the minimum value of the function is given by
TSEDUHUB
www.tseduhub.blogspot.com

JNTUWORLD
MATHEMATICS-I
Curvature, Evolutes & Envelopes
Curve Tracing
I YEAR B.Tech
By
Y.Prabhaker Reddy
Asst.
Professor of Mathematics
Guru Nanak Engineering College
Ibrahimpatnam, Hyderabad.
TSEDUHUB
www.tseduhub.blogspot.com

JNTUWORLD
U
NIT-III
CURVATURE
The shape of a plane Curve is characterized by the degree of Bentness or Curvedness.
R
ADIUS OF CURVATURE
The reciprocal of the curvature of a curve is called the radius of curvature of curve.
F
ORMULAE FOR THE EVALUATION OF RADIUS OF CURVATURE
In this we have three types of problems
P

P

P
.
In Cartesian Co-ordinates
Let us consider be the given curve, then radius of curvature is given by
I
f the given equation of the curve is given as
, then the radius of curvature is given
b
y
TSEDUHUB
www.tseduhub.blogspot.com

JNTUWORLD
I
n Parametric Form
(i
.e.
in terms of other variable and in terms of other variable say
)
If the equation of the curve is given in parametric form then
I
n Polar co-ordinates
If the equation of the curve is given in polar form i.e. then
w
here,
and .
R
adius of Curvature at origin Newton?s Theorem
Suppose a curve is passing through the origin and -axis or -axis is tangent to the curve at
th
e origin. Then the following results will be useful to find
at
I
f
-axis is tangent at , then
I
f
-axis is tangent at , then
Not
e: If the given curve
passes through and neither -axis nor -axis is
ta
ngent to the curve, then
Using Maclaurin?s series expansion of
, we get
TSEDUHUB
www.tseduhub.blogspot.com

JNTUWORLD
th
en,
B
y this we can calculate
at using the formula for finding Radius of Curvature in
C
artesian Co-ordinates.
Problems on Radius of Curvature
1) Find the radius of curvature at the point o

.
S
ol: Clearly, the given equation of curve belongs to Cartesian coordinates.
We k
now that, the radius of curvature
at any point on the curve is given by
No
w, consider
Dif
ferentiate w.r.t
, we get
H
ere,
A
gain, differentiating
, w.r.t , we get
TSEDUHUB
www.tseduhub.blogspot.com

JNTUWORLD
No
w, the Radius of curvature at
is given by
numerically… since radius cannot be negative
2) If be the radius of curvature at the extremities of an chord of the cardioid
which passes through the pole, show that
S
ol: Let us consider the equation of the cardioids to be
L
et us consider
to be the extremities of the chord,
w
hose coordinates are given by
and
L
et
be the radius of curvature at the point and
r
espectively.
Let us find
:
We
know that the radius of curvature for the curve
is given by
No
w, consider
and
H
ence,
TSEDUHUB
www.tseduhub.blogspot.com

JNTUWORLD
S
imilarly,
No
w, consider L.H.S:
i.
e.
.
3
) Show that the radius of curvature at each point of the curve
,
is inversely proportional to the length of the normal intercepted between the
po
int on the curve and the
axis.
S
ol: Clearly, the equation of the curve is in parametric form
Le
t us change the problem of solving radius of curvature in parametric form to problem of
solving radius of curvature in Cartesian form.
H
ere, given
TSEDUHUB
www.tseduhub.blogspot.com

JNTUWORLD
S
imilarly,
No
w,
Ag
ain,
H
ence, the radius of curvature for the equation of the curve
at any point is given
b
y
TSEDUHUB
www.tseduhub.blogspot.com

JNTUWORLD
A
lso, we know that the length of the normal is given by
H
ence, from
we can say that, the radius of curvature at each point of the curve
is inversely proportional to the length of the normal
in
tercepted between the point on the curve and the
axis.
H
ence the result.
CENTRE OF CURVATURE
Definition : The Centre of Curvature at a point ?P? o
f a curvature is the point ?C? which lies
on the Positive direction of the normal at ?P? and is at a distance
(in magnitude) from it.
CI
RCLE OF CURVATURE
Definition:
The circle of curvature at a point ?P? of a curve is the circle whose centre is at
th
e centre of Curvature ?C? and whose radius is
in magnitude
TSEDUHUB
www.tseduhub.blogspot.com

JNTUWORLD
P
roblems on Centre of Curvature and circle of curvatu
re
1) Find the centre of curvature at the point of the curve . Find
a
lso the equation of the circle of curvature at that point.
Sol: We know that, if
are the coordinates of the centre of curvature at any point
on the curve , then
No
w, given
Dif
ferentiate w.r.t
, we get
No
w,
A
gain, Differentiate (1) w.r.t
, we get
H
ence, the coordinates of the centre of curvature are
TSEDUHUB
www.tseduhub.blogspot.com

JNTUWORLD
H
ence,
No
w, radius of curvature at the point
is given by
H
ence, the equation of circle of curvature at the given point
is given by
E
VOLUTE
Corresponding to each point on a curve we can find the curvature of the curve at that point.
Dr
awing the normal at these points, we can find Centre of Curvature corresponding to each
of these points. Since the curvature varies from point to point, centres of curvature also
differ. The totality of all such centres of curvature of a given curve will define another curve
and this curve is called the evolute of the curve.
The Locus of centres of curvature of a given curve is called the evolute of that curve.
The locus of the centre of curvature
of a variable point on a curve is called the evolute
o
f the curve. The curve itself is called involute of the evolute.
TSEDUHUB
www.tseduhub.blogspot.com

JNTUWORLD
H
ere, for different points on the curve, we get different centre of curvatures. The locus of all
these centres of curvature is called as Evolute.
The external curve which satisfies all these centres of curvature is called as Evolute. Here
Evolute is nothing but an curve equation.
To find Evolute, the following models exist.
1)If an equation of the curve is given and If we are asked to show / prove L.H.S=R.H.S,
Then do as follows.
First find Centre of Curvature
, where
A
nd then consider L.H.S: In that directly substitute
in place of and in place of
. Similarly for R.H.S. and then show that L.H.S=R.H.S
2)I
f a curve is given and if we are asked to find the evolute of the given curve, then do
as follows:
First find Centre of curvature
and then re-write as
in terms of and in terms of . and then substitute in the given curve, which
g
ives us the required evolute.
3)If a curve is given, which is in parametric form, then first find Centre of curvature,
which will be in terms of parameter. then using these values of
and eliminate
th
e parameter, which gives us evolute.
TSEDUHUB
www.tseduhub.blogspot.com

JNTUWORLD
P
roblem
1) Find the coordinates of centre of curvature at any point of the parabola
a
nd also show its evolute is given by
S
ol: Given curve is
We
know that, if
are the coordinates of the centre of curvature at any point
o
n the curve
, then
No
w,
A
lso,
C
onsider,
TSEDUHUB
www.tseduhub.blogspot.com

JNTUWORLD
C
onsider,
No
w, required to prove is
L
.H.S
R.
H.S
H
ence,
H
ence the Result.
ENVELOPE
A curve which touches each member of a given family of curves is called envelope of that
f
amily.
Procedure to find envelope for the given family of curves:

C
ase 1: Envelope of one parameter
Le
t us consider
to be the given family of curves.
S
tep 1: Differentiate w.r.t to the parameter partially, and find the value of the parameter
TSEDUHUB
www.tseduhub.blogspot.com

JNTUWORLD
S
tep 2: By Substituting the value of parameter in the given family of curves, we get
required envelope.
Special Case: If the given equation of curve is quadratic in terms of parameter
, then
envelope is given by

Case 2: Envelope of two parameter
Le
t us consider
to be the given family of curves, and a relation connecting these
tw
o parameters
Step 1: Obtain one parameter in terms of other parameter from the given relation
Step 2: Substitute in the given equation of curve, so that the problem of two parameter
converts to problem of one parameter.
Step 3: Use one parameter technique to obtain envelope for the given family of curve
Pr
oblem
1) Find the envelope of the family of straight line is the
pa
rameter.
Sol: Given equation of family of curves is
S
tep 1: Differentiate partially w.r.t the parameter
S
tep 2: Substitute the value of
in the given family of curves
TSEDUHUB
www.tseduhub.blogspot.com

