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

1. ## 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. $(D-1)^2$ would be $c1e^x + xc2e^x.$

However, what if I had $D^3?$ or $(D+1)^3?$ Judging by what happens in a square, I am guessing I am integrating? But I am unsure.

So would it be $c1e^-x + xc2e^-x + ((x^2)/2)c3e^-x?$

What do you think?

2. 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. $(D-1)^2$ would be $c1e^x + xc2e^x.$

However, what if I had $D^3?$ or $(D+1)^3?$ Judging by what happens in a square, I am guessing I am integrating? But I am unsure.

So would it be $c1e^-x + xc2e^-x + )(x^2)/2)c3e^-x?$

What do you think?
Actually for $(D+1)^3$ you would get $y=c_1e^{-t}+c_2te^{-t}+c_3t^2e^{-t}$

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

In general If you have $(D+1)^n=c_1e^{-t}+c_2te^{-t}+...c_nt^{n-1}e^{-t}$

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

3. read $csub1 , csub2, etc.$ where I have c1 and c2.. I'm just learning the code...

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

4. The original equation is:

$y^{II} + 5y^I + 6y = 2-3x+4e^{-x}$

Shouldn't I be able to use $D^2+5D+6$ , Factor:

$(D+2)(D+3)$ and use annihilators on the right side and add them to the left:

$(D-1) D^2 D$

So I would have $(D-1)D^2D(D+2)(D+3) = 0$

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

$
y=c_1e^{-3x} + c_2e^{-2x} + c_3 + xc_4 + ?$