10 Chapter 0 Ordinary Differential Equations
Root Multiplicity Contribution
m(real) 1 cemt
m(real) k (c 1 +c 2 t+···+cktk−^1 )emt
m,m(complex) 1 (acos(βt)+bsin(βt))eαt
m=α+iβ
m,m(complex) k (a^1 +a^2 t+···+aktk− (^1) )cos(βt)eαt
+(b 1 +b 2 t+···+bktk−^1 )sin(βt)eαt
Table 3 Contributions to general solution
in which the coefficientsk 1 (t),k 2 (t), etc., are given functions oft.Thetech-
niques of solution are analogous to those for second-order equations. In par-
ticular, they depend on the Principle of Superposition, which remains valid
for this equation. That principle allows us to say that the general solution
of Eq. (26) has the form of a linear combination ofnindependent solutions
u 1 (t),u 2 (t),...,un(t)with arbitrary constant coefficients,
u(t)=c 1 u 1 (t)+c 2 u 2 (t)+···+cnun(t).
Of course, we cannot solve the generalnth-order equation (26), but we can
indeed solve any homogeneous linear equation withconstantcoefficients,
u(n)+k 1 u(n−^1 )+···+kn− 1 u(^1 )+knu= 0. (27)
We must now findnindependent solutions of this equation. As in the second-
order case, we assume that a solution has the formu(t)=emt,andfindvalues
ofmfor which this is true. That is, we substituteemtforuin the differen-
tial equation (27) and divide out the common factor ofemt. The result is the
polynomial equation
mn+k 1 mn−^1 +···+kn− 1 m+kn= 0 , (28)
called thecharacteristic equationof the differential equation (27).
Each distinct root of the characteristic equation contributes as many inde-
pendent solutions as its multiplicity, which might be as high asn.Recallalso
that the polynomial equation (28) may have complex roots, which will occur
in conjugate pairs if — as we assume — the coefficientsk 1 ,k 2 , etc., are real.
When this happens, we prefer to have real solutions, in the form of an expo-
nential times sine or cosine, instead of complex exponentials. The contribution
of each root or pair of conjugate roots of Eq. (28) is summarized in Table 3.
Since the sum of the multiplicities of the roots of Eq. (28) isn, the sum of the
contributions produces a solution withnterms, which can be shown to be the
general solution.