MIT ADTU’s
MIT School of Engineering, Pune
Subject: Differential Equation and Transform Techniques
Unit-V
Numerical Solution of Ordinary Differential equations
Many ordinary differential equations can be solved by analytical methods. However majority of
differential equations appearing in physical problems cannot be solved analytically. Thus it
becomes
Imperative to discuss their solution by numerical methods. The different methods are
1) Euler’s Method
Consider the differential equation
dy
 f x, y  with the initial condition y  y0 at
dx
x  x0
Suppose we want to solve the equation we use the following approximation
y1  y0  h f x0 , y0  ,
y2  y1  h f x1 , y1  ,
And so on, In general y n1  y n  h f xn , y n , n  0,1, 2,
To obtain the solution with desired accuracy, one has to take a smaller value of h , hence
the solution is obtained very slowly. Due to this the method is rarely used. The more
accurate method will be obtained by the modified method.
ILLUSTRATIONS
Ex.1. Use Euler’s method to solve the equation
dy
 1  xy
dx
Subject to the conditions at x  0, y 1find y  0.4 
Solution: - given f  x , y  1  xy , h  0.1, x0  0, y0 1 and h 
y1  y0  h f  x0 , y0 
1  0.1 f  0,1
1  0.11  0 1  1.1
y2  y1  h f  x1 , y1 
1.1  0.1 f  0.1,1.1
30
 0.5
6
1.1  0.11  0.11.1  1.211
y3  y2  h f  x2 , y2 
1.211  0.1 f  0.2,1.211
1.211  0.11  0.2 1.211  1.3352
y4  y3  h f  x3 , y3 
1.3352  0.1 f  0.3,1.3352 
1.3352  0.11  0.3 1.3352  1.4752
y3  y2  h f  x2 , y2 
1.211  0.1 f  0.2,1.211
1.211  0.11  0.2 1.211  1.3352
2) Modified Euler’s Method
To start, we use Euler’s formula to find value of y at x  x1 ,
therefore y1  y0  h f x0 , y0 
To find more correct approximation at x  x1 , we use Euler’s modified iteration formula
h
 n 1
n 
y1
 y 0  f  x0 , y 0   f x1 , y1 , n 1, 2,
……….. (1)
2
1
n 
i.e. new y  old y  h Meanslope  , where Mean slope  f  x0 , y 0   f x1 , y1
2
1
2 
n1
n 
by making use of (1) we can obtain approximations y1 , y1 ,, y1 , y1 . This
0 




process is repeated till no significant change occurs. i e. y1
n1

 y1
n 

we call this as y1
Then y1 is more correct approximation at x  x1
To find the value of y at x  x2 , the above procedure is repeated in the interval x1 , x2 
(note x2  x1  h )
ILLUSTRATIONS
Ex.1. Use modified Euler’s method to solve
h  0.2
dy
 x  y 2 , y  0   1 to calculate y  0.4  taking
dx
Solution: - given f  x , y   x  y 2 , h  0.2, x0  0, y0 1
Step 1: to find y1   y  x 0.2
by Euler’s method
y10  y0  h f  x0 , y0   1  0.2 0  1   0.8


2
by modified Euler’s method
h
y11  y0   f  x0 , y0   f  x1 , y10  
2
 1
0.2
 f  0,1  f  0.2, 0.8  
2 
 1
0.2
 1  0.44   0.8560
2
h
y12  y0   f  x0 , y0   f  x1 , y11  
2
 1
0.2
 f  0,1  f  0.2, 0.8560  
2 
 1
0.2
 1  0.5327   0.8467
2
 y1   y  x 0.2  0.8467
Step 2: to find y1   y  x 0.4
by Euler’s method
2
y2 0  y1  h f  x1 , y1   0.8467  0.2 0.2   0.8467    0.7962