JNTUWORLD
The envelope of the given family of straight lines is an ellipse.
2
) Find the envelope of family of straight line
, where are two
pa
rameters which are connected by the relation
.
S
ol: Given equation of family of straight lines is
A
lso given,
S
ubstituting
in , we get
S
tep 1: Differentiate w.r.t
partially, we get
TSEDUHUB
www.tseduhub.blogspot.com

JNTUWORLD
No
w, substitute the value of
in
S
tep 2: Substitute the values of
in the given family of curves , we get
is the required envelope
TSEDUHUB
www.tseduhub.blogspot.com

JNTUWORLD
C
URVE TRACING
Drawing a rough sketch of the curve is called as Curve Tracing.
A
im: To find the appropriate shape of a curve for the given equation.
METHOD OF TRACING CURVES
Cartesian Co-ordinates: In order to obtain general shape of the curve from the given
e
quations, we have to examine the following properties.
1)SYMMETRY
a)If the equation contains even powers of
only, the curve is symmetrical about

Example:
b)I
f the equation contains even powers of
only, the curve is symmetrical about
Ex
ample:
c)I
f all the powers of
and in the given equation are even, the curve is symmetrical
a
bout the both the axes. i.e. about the origin.
Example:
d)I
f the equations of the curve is not changed by interchanging
and , then the curve
is
symmetrical about the line
.
Ex
ample:
e)I
f the equation of a curve remains unchanged when both
and are interchanged by
–and – respectively, then the curve is symmetrical in opposite-quadrants.
Ex
amples:
TSEDUHUB
www.tseduhub.blogspot.com

JNTUWORLD
2)ORIGIN
If the equation of a curve is satisfied by then the curve passes through
th
e origin.
Example:
3)INTERSECTION WITH CO-ORDINATE AXIS
Put in the given equation to get points of intersection with
P
ut
in the given equation to get points of intersection with
4)REGION
If possible write the given equation in the form of Give values of to make
imaginary. Let be imaginary for the values of lying between and .
T
hen, no part of the curve lies between
and . Similarly, the curve does
no
t lies between those values of
for which is imaginary.
5)TANGENTS
a)If the curve passes through the origin, the tangents at the origin are given by
equ
ating the lowest degree terms to zero.
Example: For the curve
, the tangents are given by equating the
l
owest degree terms to zero. i.e.
are tangents at origin.
b)I
f the curve is not passing through the origin, the tangents at any point are given
by finding
at that point and this indicates the direction of the tangent at that
p
oint.
Note: If there are two tangents at the origin, the origin is a double point.
A)If
the two tangents are real and coincident, the origin is a cusp.
B)If the two tangents are real and different, the origin is a node.
TSEDUHUB
www.tseduhub.blogspot.com

JNTUWORLD
C)I
f the two tangents are Imaginary, the origin is a conjugate point or Isolated
point.
6)EXTENSION OF THE CURVE TO INFINITY
Give values to for which the value of is Infinity and also give values to for which
is Infinity. These values indicate the direction in which the curve extends to
I
nfinity.
7)ASYMPTOTES
An Asymptote is a straight line which cuts a curve in two points, at an infinite
d
istance from the origin and yet is not itself wholly at Infinity.
To find Asymptotes
a)Asymptotes parallel to axis
Asymptotes parallel to -axis are obtained by equating the coefficient
o
f the highest power of
to zero.
Asymptotes parallel to -axis are obtained by equating the coefficient
o
f the highest power of
to zero.
b)Oblique Asymptotes: ( A

Let
be an asymptote. Put in the given equation of the curve.
E
quate the coefficients of highest powers of
to zero and solve for and .
TSEDUHUB
www.tseduhub.blogspot.com

JNTUWORLD
P
roblems on Curve Tracing - Cartesian Coordinates
1
) Trace the curve
S
ol: In order to trace a curve, we need to check the following properties
Sy
mmetry
Origin
Intersection with coordinate axis
Region
Tangents
Intersection of curve to
A
symptotes
Let us consider given equation of curve to be
S
ymmetry:
If we interchange and , the equation of
cu
rve is not changing.
Hence, the curve is symmetric about the line
O
rigin:
If we substitute and , the equation of
cu
rve is satisfied.
Hence, we can say that the curve passes through the
origin.
I
ntersection with coordinate axis:

P
ut
The curve meets the axis only at the Origin
A
gain
The curve meets the axis only at the Origin
Re
gion:
Consider
is positive is also positive
The curve lies in 1
st
Q
uadrant
is negative is positive
The curve lies in 2
nd
Q
uadrant
is negative & is negative
TSEDUHUB
www.tseduhub.blogspot.com

JNTUWORLD
T
hen the equation of curve is not satisfied
The curve does not lie in 3
rd
Q
uadrant
is positive is negative
The curve lies in 4
th
Q
uadrant
Tan
gents:
The given equation of curve passes through
O
rigin.
Hence, the tangents at origin are given by equating lowest degree terms to zero.
i.e.,
There exists two tangents namely,
H
ere, the two tangents at origin are real and distinct. Hence, we get a node
Now, let us check, at what points the curve meets the line
H
ence, the curve meets at
Ex
tension of the curve to Infinity:

A
s
The curve extends to infinity in the second quadrant.
A
lso, as
T
he curve extends to infinity in the fourth quadrant.
Asymptotes:

TSEDUHUB
www.tseduhub.blogspot.com

JNTUWORLD
A
symptotes parallel to axis:
No Asymptote is parallel to
No
Asymptote is parallel to
A
symptotes not parallel to axis (Oblique Asymptotes)
Let us consider
to be the required asymptote.
No
w, substitute in the given equation of the curve and solving for
and , we get
H
ence, the required asymptote is
H
ence the shape of the curve is as follows
TSEDUHUB
www.tseduhub.blogspot.com

JNTUWORLD
P
ROCEDURE FOR TRACING CURVES IN POLAR CO-ORDINATES
1)S
YMMETRY
i)If the equation does not alter by changing
to –, the curve is symmetrical about the
in
itial line or
-axis.
Example:
ii)I
f the equation does not alter by changing
to –, the curve is symmetrical about the
p
ole.
Example:
iii)I
f the equation of the curve remains unaltered when
is changed to or by changing
to –to –, the curve is symmetric about the line or -axis.

Example:
iv) I
f the equation of the curve remains unaltered when
is changed to , then the curve
is
symmetrical about the line
Example:
v)I
f the equation of the curve does not alter by changing
to then the curve is
s
ymmetrical about the line
Example: .
2)Discussion for and
G
ive certain values to
and find the corresponding values of and then plot the points.
S
ometimes it is inconvenient to find the corresponding values of
for certain values of . In
s
uch cases, a particular region for
may be considered and find out whether increases or
d
ecreases in that region.
For example: It is inconvenient to find the value of
for but it is equal to know
w
hether
increases or decreases in the region from to in which value is also
in
cluded.
3)Region
No part of the curve exists for those values ofwhich make corresponding value of
im
aginary.
TSEDUHUB
www.tseduhub.blogspot.com

JNTUWORLD
4)Tangents
Find , w

is the angle between the radius vector and the tangent at
. It will indicate the direction of the tangents at any point .
5)Asymptotes
Find the value of
which makes infinity. The curve has an asymptote in that direction.
TSEDUHUB
www.tseduhub.blogspot.com

JNTUWORLD
B
y
Y. Prabhaker Reddy
Asst.
Professor of Mathematics
Guru Nanak Engineering College
Ibrahimpatnam, Hyderabad.
MATHEMATICS-I
APPLICATIONS OF INTEGRATION
I YEAR B.Tech
TSEDUHUB
www.tseduhub.blogspot.com

