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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 87,193

That makes v1(x) = v2(x) which makes the determinant 0 that means they are linearly dependent.

Seems nonsensical.

**In mathematics, you don't understand things. You just get used to them.Of course that result can be rigorously obtained, but who cares?Combinatorics is Algebra and Algebra is Combinatorics.**

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**evinda****Member**- Registered: 2013-04-13
- Posts: 105

bobbym wrote:

That makes v1(x) = v2(x) which makes the determinant 0 that means they are linearly dependent.

Seems nonsensical.

So,what have I done wrong?? Is there an other way to do this?

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 15,544

Hi evinde

For the first part, use the fact that the Wronskian is non-zero and the fact that the determinant of the system of equations is the Wronskian at x=0.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 87,193

Why is the Wronskian non - zero?

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 15,544

Because the exponential function is always non-zero.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 87,193

The determinant is 0 because v1(x) = v2(x), therefore the Wronskian is 0.

That is why I say something is very wrong.

**In mathematics, you don't understand things. You just get used to them.Of course that result can be rigorously obtained, but who cares?Combinatorics is Algebra and Algebra is Combinatorics.**

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**evinda****Member**- Registered: 2013-04-13
- Posts: 105

bobbym wrote:

Why is the Wronskian non - zero?

Also,because of the fact that the solutions are linearly independent!!!

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 87,193

Isn't that what are you trying to prove? Using it in the proof would be circular reasoning?!

Of course that result can be rigorously obtained, but who cares?

Combinatorics is Algebra and Algebra is Combinatorics.

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**evinda****Member**- Registered: 2013-04-13
- Posts: 105

anonimnystefy wrote:

Hi evinde

For the first part, use the fact that the Wronskian is non-zero and the fact that the determinant of the system of equations is the Wronskian at x=0.

How can I use these facts??I haven't understood yet...

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 15,544

Find the determinant of the system of equations given in the second to last line of your post 1. You will see that it actually represents the Wronskian at x=0.

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Taking a new step, uttering a new word, is what people fear most. ― Fyodor Dostoyevsky, Crime and Punishment

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**evinda****Member**- Registered: 2013-04-13
- Posts: 105

anonimnystefy wrote:

Find the determinant of the system of equations given in the second to last line of your post 1. You will see that it actually represents the Wronskian at x=0.

Don't you mean these two:

, d_{1}{v_{1}}'(0)+d_{2}{v_{2}}'(0)={f}'(0) ?Can I find the determinant,although I haven't shown that there and that satisfy these conditions??

*Last edited by evinda (2013-12-11 10:41:16)*

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 15,544

Well, the point of finding the determinant is to show that there are solutions, so, yes, you can. Of course, you should treat d1 and d2 as the unknowns when finding the determinant.

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**evinda****Member**- Registered: 2013-04-13
- Posts: 105

anonimnystefy wrote:

Well, the point of finding the determinant is to show that there are solutions, so, yes, you can. Of course, you should treat d1 and d2 as the unknowns when finding the determinant.

So,I find the determinant and notice that it equals to the Wronskian,so it is

.And then?Offline

**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 15,544

Yes, but first you need to find the Wronskian. You can do this using Abel's identity.

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**evinda****Member**- Registered: 2013-04-13
- Posts: 105

anonimnystefy wrote:

Yes, but first you need to find the Wronskian. You can do this using Abel's identity.

Do we have a

we could use?*Last edited by evinda (2013-12-11 11:21:58)*

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**evinda****Member**- Registered: 2013-04-13
- Posts: 105

Do you mean that I have to use

? I tried and got W(x)=(v1(0)v2'(0)-v2(0)v1'(0))*e^(-ax).Is it right so far?and how can I continue?Offline

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 87,193

What did you get for

(v1(0)v2'(0)-v2(0)v1'(0))?

Of course that result can be rigorously obtained, but who cares?

Combinatorics is Algebra and Algebra is Combinatorics.

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**evinda****Member**- Registered: 2013-04-13
- Posts: 105

bobbym wrote:

What did you get for

(v1(0)v2'(0)-v2(0)v1'(0))?

How can I find this value?

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 87,193

What do you get for v1(x) and v2(x)?

Of course that result can be rigorously obtained, but who cares?

Combinatorics is Algebra and Algebra is Combinatorics.

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**evinda****Member**- Registered: 2013-04-13
- Posts: 105

bobbym wrote:

What do you get for v1(x) and v2(x)?

I haven't found them..how could I do this?

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 87,193

It has been stated the v1 and v2 are the 2 solutions to the DE. Do you agree?

Of course that result can be rigorously obtained, but who cares?

Combinatorics is Algebra and Algebra is Combinatorics.

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 15,544

There is an infinite nunber of solutions of the DE.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 87,193

Without fixing c1 and c2, that determinant will always be zero. That means the Wronskian will be 0. Do you agree?

Of course that result can be rigorously obtained, but who cares?

Combinatorics is Algebra and Algebra is Combinatorics.

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 15,544

Huh?

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 87,193

(v1(0)v2'(0)-v2(0)v1'(0))

Without getting the 2 constants v1 = v2, therefore the above expression will always be 0.

Of course that result can be rigorously obtained, but who cares?

Combinatorics is Algebra and Algebra is Combinatorics.

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