One possible approximation is as follows. When $\tau$ tends to infinity, the original equation is reduced to

$${\frac {d^{2}}{d{\tau}^{2}}}A \left( \tau \right) + \left( c+{

\frac {{k}^{2}}{8\tau}} \right) A \left( \tau \right) =0$$

and the corresponding analytical solution is

$$A \left( \tau \right) =C_{1}

{{\rm \bf M}\left({\frac {-i{k}^{2}}{16\sqrt {c}}},\frac{1}{2},\,2\,i\sqrt {c}\tau\right)}

+C_{2}

{{\rm \bf W}\left({\frac {-i{k}^{2}}{16\sqrt {c}}},\frac{1}{2},2\,i\sqrt {c}\tau\right)}$$

where $M$ is the Whittaker $M$ function and $W$ is the Whittaker $W$ function.

Making the change $c = -a^2$ and $a = -b$ we obtain

$$A \left( \tau \right) =C_{1}

{{\rm \bf M}\left({\frac {{k}^{2}}{16b}},\frac{1}{2} ,\,2\,b\tau\right)}+

C_{2}

{{\rm \bf W}\left({\frac {{k}^{2}}{16b}},\frac{1}{2}, \,2\,b\tau\right)}$$

Then for $k>0$ and $b>0$, the solution

$$A \left( \tau \right) =C_{1}

{{\rm \bf M}\left({\frac {{k}^{2}}{16b}},\frac{1}{2} ,\,2\,b\tau\right)}$$

gives a exponentially growing solution; and

$$A \left( \tau \right) =C_{2}

{{\rm \bf W}\left({\frac {{k}^{2}}{16b}},\frac{1}{2}, \,2\,b\tau\right)}$$

gives a exponential decaying solution.