# Help with Annihilator Method for Non-homogeneous Second Order D.Es

1. Mar 4, 2013

### Ryuzaki

I must say I'm utterly confused with the Annihilator method for solving Non-Homogeneous Constant Coefficient Second order O.D.Es. I guess it'd be better to list out my questions:-

1. Is it possible to find an annihilator for every single function out there? I mean, is it always possible to make the RHS 0 regardless of the function?

2. Is it always possible to factorise the LHS after multiplication with the annihilator? The problems I've worked with, all had neat solutions, but I'm somehow not convinced. What if the annihilator is so complex (not the mathematical sense) that it becomes incredibly hard to factorise the LHS?

3. I've sticked with this method for the reasons :- 1) that it requires a minimal memorisation of formulae (and it somehow feels more rewarding!) & 2) it is(as far as I've felt) somewhat general in application. Are there other general methods that work for a large class of functions, other than the Method of Variation of Parameters or the Method of Undetermined Coefficients?

4. Lastly, why does the Annihilator method work? Can someone please explain the theory behind this to me? I've solved so many problems by this method, yet I don't think I've understood the theory completely. I'd be really grateful if someone could explain it on a step-by-step basis?

Thank you for taking the time to read this. Any and all help is appreciated.

2. Mar 4, 2013

### lurflurf

1)It depends the usual case is differential operators with constant coefficients like 3D^17-4D^15+D^7 which can annihilate products of polynomials, trigonometric polynomials and exponential polynomials like exp(7x)[3sinh(3x)-2cosh(3x)][3sin(5x)-2cos(5x)][3x^2+2x+1]

It is possible to annihilate generalize more functions, but the operator will not have constant constant coefficients

2)factorization is easy in principle any polynomial can be factored into linear factors (x-a) over complex numbers , if we know e^(-a x) is a solution the accuracy will be limited by how accurately we know a.

3)all the methods are related, there are also integrating factors and various heuristics
The idea is if we assume the solution is restricted in some way we may be able find it more quickly, but we risk the method failing if for some cases. More general methods risk wasting effort and getting solutions in a less useful form (like a messy integral).

4)We want to solve Py=f where P is an operator and y is the unknown function and f is the known function
fin an operator Q so that
Qf=0
then
QPy=0
this is a homogeneous equation if Y is a solution it can be separated into two parts
Y=y1+y2
if we chose Py1=0 Qy2=f we have solved the original eqation
where Py1=0