# Characteristic equation (calculus)

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

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

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 $u(x) = c_{1} + c_{2}x + c_{3}x^2 + \cdots + c_{k}x^{k-1}$. 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. 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 $r = a \pm bi \,$ $y_{C}(x) = e^{ax}(c_{1} \cos bx +c_{2} \sin bx ) \,$

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