Cauchy-Euler equation

The Cauchy-Euler equation is a linear homogeneous ordinary differential equation with variable coefficients of the form
 * $$a_n y^{(n)}(x) x^n + a_{n-1} y^{(n-1)} x^{n-1} + \dots + a_0 y^{(0)}(x) x^0 = 0$$,

where $$y^{(n)}(x)$$ is defined as the $$n^{\text{th}}$$ derivative of $$y(x)$$. The solution to this differential equation can be explicitly solved in two ways.

Method 1: Solution by trial of solution
Consider the second order homogeneous differential equation
 * $$x^2 \frac{d^2y}{dx^2} + ax \frac{dy}{dx} + by = 0$$.

Assume $$y(x) = x^m$$. Differentiating our assumption gives
 * $$\frac{dy}{dx} = mx^{m-1}, \quad \frac{d^2y}{dx^2} = m(m-1)x^{m-2}$$.

Substituting our derivatives into the differential equation gives
 * $$x^2\left(m(m - 1)x^{m-2}\right) + ax\left(mx^{m-1}\right) + bx^{m} = 0$$.

Factoring out $$x^m$$ gives
 * $$x^2\left(m(m - 1) + am + b\right) = 0.$$

Assuming that $$x^2 \neq 0$$, we let
 * $$m(m - 1) + am + b = 0$$

which can be rearranged to become
 * $$m^2 + m(a - 1) + b = 0$$.

Depending on our choice of $$a$$ and $$b$$, we will end up with different forms of solutions. We shall consider each case individually.

Case 1: Distinct roots
In the first case, we will end up with two distinct solutions (we shall call them $$m_1$$ and $$m_2$$). Then the form of our solution is simply
 * $$y(x) = c_1 x^{m_1} + c_2 x^{m_2}$$.

Case 2: Repeated real root
In the second case, we have one real root. Then the form of our solution takes
 * $$y(x) = c_1 x^{m} + c_2 x^{m} \ln x$$. To see where this comes from, see here.

Case 3: Complex roots
In the last case, we have complex roots. Assume that the roots are $$m = \alpha \pm \beta i$$. Then the form of our solution will take
 * $$y(x) = x^\alpha\left(c_1 \cos\left(\beta \ln x\right) + c_2 \sin\left(\beta \ln x\right)\right)$$. To see where this comes from, see here.