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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 90,799

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 agree with you regarding the satisfaction and importance of actually computing some numbers. I can't tell you how often I see time and money wasted because someone didn't bother to run the numbers.**

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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: 15,673

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
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Why is the Wronskian non - zero?

**In mathematics, you don't understand things. You just get used to them.**

**I agree with you regarding the satisfaction and importance of actually computing some numbers. I can't tell you how often I see time and money wasted because someone didn't bother to run the numbers.**

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 15,673

Because the exponential function is always non-zero.

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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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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 90,799

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 agree with you regarding the satisfaction and importance of actually computing some numbers. I can't tell you how often I see time and money wasted because someone didn't bother to run the numbers.**

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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
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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.**

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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
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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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**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
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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

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
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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
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What did you get for

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

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

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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
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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.**

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**anonimnystefy****Real Member**- From: The Foundation
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There is an infinite nunber of solutions of the DE.

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**bobbym****Administrator**- From: Bumpkinland
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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.**

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
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Huh?

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 90,799

(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.**

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