I'm trying to use multicol to create a formula sheet with two columns. I have specified the use of such columns, and their position, but when I compile they come out one after the other rather than split vertically on each side of the page. It would be ideal to have two separate columns very close together to maximize space as this will be used during an examination. I have things like
\pagestyle{empty}
\setlength{\parindent}{0pt}
\setlength{\parskip}{0pt plus 0.5ex}
in my preamble to use as much of the page as possible, but every compilation yields the same output. I looked at a few posts with the lipsum environment but I'm not sure how that works with multicol. I realize this is a simple problem, and definitely has a simple answer, but I cannot achieve the desired output. It would also be excellent to have a framebox around each column, separated by \hspace or something of that nature.
Thanks in advance
EDIT:
\begin{multicols}{3}
2.1 a)
\begin{align*}
\ddot{x} &= \frac{1}{m}\left(F_0+ct\right)\\
\dot{x} &= \int_{0}^{t}\frac{1}{m}\left(F_0+ct\right)\,dt = \dfrac{F_0}{m}t+\dfrac{c}{2m}t^2\\
x &= \int_{0}^{t}\left(\dfrac{F_0}{m}t+\dfrac{c}{2m}t^2\right)\\
\end{align*}
\end{multicols}
Here is a sample, it's simply a strategy/reference to an assignment problem. I can post others as well as this is just a snippet, let me know.
EDIT 2
\begin{multicols}{3}
{\bf Bullet shot up with quadratic drag}
\begin{align*}
F &=-c_2v^2-mg\\
mv\dfrac{dv}{dx} &= -c_2v^2-mg\\
\int\dfrac{mv}{c_2v^2+mg}\,dv &=\int -1\,dx\\
\dfrac{m}{2c_2}\int \ln(c_2v^2+mg) &= -x+\beta\\
\beta &= \dfrac{m}{c_2}\ln(c_2v_0^2+mg)\\
v(x) &= \left((v_0^2+\frac{mg}{c_2})\exp{\dfrac{-2c_2x}{m}}-\dfrac{mg}{c_2}\right)^{1/2}\\
\end{align*}
{\bf Useful Proof}
\begin{align*}
\dfrac{d}{dt}\left[\vec{r}\cdot(\vec{v}\times\vec{a})\right] &= \vec{r}\cdot(\vec{v}\times\vec{a})\\
&= \dot{\vec{r}}\cdot(\vec{v}\times\vec{a}) + \vec{r}\cdot\dfrac{d}{dt}(\vec{v}\times\vec{a})\\
&= \vec{v}\cdot(\vec{v}\times\vec{a})+\vec{r}\cdot\left[\dot{\vec{v}}\times\vec{a}+\vec{v}\times\dot{\vec{a}}\right]\\
&= \vec{r}\cdot(\vec{v}\times\dot{\vec{a}})\\
\end{align*}
{\bf Find $V(x)$ where $F(x)=-kx+\frac{kx^3}{A^2}$,$v(t=0) = A\sqrt{\frac{k}{2m}}$ and $x(t=0) = 0$ }
\begin{align*}
F&=-\dfrac{dV}{dx}\\
k\int x-\dfrac{x^3}{A^2}\,dx &= \int dV\\
\dfrac{kx^2}{2}-\dfrac{kx^4}{4A^2} &= V(x) + C\\
V(x) &= \dfrac{kx^2}{2} -\dfrac{kx^4}{4A^2}\\
\end{align*}
{\bf Proof of successive maxima}
\begin{align*}
x(t) &= \exp{-\gamma t}(A\cos(\omega_dt+\theta_0))\\
x'(t) &= -\gamma\exp{-\gamma t}(A\cos(\omega_dt+\theta_0))+\exp{-\gamma t}(-A\omega_d\sin(\omega_dt+\theta_0))\\
0 &=A(-\gamma\cos(\omega_dt+\theta_0)-\omega_d\sin(\omega_dt+\theta_0))\\
1&=\dfrac{-\omega_d}{\gamma}\tan(\omega_dt+\theta_0)\\
\dfrac{-\gamma}{\omega_d}&=\tan(\omega_dt+\theta_0)\\
(\omega_dt_{i+1}+\theta_0) - (\omega_dt_i+\theta_0) &= 2\pi\\
\omega_d(t_{i+1}-t_i) &= 2\pi\\
t_{i+1}-t_i &= \dfrac{2\pi}{\omega_d}\\
t_{i+1}-t_i &= \dfrac{2\pi}{(\omega_0^2-\gamma^2)^{1/2}}
\end{align*}