Understanding the Euler operator While reading this book I came across a differential equation
$$t^5\frac{d^2y}{dt^2}+2t^4\frac{dy}{dt}-y=0$$
that was then rewritten in terms of the Euler operator, $\delta=t\frac{d}{dt}$, with the resulting operator being 
$$\delta^2-\delta-\frac{1}{t^3}=0.$$
I'm trying to understand why does it seem as if  $$\delta^2 = t^2\frac{d^2}{dt^2} .$$ Shouldn't
$$\delta^2=\delta(\delta)=t\frac{d}{dt}\left(t\frac{d}{dt}\right)= t\frac{d}{dt}+t^2\frac{d^2}{dt^2}.$$
Also, I'm looking for a possible generalization of this for $n^{th}$ order polynomials, that is, given a differential operator
$$A(x)\frac{d^2}{dt^2}+B(x)\frac{d}{dt}+C(x)=0$$
where $A(x), B(x)$ and $C(x)$ are $n^{th}$ degree polynomials, how can it be expressed in terms of the Euler operator
 A: I think you are right about the definition of $\delta^2$, but  the operator should be:
$\delta^2+\delta-1/t^3$, check this with your definition and  it all cancels correctly.
Then the generalisation would be:
$\frac{A(t)}{t^2}\delta^2+(B(t)-\frac{A(t)}{t})\delta+C(t)$
A: Your operator acts on sufficiently differentiable functions. Look:
Let $f$ be a sufficiently differentiable function.
The meaning of $\delta^2f$ is:
$$\delta^2f=\delta(\delta f).$$
Then:
\begin{align*}
\bigl(\delta^2f)
&=t\frac{\mathrm{d}}{\mathrm{d}t}\left(t\frac{\mathrm{d}f}{\mathrm{d}t}\right)\\
&=t\left(\frac{\mathrm{d}f}{\mathrm{d}t}+t\frac{\mathrm{d}^2f}{\mathrm{d}t^2}\right)\\
&=t\frac{\mathrm{d}f}{\mathrm{d}t}+t^2\frac{\mathrm{d}^2f}{\mathrm{d}t^2}
\end{align*}
Conclusion:
$$\delta^2=t\frac{\mathrm{d}}{\mathrm{d}t}+t^2\frac{\mathrm{d}^2}{\mathrm{d}t^2}=\delta+t^2\frac{\mathrm{d}^2}{\mathrm{d}t^2}$$
Regarding the other part of your question: Let $A$, $B$ and $C$ be any function.
Observe that
$$\frac{\mathrm{d}^2}{\mathrm{d}t^2}=\frac1{t^2}\bigl(\delta^2-\delta\bigr).$$
Then:
$$A(t)\frac{\mathrm{d}^2}{\mathrm{d}t^2}+B(t)\frac{\mathrm{d}}{\mathrm{d}t}+C(t)=\frac{A(t)}{t^2}\delta^2+\left(\frac{tB(t)-A(t)}{t^2}\right)\delta+C(t).$$
A: No, the identity $$\delta^2=\delta(\delta)=t\frac{d}{dt}\left(t\frac{d}{dt}\right)= t\frac{d}{dt}+t^2\frac{d^2}{dt^2}.$$
not correct, it should be as you said first $\delta^2=t^2\frac{d^2}{dt^2}$. Since 
\begin{align}
\delta ^2  = \delta \left( \delta  \right) = t\frac{d}{{dt}}\left( {t\frac{d}{{dt}}} \right) = t\left\{ {\frac{d}{{dt}}\left( {t\frac{d}{{dt}}} \right)} \right\} &= t\left\{ {t\frac{{d^2 }}{{dt^2 }} + \frac{d}{{dt}}\underbrace {\left( {\frac{{dt}}{{dt}}} \right)}_{ = 1}} \right\}
\\
 &= t\left\{ {t\frac{{d^2 }}{{dt^2 }} + 0} \right\} 
\\
&= t^2 \frac{{d^2 }}{{dt^2 }}
\end{align}
