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MATH
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Definitions from Second-order constant-coefficient equations

All definitions from Second-order constant-coefficient equations.

Constant-coefficient equation

A second-order constant-coefficient linear ODE has the form

ay′′+by′+cy=f(x),a≠0,ay''+by'+cy=f(x), \qquad a\neq0,

where aa, bb and cc are constants. It is homogeneous when f(x)≡0f(x)\equiv0.

The same idea extends to an nnth-order equation: each derivative has a constant coefficient and the corresponding characteristic polynomial has degree nn.

Characteristic equation

For the homogeneous equation

ay′′+by′+cy=0,ay''+by'+cy=0,

the exponential trial function y=erxy=e^{rx} gives the characteristic equation

ar2+br+c=0.ar^2+br+c=0.

The trial is useful because

y′=rerx,y′′=r2erx,y'=re^{rx},\qquad y''=r^2e^{rx},

so substitution gives

(ar2+br+c)erx=0.(ar^2+br+c)e^{rx}=0.

Since erx≠0e^{rx}\neq0, the differential equation reduces to an algebraic equation. For an nnth-order constant-coefficient equation, replace y(j)y^{(j)} by rjr^j to obtain the corresponding degree-nn polynomial.

Spring-mass-damper model

For a mass attached to a spring with spring constant kk and viscous friction coefficient cc, Newton's law gives

mx¨+cx˙+kx=0.m\ddot{x}+c\dot{x}+kx=0.

The characteristic roots determine whether the motion is non-oscillatory or oscillatory and whether its amplitude grows or decays.

With an external force, the right side becomes a forcing term. The solution then has a transient homogeneous response and a particular forced response. In the usual mechanical units, mm is mass, cc has mass/time units and kk has mass/time2^2 units.

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