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WP:CHECKWIKI error fix for #61. Punctuation goes before References. Do general fixes if a problem exists. - using AWB (9890)

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In [[mathematics]], the '''characteristic equation''' (or '''auxiliary equation'''<ref name="edwards" />) is an [[Algebraic function|algebraic]] equation of [[Degree of a polynomial|degree]] <math> n \, </math> on which depends the solutions of a given <math>n\,</math><sup>th</sup>-[[Derivative#Higher_derivatives|order]] [[differential equation]].<ref name="smith">{{cite web|url=http://etc.usf.edu/lit2go/contents/2800/2892/2892_txt.html|title=History of Modern Mathematics: Differential Equations|last=Smith|first=David Eugene|publisher=[[University of South Florida]]}}</ref> The characteristic equation can only be formed when the differential equation is [[Linear differential equation|linear]], [[Linear homogeneous differential equation|homogeneous]], and has constant [[coefficient]]s.<ref name="edwards">{{cite book|last1=Edwards |first1=C. Henry |last2=Penney |first2=David E. |others=David Calvis |title=Differential Equations: Computing and Modeling |publisher=Pearson Education |location=[[Upper Saddle River]], [[New Jersey]] |pages=156–170 |chapter=3 |isbn=978-0-13-600438-7}}</ref> Such a differential equation, with <math>y \,</math> as the [[dependent variable]] and <math>a_{n}, a_{n-1}, \ldots , a_{1}, a_{0}</math> as [[Mathematical constant|constants]],
:<math>a_{n}y^{(n)} + a_{n-1}y^{(n-1)} + \cdots + a_{1}y' + a_{0}y = 0</math>
will have a characteristic equation of the form
:<math>a_{n}r^{n} + a_{n-1}r^{n-1} + \cdots + a_{1}r + a_{0} = 0</math>
where <math>r^{n}, r^{n-1}, \ldots ,r</math> are the roots from which the [[general solution]] can be formed.<ref name="edwards" /><ref name=eFunda>{{cite web |url=http://www.efunda.com/math/ode/linearode_consthomo.cfm |title=Linear Homogeneous Ordinary Differential Equations with Constant Coefficients |last1=Chu |first1=Herman |last2=Shah |first2=Gaurav |last3=Macall |first3=Tom |publisher=eFunda |accessdate=1 March 2011}}</ref><ref name="cohen">{{cite book|last=Cohen|first=Abraham|title=An Elementary Treatise on Differential Equations|publisher=[[D. C. Heath and Company]]|year=1906}}</ref> This method of [[Integral|integrating]] linear ordinary differential equations with constant coefficients was discovered by [[Leonhard Euler]], who found that the solutions depended on an algebraic 'characteristic' equation.<ref name="smith" /> The qualities of the Euler's characteristic equation were later considered in greater detail by French mathematicians [[Augustin-Louis Cauchy]] and [[Gaspard Monge]].<ref name="smith" /><ref name="cohen" />

== Derivation ==

Starting with a linear homogeneous differential equation with constant coefficients <math>a_{n}, a_{n-1}, \ldots , a_{1}, a_{0}</math>,
:<math>a_{n}y^{(n)} + a_{n-1}y^{(n-1)} + \cdots + a_{1}y^{'} + a_{0}y = 0</math>
it can be seen that if <math>y(x) = e^{rx} \, </math>, each term would be a constant multiple of <math> e^{rx} \, </math>. This results from the fact that the derivative of the [[exponential function]] <math> e^{rx} \, </math> is a multiple of itself. Therefore, <math>y' = re^{rx} \, </math>, <math>y'' = r^{2}e^{rx} \, </math>, and <math>y^{(n)} = r^{n}e^{rx} \, </math> are all multiples. This suggests that certain values of <math> r \, </math> will allow multiples of <math> e^{rx} \, </math> to sum to zero, thus solving the homogeneous differential equation.<ref name="eFunda" /> In order to solve for <math> r \, </math>, one can substitute <math>y = e^{rx} \, </math> and its derivatives into the differential equation to get
:<math>a_{n}r^{n}e^{rx} + a_{n-1}r^{n-1}e^{rx} + \cdots + a_{1}re^{rx} + a_{0}e^{rx} = 0</math>
Since <math> e^{rx} \, </math> can never equate to zero, it can be divided out, giving the characteristic equation
:<math>a_{n}r^{n} + a_{n-1}r^{n-1} + \cdots + a_{1}r + a_{0} = 0</math>
By solving for the roots, <math> r \, </math>, in this characteristic equation, one can find the general solution to the differential equation.<ref name="edwards" /><ref name="cohen" /> For example, if <math> r \, </math> is found to equal to 3, then the general solution will be <math>y(x) = ce^{3x} \, </math>, where <math> c \, </math> is a [[Constant of integration|constant]].

