Nancyvarkey at 18:46, 1 February 2014

2014-02-01T18:46:50Z

Lavitha Pereira: Created page with ''''Title of script''': '''Solving ODEs using Scilab ode Function''' '''Author: Shamika''' '''Keywords: ODEs''' {| style="border-spacing:0;" ! Visual Cue !
2013-12-24T11:54:56Z

Created page with ''''Title of script''': '''Solving ODEs using Scilab ode Function''' '''Author: Shamika''' '''Keywords: ODEs''' {| style="border-spacing:0;" ! <center>Visual Cue</center> ! <c…'

New page
'''Title of script''': '''Solving ODEs using Scilab ode Function'''

'''Author: Shamika'''

'''Keywords: ODEs'''

{| style="border-spacing:0;"
! <center>Visual Cue</center>
! <center>Narration</center>

|-
| style="border-top:none;border-bottom:1pt solid #000000;border-left:1pt solid #000000;border-right:none;padding:0.097cm;"| Slide 1
| style="border-top:none;border-bottom:1pt solid #000000;border-left:1pt solid #000000;border-right:1pt solid #000000;padding:0.097cm;"| Dear Friends,

Welcome to the Spoken Tutorial on “'''Solving ODEs using Scilab ode Function'''”

|-
| style="border-top:none;border-bottom:1pt solid #000000;border-left:1pt solid #000000;border-right:none;padding:0.097cm;"| Slide 2 -Learning Objective Slide
| style="border-top:none;border-bottom:1pt solid #000000;border-left:1pt solid #000000;border-right:1pt solid #000000;padding:0.097cm;"| At the end of this tutorial, you will learn how to:

* '''Use Scilab ode function'''
* Solve typical examples of '''ODEs '''and
* Plot the solution

|-
| style="border-top:none;border-bottom:1pt solid #000000;border-left:1pt solid #000000;border-right:none;padding:0.097cm;"| Slide 3 -Learning Objective Slide
| style="border-top:none;border-bottom:1pt solid #000000;border-left:1pt solid #000000;border-right:1pt solid #000000;padding:0.097cm;"| The typical examples we will be solving are:

* Motion of '''simple pendulum'''
* '''Van der Pol equation'''
* '''Lorenz system'''

|-
| style="border-top:none;border-bottom:1pt solid #000000;border-left:1pt solid #000000;border-right:none;padding:0.097cm;"| Slide 4-System Requirement slide
| style="border-top:none;border-bottom:1pt solid #000000;border-left:1pt solid #000000;border-right:1pt solid #000000;padding:0.097cm;"| To record this tutorial, I am using

* '''Ubuntu 12.04''' as the operating system
* and '''Scilab 5.3.3''' version

|-
| style="border-top:none;border-bottom:1pt solid #000000;border-left:1pt solid #000000;border-right:none;padding:0.097cm;"| Slide 5- Prerequisites slide
| style="border-top:none;border-bottom:1pt solid #000000;border-left:1pt solid #000000;border-right:1pt solid #000000;padding:0.097cm;"| To practise this tutorial, a learner

* should have basic knowledge of '''Scilab '''
* and should know how to solve''' ODEs.'''

To learn '''Scilab''', please refer to the relevant tutorials available on the '''Spoken Tutorial '''website.

|-
| style="border-top:none;border-bottom:1pt solid #000000;border-left:1pt solid #000000;border-right:none;padding:0.097cm;"| Slide 6- ode Function
| style="border-top:none;border-bottom:1pt solid #000000;border-left:1pt solid #000000;border-right:1pt solid #000000;padding:0.097cm;"| The '''ode function''' is an '''ordinary differential equation solver.'''

The syntax is '''y equal to ode within paranthesis y zero, t zero, t and f'''

Here''' '''

* '''y zero''' is the initial conditon of the '''ODEs'''

* '''t zero '''is the '''initial time'''
* '''t''' is the '''time range'''
* and '''f''' is the '''function'''

|-
| style="background-color:transparent;border-top:none;border-bottom:1pt solid #000000;border-left:1pt solid #000000;border-right:none;padding:0.097cm;"| Slide 7- Motion of Simple pendulum
| style="background-color:transparent;border-top:none;border-bottom:1pt solid #000000;border-left:1pt solid #000000;border-right:1pt solid #000000;padding:0.097cm;"| Consider the motion of simple pendulum.

