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#1 2014-05-02 00:41:02

ammar
Member
Registered: 2013-12-16
Posts: 69

Modified Euler method

Use modified Euler method with (h=0.1) to solve the following differential equation:

d2y/dx2=2+y/x , Where y(1)=1 for x=1 to 1.4


How to convert second derivative to first  derivative

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#2 2014-05-02 01:03:58

bobbym
Administrator
From: Bumpkinland
Registered: 2009-04-12
Posts: 83,211

Re: Modified Euler method

Hi ammar;

The Euler method ( modified or not ) is usually for a first order ODE.


In mathematics, you don't understand things. You just get used to them.
I have the result, but I do not yet know how to get it.
All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

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#3 2014-05-02 01:10:10

ammar
Member
Registered: 2013-12-16
Posts: 69

Re: Modified Euler method

Yes,but we can solve it by  change the second derivative to first derivative but how to change this equation ?

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#4 2014-05-02 01:13:15

bobbym
Administrator
From: Bumpkinland
Registered: 2009-04-12
Posts: 83,211

Re: Modified Euler method

Yes, you can change any second order de into 2 first order or coupled ones.

I have to get offline for a bit, in the meantime watch this video. When I get back if it still is a problem then we will work on it.

http://www.youtube.com/watch?v=k2V2UYr6lYw


In mathematics, you don't understand things. You just get used to them.
I have the result, but I do not yet know how to get it.
All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

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#5 2014-05-02 01:26:51

ammar
Member
Registered: 2013-12-16
Posts: 69

Re: Modified Euler method

my problem is : can I assume dy/dx=z when this equation does not contain dy/dx

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#6 2014-05-02 04:36:20

bobbym
Administrator
From: Bumpkinland
Registered: 2009-04-12
Posts: 83,211

Re: Modified Euler method

Hi;

Is the first equation, the second is

You left out an initial condition in your original problem.


In mathematics, you don't understand things. You just get used to them.
I have the result, but I do not yet know how to get it.
All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

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#7 2014-05-02 04:42:51

ammar
Member
Registered: 2013-12-16
Posts: 69

Re: Modified Euler method

OK that's what I need to know

Thanks for help  smile   smile   smile

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#8 2014-05-02 04:43:58

bobbym
Administrator
From: Bumpkinland
Registered: 2009-04-12
Posts: 83,211

Re: Modified Euler method

What is the missing initial condition?


In mathematics, you don't understand things. You just get used to them.
I have the result, but I do not yet know how to get it.
All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

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#9 2014-05-02 05:11:51

ammar
Member
Registered: 2013-12-16
Posts: 69

Re: Modified Euler method

y0=1
x0=1

do you mean ?

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#10 2014-05-02 05:18:49

bobbym
Administrator
From: Bumpkinland
Registered: 2009-04-12
Posts: 83,211

Re: Modified Euler method

There should be a y'(x) too. But just to get the 2 equations you do not need it.


In mathematics, you don't understand things. You just get used to them.
I have the result, but I do not yet know how to get it.
All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

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#11 2014-05-02 05:38:21

ammar
Member
Registered: 2013-12-16
Posts: 69

Re: Modified Euler method

The equation doesn't have (dy/dx) ,I think I don't need it

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#12 2014-05-02 06:15:07

ammar
Member
Registered: 2013-12-16
Posts: 69

Re: Modified Euler method

Can I make Integral to converted to first derivative ?

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#13 2014-05-02 08:27:33

bobbym
Administrator
From: Bumpkinland
Registered: 2009-04-12
Posts: 83,211

Re: Modified Euler method

Can I make Integral to converted to first derivative ?

I am not following you. What do you mean?


In mathematics, you don't understand things. You just get used to them.
I have the result, but I do not yet know how to get it.
All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

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#14 2014-05-02 16:27:50

ammar
Member
Registered: 2013-12-16
Posts: 69

Re: Modified Euler method

convert (d2y/dx2=2+y/x)  to (dy/dx) by integral ??

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#15 2014-05-02 19:53:53

bobbym
Administrator
From: Bumpkinland
Registered: 2009-04-12
Posts: 83,211

Re: Modified Euler method

Hi;

Can you copy the question exactly as it appears?


In mathematics, you don't understand things. You just get used to them.
I have the result, but I do not yet know how to get it.
All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

Offline

#16 2014-05-02 22:07:30

ammar
Member
Registered: 2013-12-16
Posts: 69

Re: Modified Euler method

∫d2y/dx2=∫(2+y/x)

this lead to

dy/dx=2x+y*ln(x) , is this correct ?

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#17 2014-05-02 22:11:58

bobbym
Administrator
From: Bumpkinland
Registered: 2009-04-12
Posts: 83,211

Re: Modified Euler method

Oh I see what you wanted now. Sorry, I did not understand. Yes, that is what I would get.


In mathematics, you don't understand things. You just get used to them.
I have the result, but I do not yet know how to get it.
All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

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#18 2014-05-02 22:17:03

ammar
Member
Registered: 2013-12-16
Posts: 69

Re: Modified Euler method

But I think it's would be dy/dx=2x+y*ln(x)+constant

and if we have a boundary condition (dy/dx) I will find the constant

but in the question the boundary is  y(1)=1, that's not work

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#19 2014-05-02 22:18:23

bobbym
Administrator
From: Bumpkinland
Registered: 2009-04-12
Posts: 83,211

Re: Modified Euler method

Yes, there would be a constant.


In mathematics, you don't understand things. You just get used to them.
I have the result, but I do not yet know how to get it.
All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

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#20 2014-05-02 22:25:50

ammar
Member
Registered: 2013-12-16
Posts: 69

Re: Modified Euler method

OK that's mean the integral to find the first derivative does't work

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#21 2014-05-02 22:29:19

bobbym
Administrator
From: Bumpkinland
Registered: 2009-04-12
Posts: 83,211

Re: Modified Euler method

That is why I asked you what the initial conditions were. You need 2, y(1) = 1 and y'(x) = something.


In mathematics, you don't understand things. You just get used to them.
I have the result, but I do not yet know how to get it.
All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

Offline

#22 2014-05-02 23:11:28

ammar
Member
Registered: 2013-12-16
Posts: 69

Re: Modified Euler method

OK thanks for help I'll try to solve it by the assumption of dy/dx=z

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#23 2014-05-03 01:08:33

bobbym
Administrator
From: Bumpkinland
Registered: 2009-04-12
Posts: 83,211

Re: Modified Euler method

Usually then you need z(0).


In mathematics, you don't understand things. You just get used to them.
I have the result, but I do not yet know how to get it.
All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

Offline

#24 2014-05-03 05:25:07

ammar
Member
Registered: 2013-12-16
Posts: 69

Re: Modified Euler method

Yes but the question like i posted , any idea for solving ?

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#25 2014-05-03 07:21:26

bobbym
Administrator
From: Bumpkinland
Registered: 2009-04-12
Posts: 83,211

Re: Modified Euler method

How can you determine the constant without another initial condition?


In mathematics, you don't understand things. You just get used to them.
I have the result, but I do not yet know how to get it.
All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

Offline

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