1Let $$x=\frac{1}{t}$$, then $\frac{d}{dx}=\frac{d}{dt}\frac{dt}{dx}=-t^{2}\frac{d}{dt}$ And \begin{align*} \frac{d^{2}}{dx^{2}} & =\frac{d}{dx}\left ( \frac{d}{dx}\right ) =\left ( -t^{2}\frac{d}{dt}\right ) \left ( -t^{2}\frac{d}{dt}\right ) =-t^{2}\frac{d}{dt}\left ( -t^{2}\frac{d}{dt}\right ) \\ & =-t^{2}\left ( -2t\frac{d}{dt}-t^{2}\frac{d^{2}}{dt^{2}}\right ) \\ & =2t^{3}\frac{d}{dt}+t^{4}\frac{d^{2}}{dt^{2}} \end{align*}

The original ODE becomes \begin{align*} \left ( 2t^{3}\frac{d}{dt}+t^{4}\frac{d^{2}}{dt^{2}}\right ) y+\left . p\left ( x\right ) \right \vert _{x=\frac{1}{2}}\left ( -t^{2}\frac{d}{dt}\right ) y+\left . q\left ( x\right ) \right \vert _{x=\frac{1}{t}}y & =0\\ \left ( 2t^{3}y^{\prime }+t^{4}y^{\prime \prime }\right ) -t^{2}p\left ( t\right ) y^{\prime }+q\left ( t\right ) y & =0\\ t^{4}y^{\prime \prime }+\left ( -t^{2}p\left ( t\right ) +2t^{3}\right ) y^{\prime }+q\left ( t\right ) y & =0\\ y^{\prime \prime }+\frac{\left ( -p\left ( t\right ) +2t\right ) }{t^{2}}y^{\prime }+\frac{q\left ( t\right ) }{t^{4}}y & =0 \end{align*}