Solution of ${x}^{2} \dfrac {d^{2}y}{dx^{2}}+ x \dfrac{dy}{dx}+y=0 $ 
Find the general solution of $${x}^{2} \dfrac {d^{2}y}{dx^{2}}+ x \dfrac{dy}{dx}+y=0 $$

$$$$
This is an example of a Cauchy-Euler equation. However I'm interested in solving it without assuming a solution of the form $y=x^m$
$$$$
I tried as follows:
Let $D=\dfrac{dy}{dx}$. $$$$Thus
$$x^2\dfrac{dD}{dx}+xD+y=0$$
Differentiating wrt $x$,
$$x^2\dfrac{d^2D}{dx^2}+2x\dfrac{dD}{dx}+D+x\dfrac{dD}{dx}+D=0$$
$$x^2\dfrac{d^2D}{dx^2}+3x\dfrac{dD}{dx}+2D=0$$
Could somebody please suggest how I could go about solving this? 
 A: Let $t=\ln x$ so $\partial_t =x\partial_x$. You should find your equation reduces to $\partial_t^2 y=-y$, so constants $A,\,B$ exist with $y=A\cos\ln x +B\sin\ln x$.
A: 
The other method you are trying to employ is factorizing the differential
  operator as follows:

${{x}^{2}}{{y}'}'+x{y}'+y=\left( {{x}^{2}}D+xD+1 \right)y=0$
Assume$\left( {{x}^{2}}D+xD+1 \right)=\left( xD+a \right)\left( xD+b \right)$
Expanding the operators on the R.H.S yields:
$\begin{align}
  & \left( {{x}^{2}}D+xD+1 \right)=\left( xD+a \right)\left( xD+b \right) \\ 
 & \quad \quad \quad \quad \quad \quad =xD\left( xD+b \right)+a\left( xD+b \right) \\ 
 & \quad \quad \quad \quad \quad \quad =xD\left( xD \right)+bxD+axD+ab \\ 
 & \quad \quad \quad \quad \quad \quad =x\left( x{{D}^{2}}+D \right)+bxD+axD+ab \\ 
 & \quad \quad \quad \quad \quad \quad ={{x}^{2}}D+xD+bxD+axD+ab \\ 
 & \quad \quad \quad \quad \quad \quad ={{x}^{2}}D+x\left( 1+b+a \right)D+ab \\ 
\end{align}
$
Equating the coefficients imply that $a+b=0,\ ab=1$. At this time we are not interested in finding the valves of $a$ and $b$, so
$\left( xD+a \right)\left( xD+b \right)y=0$.
Hence $\left( xD+a \right)z=0$   if we set $\left( xD+b \right)y=z$.
The O.D.E $\left( xD+a \right)z=0$ is Separable and has a solution $z={{c}_{1}}\ {{x}^{-a}}$ for some constant ${{c}_{1}}\ $
Finally,
$\left( xD+b \right)y={{c}_{1}}\ {{x}^{-a}}$
is a Linear first order D.E with solution $y={{c}_{1}}\ {{x}^{-a}}+{{c}_{2}}\ {{x}^{-b}}$
which is the general solution to our D.E. Unfortunately the solution ${{x}^{m}}$ appears again.
A: I don't think this approach has any advantage. Note that:


*

*the equation you got in $D(x)$ is very similar to the original equation (except for the $3$ and the $2$), and so the same method of assuming a solution which is a power of $x$ would work;

*but now you know that $y(x)$ is a solution if and only if a $D(x)$ and a $y(x)$ (such that $y'(x)\equiv D(x)$) verify
$$x^2D'(x)+xD(x)+y(x)=0,$$
and this implies that
$$x^2D''(x)+3xD'(x)+2D(x)=0,$$
but it is not the case (*) that every such $y(x)$ will be a solution;

*you would have to perform integration of $D$ and elimination of an unnecessary constant as extra steps of the procedure.


In any case, very good for you trying alternative solutions or methods; as long as you understand why your approach did not work, this is very important for deepening your understanding of the problem. I even find it useful for myself and other people.
Regarding the idea of substitution by $t=\ln(x)$, it is useful because it leads to another equation, now for $Y(t)=y(e^t)$ instead of $y(x)$, which is linear with constant coefficients. And that allows for an easier solution (specially for this case, where you have an homogeneous equation).
But besides that, it's motivation is as intuitive as for the idea of assuming a solution of the form $y(x)=x^n$ for some $n\in \mathbb N$: someone else has already worked with this type of equation and he/she knows that this idea will work.

(*) Since derivation is not an injection, that is, $f'(x) \equiv g'(x)$ does not imply $f(x) \equiv g(x)$ but only that there is a $C\in \mathbb R$ such that $f(x) \equiv g(x)+C$.
