# Characteristic equation (calculus)

In mathematics, the characteristic equation (or auxiliary equation[1]) is an algebraic equation of degree ${\displaystyle n\,}$ on which depends the solutions of a given ${\displaystyle n\,}$th-order differential equation.[2] The characteristic equation can only be formed when the differential equation is linear, homogeneous, and has constant coefficients.[1] Such a differential equation, with ${\displaystyle y\,}$ as the dependent variable and ${\displaystyle a_{n},a_{n-1},\ldots ,a_{1},a_{0}}$ 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

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

where ${\displaystyle r^{n},r^{n-1},\ldots ,r}$ are the roots from which the general solution can be formed.[1][3][4] 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.[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][4]

## Derivation

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

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

it can be seen that if ${\displaystyle y(x)=e^{rx}\,}$, each term would be a constant multiple of ${\displaystyle e^{rx}\,}$. This results from the fact that the derivative of the exponential function ${\displaystyle e^{rx}\,}$ is a multiple of itself. Therefore, ${\displaystyle y'=re^{rx}\,}$, ${\displaystyle y''=r^{2}e^{rx}\,}$, and ${\displaystyle y^{(n)}=r^{n}e^{rx}\,}$ are all multiples. This suggests that certain values of ${\displaystyle r\,}$ will allow multiples of ${\displaystyle e^{rx}\,}$ to sum to zero, thus solving the homogeneous differential equation.[3] In order to solve for ${\displaystyle r\,}$, one can substitute ${\displaystyle y=e^{rx}\,}$ 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 ${\displaystyle e^{rx}\,}$ can never equate to 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, ${\displaystyle r\,}$, in this characteristic equation, one can find the general solution to the differential equation.[1][4] For example, if ${\displaystyle r\,}$ is found to equal to 3, then the general solution will be ${\displaystyle y(x)=ce^{3x}\,}$, where ${\displaystyle c\,}$ is an arbitrary constant.

## Formation of the general solution

Example

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

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

one can see that the solutions for ${\displaystyle r\,}$ are the distinct single root ${\displaystyle r_{1}=3\,}$ and the double complex root ${\displaystyle r_{2,3,4,5}=-1\pm i}$. This corresponds to the real-valued general solution with constants ${\displaystyle c_{1},\ldots ,c_{5}}$ of

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

Solving the characteristic equation for its roots, ${\displaystyle r_{1},\ldots ,r_{n}}$, 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, ${\displaystyle h\,}$ repeated roots, and/or ${\displaystyle k\,}$ complex roots corresponding to general solutions of ${\displaystyle y_{D}(x)\,}$, ${\displaystyle y_{R_{1}}(x),\ldots ,y_{R_{h}}(x)}$, and ${\displaystyle y_{C_{1}}(x),\ldots ,y_{C_{k}}(x)}$, respectively, then the general solution to the differential equation is

${\displaystyle y(x)=y_{D}(x)+y_{R_{1}}(x)+\cdots +y_{R_{h}}(x)+y_{C_{1}}(x)+\cdots +y_{C_{k}}(x)}$

### Distinct real roots

The superposition principle for linear homogeneous differential equations with constant coefficients says that if ${\displaystyle u_{1},\ldots ,u_{n}}$ are ${\displaystyle n\,}$ linearly independent solutions to a particular differential equation, then ${\displaystyle c_{1}u_{1}+\cdots +c_{n}u_{n}}$ is also a solution for all values ${\displaystyle c_{1},\ldots ,c_{n}}$.[1][5] Therefore, if the characteristic equation has distinct real roots ${\displaystyle r_{1},\ldots ,r_{n}}$, then a general solution will be of the form

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

### Repeated real roots

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

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

The fact that ${\displaystyle y_{p}(x)=c_{1}e^{r_{1}x}}$ is one solution allows one to presume that the general solution may be of the form ${\displaystyle y(x)=u(x)e^{r_{1}x}\,}$, where ${\displaystyle u(x)\,}$ is a function to be determined. Substituting ${\displaystyle ue^{r_{1}x}\,}$ gives

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

when ${\displaystyle k=1\,}$. By applying this fact ${\displaystyle k\,}$ 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 ${\displaystyle e^{r_{1}x}\,}$, it can be seen that

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

However, this is the case if and only if ${\displaystyle u(x)\,}$ is a polynomial of degree ${\displaystyle k-1\,}$, so that ${\displaystyle u(x)=c_{1}+c_{2}x+c_{3}x^{2}+\cdots +c_{k}x^{k-1}}$.[4] Since ${\displaystyle y(x)=ue^{r_{1}x}\,}$, the part of the general solution corresponding to ${\displaystyle r_{1}}$ is

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

### Complex roots

If the characteristic equation has complex roots of the form ${\displaystyle r_{1}=a+bi}$ and ${\displaystyle r_{2}=a-bi}$, then the general solution is accordingly ${\displaystyle y(x)=c_{1}e^{(a+bi)x}+c_{2}e^{(a-bi)x}\,}$. However, by Euler's formula, which states that ${\displaystyle e^{i\theta }=\cos \theta +i\sin \theta \,}$, this solution can be rewritten as follows:

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

where ${\displaystyle c_{1}\,}$ and ${\displaystyle c_{2}\,}$ are constants that can be complex.[4]

Note that if ${\displaystyle c_{1}=c_{2}={\tfrac {1}{2}}}$, then the particular solution ${\displaystyle y_{1}(x)=e^{ax}\cos bx\,}$ is formed.

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

## References

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