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## #26 2013-12-12 09:21:09

bobbym

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### Re: Wronskian use identities !

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.
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.

## #27 2013-12-12 09:24:04

evinda
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### Re: Wronskian use identities !

#### 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?

## #28 2013-12-12 09:24:41

anonimnystefy
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### Re: Wronskian use identities !

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.

The limit operator is just an excuse for doing something you know you can't.
“It's the subject that nobody knows anything about that we can all talk about!” ― Richard Feynman
“Taking a new step, uttering a new word, is what people fear most.” ― Fyodor Dostoyevsky, Crime and Punishment

## #29 2013-12-12 09:25:52

bobbym

Online

### Re: Wronskian use identities !

Why is the Wronskian non - zero?

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.

## #30 2013-12-12 09:26:41

anonimnystefy
Real Member

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### Re: Wronskian use identities !

Because the exponential function is always non-zero.

The limit operator is just an excuse for doing something you know you can't.
“It's the subject that nobody knows anything about that we can all talk about!” ― Richard Feynman
“Taking a new step, uttering a new word, is what people fear most.” ― Fyodor Dostoyevsky, Crime and Punishment

## #31 2013-12-12 09:28:27

bobbym

Online

### Re: Wronskian use identities !

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.
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.

## #32 2013-12-12 09:28:46

evinda
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### Re: Wronskian use identities !

#### bobbym wrote:

Why is the Wronskian non - zero?

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

## #33 2013-12-12 09:29:37

bobbym

Online

### Re: Wronskian use identities !

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

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.

## #34 2013-12-12 09:29:57

evinda
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### Re: Wronskian use identities !

#### 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...

## #35 2013-12-12 09:32:01

anonimnystefy
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### Re: Wronskian use identities !

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.

The limit operator is just an excuse for doing something you know you can't.
“It's the subject that nobody knows anything about that we can all talk about!” ― Richard Feynman
“Taking a new step, uttering a new word, is what people fear most.” ― Fyodor Dostoyevsky, Crime and Punishment

## #36 2013-12-12 09:37:32

evinda
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### Re: Wronskian use identities !

#### 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 (2013-12-12 09:41:16)

## #37 2013-12-12 09:40:43

anonimnystefy
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### Re: Wronskian use identities !

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.

The limit operator is just an excuse for doing something you know you can't.
“It's the subject that nobody knows anything about that we can all talk about!” ― Richard Feynman
“Taking a new step, uttering a new word, is what people fear most.” ― Fyodor Dostoyevsky, Crime and Punishment

## #38 2013-12-12 09:43:29

evinda
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### Re: Wronskian use identities !

#### 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?

## #39 2013-12-12 09:50:29

anonimnystefy
Real Member

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### Re: Wronskian use identities !

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

The limit operator is just an excuse for doing something you know you can't.
“It's the subject that nobody knows anything about that we can all talk about!” ― Richard Feynman
“Taking a new step, uttering a new word, is what people fear most.” ― Fyodor Dostoyevsky, Crime and Punishment

## #40 2013-12-12 10:21:16

evinda
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### Re: Wronskian use identities !

#### 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 (2013-12-12 10:21:58)

## #41 2013-12-13 00:23:26

evinda
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### Re: Wronskian use identities !

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?

## #42 2013-12-13 00:39:40

bobbym

Online

### Re: Wronskian use identities !

What did you get for

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

## #43 2013-12-13 00:45:13

evinda
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### Re: Wronskian use identities !

#### bobbym wrote:

What did you get for

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

How can I find this value?

## #44 2013-12-13 00:49:14

bobbym

Online

### Re: Wronskian use identities !

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

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.

## #45 2013-12-13 01:00:19

evinda
Full Member

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### Re: Wronskian use identities !

#### bobbym wrote:

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

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

## #46 2013-12-13 01:06:41

bobbym

Online

### Re: Wronskian use identities !

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

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.

## #47 2013-12-13 03:33:18

anonimnystefy
Real Member

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### Re: Wronskian use identities !

There is an infinite nunber of solutions of the DE.

The limit operator is just an excuse for doing something you know you can't.
“It's the subject that nobody knows anything about that we can all talk about!” ― Richard Feynman
“Taking a new step, uttering a new word, is what people fear most.” ― Fyodor Dostoyevsky, Crime and Punishment

## #48 2013-12-13 05:24:52

bobbym

Online

### Re: Wronskian use identities !

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

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.

## #49 2013-12-13 05:26:54

anonimnystefy
Real Member

Offline

### Re: Wronskian use identities !

Huh?

The limit operator is just an excuse for doing something you know you can't.
“It's the subject that nobody knows anything about that we can all talk about!” ― Richard Feynman
“Taking a new step, uttering a new word, is what people fear most.” ― Fyodor Dostoyevsky, Crime and Punishment

## #50 2013-12-13 05:28:36

bobbym

Online

### Re: Wronskian use identities !

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

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