Cauchy-Euler equation

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In mathematics, a Cauchy-Euler equation (also Euler-Cauchy equation) is a linear homogeneous ordinary differential equation with variable coefficients. Because of its simple structure the equation can be replaced with an equivalent equation with constant coefficients which can then be solved explicitly.

1 Cauchy-Euler equation
- 1.1 Second order
- 1.2 Example
2 See also

[edit] Cauchy-Euler equation

Let $y (n) (x)$ be the n-derivative of the unknown function $y (x)$ , then a Cauchy-Euler equation of order n has the form

$x^n y^{(n)}(x) + a_{n-1} x^{n-1} y^{(n-1)}(x) + \ldots + a_0 y(x) = 0$

This Equation can be solved using the substitution $e u$ .

[edit] Second order

The Euler-Cauchy equation crops up in a number of engineering applications. It is given by the equation:

$x^2\frac{d^2y}{dx^2} + ax\frac{dy}{dx} + by = 0 \,$

We assume a trial solution given by:

$y = x^m \,$

Differentiating, we have:

$\frac{dy}{dx} = mx^{m-1} \,$

and

$\frac{d^2y}{dx^2} = m(m-1)x^{m-2} \,$

Substituting into the original equation, we have:

$x^2( m(m-1)x^{m-2} ) + ax( mx^{m-1} ) + b( x^m ) = 0 \,$

Or rearranging gives:

$m^2 + (a-1)m + b = 0 \,$

We then can solve for the roots of m. There are three particular cases of interest:

Case #1: Two distinct roots, m₁ and m₂
Case #2: One real double root, m
Case #3: Complex roots, α±iβ

In case #1, the solution is given by:

$y = c_{1}x^{m_{1}} + c_{2}x^{m_{2}} \,$

In case #2, the solution is given by:

$y = c_{1}x^{m}\ln(x) + c_{2}x^{m} \,$

To get to this solution, the method of reduction of order must be applied after having found one solution y = x^m.

In case #3, the solution is given by:

$y = c_{1}x^{\alpha}\cos(\beta \ln(x)) + c_{2}x^{\alpha}\sin(\beta \ln(x)) \,$

This equation also can be solved with $x = e t$ transformation.

This particular case is of no great practical importance and hence this has been left as a challenge for the reader (or another Wikipedia contributor) to solve!

[edit] Example

Given

$x^2u''-3xu'+3u=0\,$

we substitute the simple solution x^α:

$x^2(\alpha(\alpha-1)x^{\alpha-2})-3x(\alpha(x^{\alpha-1}))+3(x^\alpha)=\alpha(\alpha-1)x^\alpha-3\alpha x^\alpha+3x^\alpha\,$

$(\alpha(\alpha-1)-3\alpha+3)x^\alpha.\,$

For this to indeed be a solution, either x=0 giving the trivial solution, or the coefficient of x^α is zero, so solving that quadratic, we get α=1,3. So, the general solution is

$u=Ax+Bx^3.\,$