Higher-Order Constant-Coefficient Equations

Definition Let $y$ be a function of $t$, with $y' = dy/dt$ and $y^{(k)}$ its $k$th derivative. A homogeneous linear equation of order $n$ with constant coefficients is

with real constants $a_0, \ldots, a_n$ and $a_n \ne 0$. Write $L[y]$ for its left side. For every constant $r$, real or complex, substituting $e^{rt}$ gives $L[e^{rt}] = P(r)\,e^{rt}$ with the characteristic polynomial

Since $a_n \ne 0$, the polynomial has degree exactly $n$, and $e^{rt}$ is a solution exactly when $P(r) = 0$. By the fundamental theorem of algebra, $P$ has $n$ complex roots when each is counted as often as its multiplicity. Writing $D$ for differentiation with respect to $t$, the equation reads $P(D)\,y = 0$. Because the coefficients are constant, a factorization $P(r) = a_n(r - r_1)\cdots(r - r_n)$ over the complex numbers gives $P(D) = a_n(D - r_1)\cdots(D - r_n)$, and these factors can be applied in any order.

Definition A root $r_0$ of $P$ has multiplicity $m$ if

for a polynomial $Q$ with $Q(r_0) \ne 0$. Equivalently, $P(r_0) = P'(r_0) = \cdots = P^{(m-1)}(r_0) = 0$ and $P^{(m)}(r_0) \ne 0$. A root of multiplicity $1$ is simple. Over the complex numbers the multiplicities of all the roots add up to the degree $n$. Multiplicity is an exact property of the polynomial. Roots that are numerically close but not equal are different roots, each with its own multiplicity, however close they are.

Definition Let $r_0$ be a root of $P$ of multiplicity $m$. Then the $m$ polynomial-exponential functions

solve $P(D)\,y = 0$. The reason is the shift identity

valid for every differentiable function $p$ of $t$, because the term $r_0e^{r_0t}p$ from the product rule cancels. For a polynomial $p$, applying $D - r_0$ a total of $m$ times gives $e^{r_0t}p^{(m)}$, which is zero whenever $p$ has degree less than $m$. Since $P(D) = Q(D)(D - r_0)^m$, each $t^ke^{r_0t}$ with $k < m$ is a solution. Collecting these functions over all the roots gives exactly $n$ functions, and they are linearly independent on every interval, so they form a fundamental set, also called a basis of solutions. For a single root the independence is clear, since dividing a vanishing combination by $e^{r_0t}$ leaves a polynomial that is zero on a whole interval, so all its coefficients are zero. A root $0$ of multiplicity $m$ contributes the powers $1, t, \ldots, t^{m-1}$. For a root that is not real these functions are complex-valued, and their real and imaginary parts give the real solutions described under repeated complex roots.

Definition Let the coefficients $a_0, \ldots, a_n$ be real, and let $\alpha \pm i\beta$ with $\beta > 0$ be a pair of complex conjugate roots of $P$. For real coefficients both roots of the pair have the same multiplicity $m$, and the pair contributes the $2m$ real solutions

They are the real and imaginary parts of the complex solutions $t^ke^{(\alpha + i\beta)t}$, and real and imaginary parts of a solution are solutions because the coefficients are real. Every power $t^k$ with $k < m$ therefore appears with both the cosine and the sine. Replacing the $2m$ complex basis functions of the pair by these $2m$ real ones still gives a fundamental set.

Definition Consider an equation of order $n$ with real constant coefficients and $a_n \ne 0$. For every point $t_0$ and all real numbers $k_0, k_1, \ldots, k_{n-1}$, the initial-value problem with

has exactly one solution, and it is defined for all real $t$. For a linear equation of order $n$ with variable coefficients the same holds on each regular interval, where the leading coefficient is never zero and the normalized coefficients are continuous. Order $n$ calls for exactly $n$ conditions, the value and the first $n - 1$ derivatives at one point. With a basis $y_1, \ldots, y_n$, the conditions become the $n$ linear equations

for $j = 0, 1, \ldots, n - 1$, and their determinant is the $n$ by $n$ Wronskian of the basis at $t_0$, which is not zero. The number of basis functions always matches the order, since each real root contributes its multiplicity, each complex conjugate pair contributes twice its multiplicity, and the total is $n$.