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\title{Laplace and the Piecewise Continuous Forcing Function}
\date{March 26, 2021}

\begin{document}
\pagestyle{empty}

\ 

\vspace{-.1in}

\maketitle

\vspace{-.2in}

\dfn{
The \textbf{unit step function} is $\displaystyle u(t) = \begin{cases}
0 &\text{ if } t < 0 \\
1 &\text{ if } t \geq 0
 \end{cases} $
}

\mtd{
The piecewise function $\displaystyle 
f(t) = \begin{cases} g(t) &\text{ if } t < t_1 \\ h(t) &\text{ if } t \geq t_1
\end{cases} $
can be expressed as $
f(t) = g(t) + u(t-t_1)\left[h(t) - g(t) \right].
$
}

\thm{
If $t_1 \geq 0$ and $L\left( g(t + t_1)\right)$ exists for $s > s_0$, then $\displaystyle \boxed{ L\left( u(t-t_1)g(t) \right) = e^{-st_1}L \left(g(t+t_1)\right)}$ for $s> s_0$.
}

\prob{ [Completion]
Find the Laplace transforms of the following functions by first re-expressing them using the unit step function the applying the theorem above.
\begin{enumerate}[a)]
\item $\displaystyle f(t) = \begin{cases}
1 &\text{ if } t < 2 \\ 
t &\text{ if } t \geq 2 
\end{cases}$

\vfill

\item $\displaystyle f(t) = \begin{cases}
e^t &\text{ if } t < \ln(2) \\ 
e^{-t} &\text{ if } t \geq \ln(2)  
\end{cases}$

\vfill

\end{enumerate}
}


\thm{[Second shifting theorem]
$\displaystyle \boxed{  L^{-1}\left[ e^{-st_1}L \left(g(t)\right) \right] = u(t-t_1)g(t-t_1)}$ 
}

\prob{[Completion]
Use the second shifting theorem to find the inverse Laplace transforms of the following functions.
\begin{enumerate}[a)]
\item $\displaystyle H(s) = \frac{e^{-s}}{s-2}$. Hint: $\frac{1}{s-2} = L(e^{2t})$.

\vfill

\item $\displaystyle H(s) = \frac{1}{s} + \frac{e^{-3s}}{s^2}$

\vfill

\end{enumerate}
}

\newpage

\prob{[Graded]
Use Laplace transforms to solve the IVP: $\displaystyle y'' - 2y' = \begin{cases} 4, & 0\leq t < 1 \\ 6, & t \geq 1 \end{cases}$, $y(0) = -6$, $y'(0) = 1$.
}

\end{document}