Cauchy-Euler DE

**Aryth** · January 19th 2010, 03:11 PM

A second order homogeneous Cauchy-Euler Equation is an equation of the type:

$ax^2y''(x) + bxy'(x) + cy(x) = 0$ , a,b,c constants, $x > 0$

Explain why, in the case of the homogeneous C-E DE, a solution can be of the form $y = x^m$

It makes sense that that is the correct Ansatz, but I cannot explain why exactly that should be the case.

**chisigma** · January 19th 2010, 09:24 PM

Setting $y=x^{m}$ the DE becomes the algebraic equation in m...

$a\cdot m\cdot (m-1) + b\cdot m + c=0$ (1)

... because the 'common term' $x^{m}$ can be 'canceled'. If $m_{1}$ and $m_{2}$ are [distinct] solutions of (1), then the solution of the DE will be...

$y= c_{1}\cdot x^{m_{1}} + c_{2}\cdot x^{m_{2}}$ (2)

Kind regards

$\chi$ $\sigma$

**HallsofIvy** · January 20th 2010, 05:21 AM

Originally Posted by Aryth

A second order homogeneous Cauchy-Euler Equation is an equation of the type:

$ax^2y''(x) + bxy'(x) + cy(x) = 0$ , a,b,c constants, $x > 0$

Explain why, in the case of the homogeneous C-E DE, a solution can be of the form $y = x^m$

It makes sense that that is the correct Ansatz, but I cannot explain why exactly that should be the case.

It can be but it isn't always in exactly the way solutions to differential equations with constant coefficients can be of the form " $e^{rx}$ " although they can also be sine or cosine, polynomials or any combinations of exponential, polynomials, and sine or cosine.

And, in fact, for exactly the same reason. The substitution t= ln(x) changes an Euler-Cauchy equation into an equation with constant coefficients.

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