For a vertical spring, the second-order linear ordinary differential equation equation is $$\frac{d^2y}{dt^2}=-\omega^2(y-l)$$ where $\omega=\sqrt{\frac{k}{m}}$ and $l=-\frac{mg}{k}$
Is there a nice solution to this equation? Wolfram Alpha gives a very long answer involving the parabolic cylinder function. How do they get that?
Absolutely, the equation can be written as $$ \frac{d^2y}{dt^2}+\omega^2y=\omega^2\ell $$ The homogeneous equation associated is $\frac{d^2y}{dt^2}+\omega^2y=0$ whose solutions are $A\cos(\omega x)+B\sin(\omega x)$. In a simple way, a particular solution of the initial equation is $\ell$, therefore all the solutions are of the form $$ A\cos(\omega x)+B\sin(\omega x)+\ell $$