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Chapter 8

NUMERICAL SOLUTION

OF

ORDINARY DIFFERENTIAL EQUATIONS

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PRELIMINARIES

= f (x, y) with an initial condition y = y0 at x = x0. dx

dyConsider

The function f (x, y) may be linear, nonlinear or table of values

When the value of y is given at x = x0 and the solution is required for x0 < x < xf then the problem is called an initial value problem. If y is given at x = xf and the solution is required for xf > x > x0 then the problem is called a boundary value problem.

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INITIAL VALUE PROBLEMS

A Solution is a curve g (x, y) in the xy plane whose slope at each point (x, y) in the specified region is

given by = f (x, y).

The initial point (x0, y0) of the solution curve g(x, y) and the slope of the curve at this point is given. We then extrapolate the values of y for the required set of values in the range (x0, xf).

dx

dy

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y2

y1

y0

x0 x1 xn

EULER’S METHOD

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EULER’S METHOD

This method uses the simplest extrapolation technique.

The slope at (x0, y0) is f (x0, y0).

Taking a small step in the direction given by the above slope, we get

y1 = y (x0 + h) = y0 + hf (x0, y0)

Similarly y2 can be obtained from y1 by taking an equal step h in the direction given by the slope

f(x1, y1).

In generalyi+1= yi + h f(xi, yi)

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Modifications

Modified Euler Method● In this method the average of the slopes at (x0, y0) and (x1, y=1

(1)) is taken instead of the slope at (x0, y0) where y1

(1) = y1 + h f (x0, y0).● In general,

Improved Modified Euler Method● In this method points are averaged instead of

slopes.

yi+1 = yi + ½ h [f (xi, yi) + f (xi + h, yi + hf (xi, yi)) ]

yi+1 = yi + hf (xi + h

2, yi +

2

h f (xi, yi) )

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Example

Find y (0.25) and y (0.5) given that = 3x2 + y, y(0) = 4 by (i) Euler Method (ii) Modified Euler Method (iii) Improved Euler Method and compare the

results.

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Solution

y - value x

Euler Modified Improved Exact

0.25 5.0000 5.1484 5.1367 5.1528

0.50 6.2969 6.7194 6.6913 6.7372

Applying Formulae

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TAYLOR SERIES METHOD

= f (x, y) with an initial condition y = y0 at x = x0. dx

dyConsider

The solution curve y(x) can be expressed in a Taylor series around x = x0 as:

dx

yd

!3

h

dx

yd

!2

h

dx

dy3

33

2

22

+…

where x = x0+h.

y (x0 + h) = y0 + h

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Example

Using Taylor series find y(0.1), y(0.2) and y(0.3) given that

= x2 - y; y(0) =1dx

dy

Solution Applying formula

y(0.1) = 0.9052y(0.2) = 0.8213 y(0.3) = 0.7492

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PICARD’S METHOD OF SUCCESSIVE APPROXIMATIONS

This is an iterative method.

= f (x, y) with an initial condition y = y0 at x = x0. dx

dyConsider

Integrating in (x0, x0 + h) y(x0 +h) = y(x0) +

hx

x

0

0

dx)y,x(f

This integral equation is solved by successive approximations.

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After n steps

This process is repeated and in the nth approximation, we get

y(n) = y0 +

hx

x

1)(n0

0

)dxyf(x,

Example

Find y(1.1) given that = x – y,

y(1) = 1, by Picard’s Method.

dx

dy

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Solution

y(1)1.1 = 1 +

1.1

1

)1( dxx

= 1.005

Successive iterations yield 1.0045, 1.0046 , 1.0046

Exact value is y (1.1) = 1.0048

Thus y (1.1) = 1.0046

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RUNGE–KUTTA METHODS

Euler Method is not very powerful in practical problems, as it requires very small step size h for reasonable accuracy.

In Taylor’s method, determination of higher order derivatives are

involved.

The Runge–Kutta methods give greater accuracy without the need to calculate higher derivatives.

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nth order R.K. Method

This method employs the recurrence formula of the form

yi+1 = yi + a1 k1 + a2 k2 + + an kn

where k1 = h f ( xi, yi)

k2 = h f (xi + p1h, yi + q11 k1)

k3 = h f (xi + p2h, yi + q21 k1 + q22 k2)

kn = h f( xi + pn-1 h, yi + q n-1,1 k1 + qn-2, 2 k2 + q(n-1), (n-1)kn)

……. …….. ………

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4th order R.K. Method

Most commonly used method

yn+1 = yn + (k1 + 2k2 + 2k3 + k4)

wherek1 = hf(xn, yn)

k2 = hf (xn + , yn + )

k3 = hf (xn + , yn + )

k4 = h f (xn +h, y3 +k3)

2

h

2

h

2

k1

2

k 2

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Example

Using R.K. Method of 4th order find y(0.1) and y(0.2).

Given that =3x + ½ y, y (0) = 1 taking h = 0.1.

Solution

dy

dx

k1 = h f (x0, y0) = 0.0500

k2 = h f ( x0 + 2

h , y0 + 2

k1 ) = 0.0663

k3 = h f ( x0 + 2

h , y0 + 2

k 2 ) = 0.0667

k4 = h f ( x0 +h, y0 + k3) = 0.0833

y1 = y (0.1) = y0 + 6

1 ( k1 + 2k2 + 2k3 + k4) = 1.0674

By similar procedure y(0.2) = 1.1682

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PREDICTOR CORRECTOR METHODS

In the Runge-Kutta methods the solution, point yn+1 is evaluated using only the solution point yn and none of the previous points yn-1, yn-2 etc. on the solution curve. The behaviour of the solution in the past are ignored and each step starts afresh. This will be inefficient. The Predictor-Corrector methods take the previous solution points also into account. In these methods

● First a predictor formula is obtained and a corrector is obtained.

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Milne’s Predictor-Corrector Method This method uses the past 4 points in the solution.

Predictor formula:

yi+1=yi-3+43

h (2fi–fi-1+2fi-2)+O(h5)

Corrector formula:

yi+1=yi-1+3

h (fi+1 + 4fi + fi-1)+O(h5).

Note: If there is a significant difference between predictor and corrector value the corrector value may be taken as predictor value and a new corrector value may be calculated.

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Example

Solve using Milne’s Method given that =

y(0) = 2, y (0.2) = 2.0933, y (0.4) = 2.1755,

y (0.6) = 2.2493.

Find y (0.8).

dy

dx y x

1

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Solution for y(0.8)

yields y4,p = 2.3163

yields y4,C = 2.3164

Predictor formula:

Y4=y0+43h (2f3–f2+2fi1)

Corrector formula:

Y4=y2+3h (f4 + 4f3 + f2).

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Adams – Bashforth Predictor – Corrector Method

This method also requires information about past

4 solution points.

yi+1,p = yi + 24

h (55fi – 59fi-1 + 37fi-2 – 9fi-3).

yi+1,c = yi + 24

h (9fi+1 + 19fi – 5fi-1 + fi-2).

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Example

Given y = x2–y, y(0) = 1, y(0.1) = 0.90516, y(0.2) = 0.82127 and y (0.3) = 0.74918, find y (0.4) by Adams method.

Solution

y4,p = y3 + (55f3 – 59f2 + 37f1 – 9f0) = 0.6897

y4,c = y3 + (9f4 + 19f3 – 5f2 + f1) = 0.6897