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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.

There are infinitely many solutions.

What is the other solution?

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

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

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

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...

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

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

bobbym wrote:How do you know the two solutions are linearly independent?

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

bobbym wrote:I do not know why it should be non zero.Did you compute the Wronskian?

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!!!

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

(v1(0)v2'(0)-v2(0)v1'(0))Without getting the 2 constants v1 = v2, therefore the above expression will always be 0.