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\title{A Quick Review}
\author{ \	}
\institute{GMU}
%\subtitle{whatever}
\date{\today}

\begin{document}

\def\d#1#2{ \frac{d #1}{d #2}}
\def\dn#1#2#3{\frac{d^{#3}{#1}}{d #2^{#3}}}
\def\dd#1#2{\dn{#1}{#2}{2}}
\def\L{{\mathcal{L}}}

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%\frame{
%\frametitle{\color{blue}{Outline}}
%\frametitle{Outline}





%\begin{itemize}[<+-| alert@+>]
%%\item<1-| alert@2> The Cantor Tree
%\item First order ODEs and initial value problems
%\item Second order ODEs  and initial value problems
%\item Higher-order systems
%\item The Laplace Transform
%\item Topics in linear algebra
%\item Systems of ODEs
%\item[\ ]
%\end{itemize}




%}

\section{First order ODEs and initial value problems}


\frame{



\frametitle{First order ODEs and initial value problems}



\begin{itemize}[<+-| alert@+>]
\item  Existence and uniqueness of solutions, linear independence of solutions, the Wronskian.

\item Direction fields and qualitative behavior.

\item Solving first-order ODEs.


\begin{itemize}[<+-| alert@+>]

\item Autonomous ODEs and separable ODEs.  Solving by separating variables and integrating.

\item First order linear ODEs. $\d{y}{x} + p(x)y = q(x)$.  Solve by multiplying by the integrating factor $h(x) = e^{\int p(x) dx}$ to get $\d{yh(x)}{x} = q(x)h(x)$.	Integrate both sides with respect to $x$ and solve for $y$.

\item Exact ODEs.  Write an ODE as $g(x,y)dx + h(x,y) dy = 0$.	If	 $g_y = h_x$, there is a function $F(x,y)$ such that $F_x = g$, $F_y = h$.	 Then the solution satisfies $F(x,y) = C$.


\item Euler's method for approximating solutions to $\d{y}{x} = f(x,y)$ subject to $y(x_0) = y_0$.	Choose a (small) $h$.  Let $x_{k+1} = x_k + h$ and $y_{k+1} = y_k+f(x_k,y_k)h$.

\end{itemize}	
\item[\ ]

\end{itemize}

}

\section{Second order ODEs	and initial value problems}
\frame{

\frametitle{Second order ODEs  and initial value problems}

Solving $\dd{y}{x} + b\d{y}{x} + cy = 0$ using the characteristic polynomial $p(r) = r^2 + br + c$.

\pause

\begin{itemize}[<+-| alert@+>]

\item If there are two distinct real roots, $r_1, r_2$, the general solution is $y = C_1 e^{r_1x} + C_2e^{r_2x}$.

\item If there are two complex conjugate roots, $\alpha \pm i\beta$, the general solution is $y = C_1e^{\alpha x}\cos \beta x + C_2e^{\alpha x}\sin \beta x.$

\item It there is one repeated real root $r$, the general solution is $y = C_1e^{rx} + C_2xe^{rx}$.	
\item[\ ]

\end{itemize}



}

\frame{

\frametitle{Second order ODEs  and initial value problems}

Solving $\dd{y}{x} + b\d{y}{x} + cy = h(x)$

\pause

\begin{itemize}[<+-| alert@+>]

\item If $y_g$ is the general solution of $\dd{y}{x} + b\d{y}{x} + cy = 0$ and $y_p$ is a particular solution of $\dd{y}{x} + b\d{y}{x} + cy = h(x)$, then the general solution of $\dd{y}{x} + b\d{y}{x} + cy = h(x)$ is $y_g + y_p$.

\item  We can sometimes find a particular solution of $\dd{y}{x} + b\d{y}{x} + cy = h(x)$ using the method of undetermined coefficients.



