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

Hello!!!

Could you help me at the following exercise?

Let

and that each solution of the differential equation has the form:.

*Last edited by evinda (2013-12-11 06:37:39)*

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 87,248

Hi;

The Wronskian is

I do not see how to use v1 and v2?

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

Hi

I think evinda is referring to this:Wronskian.

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

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 87,248

I used the one in Mathematica, they should be the same thing.

**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

The Wronskian is

|u(0) v(0)|

|u'(0) v'(0) |

But...how can I show that there are constants

*Last edited by evinda (2013-12-11 09:27:13)*

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 87,248

What functions are u and v? They are not defined.

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

Well, here they are called v1 and v2 instead of u and v. Does not change much.

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

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 87,248

What is v1 and v2?

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

They are two solutions of the differential equation for which it's true that v1/v2 is not constant.

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

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 87,248

That would require solving the DE. Since when do you need the solve the DE to get the Wronskian?

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:

They are two solutions of the differential equation for which it's true that v1/v2 is not constant.

Exactly!!The Wronskian is:

| v1(0) v2(0) |

| v1'(0) v2'(0) |

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

bobbym wrote:

That would require solving the DE. Since when do you need the solve the DE to get the Wronskian?

Which DE do you mean that I have to solve??

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 87,248

When I solve the DE, I get an answer that has to arbitrary constants s1 and c2

I can not eliminate them without some more information.

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 yields a Wronskian of

Now what?

I haven't understood..Do you mean that I have to take the derivative of the Wronskian and solve the DE or which DE do you mean??

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 87,248

This DE, y''+ay'+by=0

If v1 and v2 are solutions of that DE then you have to solve the DE to get the solutions. I have already done that.

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

I found the Characteristic equation,that is

that has discriminant withSo,

Am I right so far?And how can I continue?

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

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 87,248

I am not getting that as a solution to the DE. Where is c1 and c2 in your solution?

Did you copy the question correctly?

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

Hi evinda

Those are the solutions, but I don't think you are supposed to use them.

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

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 87,248

Shouldn't you have use the roots rather than the discriminant?

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:

I am not getting that as a solution to the DE. Where is c1 and c2 in your solution?

Did you copy the question correctly?

Yes,I copied the question correcty..Should I have used

and instead of c_{1} and ?*Last edited by evinda (2013-12-11 10:14:50)*

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

anonimnystefy wrote:

Hi evinda

Those are the solutions, but I don't think you are supposed to use them.

But???How else can I start???

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

bobbym wrote:

Shouldn't you have use the roots rather than the discriminant?

Where do you mean?At the general solution?

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 87,248

This is what I am getting for a solution to that DE:

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

evinda wrote:

I found the Characteristic equation,that is

that has discriminant with

So,Am I right so far?And how can I continue?

So is she.

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

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

And how can I continue now??

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