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\begin{document}
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{\newpage
\clearpage
\samepage \begin{displaymath}Y(s)=L[y(t)](s)=\int_{0}^{\infty}e^{-st}y(t)dt
\end{displaymath}
}

{\newpage
\clearpage
\samepage \begin{displaymath}Y(s)=\int_{0}^{\infty}e^{-st}e^{t}dt=
\int_{0}^{\infty}e^{-(s-1)t}dt=\frac{1}{s-1}
\end{displaymath}
}

{\newpage
\clearpage
\samepage \begin{displaymath}Y(s)=\int_{0}^{\infty}e^{-st}\cos(3t)dt=\frac{s}{s^2+9}
\end{displaymath}
}

{\newpage
\clearpage
\samepage \begin{displaymath}|y(t)|\leq Me^{ct} \hspace{.2in} t\geq T
\end{displaymath}
}

{\newpage
\clearpage
\samepage \begin{displaymath}\int_{0}^{\infty}e^{-st}e^{t^2}dt 
\end{displaymath}
}

{\newpage
\clearpage
\samepage \begin{displaymath}L[c_1y_1(t)](s)=c_1L[y_1(t)](s)=c_1Y_1(s)
\end{displaymath}
}

{\newpage
\clearpage
\samepage \begin{displaymath}L[y_1(t)+y_2(t)](s)=L[y_1(t)](s)+L[y_2(t)](s)=Y_1(s)+Y_2(s)
\end{displaymath}
}

{\newpage
\clearpage
\samepage \begin{displaymath}L[3e^{t}](s)=3L[e^{t}](s)=3\cdot\frac{1}{s-1}=\frac{3}{s-1}
\end{displaymath}
}

{\newpage
\clearpage
\samepage \begin{displaymath}L[e^t+\cos 2t](s)=L[e^t](s)+L[\cos 2t](s)=\frac{1}{s-1}+\frac{s}{s^2+4}
\end{displaymath}
}

{\newpage
\clearpage
\samepage \begin{displaymath}L[3e^t+4\cos 2t](s)=L[3e^t](s)+L[2\cos 2t](s)=3L[e^t](s)+4L[\cos
2t](s)= \frac{3}{s-1}+\frac{4s}{s^2+4}
\end{displaymath}
}

{\newpage
\clearpage
\samepage \begin{displaymath}L[e^{at}y(t)](s)=Y(s-a),
\end{displaymath}
}

{\newpage
\clearpage
\samepage \begin{displaymath}y(t)=t \hspace{.2in} \Leftrightarrow Y(s)=\frac{1}{s^2} \hspace{.2in}
{\rm therefore}\ L[e^{2t}t](s)=Y(s-2)=\frac{1}{(s-2)^2}
\end{displaymath}
}

{\newpage
\clearpage
\samepage \begin{displaymath}L[y'(t)](s)=sY(s)-y(0)
\end{displaymath}
}

{\newpage
\clearpage
\samepage \begin{displaymath}L[y''(t)](s)=s^2Y(s)-sy(0)-y'(0)
\end{displaymath}
}

{\newpage
\clearpage
\samepage \begin{displaymath}L[y^(n)(t)](s)=s^nY(s)-s^{n-1}y(0)-s^{n-2}y'(0)-\ldots -
sy^{(n-2)}(0)-y^{(n-1)}(0)
\end{displaymath}
}

{\newpage
\clearpage
\samepage \begin{displaymath}L[t^ny(t)](s)=(-1)^nY^{(n)}(s)
\end{displaymath}
}

{\newpage
\clearpage
\samepage \begin{displaymath}L[t\sin t]=-\frac{d}{ds}[\frac{1}{s^2+1}]=\frac{2s}{(s^2+1)^2}
\end{displaymath}
}

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\end{document}