== Formation of the general solution ==
{{MetaSidebar|35%|#ffffaa|right|Example|
The linear homogeneous differential equation with constant coefficients
:<math> y^{(5)} + y^{(4)} - 4y^{(3)} - 16y'' -20y' - 12y = 0 \, </math>
has the characteristic equation
:<math> r^{5} + r^{4} - 4r^{3} - 16r^{2} -20r - 12 = 0 \, </math>
By [[Factorization|factoring]] the characteristic equation into
:<math> (r - 3)(r^{2} + 2r + 2)^{2} = 0 \, </math>
one can see that the solutions for <math> r \, </math> are the distinct single root <math> r_{1} = 3 \, </math> and the double complex root <math> r_{2,3,4,5} = -1 \pm i </math>. This corresponds to the real-valued general solution with constants <math> c_{1} , \ldots , c_{5} </math> of
: <math> 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) \, </math>}}
Solving the characteristic equation for its roots, <math> r_{1}, \ldots , r_{n} </math>, allows one to find the general solution of the differential equation. The roots may be [[real number|real]] and/or [[complex number|complex]], as well as distinct and/or repeated. If a characteristic equation has parts with distinct real roots, <math> h \, </math> repeated roots, and/or <math> k \, </math> complex roots corresponding to general solutions of <math>y_{D}(x) \, </math>, <math>y_{R_{1}}(x), \ldots , y_{R_{h}}(x) </math>, and <math>y_{C_{1}}(x), \ldots , y_{C_{k}}(x) </math>, respectively, then the general solution to the differential equation is
:<math> y(x) = y_{D}(x) + y_{R_{1}}(x) + \cdots + y_{R_{h}}(x) + y_{C_{1}}(x) + \cdots + y_{C_{k}}(x) </math>

=== Distinct real roots ===
The ''[[superposition principle]] for linear homogeneous differential equations with constant coefficients'' says that if <math> u_{1}, \ldots , u_{n} </math> are <math> n \, </math> [[linearly independent]] solutions to a particular differential equation, then <math> c_{1}u_{1} + \cdots + c_{n}u_{n} </math> is also a solution for all values <math> c_{1}, \ldots , c_{n}</math>.<ref name="edwards" /><ref name="dawkins">{{cite web|url=http://tutorial.math.lamar.edu/Classes/DE/PDETerminology.aspx|title=Differential Equation Terminology|last=Dawkins|first=Paul|work=Paul's Online Math Notes|accessdate=2 March 2011}}</ref> Therefore, if the characteristic equation has distinct [[real number|real]] roots <math> r_{1}, \ldots , r_{n} </math>, then a general solution will be of the form
:<math> y_{D}(x) = c_{1}e^{r_{1}x} + c_{2}e^{r_{2}x} + \cdots + c_{n}e^{r_{n}x} </math>

=== Repeated real roots ===
If the characteristic equation has a root <math>r_{1} \,</math> that is repeated <math> k \, </math> times, then it is clear that <math> y_{p}(x) = c_{1}e^{r_{1}x} </math> is at least one solution.<ref name="edwards" /> However, this solution lacks linearly independent solutions from the other <math> k - 1 \, </math> roots. Since <math>r_{1} \,</math> has multiplicity <math> k \, </math>, the differential equation can be factored into<ref name="edwards" />
:<math> \left ( \frac{d}{dx} - r_{1} \right )^{k}y = 0 </math>
The fact that <math> y_{p}(x) = c_{1}e^{r_{1}x} </math> is one solution allows one to presume that the general solution may be of the form <math> y(x) = u(x)e^{r_{1}x} \, </math>, where <math> u(x) \, </math> is a function to be determined. Substituting <math> ue^{r_{1}x} \, </math> gives
:<math> \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} </math>
when <math> k = 1 \, </math>. By applying this fact <math> k \, </math> times, it follows that
:<math> \left ( \frac{d}{dx} - r_{1} \right )^{k} ue^{r_{1}x} = \frac{d^{k}}{dx^{k}}(u)e^{r_{1}x} = 0</math>
By dividing out <math> e^{r_{1}x} \, </math>, it can be seen that
:<math> \frac{d^{k}}{dx^{k}}(u) = u^{(k)} = 0 </math>
However, this is the case if and only if <math> u(x) \, </math> is a polynomial of degree <math> k-1 \, </math>, so that <math> u(x) = c_{1} + c_{2}x + c_{3}x^2 + \cdots + c_{k}x^{k-1} </math>.<ref name="cohen" /> Since <math> y(x) = ue^{r_{1}x} \, </math>, the part of the general solution corresponding to <math> r_{1} </math> is
:<math> y_{R}(x) = e^{r_{1}x}(c_{1} + c_{2}x + \cdots + c_{k}x^{k-1}) </math>

=== Complex roots ===
If the characteristic equation has [[Complex number|complex]] roots of the form <math> r_{1} = a + bi </math> and <math> r_{2} = a - bi </math>, then the general solution is accordingly <math> y(x) = c_{1}e^{(a + bi)x} + c_{2}e^{(a - bi)x} \, </math>. However, by [[Euler's formula]], which states that <math> e^{i \theta } = \cos \theta + i \sin \theta \,</math>, this solution can be rewritten as follows:
:<math>\begin{align} 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{align}</math>
where <math> c_{1} \, </math> and <math> c_{2} \, </math> are constants that can be complex.<ref name="cohen" />

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

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

== References ==
{{reflist|2}}

[[Category:Ordinary differential equations]]

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en>BG19bot: WP:CHECKWIKI error fix for #61. Punctuation goes before References. Do general fixes if a problem exists. - using AWB (9890)