Solving Cauchy-Euler equations with substitution I have to solve the following equation as a Cauchy-Euler type for $1<3x$: 
$$(1-3x)^2y''-(3x-1)y'+3y=0 $$
I tried this: 
Take $$u=3x-1$$ 
then do $\frac{du}{dx}=3 $
$$du=3dx$$ 
So, $\frac{dy}{dx}= \frac{3dy}{du} $
Q. How would you do $\frac{d^2y}{du^2}$? Am I wrong? How would you proceed? 
 A: The Euler equation of the form
$$(ax+b)^2y''+p(ax+b)y'+qy=0$$
is homogeneous equation, can be solved with substitution $ax+b=e^t$ or $t=\ln(ax+b)$ convert to an equation with constant coefficients on $ax+b>0$\begin{eqnarray*}
a\frac{dy}{dt}
&=& a\frac{dy}{dx}\frac{dx}{dt} \\
&=& ay'\frac{e^t}{a}\\
&=& (ax+b)y'  \\
a^2\frac{d^2y}{dt^2}
&=& a\frac{d}{dt}(a\frac{dy}{dt}) \\
&=& a\frac{d}{dx}((ax+b)y')\frac{dx}{dt} \\
&=& a\Big[ay'+(ax+b)y''\Big]\frac{e^t}{a} \\
&=& a(ax+b)y'+(ax+b)^2y''      \\
&\implies&\left\lbrace\begin{array}{c l}(ax+b)y'=a\dfrac{dy}{dt}~~~~~~~~~~~~~~~~~~\\ (ax+b)^2y''=a^2\dfrac{d^2y}{dt^2}-a^2\dfrac{dy}{dt}.\end{array}\right.
\end{eqnarray*}
with substitution in equation we find
$$y''+(\frac{p}{a}-1)y'+\frac{q}{a^2}y=0$$
after solving this new equation, we retrieve variable to $x$ with substitution $t=\ln(ax+b)$. Here $a=3$ and $b=-1$, so try with them!
A: $$(1-3x)^2\dfrac{d^2y}{dx^2}+(1-3x)\dfrac{dy}{dx}+3y=0$$
$$e^{t}=1-3x, \dfrac{}{}$$
That will be a good starting point.
