ANNIH

The key idea of the annihilator method is to replace the problem of solving a non-homogeneous equation with the problem of solving a higher order homogeneous equation. The method is discussed in Section 2. 11 of Cottonwood's book. The more popular alternate approach is discussed in sections 5. 4-5. 5 of Trench's book. So we begin with a brief discussion of higher order linear homogeneous equations with constant coefficients. This is done in Section 2. 7 of Codington, as well as section 9. 2 of Trench, in more depth and greater detail.Such depth is not necessary for our purposes. So consider an equation of the form y (n) + an-I y+ahoy=O. Based on our experience with second order equations, we would naturally try solution of the form y = erg . If you go through the motions of differentiating and substituting into the equation you will get where urn + an-I urn-l + . +air+AAA, which is as before called the characteristic polynomial. The difficulty is that now if n > 2, the polynomial is of higher degree than before and such polynomials are hard to factor and find roots.We do not have available the quadratic formula. There are cubic formulas and quarter formulas that are known and used to appear in books, but they are rarely taught any more and no such formulas are available for polynomials of degree 5 or higher. So in practice it can be very hard to find the roots of the characteristic polynomial. Nevertheless, we can at least imagine factoring the polynomial and finding the roots. In general there would be a number of linear and irreducible quadratic factors. The quadratic factors might lead to complex roots.Any of these factors might be repeated and we would then get roots that appeared more than once. Suppose there were k distinct real roots RL , re , ; ark . For each such root, we would have a solution of the form yes = erg x . Then there might be several pairs of roots of the form a Ð’Â ± I;. These would give us pairs of real solutions of the form ex. coos (;x), ex. sin(;x). We saw in Chapter 3 that if a root occurred twice, we got an additional solution of the form Xerox . This still happens but more is true. Let me just make an authoritative statement, which I will make some explanatory comments about later.If a real root rig occurs times, then each of the functions XML erg x , for m = O, 1, ; ; , -? 1, is a solution. Similarly, if the pair a Ð’Â ± I; occurs times, then each 1 of the pairs of functions XML ex. coos(;x), XML ex. sin(;x), for m = O, 1, are solutions. Thus we can write down n solutions of the differential equation. For example, suppose in a problem we ended up with the factored characteristic polynomial p(r) = re (r – 2)3 (re + or + 3)2 . Then v' the root RL O occurs 4 times, the root re 2 occurs 3 times, and the pair of roots -?1 Ð’Â ± ii occurs 2 times.Thus we get as solutions 1, x, xx , xx , ex. , sex , xx ex. , e-x cost xx), e-x sin( xx), exe-x coos( xx), exe-x sin( xx), giving 11 solutions in all. ( Do you see how the first 4 of these solutions come from the root RL = O? ) Note that p(r) has degree 11 so the initial differential equation would have been of order 1 1 . Since the equation was assumed linear, the linearity properties would guarantee we could multiply each of these 1 1 solutions by an arbitrary constant and add to get many solutions of the original problem.We will see below that in all cases that occur, the polynomials will actually be easy to factor and we will not have any reason to be disheartened. With this preparation, we pass to a discussion of the annihilator method for constant coefficient linear differential equations. In order for the method to work, the equation to be solved must be of the form L(y) = f (x), where 1. L is linear with constant coefficients. . The non-homogeneous term f (x) is a solution Of a homogeneous differential equation M (y) = O, where M is linear with constant coefficients.So the basic idea is to do something to both sides of the g iven inhomogeneous differential equation so that the result is a homogeneous differential equation and we can do ha ewe already know how to do. Here is a motivational example: If we differentiate this equation twice, we get Clearly any solution of (1 ) is a solution of (2) (differentiating both sides of any true equation gives a true equation), but not conversely (two functions which fifer by a constant still have the same derivative). Thus the general solution of (2) will contain all solutions of (1), together with many extraneous solutions.