Chapter 0 Ordinary Differential Equations 11
Example.
Find the general solution of this fourth-order equation
u(^4 )+ 3 u(^2 )− 4 u= 0.The characteristic equation ism^4 + 3 m^2 − 4 =0, which is easy to solve because
it is a biquadratic. We find thatm^2 =−4 or 1, and thus the roots arem=± 2 i,
±1, all with multiplicity 1. From Table 3 we find thatacos( 2 t)+bsin( 2 t)cor-
responds to the complex conjugate pair,m=± 2 i,whileetande−tcorrespond
tom=1andm=−1. Thus we build up the general solution,
u(t)=acos( 2 t)+bsin( 2 t)+c 1 et+c 2 e−t. Example.
Find the general solution of the fourth-order equation
u(^4 )− 2 u(^2 )+u= 0.The characteristic equation ism^4 − 2 m^2 + 1 =0, whose roots, found as in the
preceding, are±1, both with multiplicity 2. From Table 3 we find that each of
the roots contributes a first-degree polynomial times an exponential. Thus, we
assemble the general solution as
u(t)=(c 1 +c 2 t)et+(c 3 +c 4 t)e−t.With sinh(t)=(et−e−t)/2 and cosh(t)=(et+e−t)/2, the terms of the pre-
ceding combination can be rearranged to give the general solution in a differ-
ent form,
u(t)=(C 1 +C 2 t)cosh(t)+(C 3 +C 4 t)sinh(t).
Some important equations and their solutions.1.du
dt=ku (kis constant),
u(t)=cekt.2.d^2 u
dt^2 +λ(^2) u=0,
u(t)=acos(λt)+bsin(λt).
- d
(^2) u
dt^2
−λ^2 u=0,
u(t)=acosh(λt)+bsinh(λt) or u(t)=c 1 eλt+c 2 e−λt.
4.t^2 u′′+tu′−λ^2 u=0,
u(t)=c 1 tλ+c 2 t−λ.