Editing Cauchy–Euler equation (section)

===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>