by modified Euler’s method
h
y21  y1   f  x1 , y1   f  x2 , y2 0  
2
 0.8467 
0.2
 f  0.2, 0.8467   f  0.4, 0.7962  
2 
 0.8467 
0.2
 0.5193  0.2340  0.7728
2
h
y2 2  y1   f  x1 , y1   f  x2 , y21  
2
 0.8467 
0.2
 f  0.2, 0.8467   f  0.4, 0.7728  
2 
 0.8467 
0.2
 0.5193  0.1972   0.7765
2
 y2   y  x 0.4  0.7765
Ex.2. Use modified Euler’s method to solve
dy
 x 2  y, y  0   1 to calculate y  0.1 taking
dx
h  0.05
Sol: - given f  x , y   x 2  y , h  0.05, x0  0, y0 1
Step 1: to find y1   y  x 0.05
by Euler’s method
y10  y0  h f  x0 , y0  1 0.05 0  1 1.05
by modified Euler’s method
h
y11  y0   f  x0 , y0   f  x1 , y10  
2
 1
0.05
 f  0,1  f  0.05,1.05  
2 
 1
0.05
1  1.0525  1.0513
2
h
y12  y0   f  x0 , y0   f  x1 , y11  
2
 1
0.05
 f  0,1  f  0.05,1.0513  
2 
 1
0.05
1  1.0538  1.0513
2
 y1   y  05 1.0513
Step 2: to find y2   y  x 0.1
by Euler’s method
y2 0  y1  h f  x1 , y1   1.0513  0.05  0.05   1.0513  1.1040


2
by modified Euler’s method
h
y21  y1   f  x1 , y1   f  x2 , y2 0  
2
 1.0513 
0.05
 f  0.1,1.0513  f  0.1,1.1040  
2 
 1.0513 
0.05
1.0538  1.114 1.1055
2
h
y2 2  y1   f  x1 , y1   f  x2 , y21  
2
 1.0513 
0.05
 f  0.05,1.0513   f  0.1,1.1055  
2 
 1.0513 
0.05
1.0538 1.1155 1.1055
2
 y2   y  x 0.1 1.1055
EXERCISE
Ex.1. Use Euler’s method to solve the equation
dy
 x  y, y  0   0 find y  0.4 
dx
Ex.2. Use Modified Euler’s method to find the value of y satisfying the equation
dy
 logx  y , y 1  2 for x  1.2 and x  1.4
dx
Ex.3. Use Modified Euler’s method to find the value of y satisfying the equation
d y yx

, y 0  1for x  0.1 take h  0.05
d x yx
Session -7 Numerical Solution of Ordinary Differential equations by Runge-Kutta Method
 Notes:
1) Runge-Kutta Method of fourth order
The fourth order Runge-kutta method is most commonly used.Working rule for finding
the increment k of y corresponding to an increment h of x by Runge-kutta method
dy
from
 f x, y , y x0   y0 is as follows:
dx
k 
h

Calculate successively k1  h f x0 , y 0  , k 2  h f  x0  , y0  1  ,
2
2

k 
h

k 3  h f  x0  , y 0  2  ,
2
2

1
k 4  h f x0  h, y0  k 3  Finally compute k  k1  2k 2  2k 3  k 4 
6
Which gives the required approximate value y1  y0  k
ILLUSTRATIONS
Ex.1. Use fourth order Runge- Kutta method to solve
dy

dx
x  y , y  0   1 to calculate
y at x = 0.2
Solution: - given f  x , y   x  y , h  0.2, x0  0, y0 1
Step 1: to find y1   y  x 0.2
k1  h f  x0 , y0   0.2 0  1  0.2
k 
h
0.2
0.2 


k2  h f  x0  , y0  1   0.2 f  0 
,1 

2
2
2
2 


 0.2
 0  0.1  1  0.1  0.2191
k 
h
0.2
0.2191 


k3  h f  x0  , y0  2   0.2 f  0 
,1 

2
2
2
2 


 0.2
 0  0.1  1  0.1095  0.2120
k4  h f  x0  h, y0  k3   0.2 f  0  0.2,1  0.2120 
 0.2
 0  0.1  1  0.1095  0.2120
k

