\documentclass{article}
\usepackage{hyperref}
\def\rload{R_{\rm load}}
\date{}
\begin{document}
\title{Project Executive Summary}
\author{Holden Rohrer and Nithya Jayakumar}

\maketitle
\section{Matrix Representation and Homogeneous Solution}

To determine the relevant properties of the linear system, matrix form
is useful (this form was chosen to reduce fractions' usage):
\def\x{{\bf x}}
$$\x' =
{1\over R_1C_1C_2\rload}
\pmatrix{0&-C_2\rload      &0 \cr
         0&-C_2(R_1+\rload)&C_1R_1\cr
         0&C_2R_1          &-C_1R_1}   \x +
{1\over R_1}
\pmatrix{\omega\cos(\omega t)\cr
         \omega\cos(\omega t)\cr
         0}
.$$
The characteristic polynomial is
$$-\lambda( (-\lambda-C_2(R_1+\rload))(-\lambda-C_1R_1)
- C_2C_1R_1^2).$$
Expanded,
$$-\lambda( \lambda^2 + \lambda(C_2(R_1+\rload)+C_1R_1)
+ C_1C_2R_1\rload).$$
In terms of its roots (with $b=C_2(R_1+\rload)+C_1R_1$ and
$c = C_1C_2R_1\rload,$
$$-\lambda{(\lambda-{-b+\sqrt{b^2-4c}\over2})
(\lambda-{-b-\sqrt{b^2-4c}\over 2})}$$
%For reference,
%$$b^2-4c = C_2^2(R_1+\rload)^2 + C_1^2\rload^2
%           - 2C_1C_2\rload(3R_1+\rload).$$
Let $r_1$ and $r_2$ designate these two non-zero roots.
In the original matrix $A$, this being the transformed matrix $A/c$,
$Av = cr_1$ because $A/c * v = r_1.$

The trivial zero eigenvalue corresponds to a unit x-direction vector
by inspection. The two remaining roots, in the general case of a non-%
degenerate system, which hasn't been explicitly ruled out, the middle
row can be ``ignored'' because it is linearly independent in the
following system:
$$
\pmatrix{-r&-C_2\rload &0\cr
         * &*          &*\cr
         0 &C_2R_1     &-r-C_1R_1}
$$
With $x = 1$, $\displaystyle y = -{r\over C_2\rload}$ and
$\displaystyle z = y{C_2R_1\over r+C_1R_1}.$

\def\bu{\par\leavevmode\llap{\hbox to \parindent{\hfil $\bullet$ \hfil}}}
\noindent The eigenvalues and respective eigenvectors are:

\bu $\lambda_1 = 0, v_1 = \pmatrix{1\cr0\cr0}$

\def\num#1{\pmatrix{1\cr
                    -{r_#1\over C_2\rload}\cr
                    -{R_1r_#1\over \rload(r_#1+C_1R_1)}}}

\bu $\displaystyle\lambda_2 = {r_1C_1C_2R_1\rload}, v_2 = \num1.$

\bu $\displaystyle\lambda_3 = {r_2C_1C_2R_1\rload}, v_3 = \num2.$

This corresponds to a solution of the form $f = Ce^{\lambda t}v,\cdots$
where $C\in {\bf C},$ $f\in {\bf R}\to{\bf R}$. If the non-trivial
eigenvectors are complex, %% TRY TO PROVE THIS!!
\def\re{{\rm Re}}\def\im{{\rm Im}}
their exponential solutions form, in the reals,
$g = C_1\cos{\re(\lambda)t}\re v + C_2\sin{\re(\lambda)t}\im v.$ %% DOUBLE CHECK.

\section{Nonhomogeneous System}

Extending to the nonhomogeneous system will take slightly different
paths depending on if the system has complex roots or has real roots.
But in either case, $\cos x*{\rm polynomial}+\sin x*{\rm polynomial}$
should be a particular solution. 

We can apply the method of Variation of Parameters to this. Essentially, the solution is $\bf{x} = c_1\bf{\lambda_1}(t) + c_2\bf{\lambda_2}(t) + c_3\bf{\lambda_3}(t)\bf{\lambda_p}(t)$, where the particular solution $\bf{\lambda_p}(t)$ is:
$$\bf{\lambda_p}(t) = \bf{X}(t)\int\bf{X^{-1}}(t)\bf{g}(t)dt,$$ where $\bf{X}(t)$ is the Fundamental matrix for the equation and $\bf{g}(t) = {1\over R_1}
\pmatrix{\omega\cos(\omega t)\cr
         \omega\cos(\omega t)\cr
         0}.$
\end{document}