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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 95,907

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.**

**If it ain't broke, fix it until it is.**

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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: 15,922

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.

Here lies the reader who will never open this book. He is forever dead.

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

The knowledge of some things as a function of age is a delta function.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 95,907

Why is the Wronskian non - zero?

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

**If it ain't broke, fix it until it is.**

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 15,922

Because the exponential function is always non-zero.

Here lies the reader who will never open this book. He is forever dead.

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

The knowledge of some things as a function of age is a delta function.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 95,907

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.**

**If it ain't broke, fix it until it is.**

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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: 95,907

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.**

**If it ain't broke, fix it until it is.**

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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: 15,922

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.

Here lies the reader who will never open this book. He is forever dead.

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

The knowledge of some things as a function of age is a delta function.

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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: 15,922

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.

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

The knowledge of some things as a function of age is a delta function.

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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: 15,922

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

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

The knowledge of some things as a function of age is a delta function.

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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: 95,907

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.**

**If it ain't broke, fix it until it is.**

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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: 95,907

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

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

**If it ain't broke, fix it until it is.**

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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: 95,907

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.**

**If it ain't broke, fix it until it is.**

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 15,922

There is an infinite nunber of solutions of the DE.

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

The knowledge of some things as a function of age is a delta function.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 95,907

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.**

**If it ain't broke, fix it until it is.**

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 15,922

Huh?

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

The knowledge of some things as a function of age is a delta function.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 95,907

(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.**

**If it ain't broke, fix it until it is.**

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