# Differential Equations: Undetermined Coefficients (Annihilator Method) (in Off-topic)

##
*SNK3R*
March 5 2006 5:18 PM EST

Just recently in my DE class, we went over this method. It's interesting, to say the least, but I have a question regarding more complex functions.

To start out, a basic function can easily be annihilated by seeing how many times it'll take the function to get the derivative equal to zero. For example:

8x^{3} - 5x^{2} + 1 can be annihilated to D^{4} [take the derivative four times (i.e., power + 1)].

However, when I got to exponential functions, I started not being able to "see" them quite as easily. For example:

e^{2x} can be annihilated to D - 2.

Proven the way by my professor:

(D - 2)e^{2x}

De^{2x} - 2e^{2x} = 0 [De^{2x} is the derivative of e^{2x}, which is 2e^{2x}.]

If De^{2x} = 2e^{2x}, then we can say 2e^{2x} - 2e^{2x} = 0, so it's true that D - 2 is the annihilator for the function e^{2x}.

This makes sense when it's proven, but is there a trick to see them easier than having to guess and check to see which annihilators relate to the function? For example, how would I "guess" the annihilator for the function e^{2x + 1}?

Any help is appreciated. ;)

##
*Sacredpeanut*
March 5 2006 5:35 PM EST

Not sure if this would be the "correct" method, but this is how I would do it.

1) Differentiate Original Function

eg:

F=e(2x+1)

DF=2e(2x+1)

2) Relate Differential to Original Function

eg:

DF=2e(2x+1)

DF=2F

3) Solve

eg:

DF=2F

DF-2F=0

F(D-2)=0

In other words multiplying the original function by D-2 will annihilate it to 0. Therefore e(2x+1) can be annihilated to D-2.

##
*SNK3R*
March 5 2006 5:46 PM EST

Wow, that is brilliant! That totally rocks. Thanks a **bunch**. Now I can actually solve for the annihilator instead of guessing!

why is it D - 2, instead of being D - 2F ? I'm not in DE yet, im only in calc 3. But this stuff does make sense to me, what you're talking about, lol.

ok i see you had the F(x) out front muliplying through the D-2....didnt see that far down!

##
*Sacredpeanut*
March 5 2006 5:53 PM EST

No worries :)

It's easy to write an exponential function in terms of it's derivative but is alot harder with other functions so this may not be the best method for all functions.

##
*SNK3R*
March 5 2006 5:56 PM EST

Yeah, I was looking at some of the examples later in my homework assignment that deal. For example: 6sin(x), x^{2}e^{x} + 5, and e^{x}sin(x). I guess I'll mess with them once I get to them. But this trick for exponential functions really rocks. I've already tested it on a few sample exponential functions. :)

for the 6sin(x) wouldnt it be something like F(x) + D^2 (F(x)) = 0 ?

6sin(x) - 6 sin(x) = 0

##
*Sacredpeanut*
March 5 2006 6:10 PM EST

Yes, 1+D^2 would therefore be the annihilator.

(Fx) + D^2(Fx)=0

(Fx)(1+D^2)=0

I think I did a similar paper to you last year, so if you need any help just ask :)

##
*SNK3R*
March 5 2006 6:10 PM EST

Yeah, I just did it for the problem on my homework.

D^{2} + 1 will annihilate 6sin(x):

6sin(x) * (D^{2} + 1)

D^{2}6sin(x) + 6sin(x)

-6sin(x) + 6sin(x) = 0

Good job, you'll be ready once you hit DE. ;)

##
*Stephen*
March 5 2006 6:15 PM EST

Hopefully next week he'll teach you the more general variation of parameters method for those cases where an annihilator doesn't exist

##
*SNK3R*
March 5 2006 6:21 PM EST

If you're referring to homogeneous linear ODE's with constant coefficients (problems that have 3 general solutions: 1 repeated solution, 2 different solutions, or 2 complex solutions?), we've already discussed that. ;) If not, I can't wait. I'm actually enjoyed Differential Equations more than calculus. ;)

calculus sucks, period...lol.

I'm trying to do a take home test right now....

If you want, try and help me out too....lol

heres one of the questions i cant figure out:

show that dN/ds = -kT + tB

using dT/ds = kN , and dB/ds = -tN

where:

d = derivative of

N = unit normal vector

k = kappa

T = unit tangent vector

B = Binormal vector

t = torsion

Differentials of parametrical trigonometrical equations are bad enough before I even reach Uni (glad to not be doing a Maths course when I get there).
Implicit differentiation and rate of change isn't much better although is shortest chapter in the book before we start on vectors luckily.

I used to know this.

:(

Makes me sad...

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