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MODULE IV: NUMERICAL DIFFERENTIATION & INTEGRATION
AND SOLUTIONS TO ORDINARY DIFFERENTIAL EQUATIONS (ODE)
Numerical Integration
The process of evaluating a definite integral from a set of tabulated values of integrand
𝑓(𝑥) is called numerical integration.
- Trapezoidal Rule
Let 𝐼 = ∫ 𝑓(𝑥) 𝑑𝑥
𝑏
𝑎
where 𝑓(𝑥) takes the values 𝑦₀, 𝑦₁, 𝑦₂,... , 𝑦 ₙ for 𝑥 = 𝑥₀, 𝑥₁,... , 𝑥 ₙ.
𝑏
𝑎
𝑥 0
+𝑛ℎ
𝑥 0
[(𝑦
0
𝑛
1
2
𝑛− 1
)]
- Simpson's
𝟏
𝟑
𝒓𝒅
Rule
𝑏
𝑎
[
0
𝑛
1
3
5
𝑛− 1
2
4
6
𝑛− 2
]
Number of subintervals must be even.
- Simpson's
𝟑
𝟖
𝒕𝒉
Rule
𝑏
𝑎
[
0
𝑛
1
2
4
5
7
8
𝑛− 1
3
6
9
𝑛− 3
]
Number of subintervals must be multiples of 3.
Example 1: Evaluate ∫
𝑑𝑥
1 +𝑥
2
6
0
using:
(i) Trapezoidal rule (ii) Simpson's 1/3rd rule (iii) Simpson's 3/8th rule
Solution: Divide the interval ( 0 , 6 ) into six parts of width ℎ = 1.
The value of 𝑓
1
1 +𝑥
2
is given by:
(i) Trapezoidal Rule:
2
6
0
[(𝑦
0
6
1
2
3
4
5
)]
[( 1 + 0. 027 ) + 2 ( 0. 5 + 0. 2 + 0. 1 + 0. 05884 + 0. 0385 )] = 1. 41084
(ii) Simpson's 1/
rd
Rule:
2
6
0
[(𝑦
0
6
1
3
5
2
4
)]
[( 1 + 0. 027 ) + 4 ( 0. 5 + 0. 1 + 0. 0385 ) + 2 ( 0. 2 + 0. 05884 )] = 1. 3662
(iii) Simpson's 3/
th
Rule:
2
6
0
[(𝑦
0
6
1
2
4
5
3
)]
[( 1 + 0. 027 ) + 3 ( 0. 5 + 0. 2 + 0. 05884 + 0. 0385 ) + 2 ( 0. 1 )] = 1. 3571
Example 2: Use trapezoidal rule to estimate the integral ∫ 𝑒
𝑥
2
2
0
𝑑𝑥 taking 10
subintervals.
Solution: Let 𝑦 = 𝑒
𝑥
2
, ℎ = 0. 2 and 𝑛 = 10
𝑥
2
𝑥
2
𝑥
2
2
0
[(𝑦
0
10
1
2
3
4
5
6
7
8
9
)]
Example 3: Use (𝑖) Trapezoidal (𝑖𝑖) Simpson's 1/3rd rule for ∫ 𝑒
−𝑥
2
- 6
0
𝑑𝑥 with h=0.1 (or
by taking 7 ordinates)
Solution:
(i) Trapezoidal Rule:
−𝑥
2
- 6
0
[(𝑦
0
6
1
2
3
4
5
)] = 0. 5344
(ii) Simpson's 1/
rd
Rule:
−𝑥
2
- 6
0
ℎ
3
[(𝑦
0
6
1
3
5
2
4
)]=0.
Numerical Solution of Ordinary Differential Equations
- Euler's Method
Consider the equation
𝑑𝑦
𝑑𝑥
= 𝑓(𝑥, 𝑦) given 𝑦(𝑥
0
0
. To find an approximate value of y
corresponding to 𝑥, by Euler’s method we have:
𝑖
𝑖− 1
𝑖− 1
𝑖− 1
i.e., 𝑛𝑒𝑤 𝑦 = 𝑜𝑙𝑑 𝑦 + ℎ (𝑑𝑦/𝑑𝑥)
Example 1: Using Euler's method, find 𝑦( 1 ) given
𝑑𝑦
𝑑𝑥
Solution: Divide the interval ( 0 , 1 ) into 10 subintervals with ℎ = 0. 1
The approximate value of 𝑦( 1 ) is 3.18.
