Euler Equation $ax^2y''+bxy’+cy=0$ I heard that one can solve the equation above by substituting $x$ for $e^z$.
The only way I could solve this equation was by assuming that $y(x)$ has the form $x^p$
I am kinda stuck with that new approach. Hope someone could show me how to solve it. Thanks in advance
 A: Hint: Let $x = e^t$. Then, $y(x) = y(e^t)$.
$$y'_t = \frac{dy}{dt} = \frac{dy}{dx}\frac{dx}{dt} = y' e^t \implies y' = y'_t e^{-t} \\
y'' = \frac{dy'}{dx} = \frac{dy'}{dt}\frac{dt}{dx} = \frac{dy'}{dt}\frac{1}{\frac{dx}{dt}} = (y''_t-y'_t)e^{-2t}$$
So, we have
$$\begin{cases}y' = y'_te^{-t} \\
y'' = (y''_t - y'_t)e^{-2t}\end{cases}$$
Plug in the found $y'$ and $y''$ into the equation.
$$ax^2y''+bxy’+cy=0 \\
\Downarrow \\
a\color{red}{e^{2t}}(y''_t-y'_t)\color{red}{e^{-2t}} + b\color{red}{e^t}y'_t\color{red}{e^{-t}} + cy_t = 0 \\ 
ay''_t + (b-a)y'_t + cy_t = 0$$
Now, you have constant coefficients.
A: If you let $z = \log x$, then
\begin{align*}
\frac{dy}{dx} &= \frac{1}{x}\frac{dy}{dz}\\
\frac{d^2y}{dx^2} &= \frac{1}{x^2} \left(\frac{d^2y}{dz^2}-\frac{dy}{dz}\right)
\end{align*}
with similar form for higher derivatives. In particular, this reduces an equation in Cauchy-Euler form to one with constant coefficients in terms of the new variable $z$. For instance,
$$ax^2\frac{d^2y}{dx^2}+bx\frac{dy}{dx}+cy=0$$
becomes
$$ a\frac{d^2y}{dz^2}+(b-a)\frac{dy}{dz}+cy=0$$
Solutions in terms of $z$ will be linear combinations of $e^{\lambda z}$, so solutions of the original equation will be linear combinations of $x^\lambda$. The case with repeated eigenvalues is similar.