JNTUWORLD
U
NIT-4
APPLICATIONS OF INTEGRATION
Riemann Integrals:
L
et us consider an interval
with
I
f
, then a finite set is
ca
lled as a partition of
and it is denoted by .
T
he sub intervals
are called segments (or)
sub intervals.
T
he
sub interval in this process is and its length is given by
No
te: For every interval
, it

Norm (or) Mesh of the partition:
The maximum of the lengths of the sub intervals
w.
r.t the partition
is called as Norm of the partition (or) Mesh of the partition and
it
is denoted by
Re
finement: If
and are two partitions of and if , then is called as
Re
finement of
.
L
ower and Upper Reimann Sum’s
Let is bounded and be a partition on ,
th
en
sub interval is given by and its length is given by
I
f
is bounded on , then is bounded on
l
et
and be Infimum and supremum of on , then
T
he sum
is called as lower Reimann sum and it is denoted by
T
he sum
is called as upper Reimann sum and it is denoted by
No
te: Always,
TSEDUHUB
www.tseduhub.blogspot.com

JNTUWORLD
P
roblem
1) If and be a partition of then compute
S
ol: Given
defined on and be a partition of
H
ere, let
A
nd,
L
et
and be Infimum and supremum of on , then
H
ence,
A
lso,
L
ower Reimann Integral: Let
be a bounded function and is a partition
o
f
, then supremum of is called as Lower Reimann integral on
and it is denoted by
U
pper Reimann Integral: Let
be a bounded function and is a partition
o
f
, then Infimum of is called as Upper Reimann integral on
and

Riemann Integral
If b

is a partition of and if
, then is said to be Riemann integrable on and it is denoted by
TSEDUHUB
www.tseduhub.blogspot.com

JNTUWORLD
Rec
tification: The process of finding the length of the arc of the curve is called as
Re
ctification
Length of the arc of the curve
Equation of the curve Arc Length
Cartesian Form
(i)
and
and
(ii)
and
and
Parametric Form
and
and
Polar Form
(i)
and
and
(ii)
and
and
P
roblems on length of the arc of the curve
1) Find the length of the arc of the curve f

to .
S
olution: We know that, the equation of the length of the arc of the curve
b
etween
and is g

G
iven
TSEDUHUB
www.tseduhub.blogspot.com

JNTUWORLD
T

2
) Find the perimeter of the loop of the curve
S
olution: We know that, the equation of the length of the arc of the curve
b
etween
and is given by
G
iven
TSEDUHUB
www.tseduhub.blogspot.com

JNTUWORLD
No
w,
H
ere the curve is symmetrical about the
. Hence the length of the arc will be
d
ouble that of the arc of the loop about the
.
The required length of the loop is =
TSEDUHUB
www.tseduhub.blogspot.com

JNTUWORLD
3) F
ind the perimeter of the cardioids
.
S
olution: We know that the length of the arc of the curve
and is
g
iven by
G
iven
T
he cardioids is symmetrical about the initial line and passes through the pole.
Hence the length of the arc will be double that of the arc of the loop about the
.
The required length of the loop is =
TSEDUHUB
www.tseduhub.blogspot.com

JNTUWORLD
4
)
TSEDUHUB
www.tseduhub.blogspot.com

JNTUWORLD
V
olume of solid of Revolution
Region (R) Axis
Volume of the solid
generated
Cartesian form
(i)
, the
and the lines
and
(ii)
, the
and the lines
and
(iii)
(iv)
Parametric form
(i)
(ii)
Polar form
(i)
The initial
line
(ii)
The line
(iii)
The line
P
roblems on Volume of solid of Revolution
1) Find the volume of the solid that result when the region enclosed by the curve
is revolved about the .
S
ol: We know that the volume of the solid generated by the revolution of the are
a
bounded by the curve
, the and the lines is given by
No
w, given curve
Required volume is given by
TSEDUHUB
www.tseduhub.blogspot.com

JNTUWORLD
2
) Find the volume of the solid generated by the revolution of the cardioids
about the initial line.
S
ol: We know that the volume of the solid generated by the revolution of the area
b
ounded by the curve
, the initial line and is given by
H
ere, the given cardioids is symmetrical about the initial line. The upper half of the
curve formed when
varies from to .
TSEDUHUB
www.tseduhub.blogspot.com

JNTUWORLD
S
urface area of solid of Revolution
Equation of the curve Arc Length
Cartesian Form
(i)
and
and
,
(ii)
and
and
,
Parametric Form
(i)
and
and
(ii)
and
and
Polar Form
(i)
and
and
(ii)
and
and
P
roblem
1) Find the surface area generated by the revolution of an arc of the catenary
about the
S
ol: We know that the surface area of the solid generated by the revolution of an arc
about the , is given by
TSEDUHUB
www.tseduhub.blogspot.com

JNTUWORLD
2
) Find the surface area of the solid formed by revolving the cardioid
a
bout the initial line.
Sol: We know that the surface area of the solid formed by revolving
the cardioid
, the initial line and is given by
G
iven
T
he cardioid is symmetrical about the initial line and passes through the pole.
Hence, required surface area is given by
TSEDUHUB
www.tseduhub.blogspot.com

JNTUWORLD
Let

L
ower Limit:
U
pper Limit:
Surface area –
C
hange of variables in Double Integral
Problem:
S
olution: Clearly, given coordinates are in Cartesian.
TSEDUHUB
www.tseduhub.blogspot.com

JNTUWORLD
2
)
S
olution:
TSEDUHUB
www.tseduhub.blogspot.com

JNTUWORLD
TSEDUHUB
www.tseduhub.blogspot.com

JNTUWORLD
C
hange of Order of Integration
Problem 1:
TSEDUHUB
www.tseduhub.blogspot.com

JNTUWORLD
2)
3)
TSEDUHUB
www.tseduhub.blogspot.com

JNTUWORLD
4)
TSEDUHUB
www.tseduhub.blogspot.com

JNTUWORLD
TSEDUHUB
www.tseduhub.blogspot.com

JNTUWORLD
By
Y. Prabhaker Reddy
Asst
. Professor of Mathematics
Guru Nanak Engineering College
Ibrahimpatnam, Hyderabad.
MATHEMATICS-I
DIFFERENTIAL EQUATIONS-I
I YEAR B.TECH
TSEDUHUB
www.tseduhub.blogspot.com

JNTUWORLD
SYL
LABUS OF MATHEMATICS-I (AS
PER JNTU HYD)
Name of the Unit Name of the Topic
Unit-I
Sequences and Series
1.1 Basic definition of sequences and series
1.2 Convergence and divergence.
1.3 Ratio test
1.4 Comparison test
1.5 Integral test
1.6 Cauchy?s root test
1.7 Raabe?s test
1.8 Absolute and conditional convergence
Unit-II
Functions of single variable
t.s Rolle?s theorem
t.t Lagrange?s Mean value theorem
t.u Cauchy?s Mean value theorem
2.4 Generalized mean value theorems
2.5 Functions of several variables
2.6 Functional dependence, Jacobian
2.7 Maxima and minima of function of two variables
Unit-III
Application of single variables
3.1 Radius , centre and Circle of curvature
3.2 Evolutes and Envelopes
3.3 Curve Tracing-Cartesian Co-ordinates
3.4 Curve Tracing-Polar Co-ordinates
3.5 Curve Tracing-Parametric Curves
Unit-IV
Integration and its
applications
4.1 Riemann Sum
4.3 Integral representation for lengths
4.4 Integral representation for Areas
4.5 Integral representation for Volumes
4.6 Surface areas in Cartesian and Polar co-ordinates
4.7 Multiple integrals-double and triple
4.8 Change of order of integration
4.9 Change of variable
Unit-V
Differential equations of first
order and their applications
5.1 Overview of differential equations
5.2 Exact and non exact differential equations
5.3 Linear differential equations
5.4 Bernoulli D.E
w.w Newton?s Law of cooling
5.6 Law of Natural growth and decay
5.7 Orthogonal trajectories and applications
Unit-VI
Higher order Linear D.E and
their applications
6.1 Linear D.E of second and higher order with constant coefficients
6.2 R.H.S term of the form exp(ax)
6.3 R.H.S term of the form sin ax and cos ax
6.4 R.H.S term of the form exp(ax) v(x)
6.5 R.H.S term of the form exp(ax) v(x)
6.6 Method of variation of parameters
6.7 Applications on bending of beams, Electrical circuits and simple harmonic motion
Unit-VII
Laplace Transformations
7.1 LT of standard functions
7.2 Inverse LT – first shifting property
7.3 Transformations of derivatives and integrals
7.4 Unit step function, Second shifting theorem
7.5 Convolution theorem-periodic function
7.6 Differentiation and integration of transforms
7.7 Application of laplace transforms to ODE
Unit-VIII
Vector Calculus
8.1 Gradient, Divergence, curl
8.2 Laplacian and second order operators
8.3 Line, surface , volume integrals
8.v Green?s Theorem and applications
8.5 Gauss Divergence Theorem and applications
8.x Stoke?s Theorem and applications
TSEDUHUB
www.tseduhub.blogspot.com

