Annihilator Operators and higher than second order results for aux equation

• Feb 26th 2010, 02:06 PM
eingeist
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. \$\displaystyle (D-1)^2\$ would be \$\displaystyle c1e^x + xc2e^x.\$

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

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

What do you think?
• Feb 26th 2010, 02:21 PM
TheEmptySet
Quote:

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

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

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

What do you think?

Actually for \$\displaystyle (D+1)^3\$ you would get \$\displaystyle 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 \$\displaystyle (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.
• Feb 26th 2010, 02:21 PM
eingeist
read \$\displaystyle csub1 , csub2, etc. \$ where I have c1 and c2.. I'm just learning the code...

edit: well, still learning. can't figure out subscripts(Itwasntme)
• Feb 26th 2010, 02:29 PM
eingeist
The original equation is:

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

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

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

\$\displaystyle (D-1) D^2 D \$

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

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

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