Example 2: Given
𝑑𝑦
𝑑𝑥
𝑦−𝑥
𝑦+𝑥
, 𝑦( 0 ) = 1 , find 𝑦( 0. 1 )
Solution: Divide the interval ( 0 , 0. 1 ) into 5 subintervals with ℎ = 0. 02
Thus 𝑦( 0. 1 ) ≈ 1. 0928
- Modified Euler's Method
By Euler’s Method:
𝑖
𝑖− 1
𝑖− 1
𝑖− 1
i.e., for first iteration: 𝑦 1
0
0
0
By Modified Euler’s Method:
1
is computed initially by Euler’s formula above and 𝑓(𝑥
0
0
) is replaced by
𝑓
( 𝑥
0
𝑦
0
) +𝑓
( 𝑥
1
𝑦
1
)
2
to obtain first modification 𝑦
1
( 1 )
i.e. 𝑦
1
( 1 )
0
+ ℎ [
𝑓(𝑥
0
,𝑦
0
)+𝑓(𝑥
1
,𝑦
1
)
2
]
Second modification is given by:
1
( 2 )
0
+ ℎ [
0
0
1
1
( 1 )
]
𝑡ℎ
modification is given by:
1
(𝑛)
0
+ ℎ [
0 ,
0
1
1
(𝑛− 1 )
]
The procedure is terminated when successive modification does not show any change.
Same procedure is repeated to find 𝑦 2
3
, and so on…
𝑡ℎ
modification of 𝑦
𝑘
is given by:
𝑘
(𝑛)
𝑘− 1
+ ℎ [
𝑘− 1
𝑘− 1
𝑘
𝑘
( 𝑛− 1
)
]
Example 1: Using modified Euler’s method find 𝑦( 0. 2 ) and 𝑦( 0. 4 ) for 𝑦
′
𝑥
𝑦( 0 ) = 0. Let ℎ = 0. 2
Solution: Given 𝑦
′
𝑥
0
0
i) To find 𝑦( 0. 2 )
By Euler’s Method: 𝑦
1
0
0
0
= 0. 2468 + 0. 2 [
] = 0. 6031
Fifth modification of 𝑦
2
2
( 4
)
1
+ ℎ [
1
1
2
2
( 4
)
]
= 0. 2468 + 0. 2 [
] = 0. 6031
Thus 𝑦
2
Example 2: Solve the following by Euler’s modified method
𝑑𝑦
𝑑𝑥
= log(𝑥 + 𝑦),
= 2 , find 𝑦
considering ℎ = 0. 2
Solution: Given 𝑦
′
= log(𝑥 + 𝑦), 𝑥
0
0
By Euler’s Method: 𝑦 1
0
0
0
First modification of 𝑦 1
1
( 1 )
0
+ ℎ [
0
0
1
1
] = 2 + 0. 2 [
] = 2. 0655
Second modification of 𝑦
1
1
( 2 )
0
+ ℎ [
0
0
1
1
( 1
)
] = 2 + 0. 2 [
] = 2. 0656
Third modification of 𝑦 1
1
( 3 )
0
+ ℎ [
0
0
1
1
( 2
)
] = 2 + 0. 2 [
] = 2. 0656
Thus 𝑦 1
Exercise:
- Using modified Euler’s method, find an approximate value of 𝑦 when 𝑥 = 0. 3 given
that
𝑑𝑦
𝑑𝑥
= 𝑥 + 𝑦 and 𝑦( 0 ) = 1 Ans: y(0.3) ≈ 1.
- Runge-Kutta Method
Consider the differential equation
𝑑𝑦
𝑑𝑥
0
0
Let ℎ be increment in 𝑥 and 𝑘 be the increment in 𝑦 i.e.,
0
0
0
0
Then for finding the increment 𝑘 by Runge-Kutta method, calculate:
1
0
0
2
0
0
1
3
0
0
2
4
0
0
3
1
2
3
4
And 𝑦
1
0
The above method is also called fourth-order Runge Kutta method and is applicable to
linear or non-linear differential equation
Example 1: Apply Rune-Kutta fourth order method, to find an approximate value of
𝑦( 0. 2 ) given that
𝑑𝑦
𝑑𝑥
= 𝑥 + 𝑦 and 𝑦( 0 ) = 1.
Solution:
Given 𝑥 0
0
= 1 , ℎ = 0. 2. To find 𝑘:
1
0
0
= 0. 2 × 𝑓( 0 , 1 ) = 0. 2 × 1 = 0. 2000
2
0
0
1
) = 0. 2 × 𝑓( 0. 1 , 1. 1 ) = 0. 2 × 1. 2 = 0. 2400
3
0
0
2
) = 0. 2 × 𝑓( 0. 1 , 1. 12 ) = 0. 2 × 1. 22 = 0. 2440
4
0
0
3
) = 0. 2 × 𝑓( 0. 2 , 1. 244 ) = 0. 2888
1
2
3
4
Thus, 𝑦
0
Example 2: Using Runge-Kutta method of fourth order, solve
𝑑𝑦
𝑑𝑥
𝑦
2
−𝑥
2
𝑦
2
+𝑥
2
, with 𝑦( 0 ) = 1 ,
find 𝑦( 0. 2 ).
Solution:
Given 𝑥 0
0
= 1 , ℎ = 0. 2. To find 𝑘:
1
0
0
) = 0. 2 × 𝑓( 0 , 1 ) = 0. 2000
2
0
0
1
) = 0. 2 × 𝑓( 0. 1 , 1. 1 ) = 0. 19672