JNTUWORLD
CONTE
NTS

U
NIT-5
Di
fferential Equations-I
Overview of differential equations
Ex
act and non exact differential equations
Linear differential equations
Bernoulli D.E
Orthogonal Trajectories and applications
Newton’s Law of cooling
Law of Natural growth and decay
TSEDUHUB
www.tseduhub.blogspot.com

JNTUWORLD
D
IFFERENTIAL EQUATIONS
Differentiation: The rate of change of a variable w.r.t the other variable is called as a
Dif
ferentiation.
In this case, changing variable is called Dependent variable and other variable is called as an
Independent variable.
Example:
is a Differentiation, Here is dependent variable and is Independent variable.
DI
FFERENTIAL EQUATION: An equation which contains differential coefficients is called as a D.E.
E
xamples: 1)
2)++1=0
.
Differential Equations are separated into two types
Ordinary D.E: I
n a D.E if there exists single Independent variable, it is called as Ordinary D.E
Ex
ample: 1)
+ is a Ordinary D.E 2
)
+ is a Ordinary D.E
Pa
rtial D.E: In a D.E if there exists more than one Independent variables then it is called as Partial
D.E
Ex
ample: 1)
++1=0

are two Independent variables.

2)
+ 1 =0 is a Partial D.E, since are two Independent variables
ORDER
OF D.E: The highest derivative in the D.E is called as Order of the D.E
Ex
ample: 1) Order of
+ 2 is one.
2) O
rder of
is Five.
DE
GREE OF D.E: The Integral power of highest derivative in the D.E is called as degree of the D.E
Exam
ple: 1) The degree of
is One.
2) The degree of
is Two.
NO
TE: Degree of the D.E does not exist when the Differential Co-efficient Involving with
exponential functions, logarithmic functions, and Trigonometric functions.
Example: 1) There is no degree for the D.E
2) T
here is no degree for the D.E
3) There is no degree for the D.E .
Differential Equations
Ordinary D.E Partial D.E
TSEDUHUB
www.tseduhub.blogspot.com

JNTUWORLD
N
OTE: 1) The degree of the D.E is always a +ve Integer, but it never be a negative (or) zero (or) fraction.
2) Dependent variable should not include fraction powers. It should be perfectly Linear.

Ex:

Degree does not exist.
F
ORMATION OF DIFFERNTIAL EQUATION
A D.

Differentiation Concept. If the given equation contains ?n? arbitrary constants then differentiating
it ?n? times successively and eliminating ?n? arbitrary constants we get the corresponding D.E.
NOTE: If the given D.E contains ?n? arbitrary constants , then the order of its corresponding D.E is ?n? .
NOTE: For
then the corresponding D.E is given by
In
general, if
then its D.E is given by
, where
Special
Cases: If
then D.E is given by where .
In
general, if
then the corresponding D.E is given by
N
OTE: For
then D.E is .
WR
ANSKIAN METHOD
Let
be the given equation then its corresponding D.E is given by
D
IFFERENTIAL EQUATIONS OF FIRST ORDER AND FIRST DEGREE
A D.E of the form is called as a First Order and First Degree D.E in terms of
d
ependent variable
and independent variable .
I
n order to solve above type of Equation?s, following methods exists.
1)Variable Separable Method.
2)Homogeneous D.E and Equations reducible to Homogeneous.
3)Exact D.E and Equations made to exact.
4)Linear D.E and Bernoulli?s Equations.
This method is applicable when there
are two arbitrary constants only.
TSEDUHUB
www.tseduhub.blogspot.com

JNTUWORLD
M
ethod-I:
VARIABLE SEPERABLE METHOD
First Form: Let us consider given D.E
If then proceed as follows
is the required general solution.
Second F
orm:
If then proceed as follows
Let
By
using variable separable method we can find its general solution.
Let it be
. But
.
M
ethod-2:
HOMOGENEOUS DIFFERENTIAL EQUATION METHOD
Homogeneous Function:
A Function
is said to be homogeneous function of degree ?n? if
Ex
ample: 1) If
is a homogeneous function of degree ?r?
2) If
is not a homogeneous function.
H
omogeneous D.E: A D.E of the form
is said to be Homogeneous D.E of first order
an
d first degree in terms of dependent variable ?y? and independent variable ?x? if
is a
hom
ogeneous function of degree ?r?.
Ex: 1)
is a homogeneous D.E
2)
is a homogeneous D.E
3)
is a homogeneous D.E
4
)
is n

5)
is not a homogeneous D.E

Working Rule: Let us consider given homogeneous D.E
Substituting we get
By
using variable separable method we can find the General solution of it
Let it be
. But
be the required general solution.
TSEDUHUB
www.tseduhub.blogspot.com

JNTUWORLD
N
ON-HOMOGENEOUS DIFFERENTIAL EQUATION
A D.E of the form is called as a Non-Homogeneous D.E in terms of independent
v
ariable
and dependent variable , where are real constants.
C
ase (I): If
then procedure is as follows
Let us choose constants h & k in such a way that
Let and also then above relation becomes
(From I)

Which is a Homogeneous D.E of first order and of first degree in terms of
and .

By using Homogeneous method, we can find the General solution of it. Let it be
.
Bu
t
,
is the

Case (II): If , then By Using Second form of Variable Seperable method we can
f
ind the General Solution of the given equation.
Method-3:
EXACT DIFFERENTIAL EQUATION
A D.E of the form is said to be exact D.E if .
I
ts general solution is given by
N
ON EXACT DIFFERENTIAL EQUATION
A D.E of the form is said to be Non-Exact D.E if
I
n order to make above D.E to be Exact we have to multiply with
which is known as
an

Integrating Factor.
I
TSEDUHUB
www.tseduhub.blogspot.com

JNTUWORLD
To Solve such a type of problems, we have following methods.
1)I
nspection Method
2)Method to find Integrating Factor I.F
3)M
ethod to find Integrating Factor I.F
4)M
ethod to find Integrating Factor I.F
5)M
ethod to find Integrating Factor I.F
M
ethod 1: INSPECTION METHOD
Some Formulae:
Hi
nts while solving the problems using Inspection Method
I

term is there then select another term.
A

combination with .
M
ethod-2:
Method to find Integrating Factor
If given D.E is is Non-Exact and it is Homogeneous and also
T
hen
is t

Method-3:
Method to find Integrating Factor
Let the given D.E is is Non

A

then is an I.F
: H

there shou

combination (with constants also)
i.
e. With same powers
.
This Method is used for both
exact and non-exact D.E
99% of the problems
can be solved using
Inspection method
TSEDUHUB
www.tseduhub.blogspot.com

JNTUWORLD
M
ethod-4:
Method to find the I.F
Let be the Non-Exact D.E. If
Wh
ere
= Function of x-alone or constant then I.F is
N
OTE: In this case number of terms in M is greater than or equal to number of terms in N i.e. M≥N
Method-5: Method to find the I.F
Let be the Non-Exact D.E. If
Wh
ere
=Function of y-alone or constant then I.F is
N
OTE: )n this case number of terms in N is greater than or equal to number of items in M i.e. N≥M
LINEAR DIFFERENTIAL EQUATION
A D.