1
 k1  2k2  2k3  k4 
6
1
 0.2  2  0.2191  2  0.2120  0.2377 
6
 0.2167
 y1   y  x 0.2  y0  k  1.2167
Ex.2. Use fourth order Runge- Kutta method to solve
dy
1

, y  0  1 to find
dx x  y
y at x = 0.4 taking h=0.2
Solution: - given f  x , y  
1
, h  0.2, x0  0, y0 1
x y
Step 1: to find y1   y  x 0.2
 1 
k1  h f  x0 , y0   0.2 
  0.2
 0 1 
k 
h
0.2
0.2 


k2  h f  x0  , y0  1   0.2 f  0 
,1 

2
2
2
2 


 1 
 0.2 
  0.167
 0.1  1.1 
k 
h
0.2
0.167 


k3  h f  x0  , y0  2   0.2 f  0 
,1 

2
2
2
2 


1


 0.2 
  0.169
 0.1  1.0835 
k4  h f  x0  h, y0  k3   0.2 f  0  0.2,1  0.169 
1


 0.2 
  0.1461
 0.2  1.169 
k

1
 k1  2k2  2k3  k4 
6
1
 0.2  2  0.167  2  0.169  0.1461
6
 0.1697
 y1   y  x 0.2  y0  k  1.1697
Step 2: to find y2   y  x 0.4
1


k1  h f  x1 , y1   0.2 
  0.146
0.2

1.1697


k 
h
0.2
0.146 


k2  h f  x1  , y1  1   0.2 f  0.2 
,1.1697 

2
2
2
2 


1


 0.2 
  0.1296
 0.3  1.2497 
k 
h
0.2
0.1296 


k3  h f  x1  , y1  2   0.2 f  0.2 
,1.1697 

2
2
2
2 


1


 0.2 
  0.1303
 0.3  1.2345 
k4  h f  x1  h, y1  k3   0.2 f  0.2  0.2,1.1697  1.1303
1


 0.2 
  0.1176
 0.4  1.3 
k

1
 k1  2k2  2k3  k4 
6
1
 0.146  2  0.1296  2  0.1303  0.1176 
6
 0.1306
 y2   y  x0.4  y1  k  1.3003
EXERCISE
Ex.1. Use fourth order Runge- Kutta method to solve
y 1.2  and y 1.4 
dy
 log  x  y  , y 1  2 to find
dx
Ex.2. Use fourth order Runge- Kutta method to solve
dy
 x 2  y 2 , y 1 1.5 to find
dx
y 1.2  with h  0.1
Introduction:Theory of probabilityhad its origin in mid-eighteenth century studies in games of chance
related to dice throw gambling. Almost every human activity involves some kind of chance elements and
the role for thetheory of probability to play. As time progressed, probability theory found its way into
many applications not only in Engineering and Science but also in the fields like Actuarial Science,
Agriculture, Commerce, Medicine and Psychology.
Theorems on Probability:
A) If A and B are any two events then
P ( A  B )= P( A)  P( B)  P ( A  B)
B) P( A  B)  P( A) P( B / A) and P ( A  B )  P ( B ) P ( A / B )
C) Baye’sTheorem: If A1 , A2 , ..., An are mutually exclusive events whose union is the sample
space and A
is any event, then
P( Ak / A) 
P( Ak ) P( A / Ak )
n
 P( A ) P( A / A )
k l
k
k
Ex. 1: Among six books, there are two volume of one book. These books are arranged in a random order
on a shelf.
Find the probability that the two volumes are always together.
Solution:Six books can be arranged on a shelf in 6! = 720 ways, thus n =720
The probability that the two books are always together is
P=
=
240 1

