Characteristic Equation for a Differential Equation¶

Definition¶

Consider the homogeneous linear differential equation with constant real coefficients

x^{(n)}(t)+a_{n-1}x^{(n-1)}(t)+\dots +a_1x'(t)+a_0x(t)=0,

where $x'(t)=\dfrac{\mathrm{d}x}{\mathrm{d}t}(t)$ and $x^{(k)}$ is the $k$ -th derivative. Its characteristic polynomial is

p(r)=r^n+a_{n-1}r^{n-1}+\dots+a_1r+a_0,

and $p(r)=0$ is its characteristic equation. The equation is obtained by substituting the trial solution $x(t)=\mathrm{e}^{rt}$ $e^{r t}$ : since $\frac{\mathrm{d}^k}{\mathrm{d}t^k}\mathrm{e}^{rt}=r^k\mathrm{e}^{rt}$ $r^{k} e^{r t}$ , the left-hand side becomes $p(r)\mathrm{e}^{rt}$ , which vanishes for all $t$ if and only if $p(r)=0$ $0$ . For a second-order differential equation $x''+ax'+bx=0$ the characteristic equation is $r^2+ar+b=0$ $0$ .

Interpretation¶

The substitution turns a differential equation into an algebraic one: each root of the polynomial $p$ gives an exponential "mode" $\mathrm{e}^{rt}$ of the exponential function. Writing the equation as a first-order system with a companion matrix, $p$ is exactly the characteristic polynomial of that matrix, so the roots are its eigenvalues.

Left, the roots -1 + 2i and -1 - 2i of r^2 + 2r + 5 = 0 in the complex plane; right, the real solution e^(-t) cos 2t oscillating inside the dashed envelope plus or minus e^(-t).

The figure follows the complex-roots example $x''+2x'+5x=0$ $0$ . Its characteristic equation $r^2+2r+5=0$ has the roots $r=-1\pm2\mathrm{i}$ $2 i$ , marked on the left in the complex plane. Through $\mathrm{e}^{rt}$ and Euler's formula they give the real mode $\mathrm{e}^{-t}\cos2t$ on the right: the real part $-1$ fixes the shrinking envelope $\pm \mathrm{e}^{-t}$ , and the imaginary part $2$ fixes how fast the solution oscillates inside it.

The discrete-time analogue is the characteristic polynomial for a difference equation, where the trial solution is $\lambda^t$ instead of $\mathrm{e}^{rt}$ . The two are linked by $\lambda=\mathrm{e}^{r}$ $e^{r}$ : sampling $\mathrm{e}^{rt}$ at integer times gives $(\mathrm{e}^{r})^t$ .

Common errors: forgetting that a repeated root needs the extra factor $t$ ; writing $\mathrm{e}^{(\alpha\pm\mathrm{i}\beta)t}$ as the final real answer instead of converting it with Euler's formula; and applying the method to equations with non-constant coefficients, where $\mathrm{e}^{rt}$ is generally not a solution.

Key Results¶

By the fundamental theorem of algebra, $p$ has $n$ complex roots counted with multiplicity.
A real root $r$ of multiplicity $m$ gives the $m$ solutions $\mathrm{e}^{rt},\,t\mathrm{e}^{rt},\,\dots,\,t^{m-1}\mathrm{e}^{rt}$ .
Since the coefficients are real, nonreal roots come in pairs $\alpha\pm\mathrm{i}\beta$ ( $β \neq =$ $\beta\neq0$ $0$ ) by the complex conjugate root theorem. Each pair gives the real solutions $\mathrm{e}^{\alpha t}\cos\beta t$ and $\mathrm{e}^{\alpha t}\sin\beta t$ , because $\mathrm{e}^{(\alpha+\mathrm{i}\beta)t}=\mathrm{e}^{\alpha t}(\cos\beta t+\mathrm{i}\sin\beta t)$ $i sin β t)$ ; a pair of multiplicity $m$ gives these two functions multiplied by $1,t,\dots,t^{m-1}$ .
The $n$ functions obtained this way form a basis of the solution space, so the general solution is their linear combination.
For $n=2$ $2$ , $r^2+ar+b=0$ has discriminant $a^2-4b$ $4 b$ : positive gives two real roots, zero a double root $-a/2$ , negative a complex pair $-\frac{a}{2}\pm\mathrm{i}\frac{\sqrt{4b-a^2}}{2}$ $i \frac{4 b - a ^{2}}{2}$ .

Examples¶

Distinct real roots. $x''-x'-6x=0$ $0$ : $r^2-r-6=(r-3)(r+2)$ $2)$ , so $x(t)=c_1\mathrm{e}^{3t}+c_2\mathrm{e}^{-2t}$ $c_{2} e^{- 2 t}$ .

Complex roots. $x''+2x'+5x=0$ $0$ : $r=\frac{-2\pm\sqrt{4-20}}{2}=-1\pm2\mathrm{i}$ $2 i$ , so $x(t)=\mathrm{e}^{-t}(c_1\cos2t+c_2\sin2t)$ $c_{2} sin 2 t)$ , a damped oscillation.

Double root, checked. $x''+2x'+x=0$ $0$ : $(r+1)^2=0$ $0$ . For $x=t\mathrm{e}^{-t}$ $t e^{- t}$ : $x'=(1-t)\mathrm{e}^{-t}$ $t) e^{- t}$ , $x''=(t-2)\mathrm{e}^{-t}$ $2) e^{- t}$ , and $x''+2x'+x=(t-2+2-2t+t)\mathrm{e}^{-t}=0$ $0$ .

Triple root. $x'''-3x''+3x'-x=0$ $0$ : $r^3-3r^2+3r-1=(r-1)^3$ $1)^{3}$ , so $x(t)=(c_1+c_2t+c_3t^2)\mathrm{e}^{t}$ $c_{3} t^{2}) e^{t}$ .

Prerequisites¶

Sources¶

Math V: Advanced Linear Algebra · Week 4

Aliases: auxiliary equation, characteristic polynomial of an ODE, characteristic equation of a linear ODE Keywords: e^{rt}, characteristic roots, repeated root, complex roots, e^{alpha t} cos(beta t) Symbols: alpha, beta, lambda, square root

Last updated: 22 October 2026 19:04