is called as a First order and First degree D.E in terms of
d
ependent variable
and independent variable where functions of x-alone (or)
consta
nt.
Working Rule: Given that
-------- (1)
I.F is given by
M
ultiplying with this I.F to (1), it becomes
Now Integrating both sides we get
is the required General Solution.
ANO
THER FORM
A D.E of the form
is also called as a Linear D.E where functions of
y
-alone. Now I.F in this case is given by I.F=
and General Solution is given by
E
quations Reducible to Linear Form
An Equation of the form is called as an Equation Reducible to Linear
F
orm
TSEDUHUB
www.tseduhub.blogspot.com

JNTUWORLD
Worki
ng Rule:
Given that
( I )
L
et
( I
)
which is Linear D.E in terms of
By
using Linear Method we can find its General Solution.
Let it be
But
is the required solution
B
ERNOULLIS EQUATION
A D.E of the form is called as Bernoulli?s equation in terms of dependent
v
ariable y and independent variable x. where P and Q are functions of x-alone (or) constant.
Working Rule:
Giv
en that
1
2
Let
Dif
ferentiating with respect to x, we get
(From 2), which is a Linear D.E in terms of
By
using Linear Method we can find general solution of it.
Let it be
which is general solution of the given equation.
O
rthogonal Trajectories
Trajectory: A Curve which cuts given family of curves according to some special law is called as a
Tr
ajectory.
Orthogonal Trajectory: A Curve which cuts every member of given family of curves at
is
called
as an Orthogonal Trajectory.
Hint: First make coefficient
as 1, and then make R.H.S term
purely function of alone
TSEDUHUB
www.tseduhub.blogspot.com

JNTUWORLD
O
rthogonal Trajectory in Cartesian Co-ordinates
Let be given family of curves in Cartesian Co-ordinates.
Dif
ferentiating it w.r.t
, we get .
S
ubstituting
, we get . By using previous

methods we can find general solution of it. Let it be
,
w
hich is Orthogonal Trajectory of the given family of curves.
Orthogonal Trajectory in Polar Co-ordinates
Let be given family of curves in Polar Co-ordinates.
Dif
ferentiating it w.r.t
, we get .
S
ubstituting
, we get . By using previous

methods we can find general solution of it. Let it be
, which is Orthogonal Trajectory
of
the given family of curves.
Self Orthogonal: If the Orthogonal Trajectory of given family of Curves is family of curves itself
then it
is called as Self Orthogonal.
Mutual Orthogonal: Given family of curves
are said to be Mutually
Or
thogonal if Orthogonal Trajectory of one given family of curves is other given family of curves.
NEWTON?S LAW OF COOL)NG
Statement: The rate of the temperature of a body is proportional to the difference of the
te
mperature of the body and that of the surrounding medium.
Let
be the temperature of the body at the time and be the temperature of its surrounding
med
ium air. By the Newton?s Law of cooling, we have
,

is a positive constant
For O.T,
because, two lines are
if product of
.
TSEDUHUB
www.tseduhub.blogspot.com

JNTUWORLD
Integrating, we get
P
roblem
A body is originally at and cools down to in 20 minutes. If the temperature of the air is
, find the temperature of the body after 40 minutes.
Sol: Let be the temperature of the body at a time
We
know that from Newton?s Law of cooling
,

is a positive constant
Giv
en temperature of the air
I
ntegrating, we get
I
Now
, given at
I
S
ubstituting this value of
in I

II
A
gain, given at

II
S
ubstituting this value of
in II we get
III
0
20
40
TSEDUHUB
www.tseduhub.blogspot.com

JNTUWORLD
Again, when

III
At
L
AW OF NATURAL GROWTH (Or) DECAY
If be the amount of substance at time , then the rate of change of amount of a
ch
emically changing substance is proportional to the amount of the substance available at that
time.
w
here
is a proportionality constant.
Note:
If as
increases, increases we can take , and if as increases, decreases
w
e can take
R
ATE OF DECAY OF RADIOACTIVE MATERIALS
If is the amount of the material at any time , then , where is any constant.
TSEDUHUB
www.tseduhub.blogspot.com

JNTUWORLD
By
Y. Prabhaker Reddy
Asst
. Professor of Mathematics
Guru Nanak Engineering College
Ibrahimpatnam, Hyderabad.
MATHEMATICS-I
DIFFERENTIAL EQUATIONS-II
I YEAR B.TECH
TSEDUHUB
www.tseduhub.blogspot.com

JNTUWORLD
SYL
LABUS OF MATHEMATICS-I (AS
PER JNTU HYD)
Name of the Unit Name of the Topic
Unit-I
Sequences and Series
1.1 Basic definition of sequences and series
1.2 Convergence and divergence.
1.3 Ratio test
1.4 Comparison test
1.5 Integral test
1.6 Cauchy’s root test
1.7 Raabe’s test
1.8 Absolute and conditional convergence
Unit-II
Functions of single variable
t.s Rolle’s theorem
t.t Lagrange’s Mean value theorem
t.u Cauchy’s Mean value theorem
2.4 Generalized mean value theorems
2.5 Functions of several variables
2.6 Functional dependence, Jacobian
2.7 Maxima and minima of function of two variables
Unit-III
Application of single variables
3.1 Radius , centre and Circle of curvature
3.2 Evolutes and Envelopes
3.3 Curve Tracing-Cartesian Co-ordinates
3.4 Curve Tracing-Polar Co-ordinates
3.5 Curve Tracing-Parametric Curves
Unit-IV
Integration and its
applications
4.1 Riemann Sum
4.3 Integral representation for lengths
4.4 Integral representation for Areas
4.5 Integral representation for Volumes
4.6 Surface areas in Cartesian and Polar co-ordinates
4.7 Multiple integrals-double and triple
4.8 Change of order of integration
4.9 Change of variable
Unit-V
Differential equations of first
order and their applications
5.1 Overview of differential equations
5.2 Exact and non exact differential equations
5.3 Linear differential equations
5.4 Bernoulli D.E
w.w Newton’s Law of cooling
5.6 Law of Natural growth and decay
5.7 Orthogonal trajectories and applications
Unit-VI
Higher order Linear D.E and
their applications
6.1 Linear D.E of second and higher order with constant coefficients
6.2 R.H.S term of the form exp(ax)
6.3 R.H.S term of the form sin ax and cos ax
6.4 R.H.S term of the form exp(ax) v(x)
6.5 R.H.S term of the form exp(ax) v(x)
6.6 Method of variation of parameters
6.7 Applications on bending of beams, Electrical circuits and simple harmonic motion
Unit-VII
Laplace Transformations
7.1 LT of standard functions
7.2 Inverse LT – first shifting property
7.3 Transformations of derivatives and integrals
7.4 Unit step function, Second shifting theorem
7.5 Convolution theorem-periodic function
7.6 Differentiation and integration of transforms
7.7 Application of laplace transforms to ODE
Unit-VIII
Vector Calculus
8.1 Gradient, Divergence, curl
8.2 Laplacian and second order operators
8.3 Line, surface , volume integrals
8.v Green’s Theorem and applications
8.5 Gauss Divergence Theorem and applications
8.x Stoke’s Theorem and applications
TSEDUHUB
www.tseduhub.blogspot.com

JNTUWORLD
CONTE
NTS

U
NIT-6
Di
fferential Equations-II
Linear D.E of second and higher order with constant coefficients
R.
H.S term of the form exp(ax)
R.H.S term of the form sin ax and cos ax
R.H.S term of the form exp(ax) v(x)
Method of variation of parameters
TSEDUHUB
www.tseduhub.blogspot.com

JNTUWORLD
L
INEAR DIFFERENTIAL EQUATIONS OF SECOND AND HIGHER OR
DER
A D.E of the form is called as a Linear
Dif
ferential Equation of order
with constant coefficients, where are Real constants.
L
et us denote
, then above equation becomes
which is in the form of , where
.
T
he General Solution of the above equation is