720 3
Ex.2: Two cards are drawn from a well shuffled pack of 52 cards. Find the probability that they are both
kings if
i) the first card drawn is replaced, ii)first card drawn is not replaced.
Ans : 1/169, 1/221
Exercise
Ex.1:Three coins are tossed simultaneously . Find the probability of getting at least two heads.
Ans :1/2
Ex.2:A box contains 3 red and 2 blue marbles. The box B contains 2 red and 8 blue marbles. A fair coin
is tossed. If
the coin shows Head , a marble is chosen from the box A, if it shows tail, a marble is chosen from box B.
find
the probability that a red is chosen.
Ans : 2/5
Ex.3: Urn I contains 6 white and 4 black balls and urn II contains 4 white and 5 black balls. From urn I,
two balls
aretransferred to urn II without noticing the colour. Sample of size 2 is then drawn without replacement
from
urn II. What is the probability that the sample contains exactly one white ball.
Ans :4/5
-------------------------------------------------------------------------------------------------------------------------------
Session -7
Probability Distributions:
Binomial Probability Distribution:
It is used when the experiment has only two outcomes. Then probability of ‘r’ successes in ‘n’ trials is
given by
n
r
B(n,p,r)= p(r)= Cr p q
nr
; r= 0, 1, 2, …
Where p=probability of success
q =probability of failure
Remark :A) p+q =1
B) Mean of a Binomial distribution is m=np
C) Standard deviation of a Binomial distribution is   npq
Ex.1: An unbiased coin is thrown 10 times. Find the probability of getting exactly 6 heads, at least 2
heads.
Solution:Here n=10, p=0.5, q=0.5
10
6
i)Probability of getting exactly 6 heads = P(6)= C6 p q
4
=
10
C6 (0.5) 6 (0.5) 4 = 10C6 (0.5)10 = 0.2051
ii)Probability of getting at least 2 heads = p(r  2)= p(2)+p(3)+…+p(10)=1- p(0)-p(1)
= 1 - 10C0 (0.5) 0 (0.5)10 -
10
C1 (0.5)1 (0.5) 9
= 1 – 0.000977 – 0.00977 = 0.9893
Ex.2 : The mean and variance of a binomial distribution are 6 and 2 respectively. Find p(r  1).
Solution :Given mean = 6 and variance = 2
Which implies np=6 and npq = 2
Solving we get q=1/3, p=1-q = 2/3 , n = 9
 p(r  1) = 1 – p(0) = 0.999949
Exercise
Ex.1: A random variable is X with B(n=6, p). Find p if 9P(R=4)=P(R=2).
Ans : p=1/4
Ex.2:A die is rolled six times. Find the probability that at least two 4’s appear.Ans : p = 0.6 (approx)
-------------------------------------------------------------------------------------------------------------------------------
Session-08
Poisson Distribution:
It is a limiting case of a Binomial distribution with n  , p  0 and mean that is np remaining
finite. Therefore the ‘ r ‘ successes in a large number of trials ‘ n’ with small number of probability
success p in eachtrial is given by
mr em
p(r)=
;
r!
r= 0, 1, 2, …
Remark :A) p+q =1
B) Mean of a Poisson distribution is m=np
C)Standard deviation of a Poisson distribution is   np
Ex.1: Assuming that the probability of an indivisual coal minor being killed in a mine accidents during a
year is
1/2400. Calculate the probability in a mine employing 200 miners there will be at least one fatal
accident in a
year.
Solution : Given
p=1/2400 , n=200
 m = np = 0.083
p(r)=
mr em
;
r!
r= 0, 1, 2, …
(0.083) r e  ( 0.083)
=
r!
The probability in a mine employing 200 miners there will be at least one fatal accident in a year is
p(r  1) = 1 – p(0)
= 1 – 0.92
p(r  1) = 0.08
Ex.2 : In a Poisson distribution if P(r=1)=2P(r=2), find P(r=3).
Solution :InPoisson distribution probability of r successes is given by
p(r)=
mr em
;
r!
p(r=1)= me
m
r= 0, 1, 2, …
and
m 2em
p(r=2)=
2

