The characteristic-equation method from distinct real roots, complex roots and repeated roots works for any order. An $n$th-order linear equation with constant coefficients needs $n$ independent solutions, one for each root counted with multiplicity.
Idea clave
For $a_ny^{(n)} + \dots + a_1y' + a_0y = 0$, solve $a_nr^{n} + \dots + a_1r + a_0 = 0$. A real root $r$ of multiplicity $m$ gives $e^{rt}, te^{rt}, \dots, t^{m-1}e^{rt}$. A complex pair $\alpha \pm \beta i$ of multiplicity $m$ gives $t^{j}e^{\alpha t}\cos\beta t$ and $t^{j}e^{\alpha t}\sin\beta t$ for $j = 0, \dots, m - 1$.
Por qué funciona
With $D = \frac{d}{dt}$, the equation is $a_n(D - r_1)(D - r_2)\cdots(D - r_n)y$ $= 0$: the operator factors just like the polynomial, because constant-coefficient operators commute. Each factor $(D - r)$ kills $e^{rt}$, and $(D - r)^{m}$ kills $t^{j}e^{rt}$ for $j < m$ (each application lowers the power of $t$ by one, as in the repeated-root lesson). Counting with multiplicity, there are exactly $n$ such solutions, and they are independent.
Ejemplo resuelto: Three distinct roots
Find the general solution of $y''' - 2y'' - y' + 2y = 0$.
- $r^{3} - 2r^{2} - r + 2$ $= r^{2}(r - 2) - (r - 2)$ $= (r - 2)(r - 1)(r + 1)$.
- Roots $2, 1, -1$.
Respuesta: $y = c_1e^{2t} + c_2e^{t} + c_3e^{-t}$
Ejemplo resuelto: Real and imaginary roots
Find the general solution of $y^{(4)} - 16y = 0$.
- $r^{4} - 16 = (r^{2} - 4)(r^{2} + 4)$: roots $\pm 2$ and $\pm 2i$.
Respuesta: $y$ $= c_1e^{2t} + c_2e^{-2t} + c_3\cos 2t$ $+ c_4\sin 2t$
Ejemplo resuelto: A double root with initial values
Solve $y''' - 3y' + 2y = 0$, $y(0) = 0$, $y'(0) = 0$, $y''(0) = 9$.
- $r = 1$ is a root; dividing gives $r^{3} - 3r + 2 = (r - 1)^{2}(r + 2)$.
- $y = (c_1 + c_2t)e^{t} + c_3e^{-2t}$.
- Conditions: $c_1 + c_3 = 0$; $c_1 + c_2 - 2c_3 = 0$; $c_1 + 2c_2 + 4c_3 = 9$.
- From the first two, $c_3 = -c_1$ and $c_2 = -3c_1$; the third gives $-9c_1 = 9$.
Respuesta: $c_1 = -1$, $c_2 = 3$, $c_3 = 1$: $y = (3t - 1)e^{t} + e^{-2t}$
Error común
Stopping with too few solutions: a cubic has three roots counted with multiplicity, so the general solution needs three constants. Find one root by trying divisors of the constant term, divide it out, and factor what is left. For a repeated root, include the extra powers of $t$.
Higher-order equations appear naturally. Two masses joined by springs give a pair of second-order equations that combine into one fourth-order equation, and the bending of a loaded beam obeys $EIy^{(4)} = w(x)$. To find the roots of the characteristic polynomial by hand, list the candidates from the rational root test (factors of $a_0$ divided by factors of $a_n$), test them, and divide out each root you find. Undetermined coefficients and variation of parameters extend to any order in the same way.