Higher Order Linear Equations

A higher order linear equation is simply one with a third, fourth or even higher derivative in it, where $y$ and its derivatives only get multiplied by numbers, never by each other. ("Order" means the highest derivative that appears, and "linear" means no $y^2$, $yy'$ or anything like that.) They come up in beams, which need four derivatives, and in systems of several linked springs or circuits. The equation $a_ny^{(n)} + \cdots + a_1y' + a_0y = 0$ looks long, but it's just a list of terms: $y^{(n)}$ (the $n$th derivative, $y$ differentiated $n$ times), each multiplied by a constant $a$, all adding up to zero. Swap each derivative for a power of $r$ and you get a polynomial whose roots tell you the solutions, exactly as in second order. More terms, same idea; bear with it.

Nothing about the $e^{rt}$ trick cared that the equation was second order. Plug $y = e^{rt}$ into $a_ny^{(n)} + \cdots + a_1y' + a_0y = 0$: each derivative brings down one more factor of $r$, so the left side becomes the characteristic polynomial times $e^{rt}$. Since $e^{rt}$ is never zero, $r$ has to be a root of that polynomial. A degree-$n$ polynomial has $n$ roots counted with multiplicity, and that's exactly how many independent solutions an $n$th-order equation needs.

Building the general solution:

• Each distinct real root $r$: $e^{rt}$.
• A root repeated $k$ times: $e^{rt}, te^{rt}, \dots, t^{k-1}e^{rt}$.
• Each complex pair $\lambda \pm \mu i$: $e^{\lambda t}\cos\mu t$ and $e^{\lambda t}\sin\mu t$ (with extra powers of $t$ if repeated).

A common mistake. For a triple root it's tempting to stop at $e^{rt}$ and $te^{rt}$, because that's what a repeated root gave you in second order. It isn't enough: a root repeated $k$ times has to supply $k$ solutions, so a triple root gives $e^{rt}$, $te^{rt}$ and $t^2e^{rt}$. The pattern stops at $t^{k-1}$, never $t^k$.

Finding the roots of cubics and quartics is the hard part. Try integer roots that divide the constant term, then factor out each one you find and deal with what's left.

An $n$th-order IVP needs $n$ initial conditions: $y, y', \dots, y^{(n-1)}$ at $t_0$. It makes sense that you need that many, since the general solution has $n$ constants to pin down.

There's an animation further down of a beam clamped at both ends, which is a fourth order problem with four conditions.

Beam deflection (a fourth-order equation), coupled oscillators and multi-stage filters lead to higher-order characteristic polynomials.