Let '''theta t '''be the angle made by the '''penndulum''' with the '''vertical''' at time '''t.'''

We are given the initial conditions

* '''theta zero '''is equal to''' pi by four '''and
* '''theta dash of zero '''is equal to '''zero'''.

|-
| style="background-color:transparent;border-top:none;border-bottom:1pt solid #000000;border-left:1pt solid #000000;border-right:none;padding:0.097cm;"| Slide 8- Motion of Simple pendulum
| style="background-color:transparent;border-top:none;border-bottom:1pt solid #000000;border-left:1pt solid #000000;border-right:1pt solid #000000;padding:0.097cm;"| Then the position of the '''pendulum''' is described by

'''theta double dash t minus g by l into sin of theta t equal to zero'''.

Here

* '''g equal to 9.8 m per second square''' is the '''acceleration due to gravity '''and
* '''l equal to zero point five meter''' is the '''length '''of the '''pendulum.'''

|-
| style="background-color:transparent;border-top:none;border-bottom:1pt solid #000000;border-left:1pt solid #000000;border-right:none;padding:0.097cm;"| Slide 9- Motion of Simple pendulum
| style="background-color:transparent;border-top:none;border-bottom:1pt solid #000000;border-left:1pt solid #000000;border-right:1pt solid #000000;padding:0.097cm;"| For the given initial conditions, we have to solve the '''ODE''' within the '''time range zero less than equal to t less than equal to five'''.

We also have to '''plot''' the solution.

|-
| style="border-top:none;border-bottom:1pt solid #000000;border-left:1pt solid #000000;border-right:none;padding:0.097cm;"| Open Pendulum.sci on Scilab Editor
| style="border-top:none;border-bottom:1pt solid #000000;border-left:1pt solid #000000;border-right:1pt solid #000000;padding:0.097cm;"| Let us look at the code for solving this problem.

Open '''pendulum dot sci''' on''' Scilab editor.'''

|-
| style="border-top:none;border-bottom:1pt solid #000000;border-left:1pt solid #000000;border-right:none;padding:0.097cm;"| Highlight

y0<nowiki>=</nowiki><nowiki>[</nowiki>%pi/4 0]'<nowiki>;</nowiki>

t0<nowiki>=</nowiki>0<nowiki>;</nowiki>

t<nowiki>=</nowiki>0:1:5

| style="border-top:none;border-bottom:1pt solid #000000;border-left:1pt solid #000000;border-right:1pt solid #000000;padding:0.097cm;"| The first line of the code defines the initial conditions of the '''ODE'''.

Then we define the intial time value. And we provide the '''time range'''.

|-
| style="border-top:none;border-bottom:1pt solid #000000;border-left:1pt solid #000000;border-right:none;padding:0.097cm;"| Highlight

function '''dy'''<nowiki>=</nowiki>Pendulum('''t''', '''y''')

'''dy'''(1) <nowiki>=</nowiki> '''y'''(2)<nowiki>;</nowiki>

'''dy'''(2) <nowiki>=</nowiki> (9.8/0.5)<nowiki></nowiki>sin('''y'''(1))<nowiki>;</nowiki>

endfunction

| style="border-top:none;border-bottom:1pt solid #000000;border-left:1pt solid #000000;border-right:1pt solid #000000;padding:0.097cm;"| Next, we convert the given equation to a system of '''first order ODEs'''.

We substitute the values of '''g''' and '''l'''.

Here we take '''y '''to be the given '''variable theta''' and '''y dash''' to be '''theta dash'''.

|-
| style="border-top:none;border-bottom:1pt solid #000000;border-left:1pt solid #000000;border-right:none;padding:0.097cm;"| Highlight

y<nowiki>=</nowiki>ode(y0,t0,t,Pendulum)

| style="border-top:none;border-bottom:1pt solid #000000;border-left:1pt solid #000000;border-right:1pt solid #000000;padding:0.097cm;"| Then we call the '''ode function''' with '''arguments y zero, t zero, t '''and the '''function Pendulum'''.

|-
| style="border-top:none;border-bottom:1pt solid #000000;border-left:1pt solid #000000;border-right:none;padding:0.097cm;"| Highlight

plot(t,y(1,:),'-',t,y(2,:),'-')

| style="border-top:none;border-bottom:1pt solid #000000;border-left:1pt solid #000000;border-right:1pt solid #000000;padding:0.097cm;"| The solution to the '''equation''' will be a '''matrix''' with two '''rows'''.

