Discussion about math, puzzles, games and fun. Useful symbols: ÷ × ½ √ ∞ ≠ ≤ ≥ ≈ ⇒ ± ∈ Δ θ ∴ ∑ ∫ π -¹ ² ³ °

You are not logged in.

- Topics: Active | Unanswered

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

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

Offline

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 84,605

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.**

Offline

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

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

Offline

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 84,605

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.

**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

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

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

Offline

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 84,605

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.**

Offline

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

OK that's what I need to know

Thanks for help

Offline

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 84,605

What is the missing initial condition?

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

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

y0=1

x0=1

do you mean ?

Offline

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 84,605

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

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

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

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

Offline

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

Can I make Integral to converted to first derivative ?

Offline

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 84,605

Can I make Integral to converted to first derivative ?

I am not following you. What do you mean?

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

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

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

Offline

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 84,605

Hi;

Can you copy the question exactly as it appears?

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

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

∫d2y/dx2=∫(2+y/x)

this lead to

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

Offline

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 84,605

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

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

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

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

Offline

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 84,605

Yes, there would be a constant.

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

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

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

Offline

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 84,605

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

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

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

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

Offline

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 84,605

Usually then you need z(0).

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

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

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

Offline

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 84,605

How can you determine the constant without another initial condition?

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