\item[\ ]

\end{itemize}




}

\section{Higher-order systems}

\frame{
\frametitle{Higher-order systems}

\begin{itemize}[<+-| alert@+>]

\item Existence and uniqueness of solutions of higher-order linear ODEs and the Wronskian

\item Solutions of $\dn{y}{x}{n} + a_1 \dn{y}{x}{n-1}  + \cdots + a_n y = h(x)$.

\begin{itemize}


\item Solutions of $\dn{y}{x}{n} + a_1 \dn{y}{x}{n-1}  + \cdots + a_n y = 0$ using the characteristic polynomial.	 Everything is analogous to the second-order case.

\item Finding particular solutions of	$\dn{y}{x}{n} + a_1 \dn{y}{x}{n-1}	+ \cdots + a_n y = h(x)$ using the method of undetermined coefficients.




\end{itemize}	

\item[\ ]


\end{itemize}



}


\section{The Laplace Transform}

\frame{

\frametitle{The Laplace Transform}


\begin{itemize}[<+-| alert@+>]

\item $\L(f(t))(s) = \int_0^\infty f(t)e^{-st}dt$.

\begin{itemize}

\item Computing $\L(f)$ from the definition.

\item Finding $\L(f)$ from the formula $\L(f^{(n)}) = s^n\L(f) - s^{n-1}f(0) - \cdots - sf^{(n-2)}(0) - f^{(n-1)}(0)$.

\item Finding inverse Laplace transforms from partial fraction expansions using linearity and	tables


\end{itemize}

\item Unit step functions $u_c$ and expressing piecewise functions in terms of unit step functions.

\begin{itemize}

\item Using the formulas $\L(u_c(t)f(t -c))(s) = e^{-cs}\L(f(t))(s)$ and $\L(e^{ct}f(t)) = \L(f(t))(s - c)$ to find Laplace transforms and inverse Laplace transforms.

\item Using Laplace transforms to solve initial-value problems with piecewise defined forcing functions.


\end{itemize}

\item[\ ]


\end{itemize}






}


\section{Topics in linear algebra}

\frame{
\frametitle{Topics in linear algebra}

\begin{itemize}[<+-| alert@+>]


\item Matrix operations.

\item Eigenvalues and eigenvectors.

\item Diagonalizability.  If the $n \times n$ matrix $A$ has $n$ distinct eigenvalues $\lambda_1, \cdots , \lambda_n$, then $P^{-1}AP = D$ where $D$ is the diagonal matrix having $\lambda_1, \cdots , \lambda_n$ along the diagonal and $P$ is the matrix whose columns are eigenvectors corresponding to $\lambda_1, \cdots , \lambda_n$.	 Therefore, $A = PDP^{-1}$.

\item The exponential of a matrix.	If $D$ is a diagonal matrix, $exp(A)$ is the diagonal matrix whose $ii^{th}$ entry is $e^{D_{ii}}$.

\item If $A = PDP^{-1}$, then $exp(A) = P exp(D) P^{-1}$.


\item[\ ]


\end{itemize}

}


\section{Systems of ODEs}

\frame{
\frametitle{Systems of ODEs}

\begin{itemize}[<+-| alert@+>]

\item Existence and uniqueness of solutions of systems of first-order linear ODEs. Linear independence of functions and the Wronskian of a set of functions.

\item A system of $n$ first-order linear ODEs can be written as $\mathbf{x'} = A(t)\mathbf{x} + \mathbf{g(x)}$.	 If $\mathbf{g(x)} = \mathbf{0}$, the system is homogeneous.

\item An $n^{th}$-order homogeneous linear ODE can be expressed as a homogeneous system of $n$ first-order ODEs.

\item If the $n \times n$ matrix $A$ has $n$ distinct eigenvalues, then the general solution of the system $\mathbf{x'}(t) = A\mathbf{x}(t)$ is $\mathbf{x} = exp(tA)\mathbf{C}$, where $\mathbf{C}$ is a constant vector $\mathbf{C} = \left(\begin{array}{c} C_1 \cr \vdots \cr C_n \end{array}\right)$.


\item[\ ]


\end{itemize}


}






\end{document} 