# Characteristic equation (calculus)

In mathematics, the characteristic equation (or auxiliary equation) is an algebraic equation of degree $n\,$ on which depends the solutions of a given $n\,$ th-order differential equation. The characteristic equation can only be formed when the differential equation is linear, homogeneous, and has constant coefficients. Such a differential equation, with $y\,$ as the dependent variable and $a_{n},a_{n-1},\ldots ,a_{1},a_{0}$ as constants,

$a_{n}y^{(n)}+a_{n-1}y^{(n-1)}+\cdots +a_{1}y'+a_{0}y=0$ will have a characteristic equation of the form

$a_{n}r^{n}+a_{n-1}r^{n-1}+\cdots +a_{1}r+a_{0}=0$ where $r^{n},r^{n-1},\ldots ,r$ are the roots from which the general solution can be formed. This 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. The qualities of the Euler's characteristic equation were later considered in greater detail by French mathematicians Augustin-Louis Cauchy and Gaspard Monge.

## Derivation

Starting with a linear homogeneous differential equation with constant coefficients $a_{n},a_{n-1},\ldots ,a_{1},a_{0}$ ,

$a_{n}y^{(n)}+a_{n-1}y^{(n-1)}+\cdots +a_{1}y^{'}+a_{0}y=0$ $a_{n}r^{n}e^{rx}+a_{n-1}r^{n-1}e^{rx}+\cdots +a_{1}re^{rx}+a_{0}e^{rx}=0$ Since $e^{rx}\,$ can never equate to zero, it can be divided out, giving the characteristic equation

$a_{n}r^{n}+a_{n-1}r^{n-1}+\cdots +a_{1}r+a_{0}=0$ By solving for the roots, $r\,$ , in this characteristic equation, one can find the general solution to the differential equation. For example, if $r\,$ is found to equal to 3, then the general solution will be $y(x)=ce^{3x}\,$ , where $c\,$ is an arbitrary constant.

## Formation of the general solution

Solving the characteristic equation for its roots, $r_{1},\ldots ,r_{n}$ , allows one to find the general solution of the differential equation. The roots may be real and/or complex, as well as distinct and/or repeated. If a characteristic equation has parts with distinct real roots, $h\,$ repeated roots, and/or $k\,$ complex roots corresponding to general solutions of $y_{D}(x)\,$ , $y_{R_{1}}(x),\ldots ,y_{R_{h}}(x)$ , and $y_{C_{1}}(x),\ldots ,y_{C_{k}}(x)$ , respectively, then the general solution to the differential equation is

$y(x)=y_{D}(x)+y_{R_{1}}(x)+\cdots +y_{R_{h}}(x)+y_{C_{1}}(x)+\cdots +y_{C_{k}}(x)$ ### Distinct real roots

The superposition principle for linear homogeneous differential equations with constant coefficients says that if $u_{1},\ldots ,u_{n}$ are $n\,$ linearly independent solutions to a particular differential equation, then $c_{1}u_{1}+\cdots +c_{n}u_{n}$ is also a solution for all values $c_{1},\ldots ,c_{n}$ . Therefore, if the characteristic equation has distinct real roots $r_{1},\ldots ,r_{n}$ , then a general solution will be of the form

$y_{D}(x)=c_{1}e^{r_{1}x}+c_{2}e^{r_{2}x}+\cdots +c_{n}e^{r_{n}x}$ ### Repeated real roots

If the characteristic equation has a root $r_{1}\,$ that is repeated $k\,$ times, then it is clear that $y_{p}(x)=c_{1}e^{r_{1}x}$ is at least one solution. However, this solution lacks linearly independent solutions from the other $k-1\,$ roots. Since $r_{1}\,$ has multiplicity $k\,$ , the differential equation can be factored into

$\left({\frac {d}{dx}}-r_{1}\right)^{k}y=0$ $\left({\frac {d}{dx}}-r_{1}\right)ue^{r_{1}x}={\frac {d}{dx}}(ue^{r_{1}x})-r_{1}ue^{r_{1}x}={\frac {d}{dx}}(u)e^{r_{1}x}+r_{1}ue^{r_{1}x}-r_{1}ue^{r_{1}x}={\frac {d}{dx}}(u)e^{r_{1}x}$ $\left({\frac {d}{dx}}-r_{1}\right)^{k}ue^{r_{1}x}={\frac {d^{k}}{dx^{k}}}(u)e^{r_{1}x}=0$ ${\frac {d^{k}}{dx^{k}}}(u)=u^{(k)}=0$ $y_{R}(x)=e^{r_{1}x}(c_{1}+c_{2}x+\cdots +c_{k}x^{k-1})$ ### Complex roots

${\begin{array}{rcl}y(x)&=&c_{1}e^{(a+bi)x}+c_{2}e^{(a-bi)x}\\&=&c_{1}e^{ax}(\cos bx+i\sin bx)+c_{2}e^{ax}(\cos bx-i\sin bx)\\&=&(c_{1}+c_{2})e^{ax}\cos bx+i(c_{1}-c_{2})e^{ax}\sin bx\end{array}}$ Similarly, if $c_{1}={\tfrac {1}{2i}}$ and $c_{2}=-{\tfrac {1}{2i}}$ , then the independent solution formed is $y_{2}(x)=e^{ax}\sin bx\,$ . Thus by the superposition principle for linear homogeneous differential equations with constant coefficients, the part of a differential equation having complex roots $r=a\pm bi\,$ will result in the following general solution: $y_{C}(x)=e^{ax}(c_{1}\cos bx+c_{2}\sin bx)\,$ 