Cauchy-Euler equation – wronskian and $e^u$ substitution

Question

Wikipedia says te following about Cauchy-Euler equation:

$$a_{n} x^n y^{(n)}(x) + a_{n-1} x^{n-1} y^{(n-1)}(x) + \cdots + a_0 y(x) = 0$$
is called Cauchy-Euler equation. Substitution $x=e^u$ (where $u = ln(x)$; for $x < 0$ we replace $x$ with $|x|$ to extend the domain to $\mathbb{R}_0$) may be used to reduce this equation to a linear differential equation with constant coefficients.

How does it work? I tried plugging $e^{lnx}$ to it:

$$
a_ne^{nlnx}y^{(n)}(e^{lnx}) + \dots + a_1e^{lnx} y'(e^{lnx}) + a_0y(e^{lnx})=0 \\
a_nxe^ny^{(n)}(x) + a_{n-1}xe^{(n-1)}y^{n-1}(x) + \dots + a_1e^{lnx} y'(x) + a_0y(x)=0
$$
but I don't see how this would lead me anywhere.

Another question popped up – how does Wronskian of such equation would look like in general? All I can find are Wronskians of concrete examples.

Answer 1

Define $z(u)=y(e^u)$. Then $$z'(u)=y'(e^u)e^u=xy'(x).$$ Similarly for higher-order derivatives.

For example, the equation $$xy'(x)+y(x)=0$$ becomes $$z'(u)+z(u)=0.$$