1. Computational
Techniques
Presentation
By Rafi Dar
student of
University of Kashmir

2. Computational Techniques
 Runge Kutta 2nd order Methods
 Runge Kutta 4th order Methods
 Eular Method
 Modified Eular Method

3. Runge Kutta Method
 Runge Kutta is the Family of Methods.
 for the approximation of solutions of ordinary
differential equations.
 2nd order and 4th order methods are widely
applied.
Steps:
1. First slope at some of the intermediate points is
computed.
2. Weighted average of the slope is used to
extrapolate the next solution point.

4. Runge Kutta 2nd order method:
the interval [x1 , xf] is divided into subintervals
and a weighted average of derivatives (slopes)
at these intervals is used to determine the value
of next dependent variable.
Advantage: of using Runge kutta method is that it
is one step method.

5.  Consider following differential equation
dy/dx = f(x,y).
With an initial condition; y =y1, x= x1.
 At starting point compute slope of the
curve as f(x1, y1) call it s1.
 Now compute slope of curve at point
(x2, y1+S1h) as f(x2, y1+ s1h) Call it s2.
 Find the average of these slopes and then
compute value of dependent variable y from
the following equation.
y2 = y1 + Sh ;
where S = (s1 + s2 ) / 2.
S1 = f(x1, y2) and s2 = f(x2, y1+ s1h).

6.  Hence starting from first point we can
find second and then third and so on.
 In general the value of y for (i+1)th
point on the solution curve is obtained
from the ith solution point using formula:
yi+1 = yi + hS
Where S = (si + si+1 ) /2 and
Si = f(xi,yi)
si+1 = f(xi+h, yi+hsi)
This formula for Runge Kutta 2nd Order is
called as Heun’s method.

7. Runge Kutta 4th Order Method
 Error in 2nd order method is O(h³) per step.
 If more precision is required we use 4th order
R.K method. In which error is O(h5) per step.
 In R.K 4th Order method, the slope at 4 points
including the starting point is computed, and
the average weight of these slopes is computed
as :
S =1/6(s1 + 2s2+ 2s3 +s4)
where s1 = f(x1 ,y1)
s2 = f(x1+h/2, y1 +h/2 s1)
s3 = f(x1+h/2, y1+h/2 s2)
s4 = f(x1+h ,y1+h s3)

8.  the value of dependent variable y is computed as :
y2 = y1 + hS
In the similar manner, starting from second solution
point we can compute the third point. The process is
repeated till we find the solution in the desired interval.
In general, the (i+1)th point of the solution curve is
obtained from the ith point using the following equation:
yi+1 =yi+ hs
Where S = (s1 + 2s2 + 2s3 + s4) /6 and
s1 = f(xi,yi)
s2 = f(xi+ h/2, yi+ h/2 s1)
s3 = f(xi+ h/2, yi+h/2 s2)
s4 = f(xi+ h, yi+ hs3)

9. Euler Method
 The Euler method can be described as piecewise
linear approximation technique to arrive at the
solution.
In Euler method we are given two starting values
(x1,y2) and slope of the curve.
 Consider an ordinary different equation dy/dx =
f(x,y).
Replacing x, y by initial value x1,y1.
dy/dx = f(x1,y1) or y1(x1) = f(x1,y1)
 Recall the mean value theorem:
If a function is continuous and differentiable
between two points (x1,y1) and (x2,y2). Then the
slope of the line joining the points is derivative
at least at one point call it (c, d) of the
function.
y1(c) = y(x2) –y(x1) / (x2 –x1)

10. Replacing x2 –x1 by h and c by x1.
We get :
y1(x1) = y(x2) –y(x1) / (x2 –x1) or
f(x1,y1) = y(x2) –y(x1) / h
y(x2) = h f(x1,y1) + y(x1)
y2 = y1+h f(x1,y1)
Using this equation we can compute the
second point on the solution curve as
(x2,y2).
Similarly taking (x2,y2) as starting point
we can determine : y3 = y2 + h f(x2,y2)
In general the (i+1)th point of the
solution curve is obtained from ith point
using following formula:
yi+1 = yi + h f(xi, yi)

11. Modified Eular Method
The modified Eular method is a multi step method,
which utilizes information from more than one
previous step to extrapolate solution curve.
the method works correctly only in case of linear
function. Alternatively we can calculate the average
slope within the interval.
Suppose we use interval bound by x =xi and x = xi+1
Average slope =(y1(xi) + y1(xi+1) )/2.
Then according to the basic Eular formula:
Yi+1 = yi + h/2[y1(xi) + y(xi+1)]
Yi+1 = yi + h/2[f(xi,yi) +f(xi+1,yi+1)]
This is an improved estimate fo yi+1 at xi+1. but we can’t
use it directly since yi+1 is unknown.

12. The value of yi+1 is predicted using
basic Eular equation as:
ypi+1 = yi + h f(xi, yi)
This is called as predictor formula.
Using this predicted value, a more
accurate value is computed by equation
called as corrector formula as follows:
yci+1 = yi + h/2[f(xi,yi ) + f (xi+1,ypi+1)].
In general we may conclude Eular method
is a two step method:
1. Predict yi+1solution point by using
predictor formula.
2. Correct the yi+1solution point using
corrector formula.

13. Thank you.
reach out to me :
rafi.rey@gmail.com
or
https://www.facebook.com/rafi.dar.98