How do I solve the following equation:
$$ F''(r) + \frac{1}{2}rF'(r) - \frac{1}{2}kF(r) = 0 $$
This equation is found when substituting the solution $u = t^{\frac{k}{2}}F(r)$ into the heat equation : $u_{xx} = u_t$.
The solution for $u$ is found by solving the characteristic equation: $$ \frac{dx}{x} = \frac{dt}{2t} = \frac{du}{ku} $$ Where I get $$ r = xt^{\frac{-1}{2}} \\ v = ut^{\frac{-1}{2}} \\ \text{where } v = F(r) \\ u = t^{\frac{k}{2}}F(xt^{\frac{-1}{2}}) $$
The general solution given is: $$ F(r) = c_1U(k+\frac{1}{2}, 2^{\frac{-1}{2}}r) + c_2V(k+\frac{1}{2}, 2^{\frac{-1}{2}}r) $$
Where $U(p,z)$ and $V(p,z)$ are parabolic cylinder functions.
How do I arrive at the given solution?
There is at least anoother solution for $$F''(r) + \frac{1}{2}rF'(r) - \frac{1}{2}kF(r) = 0$$ which looks very close to Hermite differential equation.
Let $F(r)=e^{-\frac{r^2}{4}}G(r)$ to make $$2 G''(r)-r G'(r)-(k+1) G(r)=0$$which does not look very different but easier to integrate making $r=2 x $.
So, using the solution given in the linked page, you should end with $$G(r)=c_1 H_{-(k+1)}\left(\frac{r}{2}\right)+c_2 \, _1F_1\left(\frac{k+1}{2};\frac{1}{2};\frac{r^2}{4}\right)$$ where appear Hermite polynomial and Kummer hypergeometric function.
Multiplied by $e^{-\frac{r^2}{4}}$, you arrive to the parabolic cylinder functions (have a look here).