The Characteristic Polynomial 1. The General Second Order Case and the Characteristic Equation For m, b, k constant, the homogeneous equation .. . mx + bx + kx = 0. (1) . is a lot like x + kx = 0, which has as solution x = e−kt . We’ll be optimistic and try for exponential solutions, x (t) = ert , for some as yet undetermined constant r. To see which values of r might work, plug x (t) = ert into (1). Organize the calculation: the k] , b] , m] are ﬂags indicating that we should multiply the corresponding line by this number. k] b] m] .. x = ert . x = rert .. x = r2 ert . ⇒ mx + bx + kx = (mr2 + br + k)ert = 0. An exponential is never zero, so we can divide this equation by ert . We have found that ert is a solution to (1) exactly when r satisﬁes the characteristic equation mr2 + br + k = 0. The left hand side is a polynomial called, naturally enough, the character istic polynomial and usually denoted p(r ). (You will often also see s used as the variable instead of r. With this notation the characteristic polynomial is p(s) = ms2 + bs + k.) .. . Example. Find all the solutions to x + 8x + 7x = 0. Solution. The characteristic polynomial is r2 + 8r + 7 . We want the roots. One reason we wrote out the polynomial was to remind you that you can ﬁnd roots by factoring it. This one factors as (r + 1)(r + 7) so the roots are r = −1 and r = −7, with corresponding exponential solutions are x1 (t) = e−t and x2 (t) = e−7t . By superposition, the linear combination of independent solutions gives the general solution: x ( t ) = c 1 e − t + c 2 e −7t . The Characteristic Polynomial OCW 18.03SC . Suppose that we have initial conditions x (0) = 2 and x (0) = −8 then . we can solve for c1 and c2 . Use x (t) = −c1 e−t − 7c2 e−7t and substitute t = 0 to get c1 + c2 = 2 x (0) = . x (0) = −c1 − 7c2 = −8 Adding these two equations yields −6c2 = −6, so c2 = 1 and c1 = 1. The solution to our DE with the given initial conditions is then x (0) = 2, . x (0) = −8 is x (t) = e−t + e−7t . 2. The General nth Order Case In the same way we can take the homogeneous constant coefﬁcient lin ear equation of degree n . a n x ( n ) + · · · + a1 x + a0 x = 0 and get its characteristic polynomial, p (r ) = a n r n + · · · + a1 r + a0 The exponential x (t) = ert is a solution of the homogeneous DE if and only if r is a root of p(r ), i.e. p(r ) = 0. By superposition, any linear combination of these exponentials is also a solution. 2