How to solve $tx''-x'-4t^3x=0$ 
Solve $$tx''-x'-4t^3x=0$$

I came across this example and it seems difficult to me. Any hints on how to start solving it?
 A: $\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle}
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Use $\ds{\tau \equiv t^{2}\ \imp\ t = \tau^{1/2}}$:

\begin{align}
\totald{}{t}&=\totald{\tau}{t}\,\totald{}{\tau}=2t\,\totald{}{\tau}
=2\tau^{1/2}\,\totald{}{\tau}
\\[5mm]
\totald[2]{}{t}&=4\tau^{1/2}\,\totald{}{\tau}\pars{\tau^{1/2}\,\totald{}{\tau}}
=4\tau^{1/2}\pars{\half\,\tau^{-1/2}\,\totald{}{\tau}
+\tau^{1/2}\,\totald[2]{}{\tau}}
=2\,\totald{}{\tau} + 4\tau\,\totald[2]{}{\tau}
\end{align}

The equation becomes
\begin{align}
0&=\tau^{1/2}\pars{2\,\totald{}{\tau} + 4\tau\,\totald[2]{}{\tau}}x
-\pars{2\tau^{1/2}\,\totald{}{\tau}}x - 4\tau^{3/2}x
\end{align}

\begin{align}
0&=4\tau^{3/2}\,\totald[2]{x}{\tau} - 4\tau^{3/2}x
\quad\imp\color{#66f}{\large\quad\totald[2]{x}{\tau} - x = 0}
\end{align}

Solutions are linear combinations of $\ds{\expo{\pm\tau}}$. Namely,
linear combinations of $\ds{\expo{\pm t^{2}}}$.
A: Let $x(t) = y(t^2)$. Then, $x'(t) = 2ty'(t^2)$ and $x''(t) = 2y'(t^2)+4t^2y''(t^2)$. 
$tx''(t)-x'(t)-4t^3x(t) = 0$ 
$t[2y'(t^2)+4t^2y''(t^2)]-[2ty'(t^2)]-4t^3[y(t^2)] = 0$ 
$4t^3[y''(t^2)-y(t^2)] = 0$
$y'' - y = 0$
Can you solve this ODE?
