[SOLVED] How to find differential operators/annihilators?

**horan** · November 3rd 2007, 11:23 AM

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:
$x+cos(x)$

Thanks in advance!

**topsquark** · November 3rd 2007, 04:22 PM

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:
$x+cos(x)$

Thanks in advance!

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

**horan** · November 4th 2007, 08:33 AM

Thank you so much!

**topsquark** · November 4th 2007, 07:04 PM

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

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