Editing Cauchy–Euler equation (section)

==The equation==
Let {{math|''y''{{i sup|(''n'')}}(''x'')}} be the ''n''th derivative of the unknown function&nbsp;{{math|''y''(''x'')}}. Then a Cauchy–Euler equation of order ''n'' has the form
<math display="block">a_{n} x^n y^{(n)}(x) + a_{n-1} x^{n-1} y^{(n-1)}(x) + \dots + a_0 y(x) = 0.</math>

The substitution <math>x = e^u</math> (that is, <math>u = \ln(x)</math>; for <math>x < 0</math>, in which one might replace all instances of <math>x</math> by <math>|x|</math>, extending the solution's domain to <math>\reals \setminus \{0\}</math>) can be used to reduce this equation to a linear differential equation with constant coefficients. Alternatively, the trial solution <math>y = x^m</math> can be used to solve the equation directly, yielding the basic solutions.<ref name="kreyszig">{{cite book|last=Kreyszig|first=Erwin|title=Advanced Engineering Mathematics | publisher=Wiley|date=May 10, 2006|isbn=978-0-470-08484-7}}</ref>

===Second order – solving through trial solution===
[[File:Euler-Cauchy equation solution curves real roots.svg|thumb|400px|right|Typical solution curves for a second-order Euler–Cauchy equation for the case of two real roots]]
[[File:Euler-Cauchy equation solution curves double root.svg|thumb|400px|right|Typical solution curves for a second-order Euler–Cauchy equation for the case of a double root]]
[[File:Euler-Cauchy_equation_solution_curves_complex_roots.svg|thumb|400px|right|Typical solution curves for a second-order Euler–Cauchy equation for the case of complex roots]]
The most common Cauchy–Euler equation is the second-order equation, which appears in a number of physics and engineering applications, such as when solving [[Laplace's equation]] in polar coordinates. The second order Cauchy–Euler equation is<ref name="kreyszig" /><ref>{{Cite book|title=Elementary Differential Equations and Boundary Value Problems|last1=Boyce|first1=William E.| pages=272–273|last2=DiPrima|first2=Richard C.|editor-last=Rosatone|editor-first=Laurie|edition=10th|year=2012| isbn=978-0-470-45831-0}}</ref>

<math display="block">x^2\frac{d^2y}{dx^2} + ax\frac{dy}{dx} + by = 0.</math>

We assume a trial solution<ref name="kreyszig" /> <math display="block">y = x^m.</math>

Differentiating gives <math display="block">\frac{dy}{dx} = mx^{m-1} </math> and <math display="block">\frac{d^2y}{dx^2} = m\left(m-1\right)x^{m-2}. </math>

Substituting into the original equation leads to requiring that
<math display="block">x^2\left( m\left(m-1 \right)x^{m-2} \right) + ax\left( mx^{m-1} \right) + b\left( x^m \right) = 0</math>

Rearranging and factoring gives the indicial equation
<math display="block">m^2 + \left(a-1\right)m + b = 0.</math>

We then solve for ''m''. There are three cases of interest:

* Case 1 of two distinct roots, {{math|''m''<sub>1</sub>}} and {{math|''m''<sub>2</sub>}};
* Case 2 of one real repeated root, {{mvar|m}};
* Case 3 of complex roots, {{math|''α'' ± ''βi''}}.

In case 1, the solution is <math display="block">y = c_1 x^{m_1} + c_2 x^{m_2}</math>

In case 2, the solution is <math display="block">y = c_1 x^m \ln(x) + c_2 x^m </math>

To get to this solution, the method of [[reduction of order]] must be applied, after having found one solution {{math|1=''y'' = ''x''{{i sup|''m''}}}}.

In case 3, the solution is 
<math display="block">y = c_1 x^\alpha \cos(\beta \ln(x)) + c_2 x^\alpha \sin(\beta \ln(x)) </math>
<math display="block">\alpha = \operatorname{Re}(m)</math>
<math display="block">\beta = \operatorname{Im}(m)</math>

For <math>c_1, c_2 \isin \R</math>.

This form of the solution is derived by setting {{math|1=''x'' = ''e''{{i sup|''t''}}}} and using [[Euler's formula]].

===Second order – solution through change of variables===
<math display="block">x^2\frac{d^2y}{dx^2} +ax\frac{dy}{dx} + by = 0 </math>

We operate the variable substitution defined by

<math display="block">t = \ln(x). </math>
<math display="block">y(x) = \varphi(\ln(x)) = \varphi(t). </math>

Differentiating gives
<math display="block">\frac{dy}{dx}=\frac{1}{x}\frac{d\varphi}{dt}</math>
<math display="block">\frac{d^2y}{dx^2}=\frac{1}{x^2}\left(\frac{d^2\varphi}{dt^2}-\frac{d\varphi}{dt}\right).</math>

Substituting <math>\varphi(t)</math> the differential equation becomes
<math display="block">\frac{d^2\varphi}{dt^2} + (a-1)\frac{d\varphi}{dt} + b\varphi = 0.</math>

This equation in <math>\varphi(t)</math> is solved via its characteristic polynomial
<math display="block">\lambda^2 + (a-1)\lambda + b = 0.</math>

Now let <math>\lambda_1</math> and <math>\lambda_2</math> denote the two roots of this polynomial. We analyze the case in which there are distinct roots and the case in which there is a repeated root:

If the roots are distinct, the general solution is <math display="block">\varphi(t)=c_1 e^{\lambda_1 t} + c_2 e^{\lambda_2 t},</math> where the exponentials may be complex.

