Lecture Notes of Numerical Solution of Ordinary Differential Equations Using Improved Euler s Method I

University of Babylon
College of Engineering
Department of Environmental Engineering
Engineering Analysis I (ENAN 103)







Numerical Solution of Ordinary Differential Equations

Undergraduate Level, 3th Stage



Mr. Waleed Ali Tameemi
College of Engineering/ Babylon University
M.Sc. Civil Engineering/ the University of Kansas/ USA



2016-2017
Improved Euler’s Method
This method which, requires an initial condition (y_0,x_0), is utilized in solving ordinary deferential equation numerically. It is an improved approach of Euler’s Method.
¬The following steps are required in solving ordinary differential equations using Improved Euler’s Method:

The ordinary deferential equation (ODE) dy/dx=f(x,y)
Initial condition (x_0,y_0)
Required y value at a given x value ?(x_n,y_n)
Step size ?_x=x_(i+1)-x_i
Number of steps n=(x_n-x_0)/?_x
Initial estimation of y_(i+1)^*
y_(i+1)^*=y_i+?_x×y_i
f(x_(i+1),y_(i+1) )=y_(i+1)=y_i+[(f(x_i,y_i )+f(x_(i+1),y_(i+1)^* ))/2]×?_x
Those steps may be summarized in the following table
i x_i y_i f(x_i,y_i ) y_(i+1)^* f(x_(i+1),y_(i+1)^* ) y_(i+1)
0 x_0 y_0 f(x_0,y_0 ) y_1^*=y_0+?_x×y_0 f(x_1,y_1^* ) y_1=y_0+[(f(x_0,y_0 )+f(x_1,y_1^* ))/2]×?_x
1 x_1=x_0+?_x y_1 f(x_1,y_1 ) y_2^*=y_1+?_x×y_1 f(x_2,y_2^* ) y_2=y_1+[(f(x_1,y_1 )+f(x_2,y_2^* ))/2]×?_x
2 x_2=x_1+?_x y_2 f(x_2,y_2 ) y_3^*=y_2+?_x×y_2 f(x_3,y_3^* ) y_3=y_2+[(f(x_2,y_2 )+f(x_3,y_3^* ))/2]×?_x
… … … … …

Ex1: Find the value of the given ODE at f(0.2)
x+y-y`=0
Initial condition: y(0)=1
Step size ?_x=0.1

Solution:
The ordinary deferential equation (ODE) f(x,y)=dy/dx=x+y
Initial condition (0,1)
Required y value at a given x value ?f(0.2)
Step size ?_x=0.1
Number of steps n=(0.2-0)/0.1=2 steps
Initial estimation of y_(i+1)^*
y_(i+1)^*=y_i+?_x×y_i
f(x_(i+1),y_(i+1) )=y_(i+1)=y_i+[(f(x_i,y_i )+f(x_(i+1),y_(i+1)^* ))/2]×?_x
Those steps may be summarized in the following table
i x_i y_i f(x_i,y_i ) y_(i+1)^* f(x_(i+1),y_(i+1)^* ) y_(i+1)
0 0 1 0+1=1 y_1^*=1+0.1×1=1.1 0.1+1.1=1.2 y_1=y_0+[(f(x_0,y_0 )+f(x_1,y_1^* ))/2]×?_x=1+[(1+1.2)/2]×0.1=1.11
1 x_1=0+0.1=0.1 1.11 0.1+1.11=1.21 y_2^*=1.11+0.1×1.11=1.221 0.2+1.221=1.421 y_2=1.11+[(1.21+1.421)/2]×0.1=1.2416
2 0.2 1.2416
f(0.2)=1.2416

Ex2: Find the value of the given ordinary differential equation at f(1.6)
y`=(x^2+y^2)/(x+y)
Initial condition: y(1)=0
Step size ?_x=0.2

Solution:
The ordinary deferential equation (ODE) f(x,y)=dy/dx=(x^2+y^2)/(x+y)
Initial condition (1,0)
Required y value at a given x value ?f(1.6)
Step size ?_x=0.2
Number of steps n=(1.6-1)/0.2=3 steps
Initial estimation of y_(i+1)^*
y_(i+1)^*=y_i+?_x×y_i
f(x_(i+1),y_(i+1) )=y_(i+1)=y_i+[(f(x_i,y_i )+f(x_(i+1),y_(i+1)^* ))/2]×?_x