\documentclass[12pt]{article}
\usepackage{fullpage,amsmath,amsfonts,graphicx}

\def\d{{\mathrm d}}
\def\e{{\mathrm e}}
\def\i{{\mathrm i}}
\def\R{{\mathbb R}}
\def\x{\boldsymbol{x}}

\begin{document}

\title{Applied Differential Equations and Modeling}
\author{Homework 7}
\date{Due in class Thursday, May 2}
\maketitle

\begin{enumerate}

\item Find the Laplace transform of the given function.
\begin{enumerate}
\item $\displaystyle f(t) = t^{10}$
\item $\displaystyle f(t) = \e^{2t} \, \cos 3 t$
\item $\displaystyle f(t) =
\begin{cases}
t & \text{for } 0 \leq t \leq 1 \\
1 & \text{for } t > 1
\end{cases}
$
\item $\displaystyle f(t) = t^n \, \e^{a t}$
\end{enumerate}

\item Show that the Laplace transform $\mathcal L$ satisfies
\[
  \mathcal L \int_0^t f(\tau) \, \d \tau
  = \frac1s \, \mathcal L(f)
\]
for all functions $f$ such that the transforms on the left and on the
right hand sides are well defined.

\item Compute the inverse Laplace transform of the given function.
\begin{enumerate}
\item $\displaystyle F(s) = \frac2{s^2 + 3s - 4}$
\item $\displaystyle F(s) = \frac{8s^2 - 4s +12}{s(s^2+4)}$
\item $\displaystyle F(s) = \frac{s^3-2s^2-6s-6}{(s^2+2s+2) \, s^2}$
\end{enumerate}

\item Use the Laplace transform to solve the given initial value
problem.

\begin{enumerate}
\item $\displaystyle y'' - y' - 6y = 0$

with $y(0)=1$, $y'(0)=-1$

\item $\displaystyle y'' + \omega^2 \, y = \cos 2t$

for $\omega^2 \neq 4$ with $y(0)=1$, $y'(0)=0$


\item $\displaystyle y''''- 4y = 0$

with $y(0)=1$, $y'(0)=0$, $y''(0)=-2$, $y'''(0)=0$

\end{enumerate}

\item Find the Laplace transform of the given periodic function.

\begin{enumerate}
\item $\displaystyle f(t) =
\begin{cases}
  1 & \text{for } 0 \leq t < 1 \\
  -1 & \text{for } 1 \leq t < 2
\end{cases}$

and $f$ has period 2.

\item $f(t) = t$ for $0 \leq t < 1$ and $f$ has period 1.

\end{enumerate}

\item For a function $f(t)$, write $F(s)$ to denote the Laplace
transform.  Prove the following.
\begin{enumerate}
%\item $\displaystyle \mathcal L \biggl( \frac{f(t)}t \biggr) = \int_s^\infty F(u) \,
%\d u$
\item $\displaystyle \mathcal L (\e^{ct} \, f(t)) =  F (s-c)$
\item $\displaystyle \mathcal L (f(ct)) = \frac1c \, F \Bigl(
\frac{s}c \Bigr)$
\end{enumerate}

% \item Find the solution of the given initial value problem and draw
% its graph.
% \begin{enumerate}
% \item $\displaystyle y'' +2 y' + 2y = \delta(t-\pi)$

% with $y(0)=1$, $y'(0)=0$

% \item $\displaystyle y'' + y = \delta(t-2 \pi) \, \cos t$

% with $y(0)=0$, $y'(0)=1$

% \end{enumerate}



\end{enumerate}
\end{document}
