1. Jared Rosa Math 480 Spring 2013

2. L(y) is the differential operator on y. E.g. y”’ - 3y” + 2y = x ; can be rewritten as L(y) = x. P c (r) is the characteristic polynomial of an ordinary differential equation. E.g. P c (r) = r 3 – 3r 2 + 2, is the characteristic polynomial of the ode y”’ - 3y” + 2y. Some Notations & Definitions

3. The Annihilator method is a faster method than the variation of constants used to solve non- homogenous differential equations with constant coefficients. So what is the Annihilator Method?

4. The annihilator method uses a second differential operator denoted M, where the solution of the ode M(b) = O is the right hand side of the first differential operator L(y). Then using the Pc(r) of M(L(y)) we can determine the solutions of L(y). E.G. L(y) = b(x) & M(b) = O So we have M(L(y)) = M(b) = O Hence, M annihilates L. How does the method work?

5. The annihilator method only has two simple requirement: 1)The differential operator that you want to annihilate has constant coefficients. E.g. L(y) = ay” + by’ + cy, where a, b, and c are constants 2)There exists a second differential operator M, such that if L(y) = f(x), then M(f) = O. Or said another way, the solution to the homogenous equation M(y) is f(x). When Can I use this Method?

6. FunctionP c (r) of an annihilator r – a (r-a) k+1 r 2 + a 2 (r 2 + a 2 ) k+1 e ax x k e ax sin(ax), cos(ax) (a real) x k sin(ax), x k cos(ax) (a real) List of Annihilators

7. Annihilator MethodVariation Of Constants Method L(y) = y”’ + y” + y’ + y = 1 y(0) = 0, y’(o) = 1, y”(0) = 0 First set L(y) = 0 to get P c (r): P c (r) = r 3 + r 2 + r + 1 = (r 2 + 1)(r + 1) So, y 1 = cos(x), y 2 = sin(x), y 3 = e -x and y p = U 1 y 1 + U 2 y 2 + U 3 y 3, where U 1 = ½(cos(x) – sin(x)) U 2 = ½(sin(x) + cos(x)) U 3 = ½e -x So y g = 1 + c 1 sin(x) + c 2 cos(x) + c 3 e -x And y = = 1 – ½ e -x + ½ sin(x) – ½ cos(x) L(y) = y”’ + y” + y’ + y = 1 y(0) = 0, y’(o) = 1, y”(0) = 0 M(y) = y’ = 0 M(L(y)) = y (4) + y”’ + y” + y’ = 0 P c (r) = r 4 + r 3 + r 2 +r 0 = r 4 + r 3 +r 2 + r = r(r 3 + r 2 + r + 1) 0 = r(r 2 + 1)(r + 1) y g = a + be -x + csin(x) + dcos(x) We assume a=1 since a has to be a solution of L(y) to get: y = 1 – ½ e -x + ½ sin(x) – ½ cos(x) The Annihilator Vs. VOC

8. Coddington, Earl A. An introduction to ordinary differential equations. Dover Publications, inc. New York: 1961. pg 86 – 92. Source