(or
)

Now
, to find Complementary Function
, we have to find Auxillary Equation
Aux
illary Equation: An equation of the form
is called as an Auxillary Equation.
S
ince
is a polynomial equation, by solving this we get roots. Depending upon these
r
oots we will solve further.
Complimentary Function: The General Solution of
is called as Complimentary
F
unction and it is denoted by

Dep
ending upon the Nature of roots of an Auxillary equation we can define
C
ase I:
If the Roots of the A.E are real and distinct, then proceed as follows
I
f
are two roots which are real and distinct (different) then complementary function is
g
iven by
G
eneralized condition: If
are real and distinct roots of an A.E then
C
ase II:
If the roots of A.E are real and equal then proceed as follows
I
f
then
G
eneralized condition: If
then
C
ase III: If roots of A.E are Complex conjugate i.e.
then
(Or)
(Or)
C.F= Complementary Function
P.I= Particular Function
TSEDUHUB
www.tseduhub.blogspot.com

JNTUWORLD
Not
e: For repeated Complex roots say,
C
ase IV: If roots of A.E are in the form of Surds i.e.
, where is not a perfect square
then
,
(Or)
(Or)
Not
e: For repeated roots of surds say,
P
articular Integral
The evaluation of is called as Particular Integral and it is denoted by
i.
e.
Not
e: The General Solution of
is called as Particular Integral and it is denoted by
M
ethods to find Particular Integral
Me
thod 1: Method to find P.I of
where , where is a constant.
We know that
if
if
Dep
ending upon the nature of
we can proceed further.
Not
e: while solving the problems of the type
, where Denominator =0, Rewrite the
Denomina
tor quantity as product of factors, and then keep aside the factor which troubles us.
I.e the term which makes the denominator quantity zero, and then solve the remaining quantity.
finally substitute
in place of .
Taking outside the operator by replacing with
Directly substitute in place of
TSEDUHUB
www.tseduhub.blogspot.com

JNTUWORLD
M
ethod 2:
Method to find P.I of where , a is constant
We know that
L
et us consider
, then the above equation becomes
Now
Substitute
if
I
f
then i.e.
T
hen
respectively.
M
ethod 3: Method to find P.I of
where
We know that
Now
taking Lowest degree term as common in
, above relation becomes
E
xpanding this relation upto
derivative by using Binomial expansion and hence get
Impo
rtant Formulae:
1)
2)

3)
4)
5)
6)
M
ethod 4: Method to find P.I of
where , where is a function
o
f
and is constant
We know that
I
n such cases, first take
term outside the operator, by substituting in place of .
Dep
ending upon the nature of
we will solve further.
TSEDUHUB
www.tseduhub.blogspot.com

JNTUWORLD
M
ethod 5: Method to find P.I of
where , where , is any
f
unction of
( i.e. )
We know that
C
ase I: Let
, then
C
ase II: Let
and
We k
now that
By
using previous methods we will solve
further
Finally substitute
L
et
and
We k
now that
By
using previous methods we will solve
further
Finally substitute
Ge
neral Method
To find P.I of where is a function of
We k
now that
L
et
then
S
imilarly,
then
Note:
The above method is used for the problems of the following type
TSEDUHUB
www.tseduhub.blogspot.com

JNTUWORLD
C
auchy’s Linear Equations or (omogeneous Linear Equ
ations
A Differential Equation of the form where
is called as or

and
in
dependent variable
, where are Real constants and .
S
ubstitute
and
T
hen above relation becomes
, w

using previous methods, we can find Complementary Function and Particular Integral of it, and
hence by replacing
with we get the required General Solution of Cauchy’s Linear Equation.
L
egendre’s Linear Equation
An D.E of the form is
called
as Legendre’s Linear Equation of order , where
are Real constants.
Now
substituting,
T
hen, above relation becomes
which is a Linear D.E with constant coefficients. By
using prev
ious methods we can find general solution of it and hence substituting
w
e get the general solution of Legendre’s Linear Equation.
Method of Variation of Parameters
To find the general solution of
L
et us consider given D.E
( I )
L
et the Complementary Function of above equation is
L
et the Particular Integral of it is given by
, where
TSEDUHUB
www.tseduhub.blogspot.com

JNTUWORLD
MATHEMATICS-I
LAPLACE TRANSFORMS
I YEAR B.Tech
By
Y. Prabhaker Reddy
Asst
. Professor of Mathematics
Guru Nanak Engineering College
Ibrahimpatnam, Hyderabad.
TSEDUHUB
www.tseduhub.blogspot.com

JNTUWORLD
S
YLLABUS OF MATHEMATICS-I ( AS
PER JNTU HYD)
Name of the Unit Name of the Topic
Unit-I
Sequences and Series
1.1 Basic definition of sequences and series
1.2 Convergence and divergence.
1.3 Ratio test
1.4 Comparison test
1.5 Integral test
1.6 Cauchy’s root test
1.7 Raabe’s test
1.8 Absolute and conditional convergence
Unit-II
Functions of single variable
2.1 Rolle’s theorem
2.2 Lagrange’s Mean value theorem
2.3 Cauchy’s Mean value theorem
2.4 Generalized mean value theorems
2.5 Functions of several variables
2.6 Functional dependence, Jacobian
2.7 Maxima and minima of function of two variables
Unit-III
Application of single variables
3.1 Radius , centre and Circle of curvature
3.2 Evolutes and Envelopes
3.3 Curve Tracing-Cartesian Co-ordinates
3.4 Curve Tracing-Polar Co-ordinates
3.5 Curve Tracing-Parametric Curves
Unit-IV
Integration and its applications
4.1 Riemann Sum
4.3 Integral representation for lengths
4.4 Integral representation for Areas
4.5 Integral representation for Volumes
4.6 Surface areas in Cartesian and Polar co-ordinates
4.7 Multiple integrals-double and triple
4.8 Change of order of integration
4.9 Change of variable
Unit-V
Differential equations of first
order and their applications
5.1 Overview of differential equations
5.2 Exact and non exact differential equations
5.3 Linear differential equations
5.4 Bernoulli D.E
5.5 Newton’s Law of cooling
5.6 Law of Natural growth and decay
5.7 Orthogonal trajectories and applications
Unit-VI
Higher order Linear D.E and
their applications
6.1 Linear D.E of second and higher order with constant coefficients
6.2 R.H.S term of the form exp(ax)
6.3 R.H.S term of the form sin ax and cos ax
6.4 R.H.S term of the form exp(ax) v(x)
6.5 R.H.S term of the form exp(ax) v(x)
6.6 Method of variation of parameters
6.7 Applications on bending of beams, Electrical circuits and simple harmonic motion
Unit-VII
Laplace Transformations
7.1 LT of standard functions
7.2 Inverse LT –first shifting property
7.3 Transformations of derivatives and integrals
7.4 Unit step function, Second shifting theorem
7.5 Convolution theorem-periodic function
7.6 Differentiation and integration of transforms
7.7 Application of laplace transforms to ODE
Unit-VIII
Vector Calculus
8.1 Gradient, Divergence, curl
8.2 Laplacian and second order operators
8.3 Line, surface , volume integrals
8.4 Green’s Theorem and applications
8.5 Gauss Divergence Theorem and applications
8.6 Stoke’s Theorem and applications
TSEDUHUB
www.tseduhub.blogspot.com

JNTUWORLD
CONTE
NTS

U
NIT-7
LA
PLACE TRANSFORMS
La
place Transforms of standard functions
Inverse LT- First shifting Property
Transformations of derivatives and integrals
Unit step function, second shifting theorem
Convolution theorem – Periodic function
Differentiation and Integration of transforms
Application of Laplace Transforms to ODE
TSEDUHUB
www.tseduhub.blogspot.com

