Homogeneous Euler-Cauchy equation

#math
If the ODE is of the form $$ax^2y'' + bxy' + cy = 0 \tag{8}$$
then the characteristic equation is given by $$am^2 + (b-a)m + c = 0 \tag{9}$$We consider the case $x > 0$ ;

If the roots of $(9)$ are real and distinct, $m_{1}, m_{2}$ , then the two LI solutions are $x^{m_{1}}$ and $x^{m_{2}}$
If they are repeated roots, $m_{1} = m_{2} = m$ then the two LI solutions are $x^{m}$ and $x^{m} \ln x$
If the two roots are complex conjugate, $α \pm i β$ then the two LI solutions are $x^{α} \cos (β \ln x)$ and $x^{α} \sin (β \ln x)$

Proof:

Similar to Homogeneous second order differential equation with basis of the type $e^{m x}$ , the proof is similar. Instead of considering our trial solution as $e^{m x}$ , we take $x^{m}$ , since we notice the increasing powers of $x$ in the equation that exactly match differentiation pattern of $x^{m}$ .

Trivial
Follow same method of $p (m) = p^{'} (m) = 0$ for finding the other term in the basis, other than $x^{m}$ . $\frac{\partial}{\partial m} x^{m} = x^{m} \ln x$ , which is the other LI solution.
$Y_{1} = x^{α + i β} = x^{α} \times e^{i β \ln x}$ and $Y_{2} = x^{α} e^{- i β \ln x}$ . Consider their averages, $\frac{Y_{1} + Y_{2}}{2}$ and $\frac{Y_{1} - Y_{2}}{2 i}$ , to give $x^{α} \cos (β \ln x)$ and $x^{α} \sin (β \ln x)$ .

Comments

The solution for $x < 0$ can be obtained by replacing $x$ with $- x$ everywhere.
Homogeneous Euler-Cauchy equation can be transformed into linear constant coefficient homogeneous equation by changing independent variable to $t = \ln x$ for $x > 0$
This equation type can be generalized to equation of the form $$a(\gamma x + \delta)^2y'' + b(\gamma x + \delta)y' + cy = 0$$by considering $(γ x + δ)^{m}$ as the trial solution. (It must be linear, will not work for any $f (x)$ )