MHB Linear ODE for a fundamental solution set

[rishadjb]

Question:
For the interval x > 0 and the function set S = { 3ln(x), ln2, ln(x), ln(5x)}, construct a linear ODE of the lowest order.

My work:

Taking the wronskian for this solution set, I get it as 0. Doesn't that mean that a linear ODE for this set cannot be found?

I'm very confused here, and any help is appreciated. Thanks

 

[Opalg]

Question:
For the interval x > 0 and the function set S = { 3ln(x), ln2, ln(x), ln(5x)}, construct a linear ODE of the lowest order.

My work:

Taking the wronskian for this solution set, I get it as 0. Doesn't that mean that a linear ODE for this set cannot be found?

I'm very confused here, and any help is appreciated. Thanks

Notice that the four functions in the set S are not linearly independent. They are all of the form $A + B\ln x$ (where A and B are constants). So for example you could replace S by the smaller set $\{\ln 2, \ln x\}$.

 

[rishadjb]

Ok I see, so since the terms are linearly dependant, we need to rewrite the fundamental set? So how do you come to the conclusion that we can use the smaller set of {ln2, ln x}. Is it because these two terms are lin. independent?

Could I use {3lnx, ln2} ?

 

[HallsofIvy]

Ok I see, so since the terms are linearly dependant, we need to rewrite the fundamental set? So how do you come to the conclusion that we can use the smaller set of {ln2, ln x}. Is it because these two terms are lin. independent?

Could I use {3lnx, ln2} ?

Yes, Opalg told you that all of those are of the form Aln(x)+ B for some A and B. You could, just as easily write them as A'(3 ln(x))+ B where A'= A/3.

Of your original set, 3ln(x), ln(2), ln(x), and ln(5x), note that ln(2) is a constant 3ln(x) is just 3 times ln(x), and ln(5x)= ln(x)+ ln(5). So all of them are of the form "a multiple of ln(x)" plus a constant. That is Aln(x)+ B.

 

[rishadjb]

Question solved, thanks guys :)