If the roots are equal, the general solution is <math display="block">\varphi(t)=c_1 e^{\lambda_1 t} + c_2 t e^{\lambda_1 t}.</math>

In both cases, the solution <math>y(x)</math> can be found by setting <math>t = \ln(x)</math>.

Hence, in the first case, <math display="block">y(x) = c_1 x^{\lambda_1} + c_2 x^{\lambda_2},</math> and in the second case, <math display="block">y(x) = c_1 x^{\lambda_1} + c_2 \ln(x) x^{\lambda_1}.</math>

===Second order - solution using differential operators===

Observe that we can write the second-order Cauchy-Euler equation in terms of a linear [[differential operator]] <math> L </math> as <math display="block">Ly = (x^2 D^2 + axD + bI)y = 0,</math> where <math> D = \frac{d}{dx} </math> and <math> I </math> is the identity operator.

We express the above operator as a polynomial in <math> xD </math>, rather than <math> D </math>. By the product rule, <math display="block"> (x D)^2 = x D(x D) = x(D + x D^2) = x^2D^2 + x D.</math> So, <math display="block"> L = (xD)^2 + (a-1)(xD) + bI.</math>

We can then use the quadratic formula to factor this operator into linear terms. More specifically, let <math> \lambda_1, \lambda_2 </math> denote the (possibly equal) values of <math display="block">-\frac{a-1}{2} \pm \frac{1}{2}\sqrt{(a-1)^2 - 4b}. </math> Then, <math display="block">L = (xD - \lambda_1 I)(xD - \lambda_2 I).</math>

It can be seen that these factors commute, that is <math>(xD - \lambda_1 I)(xD - \lambda_2 I) = (xD - \lambda_2 I)(xD - \lambda_1 I)</math>. Hence, if <math> \lambda_1 \neq \lambda_2 </math>, the solution to <math> Ly = 0 </math> is a linear combination of the solutions to each of <math> (xD - \lambda_1 I)y = 0 </math> and <math> (xD - \lambda_2 I)y = 0 </math>, which can be solved by [[separation of variables]].

Indeed, with <math> i \in \{1,2\} </math>, we have <math> (xD - \lambda_i I)y = x\frac{dy}{dx} - \lambda_i y = 0 </math>. So, <math display="block">\begin{align} x\frac{dy}{dx} &= \lambda_i y\\ \int \frac{1}{y}\, dy &= \lambda_i \int \frac{1}{x}\, dx\\ \ln y &= \lambda_i \ln x + C\\ y &= c_i e^{\lambda_i \ln x} = c_i x^{\lambda_i}.\end{align}</math> Thus, the general solution is <math> y = c_1 x^{\lambda_1} + c_2 x^{\lambda_2} </math>.

If <math> \lambda = \lambda_1 = \lambda_2 </math>, then we instead need to consider the solution of <math>(xD - \lambda I)^2y = 0 </math>. Let <math> z = (xD-\lambda I)y </math>, so that we can write <math display="block"> (xD - \lambda I)^2y = (xD - \lambda I)z = 0.</math> As before, the solution of <math> (xD- \lambda I)z = 0 </math> is of the form <math> z = c_1x^\lambda </math>. So, we are left to solve <math display="block"> (xD - \lambda I)y = x\frac{dy}{dx} - \lambda y = c_1x^\lambda.</math> We then rewrite the equation as <math display="block"> \frac{dy}{dx} - \frac{\lambda}{x} y = c_1x^{\lambda-1},</math> which one can recognize as being amenable to solution via an [[integrating factor]].

Choose <math> M(x) = x^{-\lambda} </math> as our integrating factor. Multiplying our equation through by <math> M(x) </math> and recognizing the left-hand side as the derivative of a product, we then obtain <math display="block">\begin{align} \frac{d}{dx}(x^{-\lambda} y) &= c_1x^{-1}\\ x^{-\lambda} y &= \int c_1x^{-1}\, dx\\ y &= x^\lambda (c_1\ln(x) + c_2)\\ &= c_1\ln(x)x^\lambda +c_2 x^\lambda.\end{align}</math>

===Example===
Given
<math display="block">x^2 u'' - 3xu' + 3u = 0\,,</math>
we substitute the simple solution {{math|''x''{{i sup|''m''}}}}:
<math display="block">x^2\left(m\left(m-1\right)x^{m-2}\right)-3x\left(m x^{m-1}\right) + 3x^m = m\left(m-1\right)x^m - 3m x^m+3x^m = \left(m^2 - 4m + 3\right)x^m = 0\,.</math>

For {{math|''x''{{i sup|''m''}}}} to be a solution, either {{math|1=''x'' = 0}}, which gives the [[Trivial (mathematics)|trivial]] solution, or the coefficient of {{math|''x''{{i sup|''m''}}}} is zero. Solving the quadratic equation, we get&nbsp;{{math|1=''m'' = 1, 3}}. The general solution is therefore

: <math>u=c_1 x+c_2 x^3\,.</math>