The first '''row''' will contain the values of '''y''' in the given '''time range.'''

The second '''row''' will contain the values of '''y dash '''within the '''time range'''.

Hence we plot both the '''rows''' with respect to '''time'''.

|-
| style="border-top:none;border-bottom:1pt solid #000000;border-left:1pt solid #000000;border-right:none;padding:0.097cm;"| Click on Execute and select Save and Execute
| style="border-top:none;border-bottom:1pt solid #000000;border-left:1pt solid #000000;border-right:1pt solid #000000;padding:0.097cm;"| Save and execute the file '''Pendulum dot sci'''

|-
| style="border-top:none;border-bottom:1pt solid #000000;border-left:1pt solid #000000;border-right:none;padding:0.097cm;"| Show plot
| style="border-top:none;border-bottom:1pt solid #000000;border-left:1pt solid #000000;border-right:1pt solid #000000;padding:0.097cm;"| The plot shows how the values of '''y''' and '''y dash''' vary with '''time'''.

|-
| style="border-top:none;border-bottom:1pt solid #000000;border-left:1pt solid #000000;border-right:none;padding:0.097cm;"| Switch to Scilab console
| style="border-top:none;border-bottom:1pt solid #000000;border-left:1pt solid #000000;border-right:1pt solid #000000;padding:0.097cm;"| Switch to '''Scilab console'''

|-
| style="border-top:none;border-bottom:1pt solid #000000;border-left:1pt solid #000000;border-right:none;padding:0.097cm;"| Type y

Press Enter
| style="border-top:none;border-bottom:1pt solid #000000;border-left:1pt solid #000000;border-right:1pt solid #000000;padding:0.097cm;"| If you want to see the values of '''y''', type '''y''' on the '''console '''and press '''Enter'''.

The values of '''y '''and '''y dash '''are displayed.

|-
| style="border-top:none;border-bottom:1pt solid #000000;border-left:1pt solid #000000;border-right:none;padding:0.097cm;"| Slide 10- Van der Pol Equation
| style="border-top:none;border-bottom:1pt solid #000000;border-left:1pt solid #000000;border-right:1pt solid #000000;padding:0.097cm;"| Let us solve '''Van der Pol equation''' using the '''ode function'''.

We are given the '''equation '''

'''v double dash of t plus epsilon into v of t square minus one into v dash of t plus v of t equal to zero'''.

The initial '''conditions''' are '''v of two equal to one '''and '''v dash of two equal to zero'''.

Assume '''epsilon is equal to zero point eight nine seven'''.

We have to find the solution within the '''time range two less than t less than ten '''and then '''plot '''the solution.

|-
| style="border-top:none;border-bottom:1pt solid #000000;border-left:1pt solid #000000;border-right:none;padding:0.097cm;"| Open Vanderpol.sci on scilab editor
| style="border-top:none;border-bottom:1pt solid #000000;border-left:1pt solid #000000;border-right:1pt solid #000000;padding:0.097cm;"| Let us look at the code for '''Van der Pol equation'''.

Switch to '''Scilab editor '''and open '''van der pol dot sci'''.

|-
| style="border-top:none;border-bottom:1pt solid #000000;border-left:1pt solid #000000;border-right:none;padding:0.097cm;"| Highlight

y0<nowiki>=</nowiki><nowiki>[</nowiki>1 0]'<nowiki>;</nowiki>

t0<nowiki>=</nowiki>0<nowiki>;</nowiki>

t<nowiki>=</nowiki>2:1:10<nowiki>;</nowiki>

| style="border-top:none;border-bottom:1pt solid #000000;border-left:1pt solid #000000;border-right:1pt solid #000000;padding:0.097cm;"| We define the initial conditions of the '''ODEs''' and '''time''' and then define the '''time range'''.

Since the '''inital time value '''is given as''' two''', we start the '''time range '''at''' two. '''

|-
| style="border-top:none;border-bottom:1pt solid #000000;border-left:1pt solid #000000;border-right:none;padding:0.097cm;"| Highlight

function '''dy'''<nowiki>=</nowiki>Vanderpol('''t''', '''y''')

'''dy'''(1) <nowiki>=</nowiki> '''y'''(2)<nowiki>;</nowiki>

'''dy'''(2) <nowiki>=</nowiki> -0.897<nowiki></nowiki>('''y'''(1).^2 - 1)<nowiki></nowiki>'''y'''(2) - '''y'''(1)<nowiki>;</nowiki>

endfunction

| style="border-top:none;border-bottom:1pt solid #000000;border-left:1pt solid #000000;border-right:1pt solid #000000;padding:0.097cm;"| Then we define the '''function van der pol '''and construct a system of '''first order ODEs'''.