P(r=1)=2P(r=2)

me  m = 2

m =1

P(r=3) =
m 2em
2
1e 1
= 0.0613
3!
Exercise
Q.1) In a certain factory turning out razor blades, there is a small chance of 1/500 for any blade to be
defective. The blades are supplied in a packets of 10. Use Poisson distribution to calculate the
approximate number of packets containing no defective and two defective blades, in a consignment
of 10,000 packets.
Ans : 9802, 2.
Q.2)The accidents per shift in a factory are given by the table :
Accidents per
shift
Frequency
0
1
2
3
4
5
142
158
67
27
5
1
Fit a Poisson distribution to the above table and calculate the frequencies as per Poisson fit.
Ans : p(0)=0.3697, p(1)=0.36785, p(2)=0.183, p(3)=0.0607, p(4)=0.0151, p(5)=0.003
Corresponding theoretical frequencies are 148, 147, 73, 24, 6, 1
respectively.
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Session-9
Normal Distribution :
Normal distribution curve is given by the equation
y
2
2
1
e ( x ) / 2
 2y
Normal variable z is defined as ,
When
x  x1 ,
z=
x

x ,
z=0
z=z1 (say)

P(   x  x1 ) = P(0  z  z1)
…….
(probability of z between
z=0 and z=z1 )
z1
=
 f ( z )dz ……….
(normal probability
0
integral)
= area under standard normal curve between z=0 and z=z1
Remark :A) The total area under the normal curve ( which is symmetric about x   ) is given by

 y dx  1

Ex.1:In a certain examination test, 2000 students appeared in a subject of statistics. Average marks
obtained were
50% with standard deviation 5%. How many students do you expect to obtain more than 60% of
marks,
supposing that marks are distributed normally? (z=2, A=0.4772)
 = 5%=5/100=0.05
Solution :  =50%=50/100= 0.5,
x1  60% = 60/100=0.6
z1 
x1  

=
0.6  0.5
=2
0.05
 P(0  z  2)=0.4772
A(z=2)=0.4772
 P( x  0.6 )=P( z  2 )
=0.5  P(0  z  2)
= 0.5  0.4772
P( x  0.6 ) = 0.0228
Number of students expected to get more than 60% marks = 0.0228
approximately.
200 = 46 students
Ex.2: In a distribution exactly normal , 7% of the items are under 35 and 89% are under 63. Find the
mean and
standard deviation of the distribution.
Solution :
Given that A(<35)=0.07 and A(>63)=0.11
That is p(x<35)=0.07 and p(x>63)=0.11
When
x=35 , z 
When x=63 , z 
35  

63  

  z1
 z2
 Area A1 = p(0<z< z1 )= 0.43
andArea A2 = p(0<z< z 2 )
Therefore
35  

corresponds to
z1= 1.48
= 0.39 corresponds to
z2= 1.23
  z1 = - 1.48
Solving these two equations we get
 =50.3,  =10.33
and
63  

 z 2 = 1.23
Exercise
Q.1) X is a normal variate with mean 30 and S.D. 5, find the probabilities that
i)
ii)
|
iii)|
[Given : A(z=0.8)=0.2881, A(z = 1) = 0.3413, A(z =2)=0.4772, A(z =3)= 0.4986]
Ans : i) 0.7653
ii) 0.0014
iii) 0.3174
Q.2)In a distribution normal, 31% of the items are under 45 and 8% are over 64. Find the mean and S.D.
of the
Distribution.[Given :
]
Ans : mean = 50, S.D.= 6
Q.3) In a test on 2000 electric bulbs, it was found that the life of a particular make, was normally
distributed with an
Average life of 2040 hours and S.D. of 60 hours. Estimate the number bulbs likely to burn for
i)more than 2150 hours
ii)less than 1950 hours
iii)more than 1920 and less than 2160 hours.
Given :A(z = 1.833) = 0.4664, A(z = 1.5) = 0.4332, A(z = 2) = 0.4772
Ans : i)67
ii )137
iii)1909
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