JNTUWORLD
L
APLACE TRANSFORMATION
INTRODUCTION
L
aplace Transformations were introduced by Pierre Simmon Marquis De Laplace (1749-1827), a
Fr
ench Mathematician known as a Newton of French.
Laplace Transformations is a powerful Technique; it replaces operations of calculus by
operations of Algebra.
Suppose an Ordinary (or) Partial Differential Equation together with Initial conditions is reduced
to a problem of solving an Algebraic Equation.
Definition of Laplace Transformation: Let
be a given function defined for all , then
the L
aplace Transformation of
is defined as
H
ere,
is called Laplace Transform Operator. The function is known as determining
f
unction, depends on
. The new function which is to be determined (i.e. F) is called
g
enerating function, depends on
.
H
ere
NO
TE: Here Question will be in
and Answer will be in .
La
place Transformation is useful since
Particular Solution is obtained without first determining the general solution
Non-Homogeneous Equations are solved without obtaining the complementary Integral
Solutions of Mechanical (or) Electrical problems involving discontinuous force functions
(R.H.S function
) (or) Periodic functions other than and are obtained easily.
T
he Laplace Transformation is a very powerful technique, that it replaces operations of
calculus by operations of algebra. For e.g. With the application of L.T to an Initial value
problem, consisting of an Ordinary( or Partial ) differential equation (O.D.E) together with
Initial conditions is reduced to a problem of solving an algebraic equation ( with any given
Initial conditions automatically taken care )
APPLICATIONS
Laplace Transformation is very useful in obtaining solution of Linear D.E?s, both Ordinary and
Pa
rtial, Solution of system of simultaneous D.E?s, Solutions of )ntegral equations, solutions of
Linear Difference equations and in the evaluation of definite Integral.
TSEDUHUB
www.tseduhub.blogspot.com

JNTUWORLD
T
hus, Laplace Transformation transforms one class of complicated functions
to
p
roduce another class of simpler functions
.
A
DVANTAGES
Wit

without the necessity of first determining general solution and then obtaining the particular
solution (by substitution of Initial Conditions).
L.T solves non-homogeneous D.E without the necessity of first solving the corresponding
homogeneous D.E.
L.T is applicable not only to continuous functions but also to piece-wise continuous
functions, complicated periodic functions, step functions, Impulse functions.
L.T of various functions are readily available.
The symbol ?? denotes the L.T operator, when it operated on a f
unction
, it transforms
in
to a function
of complex variable . We say the operator transforms the function in
the domain (usually called time domain) into the function in the domain (usually called
com
plex frequency domain or simply the frequency domain)
Because the Upper limit in the Integral is Infinite,
the domain of Integration is Infinite. Thus the
Integral is an example of an Improper Integral.
T
he Laplace Transformation of
is said to exist if the Integral Converges for
som
e values of
, Otherwise it does not exist.
Domain
(time domain)
Domain
(frequency domain)
Original Equation
(In terms of t)
Laplace Transform
Final Equation
(In terms of s)
TSEDUHUB
www.tseduhub.blogspot.com

JNTUWORLD
Defini
tion: A function
is said to be piece wise Continuous in any Interval , if it is defined
on
that Interval and is such that the Interval can be broken up into a finite number of sub-
Intervals in each of which
is Continuous.
I
n Mathematics, a transform is usually a device that converts one type of problem into another
type.
The main application of D.E using Laplace Transformation and Inverse Laplace Transformation is
that, By solving D.E directly by using Variation of Parameters, etc methods, we first find the
general solution and then we substitute the Initial or Boundary values. In some cases it will be
more critical to find General solution.
By suing Laplace and Inverse Laplace Transformation, we will not going to find General solution
and in the middle we substitute the Boundary conditions, so the problem may becomes simple.
Note: Some Problems will be solved more easier in Laplace than by doing using Methods
(variation of Parameter etc) and vice-versa.
PROPERTIES OF LAPLACE TRANSFORMATION
L

is
Sol:
We know that
T

is
Sol
: We know that
Definition of Gama Function
(O
R)
.
Note:
i)

ii)
TSEDUHUB
www.tseduhub.blogspot.com

JNTUWORLD
T

, where is a non-negative Real number.
Sol
: We know that

Put
As
Problems
1)Find the Laplace Transformation of
Sol
: We know that
2)Find the Laplace Transformation of
Sol
: We know that
SE
CTIONALLY CONTINUOUS CURVES (Or) PIECE-WISE CONTINUOUS
A function is said to be Sectionally Continuous (or) Piece-wise Continuous in any Interval
, if it is continuous and has finite Left and Right limits in any Sub-Interval of .
I
n the above case Laplace Transformation holds good.
TSEDUHUB
www.tseduhub.blogspot.com

JNTUWORLD
F
UNCTIONS OF EXPONENTIAL ORDER
A function is said to be exponential of order as if is a finite
v
alue.
Example: Verify
is an exponential order (or) not?
Sol
:
, a
.
is an exponential order.
S
ufficient conditions for the Existence of Laplace Tra
nsformation
The Laplace Transformation of exists i.e. The Improper Integral of
C
onverges (finite value) when the following conditions are satisfied.
1)
is a piece-wise continuous
2) is an exponential of order .
PROPE
RTIES OF LAPLACE TRANSFORMATION
LINEAR PROPERTY
Statement:
If
, then
P
roof: Given that
and
L.
H.S:
=R.H.S
F
IRST SHIFTING PROPERTY (or) FIRST TRANSLATION PROPERTY

Statement:
If
then
P
roof: We know that
(Her
e we have taken exponential quantity as
negative, but not positive, because
as
)
Put
He
nce, If
then
w

, f

which is equal to an

evaluate
, w

in place of in
TSEDUHUB
www.tseduhub.blogspot.com

JNTUWORLD
P
roblem
F

S
ol: To find
first we shall evaluate

Now,
CH
ANGE OF SCALE PROPERTY
Statement: If then
P
roof: We know that
Put
Put then,
TSEDUHUB
www.tseduhub.blogspot.com

JNTUWORLD
P
roblems
1)F

Sol
: We know that
an
d

Now,
2)Find the Laplace Transformation of
Sol
: We know that
an
d
Now
,
3)Find
Sol
: We know that
I
But
II
From I &II,
Equating the corresponding coefficients, we get
,
and
L
APLACE TRANSFORMATION OF DERIVATIVES
Statement:
If
, then
P
roof: We know that

Let us consider
I
n view of this,
Now
, put
S
ince,
TSEDUHUB
www.tseduhub.blogspot.com

JNTUWORLD
Gene
ralizing this, we get finally
P
roblem: Find the
by using derivatives method.
we know that
Giv
en that
D
IVISION BY
?t? METHOD (or) Laplace Integrals
Statement:
If
, then
P
roof: Let us consider
Now
, Integrate on both sides w.r.t
by taking the Limits from then
Since and are Independent variables, by Interchanging the order of Integration,
P
roblems
F

S
ol: Here

We know that
v
TSEDUHUB
www.tseduhub.blogspot.com

JNTUWORLD
I

, then .
Sol: L
et us consider
and
Now
,
Gene
ralization of above one:
n
-times
Multiplication of
Statement: If , then
P
roof: We know that
–
Gene
ralization:
P
roblem
F

Sol
: we know that
Here
TSEDUHUB
www.tseduhub.blogspot.com

JNTUWORLD
I
NVERSE LAPLACE TRANSFORMATION
Definition:

, then is known as Inverse Laplace Transformation of and it
is
denoted by
, where is known as Inverse Laplace Transform operator and is
suc
h that
.
I
nverse Elementary Transformations of Some Elementary Fu
nctions
P
roblems based on Partial Fractions
A fraction of the form in which both powers and are positive numbers
is ca
lled rational algebraic function.
When the degree of the Numerator is Lower than the degree of Denominator, then the fraction is
called as Proper Fraction.
To Resolve Proper Fractions into Partial Fractions, we first factorize the denominator into real
factors. These will be either Linear (or) Quadratic and some factors may be repeated.
From the definitions of Algebra, a Proper fraction can be resolved into sum of Partial fractions.
S.No Factor of the Denominator Corresponding Partial Fractions
1.
Non-Repeated Linear Factor
Ex: , [
occurs only
one time]
2.
Repeated Linear Factor,
repeated ?r? times
Ex:
3.
Non-repeated Quadratic
Expression
Ex:
, Here atleast one of
4.
Repeated Quadratic
Expression, repeated ?r? times
Ex:
TSEDUHUB
www.tseduhub.blogspot.com

