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But the Wronkian should be nonzero..So,what do I have to do??
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I do not know why it should be non zero.
Did you compute the Wronskian?
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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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How do you know the two solutions are linearly independent?
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
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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,Offline
That is true but have you used the solutions to compute the wronskian?
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
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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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Of course you do, but you already got the solution to DE earlier.
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
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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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That is what I am saying. If there is only one solution. Is there another solution?
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
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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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What is the other solution?
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
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There are infinitely many solutions.
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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.
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
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