We substitute the value of '''epsilon '''with '''zero point eight nine seven'''.

Here '''y '''refers to the '''voltage v'''.

|-
| style="border-top:none;border-bottom:1pt solid #000000;border-left:1pt solid #000000;border-right:none;padding:0.097cm;"| Highlight

y<nowiki>=</nowiki>ode(y0,t0,t,Vanderpol)

| style="border-top:none;border-bottom:1pt solid #000000;border-left:1pt solid #000000;border-right:1pt solid #000000;padding:0.097cm;"| Then we call '''ode function''' and solve the system of '''equations'''.

|-
| style="border-top:none;border-bottom:1pt solid #000000;border-left:1pt solid #000000;border-right:none;padding:0.097cm;"| Highlight

plot(t,y(1,:),'-',t,y(2,:),'--')

| style="border-top:none;border-bottom:1pt solid #000000;border-left:1pt solid #000000;border-right:1pt solid #000000;padding:0.097cm;"| Finally we plot '''y '''and''' y dash versus t.'''

|-
| style="border-top:none;border-bottom:1pt solid #000000;border-left:1pt solid #000000;border-right:none;padding:0.097cm;"| Click on Execute and select Save and Execute
| style="border-top:none;border-bottom:1pt solid #000000;border-left:1pt solid #000000;border-right:1pt solid #000000;padding:0.097cm;"| Save and execute the file '''van der pol dot sci. '''

|-
| style="border-top:none;border-bottom:1pt solid #000000;border-left:1pt solid #000000;border-right:none;padding:0.097cm;"| Show plot
| style="border-top:none;border-bottom:1pt solid #000000;border-left:1pt solid #000000;border-right:1pt solid #000000;padding:0.097cm;"| The '''plot''' showing '''voltage versus time''' is shown.

Let's move onto '''Lorenz system of equations.'''

|-
| style="border-top:none;border-bottom:1pt solid #000000;border-left:1pt solid #000000;border-right:none;padding:0.097cm;"| Slide 11, 12- Lorenz system

| style="border-top:none;border-bottom:1pt solid #000000;border-left:1pt solid #000000;border-right:1pt solid #000000;padding:0.097cm;"| The '''Lorenz system''' '''of''' '''equations''' is given by

* '''x one dash equal to sigma into x two minus x one, '''
* '''x two dash equal to one plus r minus x three into x one minus x two '''and''' '''
* '''x three dash equal to x one into x two minus b into x three. '''

The initial conditions are '''x one zero''' equal to '''minus ten, x two zero equal to ten '''and''' x three zero equal to twenty five'''.

Let '''sigma '''be''' equal to ten, r '''be '''equal to twenty eight '''and''' b equal to eight by three. '''

|-
| style="border-top:none;border-bottom:1pt solid #000000;border-left:1pt solid #000000;border-right:none;padding:0.097cm;"| Open Lorenz.sci on Scilab editor

| style="border-top:none;border-bottom:1pt solid #000000;border-left:1pt solid #000000;border-right:1pt solid #000000;padding:0.097cm;"| Switch to '''Scilab editor''' and open '''Lorenz dot sci'''

|-
| style="border-top:none;border-bottom:1pt solid #000000;border-left:1pt solid #000000;border-right:none;padding:0.097cm;"| Highlight

x0<nowiki>=</nowiki><nowiki>[</nowiki>-10 10 25]'<nowiki>;</nowiki>

t0<nowiki>=</nowiki>0<nowiki>;</nowiki>

t<nowiki>=</nowiki>0:1:25<nowiki>;</nowiki>

| style="border-top:none;border-bottom:1pt solid #000000;border-left:1pt solid #000000;border-right:1pt solid #000000;padding:0.097cm;"| We start by defining the initial conditions of the''' ODEs'''.

Since there are three different '''ODEs''', there are three initial conitions.

