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**Unformatted text preview: **18 Arbitrary Homogeneous Linear Equations with Constant Coefficients In chapter 16, we saw how to solve any equation of the form ay ′′ + by ′ + cy = when a , b and c are real constants. Unsurprisingly, the same basic ideas apply when dealing with any equation of the form a y ( N ) + a 1 y ( N − 1 ) + ··· + a N − 2 y ′′ + a N − 1 y ′ + a N y = when N is some positive integer and the a k ’s are all real constants. Assuming y = e rx still leads to the corresponding “characteristic equation” a r N + a 1 r N − 1 + ··· + a N − 2 r 2 + a N − 1 r + a N = , and a general solution to the differential equation can then be obtained using the solutions to the characteristic equation, much as we did in chapter 16. Computationally, the only significant difficulty is in the algebra needed to find the roots of the characteristic polynomial. So let us look at that algebra, first. 18.1 Some Algebra A basic fact of algebra is that any second-degree polynomial p ( r ) = ar 2 + br + c can be factored to p ( r ) = a ( r − r 1 )( r − r 2 ) where r 1 and r 2 are the roots of the polynomial (i.e., the solutions to p ( r ) = 0 ). These roots may be complex, in which case r 1 and r 2 are complex conjugates of each other (assuming a , b and c are real numbers). It is also possible that r 1 = r 2 , in which case the factored form of the polynomial is more concisely written as p ( r ) = a ( r − r 1 ) 2 . 375 376 Arbitrary Homogeneous Linear Equations with Constant Coefficients The idea of “factoring”, of course, extends to polynomials of higher degree. And to use this idea with these polynomials, it will help to introduce the “completely factored form” for an arbitrary N th-degree polynomial p ( r ) = a r N + a 1 r N − 1 + ··· + a N − 2 r 2 + a N − 1 r + a N . We will say that we’ve (re)written this polynomial into its completely factored form if and only if we’ve factored it to an expression of the form p ( r ) = a ( r − r 1 ) m 1 ( r − r 2 ) m 2 ··· ( r − r K ) m K (18.1) where { r 1 , r 2 , . . . , r K } is the set of all different (possibly complex) roots of the polynomial (i.e., values of r satisfying p ( r ) = 0 ), and { m 1 , m 2 , . . . , m K } is some corresponding set of positive integers. Let’s make a few simple observations regarding the above, and then look at a few exam- ples: 1. It will be important for our discussion that { r 1 , r 2 , . . . , r K } is the set of all different roots of the polynomial. If j negationslash= k , then r j negationslash= r k . 2. Each m k is the largest integer such that ( r − r k ) m k is a factor of the original polynomial. Consequently, for each r k , there is only one possible value for m k . We call m k the multiplicity of r k ....

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