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Visvesvaraya Technological University VTU
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- Numerical Solution Of Ordinary Differential Equations Of First Order And First Degree - ( 1 - 45 )
- Complex Variables - ( 46 - 77 )
- complex integration - ( 78 - 91 )
- Cauchy's integral formula - ( 92 - 105 )
- Conformal transformation - ( 106 - 117 )
- Numerical Methods And Special Functions - ( 118 - 148 )
- Probability And Joint Probability - ( 149 - 155 )
- Poisson's distribution - ( 156 - 158 )
- Exponential distribution - ( 159 - 164 )
- Mean And Variance Of Binomial Distribution - ( 165 - 171 )
- Sampling Theory And Stochastic Process - ( 172 - 177 )
- Sampling distribution - ( 178 - 183 )
- Sample variables - ( 184 - 196 )
- Stochastic process - ( 197 - 199 )

Topic:

Module – I
Engineering Mathematics - IV 2017
Numerical Solution of Ordinary Differential
Equations of First order and first degree
Introduction
Many ordinary differential equations can be solved by analytical methods
discussed earlier giving closed form solutions i.e. expressing y in terms of a
finite number of elementary functions of x. However, a majority of differential
equations appearing in physical problems cannot be solved analytically. Thus it
becomes imperative to discuss their solution by numerical methods.
Numerical methods for Initial value problem:
Consider the first order and first degree differential equations
the initial condition
problem.
(
that is
)
and
(
)with
called initial value
We discuss the following numerical methods for solving an initial value
problem.
1. Taylor’s series method
2. Modified Euler’s method
3. Runge - Kutta method of order IV
4. Milne’s Predictor - Corrector Method
5. Adams – Bashforth Predictor - Corrector Method
Type -1
Taylor’s series method
Consider
(
the
)
first
condition
order
(
)
and
first
degree
equations
.
Taylor’s series expansion of ( ) in powers of (
Dr. A.H.Srinivasa, MIT, Mysore
differential
) is
Page 1

Module – I
Engineering Mathematics - IV 2017
(
( )
)
(
(
)
)
(
(
)
)
(
(
)
)
(
)
Where
at the point(
)
Worked Examples
1.
Using Taylor’s Series method, find the value of y at x = 0.1, and x = 0.2 for
the initial value problem
( )
Solution:
Taylor’s Series expansion of ( ) about a point
(
( )
)
Here, compare
( )
(
(
)
)
( )
( )
( )
w.r.t
( )
(
w.r.t
)
Differentiate
(
( )( )]
( )
( )
(
)
(
)
(
)
and
( )
( )
[
( )
]
( )
( )( )
[
( )
]
we get,
( )
(
( )
( )
[( )( )
( ) ] [
( )
]
we get,
)
( )
)
we get,
( )
w.r.t
(
( )
( )
( )
Differentiate
)
, then
( )
( )
Differentiate
(
is given by
(
)
Substitute the values of
( )
( )
( )
(
(
( )
)
( )
[( )(
)
( )
)
[ ( )
]
( ) in equation (*)
( )
Dr. A.H.Srinivasa, MIT, Mysore
Page 2

Module – I
Engineering Mathematics - IV 2017
( )
This is called Taylors series expansion up to fourth degree term.
Put x = 0.1. x = 0.2
2.
(
)
(
)
(
)
(
)
(
)
(
)
(
)
(
)
Using Taylor’s Series method, find the value of y at x = 0.1, and x = 0.2 for
the initial value problem
( )
Solution:
Taylor’s Series expansion of ( ) about a point
( )
(
)
Here, compare
( )
(
)
(
)
( )
( )
( )
( )
( )
w.r.t
w.r.t
(
Differentiate
)( )
w.r.t
(
)( )
)
(
(
)
(
)
and
( )
(
)
)( )
( )
( )
[
( )
]
we get,
( )
Differentiate
(
)
, then
( )
( )
Differentiate
(
is given by
( )
( )
(
)(
)
( )( ) [
( )
]
we get,
( )
( )(
( )
)
( ) [
( )
( )
( )
]
( )
we get,
( )
( )( )
Dr. A.H.Srinivasa, MIT, Mysore
( )
( ) [
( )
( )
( )
( )
]
Page 3

Module – I
Engineering Mathematics - IV 2017
Substitute the values of
( )
(
)
( )
( )
( )
( )
( ) in equation (*)
( )
(
)
( )
This is called Taylors series expansion up to fourth degree term.
Put x = 0.1 and x = 0.2
3.
(
)
(
)
(
)
(
)
(
)
(
)
(
)
(
)
Using Taylor’s Series method, find the value of y at x = 0.1, and x = 0.2, for
the initial value problem
( )
Solution:
Taylor’s Series expansion of ( ) about a point
( )
(
Here, compare
( )
)
(
)
(
( )
( )
( )
( )
Differentiate
Differentiate
w.r.t
we get,
( )
( )
w.r.t
we get,
w.r.t
(
( )
]
(
)
)
(
, then
( )
( )
)
(
)
(
)
and
( )
( )
( )
[ ( )
Differentiate
)
is given by
( )
( )
( )
( )
( )
( )
( )
( )
( )
( )
(
[
( )
]
[
( )
]
( )
we get,
Dr. A.H.Srinivasa, MIT, Mysore
Page 4

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