JNTUWORLD
S
HIFTING PROPERTY OF INVERSE LAPLACE TRANSFORMATION
We know that
F
ORMULAS
I

then,
I

and then,
I
n general,
, provided
I

then,
I

then,
I

then,
C
ONVOLUTION THEOREM
(A Differential Equation can be converted into Inverse Laplace Transformation)
(I
n this the denominator should contain atleast two terms)
Convolution is used to find Inverse Laplace transforms in solving Differential Equations and
Integral Equations.
Statement: Suppose two Laplace Transformations
and are given. Let and are
t
heir Inverse Laplace Transformations respectively i.e.
T
hen,
W
here
is called Convolution. (Or) F
alting of
.
P
roof: Let
Now

T
he above Integration is within the region lying below the
line, and above
.
(Here e
quation of
is )

L
et
is taken on line and is taken on line, with as Origin.
M
TSEDUHUB
www.tseduhub.blogspot.com

JNTUWORLD
T
he axes are perpendicular to each other.
If the order of Integration is changed, the strip will be taken parallel to
. So that the limits of
are fr
om
and is taken as
Put , then Lower Limit:
Upper Limit:
T
hen, Consider
A
gain,
P
roblem
A

Sol
: Given
L
et us choose one quantity as
and other quantity as
Now
, Let
,
Now

A
gain
By Convolution Theorem,
TSEDUHUB
www.tseduhub.blogspot.com

JNTUWORLD
–
–
A
PPLICATIONS OF D.
E?s BY USING LAPLACE AND INVERSE LAPLACE TRANSFORMATIONS
Laplace Transform Method of solving Differential Equations yields particular solutions without
ne
cessity of first finding General solution and elimination of arbitrary constants.
Suppose the given D.Eq is of the form
I
is a
Linear D.Eq of order 2 with constants a, b.
Case 1: Suppose in Equation I, we assume a,b are constants and the boundary conditions are
.
We Kn
ow that
an
d
(O
r)
Here and is Second derivative
an
d
Pro
cedure: Apply Laplace Transformation to equation ( I )
i.e.
Now
, apply Inverse Laplace Transformation
TSEDUHUB
www.tseduhub.blogspot.com

JNTUWORLD
i.
e.
By
solving this, we get the required answer.
Case 2: If a, b are not constants (i.e. D.E with Variable Co-efficient)
Let
the D.E is of the form
II

Here
are some functions of , with Initial conditions
We k
now that
Now
,
A
nd,
A
pply Laplace Transformation on both sides to ( II ), we get
S
ubstituting the boundary conditions in equation II and get the values of
III
Req
uired solution is obtained by taking Inverse Laplace Transformation for equation III.
Problem
S

and
and
S
ol: Given
I
A
pply Laplace Transformation on both sides, we get
II
Now
substitute boundary conditions Immediately before solving in equation II, we get
TSEDUHUB
www.tseduhub.blogspot.com

JNTUWORLD
(By resolving into partial fractions)
is the required solution.
S

Sol
: Taking Laplace Transform on both sides, we get
H
ere, given Initial/Boundary conditions are
A
pply Inverse Laplace Transformation on both sides, we get
We k
now that
(O
r)
H
ints for solving problems in Inverse Laplace Transforma
tion
I

(i.e. resolve into partial fractions and solve further)
S

denominator quantity in the form of
etc
Note

fractions can be solved in other methods also.
TSEDUHUB
www.tseduhub.blogspot.com

JNTUWORLD
S
ome Formulas
I

then
Now
, If
, then
I

then
Now
, If
, then

Generalization: If
then
Now
, If
, then
I

then
Now
, If
, then

Generalization: If
then
Now
, If
, then
I

then
Now
, If
, then
I

then
Now
, If
, and , then

Generalization: If
then
Now
, If
, and , then

Similarly, If
then
Now
, If
, and , then
I

and Then,
Now
, If
, then
TSEDUHUB
www.tseduhub.blogspot.com

JNTUWORLD
P
roblems
F

S
ol: Here, if we observe, the denominator is in the product of factors form. So we can use partial
fr
actions method. Or we can use the following method.
We know that, If
then
Now
, If
, then I
L
et us consider
Substituting in I, we get
Find
S
ol: Here if we observe, it is possible to express denominator as product of partial fractions. So we
can
use partial fractions method also. But the method will be lengthy.
So, we go for another method, by which we can solve the problem easily.
Now, We know that If
then
Now
, If
, then I
L
et us consider
From I, we have
(Or)
If Numerator is and
den
ominator is
term,
then always use
model.
TSEDUHUB
www.tseduhub.blogspot.com

JNTUWORLD
I

and denominator is term, then always use model.
I

term, then always use model.
I

of etc. or any unknown quantity, where
w
e don?t have any direct formula, in such cases always use
model.
P
ERIODIC FUNCTIONS OF LAPLACE TRANSFORMATIONS
Periodic Function: A
Function
is said to be periodic function of the period if
E
xample:
are the periodic functions of period .
T
heorem: The Laplace Transformation of Piece-wise periodic function
with period is

Proof: Let
be the given function then
Put in the second Integral
in the third Integral
in the Integral etc
S
ince by substituting
in 1
st
I

function
T
his is a Geometric Progression
TSEDUHUB
www.tseduhub.blogspot.com

JNTUWORLD
U
nit Step Function (
eaviside?s Unit Function
The Unit Step Function or is defined as , where
is p
ositive number always.
S
econd Shifting Property (or) Second Translation Prop
erty
Statement:
If
and the shifted function
Then,
Pro
of: We know that
Put then
Not
e: The Laplace Transform of Unit Step Function (put
) is
U
nit Impulse Function (or Diract delta Function)
Suppose a large force (like Earthquake, collision of two bodies) acts on a system, produces large
eff
ect when applied for a very short interval of time. To deal with such situations, we introduce a
function called unit impulse function, which is a discontinuous function.
If a large force acts for a short time, the product of the force and time is called impulse. To deal
with similar type of problems in applied mechanics, the unit impulse is to be introduced.
1
TSEDUHUB
www.tseduhub.blogspot.com

JNTUWORLD
L
aplace Transform of Unit Step Function
Find Laplace Transformation of
S
ol: We know that
Find Laplace Transformation of
S
ol: We know that
TSEDUHUB
www.tseduhub.blogspot.com

M1 complete notes

About This Presentation

Slide Content

Tags

Categories

Download

Quick Actions

Statistics

Related Slideshows

M1 complete notes

About This Presentation

Slide Content

Slide 8

Slide 9

Slide 10

Slide 11

Slide 12

Slide 13

Slide 14

Slide 15

Slide 16

Slide 17

Slide 18

Slide 19

Slide 20

Slide 21

Slide 22

Slide 23

Slide 24

Slide 25

Slide 26

Slide 27

Slide 28

Slide 29

Slide 30

Slide 31

Slide 32

Slide 33

Slide 34

Slide 35

Slide 36

Slide 37

Slide 38

Slide 39

Slide 40

Slide 41

Slide 42

Slide 43

Slide 44

Slide 45

Slide 46

Slide 47

Slide 48

Slide 49

Slide 50

Slide 51

Slide 52

Slide 53

Slide 54

Slide 55

Slide 56

Slide 57

Slide 58

Slide 59

Slide 60

Slide 61

Slide 62

Slide 63

Slide 64

Slide 65

Slide 66

Slide 67

Slide 68

Slide 69

Slide 70

Slide 71

Slide 72

Slide 73

Slide 74

Slide 75

Slide 76

Slide 77

Slide 78

Slide 79

Slide 80

Slide 81

Slide 82

Slide 83

Slide 84