Then we define the '''inital time''' condition and next the '''time range.'''

|-
| style="border-top:none;border-bottom:1pt solid #000000;border-left:1pt solid #000000;border-right:none;padding:0.097cm;"| Highlight

function '''dx'''<nowiki>=</nowiki>Lorenz('''t''', '''x''')

sigma <nowiki>=</nowiki> 10<nowiki>;</nowiki>

r <nowiki>=</nowiki> 28<nowiki>;</nowiki>

b <nowiki>=</nowiki> 8/3<nowiki>;</nowiki>

'''dx'''(1) <nowiki>=</nowiki> sigma<nowiki></nowiki>('''x'''(2) - '''x'''(1) )<nowiki>;</nowiki>

'''dx'''(2) <nowiki>=</nowiki> ((1 + r) - '''x'''(3))<nowiki></nowiki>'''x'''(1) - '''x'''(2)<nowiki>;</nowiki>

'''dx'''(3) <nowiki>=</nowiki> '''x'''(1)<nowiki></nowiki>'''x'''(2) - b<nowiki></nowiki>'''x'''(3)<nowiki>;</nowiki>

endfunction

| style="border-top:none;border-bottom:1pt solid #000000;border-left:1pt solid #000000;border-right:1pt solid #000000;padding:0.097cm;"| We define the '''function Lorenz''' and then define the given constants '''sigma, r '''and''' b.''' Then we define the '''first order ODEs'''.

|-
| style="border-top:none;border-bottom:1pt solid #000000;border-left:1pt solid #000000;border-right:none;padding:0.097cm;"| Highlight

x<nowiki>=</nowiki>ode(x0,t0,t,Lorenz)

| style="border-top:none;border-bottom:1pt solid #000000;border-left:1pt solid #000000;border-right:1pt solid #000000;padding:0.097cm;"| Then we call the '''ode function''' to solve the '''Lorenz system of equations'''.

We equate the solution to '''x'''.

|-
| style="border-top:none;border-bottom:1pt solid #000000;border-left:1pt solid #000000;border-right:none;padding:0.097cm;"| Highlight

plot(t,x(1,:),'**',t,x(2,:),'--', t,x(3,:),'..')

| style="border-top:none;border-bottom:1pt solid #000000;border-left:1pt solid #000000;border-right:1pt solid #000000;padding:0.097cm;"| Then we '''plot x one, x two '''and''' x three versus time. '''

|-
| style="border-top:none;border-bottom:1pt solid #000000;border-left:1pt solid #000000;border-right:none;padding:0.097cm;"| Click on Execute and select Save and Execute
| style="border-top:none;border-bottom:1pt solid #000000;border-left:1pt solid #000000;border-right:1pt solid #000000;padding:0.097cm;"| Save and execute the file''' Lorenz dot sci. '''

|-
| style="border-top:none;border-bottom:1pt solid #000000;border-left:1pt solid #000000;border-right:none;padding:0.097cm;"| Show plot
| style="border-top:none;border-bottom:1pt solid #000000;border-left:1pt solid #000000;border-right:1pt solid #000000;padding:0.097cm;"| The '''plot''' of '''x one, x two''' and '''x three versus time''' is shown.

|-
| style="border-top:none;border-bottom:1pt solid #000000;border-left:1pt solid #000000;border-right:none;padding:0.097cm;"| Slide 13- Summary
| style="border-top:none;border-bottom:1pt solid #000000;border-left:1pt solid #000000;border-right:1pt solid #000000;padding:0.097cm;"| Let us summarize this tutorial.

In this tutorial we have learnt to develop '''Scilab code''' to solve an '''ODE''' using '''Scilab ode function'''.

Then we have learnt to '''plot''' the solution.

|-
| style="border-top:none;border-bottom:1pt solid #000000;border-left:1pt solid #000000;border-right:none;padding:0.097cm;"| '''Show Slide 14'''

'''Title: About the Spoken Tutorial Project'''

* Watch the video available at [http://spoken-tutorial.org/What_is_a_Spoken_Tutorial http://spoken-tutorial.org/What_is_a_Spoken_Tutorial]

* It summarises the Spoken Tutorial project

* If you do not have good bandwidth, you can download and watch it

| style="border-top:none;border-bottom:1pt solid #000000;border-left:1pt solid #000000;border-right:1pt solid #000000;padding:0.097cm;"| * About the Spoken Tutorial Project<br/>

