The document introduces Euler's method for numerically solving ordinary differential equations. It provides the formulation of Euler's method as a recurrence relation and gives examples of applying the method to solve various initial value problems by discretizing the interval and time steps. Euler's method approximates the slope of the tangent line at each step to iteratively calculate subsequent y-values.

1. Introduction
Euler’s Method
Engineering Computations
Numerical Ordinary Differential Equations
Dr.Wisam Haitham
Computer Engineering Department
College of Engineering
Al-Mustansiriyah University
July/2020
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2. Introduction
Euler’s Method
Agenda
1 Introduction
2 Euler’s Method
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3. Introduction
Euler’s Method
Introduction
Many problems in science and engineering when formulated mathe-
matically are readily expressed in terms of ordinary differential equa-
tions (ODE) with initial and boundary condition.
Consider the problem of solving the first-order differential equation
dy
dt
= f(t, y)
such that we know the rate of change of a quantity as a function
of time and the quantity itself. The problem is to obtain a solution
for y(t), possibly in a general form, but in practical problems usually
where an initial condition y(t0) = y0 is known. In many problems
this can be solved analytically. In general such an integration is not
possible, for example, the differential equation dy/dt = −y2 − t,
and we have to solve the equation numerically, which usually means
obtaining a sequence of numerical values y1, y2, ... corresponding to
specified values t1, t2, ....
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4. Introduction
Euler’s Method
The figure below shows the nature of the numerical solution of a
differential equation
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5. Introduction
Euler’s Method
Agenda
1 Introduction
2 Euler’s Method
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6. Introduction
Euler’s Method
Euler’s Method
Euler method is one of the oldest numerical methods used for inte-
grating the ordinary differential equations. Though this method is
not used in practice, its understanding will help us to gain insight
into nature of predictor-corrector method.
To find the solution of the given Differential Equation in the form
of a recurrence relation
ym+1 = ym + hf(tm, ym) Is called Euler Method
FORMULA DERIVATION
Consider the differential Equation of the first order
dy
dx = f(t, y) and y(t0) = y0
Let (t0, y0) and (t1, y1) be two points of approximation curve. Then
y1 − y0 = m(x1 − x0) ....... (i)(point Slope form)
Given That dy
dx = f(t, y)
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7. Introduction
Euler’s Method
y1 − y0 = f(t0, y0)(x1 − x0)
So y1 = y0 + f(t0, y0)(x1 − x0)
Similarly
y2 = y1 + f(t1, y1)(x2 − x1)
y3 = y2 + f(t2, y2)(x3 − x2)
. . .
. . .
. . .
ym+1 = ym + f(tm, ym)(xm+1 − xm)
ym+1 = ym + hf(tm, ym) (m = 0, 1, ..)
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8. Introduction
Euler’s Method
Example 1 :
Use Euler’s method with h = 0.1 to solve the initial value problem
dy
dx = x2 + y2 with y(0)=0 in the range 0 6 x 6 0.5.
Solution:
Here
f(x, y) = x2 + y2, x0 = 0, y0 = 0, h = 0.1
Hence
x1 = x0 + h = (0 + 0.1) = 0.1, x2 = x1 + h = (0.1 + 0.1) = 0.2,
x3 = x2 + h = 0.3, x4 = x3 + h = 0.4, x5 = x4 + h = 0.5
ym+1 = ym + hf(tm, ym)
y1 = y0 + 0.1(x2
0 + y2
0) = 0 + 0.1(0 + 0) = 0
y2 = y1 + 0.1(x2
1 + y2
1) = 0 + 0.1((0.1)2 + 0) = 0.001
y3 = y2 + 0.1(x2
2 + y2
2) = 0.001 + 0.1((0.2)2 + (0.001)2) = 0.005
y4 = y3 + 0.1(x2
3 + y2
3) = 0.005 + 0.1((0.3)2 + (0.005)2) = 0.014
y5 = y4 +0.1(x2
4 +y2
4) = 0.014+0.1((0.4)2 +(0.014)2) = 0.0300196
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9. Introduction
Euler’s Method
Hence
y(0)=0 y(0.1)=0 y(0.2)=0.001 y(0.3)=0.005
y(0.4)=0.014 y(0.5)=0.0300196
Example 2 :
Using Euler method solve the equation dy
dx = 2xy + 1 with y(0)=0,
h = 0.02, for x=0.1
Solution:
Here
f(x, y) = 2xy + 1, x0 = 0, y0 = 0, h = 0.02
Hence
x1 = x0 + h = (0 + 0.02) = 0.02,
x2 = x1 + h = (0.02 + 0.02) = 0.04, x3 = x2 + h = 0.06,
x4 = x3 + h = 0.08, x5 = x4 + h = 0.1
ym+1 = ym + hf(tm, ym)
y1 = y0 + 0.02(2x0y0 + 1) = 0 + 0.02(0 + 1) = 0.02
y2 = y1 + 0.02(2x1y1 + 1) = 0.02 + 0.02(2 ∗ 0.02 ∗ 0.02 + 1) = 0.04
y3 = y2 + 0.02(2x2y2 + 1) = 0.04 + 0.02(2 ∗ 0.04 ∗ 0.04 + 1) = 0.06
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10. Introduction
Euler’s Method
y4 = y3 +0.02(2x3y3 +1) = 0.005+0.02(2∗0.06∗0.06+1) = 0.08
y5 = y4 + 0.02(2x4y4 + 1) = 0.014 + 0.02(2 ∗ 0.08 ∗ 0.08 + 1) = 0.1
Hence
y(0)=0 y(0.02)=0.02 y(0.04)=0.04 y(0.06)=0.06
y(0.08)=0.08 y(0.1)=0.1
That is the approximate value of y(0.1) is 0.1
Example 3 :
Using Euler’s method, compute y in the range 0 6 x 6 0.5, if y
satisfies dy
dx = 3x + y2 with y(0)=1, h=0.1
Solution:
Here
f(x, y) = 3x + y2, x0 = 0, y0 = 1, h = 0.1
Hence
x1 = x0 + h = 0.1, x2 = x1 + h = 0.2, x3 = x2 + h = 0.3,
x4 = x3 + h = 0.4, x5 = x4 + h = 0.5
ym+1 = ym + hf(tm, ym)
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11. Introduction
Euler’s Method
y1 = y0 + 0.1(3x0 + y2
0) = 1.1
y2 = y1 + 0.1(3x1 + y2
1) = 1.251
y3 = y2 + 0.1(3x2 + y2
2) = 1.4675
y4 = y3 + 0.1(3x3 + y2
3) = 1.7728
y5 = y4 + 0.1(3x4 + y2
4) = 2.2071
Hence
y(0)=1 y(0.1)=1.1 y(0.2)=1.251 y(0.3)=1.4675
y(0.4)=1.7728 y(0.5)=2.2071
Example 4 :
Given dy
dx = y−x
y+x , with y=1 for x=0 . Find y approximately for
x=0.1 by Euler’s method in five steps, choosing h=0.002.
Solution:
Here
f(x, y) = y−x
y+x , x0 = 0, y0 = 1, h = 0.002
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12. Introduction
Euler’s Method
y1 = y0 + 0.002(y0−x0
y0+x0
) = 1.02
y2 = y1 + 0.002(y1−x1
y1+x1
) = 1.0392
y3 = y2 + 0.002(y2−x2
y2+x2
) = 1.0577
y4 = y3 + 0.002(y3−x3
y3+x3
) = 1.0756
y5 = y4 + 0.002(y4−x4
y4+x4
) = 1.0928
Hence
y=1.0928 and x=0.1
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