How can I find functions satisfying the Bessel equation $\frac{d^2}{dt^2}y(t)+p(t)\frac d{dt}y(t)+q(t)y(t)=0$? We are tasked with finding the functions $u(t), Q(t)$ such that the conversion $y(t) = u(t)v(t)$ takes us from
\begin{align}
\dfrac{d^2}{dt^2}y(t) + p(t)\dfrac{d}{dt}y(t) + q(t)y(t) = 0
\end{align}
to the form
\begin{align}
\dfrac{d^2}{dt^2}u(t) + Q(t)u(t)=0.
\end{align}
From then on we must use this to solve $t^2\dfrac{d^2}{dt^2}y(t) + t\dfrac{d}{dt}y(t) + \left(t^2-\dfrac{1}{4}\right)y(t) = 0$.
My take was this: I started differentiating (misusing the notation to make it easier to read)
\begin{align}
&y(t)=u(t)v(t)\\
&y'(t)=u'(t)v(t)+u(t)v'(t)\\
&y''(t)=u''(t)v(t) + 2u'(t)v'(t) + u(t)v''(t)
\end{align}
thus the original equation gives
\begin{align}
u''(t) v(t) + 2u'(t)v'(t) + u(t)v''(t)+ p(t) \left[ u'(t)v(t)+u(t)v'(t) \right] + q(t)u(t)v(t) = 0
\end{align}
dividing by $v(t)$ (we can assume it's nonzero) we have
\begin{align}
\boxed{u''(t)} + \dfrac{2u'(t)v'(t)}{v(t)} + \boxed{u(t)} \dfrac{v''(t)}{v(t)} + p(t)u'(t) + \boxed{u(t)}\dfrac{p(t)v'(t)}{v(t)} + \boxed{u(t)}q(t)=0.
\end{align}
All the boxed terms give us (by common factor) the $Q(t)$ we ask for. However the remaining non-boxed terms must be zero right? So let's ask the equation to zero this for us:
\begin{align}
\dfrac{2u'(t)v'(t)}{v(t)} + p(t)u'(t) = 0\\
\implies u'(t) \left[ 2\dfrac{v'(t)}{v(t)} + p(t) \right] = 0\\
\implies u(t) = c \ \lor \ 2\left( \ln v(t) \right)' + p(t) = 0\\
\implies u(t) = c \ \lor \ v(t) = e^{-\int \left(\dfrac{p(t)}{2}dt\right)}.
\end{align}
This is where I have arrived, and I am not sure how to proceed. Can we assume that $u(t)$ cannot be constant? What about $v(t)$? Nowhere is this asked to find right?
Any help would be appreciated.
 A: Too long for a comment
...
From then on we must solve
$$t^2\frac{d^2y}{dt^2}+t\frac{dy}{dt}+(t^2-\frac{1}{4})y=0.\tag{*}$$
Then by dividing $(*)$ by $t^2$, we get
$$\frac{d^2y}{dt^2}+\frac{1}{t}\frac{dy}{dt}+(1-\frac{1}{4t^2})y=0.$$
Now, here, $p=p(t)=\frac{1}{t}$ and $q=q(t)=1-\frac{1}{4t^2}$.
Then, we apply Algevristis's transformation. Let $y=u(t)v(t)$ where
$$v=v(t)=e^{-\int \frac{p(t)}{2}dt}=e^{-\int \frac{1}{2t}dt}=e^{-\frac{1}{2}\ln t}=t^{-\frac{1}{2}}.$$
According to Algevristis's transformation, $u=u(t)$ satisfies the equation
$$u''+(\frac{v''}{v}+\frac{v'}{v}p+q)u=0.$$
We find $v''=\frac{3}{4}t^{-\frac{5}{2}}$ and $v'=-\frac{1}{2}t^{-\frac{3}{2}}$. If we put $v, v', v'', p, q$ in the equation above, we get
$$u''+(\frac{3}{4}t^{-2}-\frac{1}{2}t^{-2}+1-\frac{1}{4}t^{-2})u=0.$$
Wow. A miracle happens! After cancellations we get clean
$$u''+u=0.\tag{**}$$
Now, we can quickly reach a solution. The solution of $(**)$ is  $u=c_1\cos t+c_2\sin t$, so the solution of $(*)$ is
$$y=c_1\frac{\cos t}{\sqrt{t}}+c_1\frac{\sin t}{\sqrt{t}}$$
because $y$ was $uv$.
My thoughts about the transformation: Is the equation $(*)$ fake Bessel equation?
