In mathematics, the characteristic equation (or auxiliary equation^[1]) is an algebraic equation of degree n upon which depends the solution of a given nth-order differential equation^[2] or difference equation.^[3][4] The characteristic equation can only be formed when the differential or difference equation is linear and homogeneous, and has constant coefficients.^[1] Such a differential equation, with y as the dependent variable, superscript (n) denoting n^th-derivative, and a_n, a_n − 1, ..., a₁, a₀ as constants,

${\displaystyle a_{n}y^{(n)}+a_{n-1}y^{(n-1)}+\cdots +a_{1}y'+a_{0}y=0,}$

will have a characteristic equation of the form

${\displaystyle a_{n}r^{n}+a_{n-1}r^{n-1}+\cdots +a_{1}r+a_{0}=0}$

whose solutions r₁, r₂, ..., r_n are the roots from which the general solution can be formed.^[1][5][6] Analogously, a linear difference equation of the form

${\displaystyle y_{t+n}=b_{1}y_{t+n-1}+\cdots +b_{n}y_{t}}$

has characteristic equation

${\displaystyle r^{n}-b_{1}r^{n-1}-\cdots -b_{n}=0,}$

discussed in more detail at Linear recurrence with constant coefficients.

The characteristic roots (roots of the characteristic equation) also provide qualitative information about the behavior of the variable whose evolution is described by the dynamic equation. For a differential equation parameterized on time, the variable's evolution is stable if and only if the real part of each root is negative. For difference equations, there is stability if and only if the modulus of each root is less than 1. For both types of equation, persistent fluctuations occur if there is at least one pair of complex roots.

The method of integrating linear ordinary differential equations with constant coefficients was discovered by Leonhard Euler, who found that the solutions depended on an algebraic 'characteristic' equation.^[2] The qualities of the Euler's characteristic equation were later considered in greater detail by French mathematicians Augustin-Louis Cauchy and Gaspard Monge.^[2][6]

Starting with a linear homogeneous differential equation with constant coefficients a_n, a_n − 1, ..., a₁, a₀,

${\displaystyle a_{n}y^{(n)}+a_{n-1}y^{(n-1)}+\cdots +a_{1}y^{\prime }+a_{0}y=0,}$

it can be seen that if y(x) = e^rx, each term would be a constant multiple of e^rx. This results from the fact that the derivative of the exponential function e^rx is a multiple of itself. Therefore, y′ = re^rx, y″ = r²e^rx, and y⁽ⁿ⁾ = rⁿe^rx are all multiples. This suggests that certain values of r will allow multiples of e^rx to sum to zero, thus solving the homogeneous differential equation.^[5] In order to solve for r, one can substitute y = e^rx and its derivatives into the differential equation to get

${\displaystyle a_{n}r^{n}e^{rx}+a_{n-1}r^{n-1}e^{rx}+\cdots +a_{1}re^{rx}+a_{0}e^{rx}=0}$

Since e^rx can never equal zero, it can be divided out, giving the characteristic equation

${\displaystyle a_{n}r^{n}+a_{n-1}r^{n-1}+\cdots +a_{1}r+a_{0}=0.}$

By solving for the roots, r, in this characteristic equation, one can find the general solution to the differential equation.^[1][6] For example, if r has roots equal to 3, 11, and 40, then the general solution will be ${\displaystyle y(x)=c_{1}e^{3x}+c_{2}e^{11x}+c_{3}e^{40x}}$, where ${\displaystyle c_{1}}$, ${\displaystyle c_{2}}$, and ${\displaystyle c_{3}}$ are arbitrary constants which need to be determined by the boundary and/or initial conditions.

Solving the characteristic equation for its roots, r₁, ..., r_n, allows one to find the general solution of the differential equation. The roots may be real or complex, as well as distinct or repeated. If a characteristic equation has parts with distinct real roots, h repeated roots, or k complex roots corresponding to general solutions of y_D(x), y_R1(x), ..., y_Rh(x), and y_C1(x), ..., y_Ck(x), respectively, then the general solution to the differential equation is

${\displaystyle y(x)=y_{\mathrm {D} }(x)+y_{\mathrm {R} _{1}}(x)+\cdots +y_{\mathrm {R} _{h}}(x)+y_{\mathrm {C} _{1}}(x)+\cdots +y_{\mathrm {C} _{k}}(x)}$

The linear homogeneous differential equation with constant coefficients

${\displaystyle y^{(5)}+y^{(4)}-4y^{(3)}-16y''-20y'-12y=0}$

has the characteristic equation

${\displaystyle r^{5}+r^{4}-4r^{3}-16r^{2}-20r-12=0}$

By factoring the characteristic equation into^{[further explanation needed]}

${\displaystyle (r-3)(r^{2}+2r+2)^{2}=0}$

one can see that the solutions for r are the distinct single root r₁ = 3 and the double complex roots r_2,3,4,5 = 1 ± i. This corresponds to the real-valued general solution

${\displaystyle y(x)=c_{1}e^{3x}+e^{x}(c_{2}\cos x+c_{3}\sin x)+xe^{x}(c_{4}\cos x+c_{5}\sin x)}$

with constants c₁, ..., c₅.

