# Thread: [SOLVED] How to find differential operators/annihilators?

1. ## [SOLVED] How to find differential operators/annihilators?

I don't understand how to find differential operators that annihilate functions... my professor just does them in lecture like they are nothing, but none of my friends (nor I) know how these work.

Some examples...

$y=x^2 exp(2x) \rightarrow(D-2)^3$
$y=x^3 exp(-x) \rightarrow(D+1)^4$
$y=exp(-x) cos(3x)\rightarrow(D+1)^2+3^2$
$y=x exp(-x) cos(3x)\rightarrow[(D+1)^2+3^2 ]^2$
$y=x^3 exp(-x) cos(3x)\rightarrow[(D+1)^2+3^2 ]^4$

But how would I, for example, annihilate:
$x+cos(x)$

2. Originally Posted by horan
I don't understand how to find differential operators that annihilate functions... my professor just does them in lecture like they are nothing, but none of my friends (nor I) know how these work.

Some examples...

$y=x^2 exp(2x) \rightarrow(D-2)^3$
$y=x^3 exp(-x) \rightarrow(D+1)^4$
$y=exp(-x) cos(3x)\rightarrow(D+1)^2+3^2$
$y=x exp(-x) cos(3x)\rightarrow[(D+1)^2+3^2 ]^2$
$y=x^3 exp(-x) cos(3x)\rightarrow[(D+1)^2+3^2 ]^4$

But how would I, for example, annihilate:
$x+cos(x)$

I've never done it quite this way, but probably the easiest method would be to find a homogeneous linear differential equation, such that $y = x + cos(x)$ is a solution.

So let
$y = x + cos(x)$

Then
$y^{\prime} = 1 - sin(x)$
and
$y^{\prime \prime} = -cos(x)$

Note that
$y^{\prime \prime} + y = x$

We aren't done yet because this is not a homogeneous equation. But
$y^{(3)} + y^{\prime} = 1$

$y^{(4)} + y^{\prime \prime} = 0$

So your annihilator would be $(D^4 + D^2)$.

-Dan

3. Thank you so much!

4. Originally Posted by horan
Thank you so much!
You are welcome. I should add that the method I used in my solution is only guaranteed to find an annihilator. It will not necessarily find the simplest annihilator.

-Dan