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Introduction to Ordinary Differential Equations: Taylor Series and Euler Method, Slides of Mathematical Methods for Numerical Analysis and Optimization

An introduction to ordinary differential equations (odes) with a focus on taylor series and the euler method. It covers the expansion of a function using taylor series, the concept of an initial value problem, and the derivation of the euler method. The document also includes examples of applying the euler method to find approximate solutions to odes.

Typology: Slides

2011/2012

Uploaded on 08/05/2012

saruy
saruy 🇮🇳

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Download Introduction to Ordinary Differential Equations: Taylor Series and Euler Method and more Slides Mathematical Methods for Numerical Analysis and Optimization in PDF only on Docsity!

Ordinary

DifferentialEquations

Ordinary

DifferentialEquations

TAYLOR’S

SERIES

METHOD

We considered aninitial value problemdescribed by

0

0

(^
dy
f t y
y t
y
dt

Noting that

f

is an implicit

function of

y

, we have

( ,

)

x^

y

y

f t y

f^

f dy

y

f^

ff

x

y dx

 

 

 

 docsity.com

Similarly

2

2

2

2

(^

)^

(^

)

2

(^

)

3

3

(^

2

)

3(

)(

)

(^

)

xx^

xy^

xy^

yy^

y^

x^

y

xx^

xy^

yy^

y^

x^

y

IV

xxx

xxx

xyy

y^

xx^

xy^

yy

x^

y^

xy^

yy

y^

x^

y

y^

f^

ff

f^

f^

ff

f^

f^

ff

f^

ff

f^

f^

f^

f^

ff

y^

f^

ff

f^

f

f^

f^

ff

f^

f

f^

ff

ff

ff

f^

f^

ff

 

^

^

^

^

 

^

^

 

^

^

 

^

 

^

 

^

 docsity.com

Euler method is one of theoldest numerical methodsused for integrating theordinary differential equations.Though this method is notused in practice, itsunderstanding will help us togain insight into nature ofpredictor-corrector method

Consider the differentialequation of first order withthe initial condition

y

( t

) = 0

y

. 0

( ,

)

dy

f t y

dt

Here we use a property that in asmall interval, a curve is nearlya straight line. Thus at (

t^0

,^

y

), 0

we approximate the curve by atangent at that point.Therefore,

0

0

0

0

0

(^ ,

)

0

(^

,^

)

t^ y

y^

y

dy

f t

y

dt

t^  t

^

^

^

^

^

^

That is,

0

0

0

0

(^
)^
(^
,^
y
y
t^
t^
f t
y

Hence, the value of ycorresponding to

t

=

t

1

is

given by

1

0

1

0

0

0

(

)

(^

,^

)

y

y

t^

t^

f t

y

Thus, we obtain ingeneral, the solution ofthe given differentialequation in the form of arecurrence relation

1

(

,^

)

m

m

m

m

y

y

hf t

y

Geometrically, this methodhas a very simple meaning.The desired function curveis approximated by apolygon train, where thedirection of each part isdetermined by the value ofthe function

f

(

t, y

) at its

starting point.

SolutionSince the number of steps arefive, we shall proceed in stepsof (0.1)/5 = 0.02.Therefore, taking step size h

= 0.02, we shall compute the value of

y

at

t^

= 0.02, 0.04, 0.06, 0.08 and 0.

Thus

1

0

0

0

(^

,^

),

y^

y^

hf t

y

^

^

where

0

0 1,^

0

y^

t ^

1

1

0

1

1

0

y^

^

^

 

Therefore,

2

1

1

1

3

2

2

2

4

3

3

3

5

4

4

4

(^

)^

(^

,^

)^

(^

,^

)^

(^

,^

)^

y^

y^

hf t

y

y^

y^

hf t

y

y^

y^

hf t

y

y^

y^

hf t

y

^

^

^

^

^

^

^

^

^

 

^

^

^

^

 

^

^

^

^

Similarly,