# Characteristic equation (calculus)

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Characteristic equation (calculus)

In mathematics, the characteristic equation (or auxiliary equation[1]) is an algebraic equation of degree $n \,$ on which depends the solutions of a given $n \,$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 $y \,$ 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.[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 $a_{n}, a_{n-1}, \ldots , a_{1}, a_{0}$,

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

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, $y' = re^{rx} \,$, $y'' = r^{2}e^{rx} \,$, and $y^{(n)} = r^{n}e^{rx} \,$ 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.[3] In order to solve for $r \,$, one can substitute $y = e^{rx} \,$ and its derivatives into the differential equation to get

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

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.[1][4] For example, if $r \,$ is found to equal to 3, then the general solution will be $y(x) = ce^{3x} \,$, where $c \,$ is a constant.

## Formation of the general solution

Example

The linear homogeneous differential equation with constant coefficients

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

has the characteristic equation

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

By factoring the characteristic equation into

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

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

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

Solving the characteristic equation for its roots, $r_{1}, \ldots , r_{n}$, 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 $c_{1}, \ldots , c_{n}$.[1][5] 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.[1] 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[1]

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

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

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

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

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

By dividing out $e^{r_{1}x} \,$, it can be seen that

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

However, this is the case if and only if $u(x) \,$ is a polynomial of degree $k \,$, so that $u(x) = c_{1} + c_{2}x + c_{3}x^2 + \cdots + c_{k}x^{k-1}$.[4] Since $y(x) = ue^{r_{1}x} \,$, the part of the general solution corresponding to r1 is

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

### Complex roots

If the characteristic equation has complex roots of the form r1 = a + bi and r2 = abi, then the general solution is accordingly $y(x) = c_{1}e^{(a + bi)x} + c_{2}e^{(a - bi)x} \,$. However, by Euler's formula, which states that $e^{i \theta } = \cos \theta + i \sin \theta \,$, this solution can be rewritten as follows:

$y(x) = c_{1}e^{(a + bi)x} + c_{2}e^{(a - bi)x} = c_{1}e^{ax}(\cos bx + i \sin bx) + c_{2}e^{ax}( \cos bx - i \sin bx ) = (c_{1} + c_{2})e^{ax} \cos bx + i(c_{1} - c_{2})e^{ax} \sin bx \,$

where $c_{1} \,$ and $c_{2} \,$ are constants that can be complex.[4] Note that if $c_{1} = c_{2} = \tfrac{1}{2}$, then the particular solution $y_{1}(x) = e^{ax} \cos bx \,$ is formed. Similarly, if $c_{1} = \tfrac{1}{2}i$ and $c_{2} = - \tfrac{1}{2}i$, 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 following general solution results for the part of a differential equation having complex roots $r = a \pm bi \,$

$y_{C}(x) = e^{ax}(c_{1} \cos bx +c_{2} \sin bx ) \,$

## References

1. ^ a b c d e f g Edwards, C. Henry; David E. Penney. "3". Differential Equations: Computing and Modeling. David Calvis. Upper Saddle River, New Jersey: Pearson Education. pp. 156–170. ISBN 978-0-13-600438-7.
2. ^ a b c Smith, David Eugene. "History of Modern Mathematics: Differential Equations". University of South Florida. Retrieved 2 March 2011.
3. ^ a b Chu, Herman; Gaurav Shah, Tom Macall. "Linear Homogeneous Ordinary Differential Equations with Constant Coefficients". eFunda. Retrieved 1 March 2011.
4. ^ a b c d e Cohen, Abraham (1906). An Elementary Treatise on Differential Equations. D. C. Heath and Company.
5. ^ Dawkins, Paul. "Differential Equation Terminology". Paul's Online Math Notes. Retrieved 2 March 2011.

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