\documentclass[letterpaper]{amsart}

\usepackage{amsmath, amssymb, amsthm, enumerate}
\usepackage[margin = 0.5in]{geometry}
\usepackage[pdflang={en-US}]{hyperref}

\newcommand{\sm}{\vspace{.1in}}

\newcommand{\header}[1]{\large{\textbf{#1}}}

\theoremstyle{definition}
\newtheorem*{theorem}{Theorem}
\newtheorem*{lemma}{Lemma}
\newtheorem*{proposition}{Proposition}
\newtheorem*{definition}{Definition}
\newtheorem{problem}{}
\newtheorem*{challenge}{Challenge}
\newtheorem*{example}{Examples}
\newtheorem*{method}{Method}
\newtheorem*{comment}{Comment}
\newtheorem{video}{Video}

\newcommand{\thm}[1]{\begin{theorem}#1\end{theorem}}
\newcommand{\prop}[1]{\begin{proposition}#1\end{proposition}}
\newcommand{\lem}[1]{\begin{lemma}#1\end{lemma}}
\newcommand{\dfn}[1]{\begin{definition}#1\end{definition}}
\newcommand{\prob}[1]{\begin{problem}#1\end{problem} \vfill}
\newcommand{\chal}[1]{\begin{challenge}#1\end{challenge}}
\newcommand{\eg}[1]{\begin{example}#1\end{example}}
\newcommand{\mtd}[1]{\begin{method}#1\end{method}}
\newcommand{\cmt}[1]{\begin{comment}#1\end{comment}}
\newcommand{\vid}[1]{\begin{video}#1\end{video}}

\newcommand{\abs}[1]{\lvert #1 \rvert}

\renewcommand{\familydefault}{\sfdefault}

\title{Solving IVPs using Laplace transforms}
\date{March 22, 2021}

\begin{document}
\pagestyle{empty}

\ 

\vspace{-.1in}

\maketitle

\vspace{-.2in}

\thm{
Suppose $f$ and $f'$ are continuous on $[0, \infty)$ and of exponential order $s_0$, and that $f''$ is piecewise continuous on $[0,\infty)$. 
Then $f$, $f'$, and $f''$ have Laplace trannsforms for $s> s_0$:
\[ \boxed{L(f') = s L(f) - f(0)} \text{ and } \boxed{L(f'') = s^2 L(f) - sf(0) - f'(0)} \]
}

\mtd{
How to use Laplace transforms to solve an IVP:
\begin{enumerate}[1)]
\item Take the Laplace transform of the differential equation
\item Use the initial conditions
\item Solve for $Y = L(y)$ (do as little algebra as possible)
\item Take the inverse Laplace transform to find $y$
\end{enumerate}
}

\noindent \hrulefill

\prob{[Completion]
Use Laplace transforms to solve the following IVPs:
\begin{enumerate}[a)]
\item $\displaystyle y'' + y' -2y = 2e^{3t}$, $y(0) = -1$, $y'(0) = 4$
\item $\displaystyle y'' - 4y' + 4y = 1$, $y(0) = 0$, $y'(0) = 1$ (be careful with the partial fractions decomposition in this one)
\end{enumerate} 
}


\end{document}