Thread: Annihilator Operators and higher than second order results for aux equation

Hello,

I am solving a DE by the Annihilator method.

I know that if you have a solution for the auxillary repeated then you must add an (x) before the next constant.

i.e. would be

However, what if I had or Judging by what happens in a square, I am guessing I am integrating? But I am unsure.

So would it be


What do you think?

Originally Posted by eingeist

Hello,

I am solving a DE by the Annihilator method.

I know that if you have a solution for the auxillary repeated then you must add an (x) before the next constant.

i.e. would be

However, what if I had or Judging by what happens in a square, I am guessing I am integrating? But I am unsure.

So would it be


What do you think?

Actually for you would get

One way to derive this is to use reduction of order.

In general If you have

By taking the Wronskian you can see that these are linearly independant.

read where I have c1 and c2.. I'm just learning the code...

edit: well, still learning. can't figure out subscripts

The original equation is:



Shouldn't I be able to use , Factor:

and use annihilators on the right side and add them to the left:



So I would have

See my problem? I have repeated D three times. So