The superposition principle for linear homogeneous says that if u₁, ..., u_n are n linearly independent solutions to a particular differential equation, then c₁u₁ + ⋯ + c_nu_n is also a solution for all values c₁, ..., c_n.^[1][7] Therefore, if the characteristic equation has distinct real roots r₁, ..., r_n, then a general solution will be of the form

${\displaystyle y_{\mathrm {D} }(x)=c_{1}e^{r_{1}x}+c_{2}e^{r_{2}x}+\cdots +c_{n}e^{r_{n}x}}$

If the characteristic equation has a root r₁ that is repeated k times, then it is clear that y_p(x) = c₁e^r₁x is at least one solution.^[1] However, this solution lacks linearly independent solutions from the other k − 1 roots. Since r₁ has multiplicity k, the differential equation can be factored into^[1]

${\displaystyle \left({\frac {d}{dx}}-r_{1}\right)^{k}y=0.}$

The fact that y_p(x) = c₁e^r₁x is one solution allows one to presume that the general solution may be of the form y(x) = u(x)e^r₁x, where u(x) is a function to be determined. Substituting ue^r₁x gives

${\displaystyle \left({\frac {d}{dx}}-r_{1}\right)\!ue^{r_{1}x}={\frac {d}{dx}}\left(ue^{r_{1}x}\right)-r_{1}ue^{r_{1}x}={\frac {d}{dx}}(u)e^{r_{1}x}+r_{1}ue^{r_{1}x}-r_{1}ue^{r_{1}x}={\frac {d}{dx}}(u)e^{r_{1}x}}$

when k = 1. By applying this fact k times, it follows that

${\displaystyle \left({\frac {d}{dx}}-r_{1}\right)^{k}ue^{r_{1}x}={\frac {d^{k}}{dx^{k}}}(u)e^{r_{1}x}=0.}$

By dividing out e^r₁x, it can be seen that

${\displaystyle {\frac {d^{k}}{dx^{k}}}(u)=u^{(k)}=0.}$

Therefore, the general case for u(x) is a polynomial of degree k − 1, so that u(x) = c₁ + c₂x + c₃x² + ⋯ + c_kx^k −1.^[6] Since y(x) = ue^r₁x, the part of the general solution corresponding to r₁ is

${\displaystyle y_{\mathrm {R} }(x)=e^{r_{1}x}\!\left(c_{1}+c_{2}x+\cdots +c_{k}x^{k-1}\right).}$

If a second-order differential equation has a characteristic equation with complex conjugate roots of the form r₁ = a + bi and r₂ = a − bi, then the general solution is accordingly y(x) = c₁e^{(a + bi )x} + c₂e^{(a − bi )x}. By Euler's formula, which states that e^iθ = cos θ + i sin θ, this solution can be rewritten as follows:

${\displaystyle {\begin{aligned}y(x)&=c_{1}e^{(a+bi)x}+c_{2}e^{(a-bi)x}\\&=c_{1}e^{ax}(\cos bx+i\sin bx)+c_{2}e^{ax}(\cos bx-i\sin bx)\\&=\left(c_{1}+c_{2}\right)e^{ax}\cos bx+i(c_{1}-c_{2})e^{ax}\sin bx\end{aligned}}}$

where c₁ and c₂ are constants that can be non-real and which depend on the initial conditions.^[6] (Indeed, since y(x) is real, c₁ − c₂ must be imaginary or zero and c₁ + c₂ must be real, in order for both terms after the last equals sign to be real.)

For example, if c₁ = c₂ = ⁠1/2⁠, then the particular solution y₁(x) = e^ax cos bx is formed. Similarly, if c₁ = ⁠1/2i⁠ and c₂ = −⁠1/2i⁠, then the independent solution formed is y₂(x) = e^ax sin bx. Thus by the superposition principle for linear homogeneous differential equations, a second-order differential equation having complex roots r = a ± bi will result in the following general solution:

${\displaystyle y_{\mathrm {C} }(x)=e^{ax}(C_{1}\cos bx+C_{2}\sin bx)}$

This analysis also applies to the parts of the solutions of a higher-order differential equation whose characteristic equation involves non-real complex conjugate roots.

• Characteristic polynomial