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

But the Wronkian should be nonzero..So,what do I have to do??

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 105,281

I do not know why it should be non zero.

Did you compute the Wronskian?

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

bobbym wrote:

I do not know why it should be non zero.

Did you compute the Wronskian?

Because there is a theorem that says that if two solutions of a differential equation are linearly independent,their Wronskian is nonzero!!!

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
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How do you know the two solutions are linearly independent?

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

bobbym wrote:

How do you know the two solutions are linearly independent?

Because the exercise says that v1,v2 are solutions of the differential equation so that

is not constant..So,.So, ...

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
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That is true but have you used the solutions to compute the wronskian?

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

bobbym wrote:

That is true but have you used the solutions to compute the wronskian?

Can't I just write that the Wronskian is equal to:

| v_{1}(0) (v_{1}(0))' |

| v_{2}(0) (v_{2}(0))' |

Do you mean that I have to solve the differential equation that is given for

and ?*Last edited by evinda (2013-12-12 08:37:26)*

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 105,281

Of course you do, but you already got the solution to DE earlier.

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

bobbym wrote:

Of course you do, but you already got the solution to DE earlier.

But then don't we find that v1(x)=v2(x) ? Or not? I haven't understood...

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**bobbym****Administrator**- From: Bumpkinland
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That is what I am saying. If there is only one solution. Is there another solution?

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

bobbym wrote:

That is what I am saying. If there is only one solution. Is there another solution?

There should be..because v1 and v2 are linearly independent

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
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What is the other solution?

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**anonimnystefy****Real Member**- From: Harlan's World
- Registered: 2011-05-23
- Posts: 16,016

There are infinitely many solutions.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 105,281

Yes but when you plug them into that determinant they are going to become 0.

We have no particular solution just a 1 general one. To continue with this route he will need 2 different general solutions.

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