Editing Cauchy–Euler equation (section)

===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]].