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## Topic review (newest first)

bobbym
2013-12-14 10:29:15

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.

anonimnystefy
2013-12-14 10:26:53

There are infinitely many solutions.

bobbym
2013-12-14 10:20:41

What is the other solution?

evinda
2013-12-14 10:18:19

#### 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

bobbym
2013-12-14 08:54:31

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

evinda
2013-12-14 08:37:07

#### 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...

bobbym
2013-12-13 09:52:48

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

evinda
2013-12-13 07:37:03

#### 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
?

bobbym
2013-12-13 07:13:08

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

evinda
2013-12-13 06:57:13

#### 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,
...

bobbym
2013-12-13 05:54:48

How do you know the two solutions are linearly independent?

evinda
2013-12-13 05:48:00

#### 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!!!

bobbym
2013-12-13 05:43:02

I do not know why it should be non zero.

Did you compute the Wronskian?

evinda
2013-12-13 05:40:43

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

bobbym
2013-12-13 05:28:36

(v1(0)v2'(0)-v2(0)v1'(0))

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