* Watch the video available given at the link shown below <br/>

* It summarises the Spoken Tutorial project<br/>

* If you do not have good bandwidth, you can download and watch it

|-
| style="border-top:none;border-bottom:1pt solid #000000;border-left:1pt solid #000000;border-right:none;padding:0.097cm;"| '''Show Slide 15'''

'''Title: Spoken Tutorial Workshops'''

The Spoken Tutorial Project Team

* Conducts workshops using spoken tutorials

* Gives certificates for those who pass an online test

* For more details, please write to contact@spoken-tutorial.org

| style="border-top:none;border-bottom:1pt solid #000000;border-left:1pt solid #000000;border-right:1pt solid #000000;padding:0.097cm;"| The Spoken Tutorial Project Team

* Conducts workshops using spoken tutorials

* Gives certificates for those who pass an online test

* For more details, please write to contact at spoken hyphen tutorial dot org

|-
| style="border-top:none;border-bottom:1pt solid #000000;border-left:1pt solid #000000;border-right:none;padding:0.097cm;"| '''Show Slide 16'''

'''Title: Acknowledgement'''

* Spoken Tutorial Project is a part of the Talk to a Teacher project

* It is supported by the National Mission on Education through ICT, MHRD, Government of India

* More information on this Mission is available at

* [http://spoken-tutorial.org/NMEICT-Intro http://spoken-tutorial.org/NMEICT-][http://spoken-tutorial.org/NMEICT-Intro Intro]

| style="border-top:none;border-bottom:1pt solid #000000;border-left:1pt solid #000000;border-right:1pt solid #000000;padding:0.097cm;"| * Spoken Tutorial Project is a part of the Talk to a Teacher project
* It is supported by the National Mission on Education through ICT, MHRD, Government of India
* More information on this Mission is available at
* spoken hyphen tutorial dot org slash NMEICT hyphen Intro <br/>

|-
| style="border-top:none;border-bottom:1pt solid #000000;border-left:1pt solid #000000;border-right:none;padding:0.097cm;"|
| style="border-top:none;border-bottom:1pt solid #000000;border-left:1pt solid #000000;border-right:1pt solid #000000;padding:0.097cm;"| This is Ashwini Patil from IIT Bombay signing off. Thanks for joining.

|}

@@ Line 74: / Line 74: @@
 |-
 | style="background-color:transparent;border-top:none;border-bottom:1pt solid #000000;border-left:1pt solid #000000;border-right:none;padding:0.097cm;"| Slide 7- Motion of Simple pendulum
-| style="background-color:transparent;border-top:none;border-bottom:1pt solid #000000;border-left:1pt solid #000000;border-right:1pt solid #000000;padding:0.097cm;"| Consider the motion of simple pendulum.
+| style="background-color:transparent;border-top:none;border-bottom:1pt solid #000000;border-left:1pt solid #000000;border-right:1pt solid #000000;padding:0.097cm;"| Consider the motion of '''simple pendulum'''.
@@ Line 305: / Line 305: @@
 * '''x three dash equal to x one into x two minus b into x three. '''
-The initial conditions are '''x one zero''' equal to '''minus ten, x two zero equal to ten '''and''' x three zero equal to twenty five'''.
+The initial conditions are '''x one zero''' equal to '''minus ten, x two zero''' equal to '''ten '''and''' x three zero''' equal to '''twenty five'''.
-Let '''sigma '''be''' equal to ten, r '''be '''equal to twenty eight '''and''' b equal to eight by three. '''
+Let '''sigma '''be equal to '''ten, r '''be equal to '''twenty eight '''and''' b''' equal to '''eight by three. '''
 |-
@@ Line 331: / Line 331: @@
-Since there are three different '''ODEs''', there are three initial conitions.
+Since there are three different '''ODEs''', there are three initial conditions.
@@ Line 357: / Line 357: @@
-| style="border-top:none;border-bottom:1pt solid #000000;border-left:1pt solid #000000;border-right:1pt solid #000000;padding:0.097cm;"| We define the '''function Lorenz''' and then define the given constants '''sigma, r '''and''' b.''' Then we define the '''first order ODEs'''.
+| style="border-top:none;border-bottom:1pt solid #000000;border-left:1pt solid #000000;border-right:1pt solid #000000;padding:0.097cm;"| We define the '''function Lorenz''' and then define the given constants '''sigma, r '''and''' b.'''
 |-

Scilab/C4/ODE-Applications/English - Revision history

Nancyvarkey at 18:46, 1 February 2014