\documentclass[12pt,fleqn]{article}
\pagestyle{empty}
\setlength{\topmargin}{-1in}
\setlength{\oddsidemargin}{-.3in}
\setlength{\textheight}{9in}
\setlength{\textwidth}{7.2in}


\usepackage{amssymb}
\usepackage{amsmath}


\newcommand{\adj}{\mathrm{Adj}}
\newcommand{\mbf}{\mathbf}
\newcommand{\R}{\mathbb{R}}
\newcommand{\ra}{\rightarrow}
\newcommand{\proj}{\mathrm{proj}}
\newcommand{\dotprod}{\bullet}
\newcommand{\sol}{\textbf{Solution}}

\begin{document}

{\Large Key to TEST 1}


\begin{enumerate}


\item  
  \begin{enumerate}
  \item  
    \[
    x \frac{dy}{dx} - y - 2x^2y=0 \qquad y(1)=1
    \]
    \sol:
    This is separable:
    \[
    \frac{dy}{y} = \frac{1+2x^2}{x} dx
    \]
    Integrating gives:
    \[
    y = Axe^{x^2}
    \]
    Solving for the constant gives $A=1/e$ and therefore the solution is:
    \[
    y = \frac{xe^{x^2}}{e}
    \]
  \item
    \[
    \frac{dy}{dx} = e^x - y \qquad y(0)=1
    \]
    \sol:
    This is linear.  We can rearrange to get:
    \[
    \frac{dy}{dx} +y = e^x 
    \]
    We need to find the integrating factor which is equal to
    \[
    IntFact = e^{\int dx} = e^x
    \]
    \[
    \frac{d}{dx} (ye^x) = e^{2x}
    \]
    We now integrate and see get
    \[
    ye^x = \frac{1}{2}e^{2x}+C
    \]
    or
    \[
    y = \frac{1}{2}e^{x}+Ce^{-x}
    \]
    Solving for the constant (using the initial condition) gives $C=1/2$ and therefore the solution is
    \[
    y = \frac{1}{2}e^{x}+\frac{1}{2}e^{-x}
    \]
  \end{enumerate}
				   
\item  
  Determine if you can guarantee that the following initial value problems have a solution.
  If you can guarantee a solution, can you guarantee that the solution is unique?
  (Justify your answers!)
  \begin{enumerate}
  \item  
    \[
    \frac{dy}{dx} = \frac{x}{x-y} \qquad y(1)=2
    \]
    \sol:
    Here we apply the existance and uniqueness theorem with $f(x,y) = x/(x-y)$.
    This function continuous in a rectangle containing $(1,2)$ so therefore 
    the theorem says a solution exists.
    
    To see if the solution is unique, we look at $f_y = x/(x-y)^2$.
    $f_y$ is continuous in a rectangle containing $(1,2)$, so therefore the solution will be unique.
  \item  
    \[
    \frac{dy}{dx} = x^3\sin y + y^{2/3} \qquad y(1)=0
    \]
    \sol:
    We apply the theorem to the funtion $f(x,y) =  x^3\sin y + y^{2/3}$ which is continuous in a rectangle
    containing $(1,0)$.
    Therefore a solution exists.

    But, $f_y = x^3\cos y +(2/3)y^{-1/3}$ is not continuous in a rectangle containing $(1,0)$.
    So, we cannot assure that the solution is unique.
  \end{enumerate}	

\item (10 points)
  Consider the linear system below:
  \[
  \begin{array}{rrrrrrr}
    x_1 & + & x_2 & - & x_3 & = 0 \\
    x_1 &  &  & + & 2x_3 & = 0 \\
    2x_1 & - & 3x_2 & + & 13x_3 & = 0
  \end{array}
  \]
  \begin{enumerate}
  \item  Write the system of equations as a matrix equation.
    \sol:  The matrix equation is $AX=B$ where $A$ is the matrix given below and
    $B$ is the column vector $[0,0,0]^T$.
  \item  Write the system equations in an augmented matrix and use Gaussian elimination on this matrix
    (you should end up with a reduced echelon matrix).
    \[
    \left[
    \begin{array}{rrr|r}
      1 & 1 & -1 & 0 \\
      1 & 0 & 2 & 0 \\
      2 & -3 & 13 & 0      
    \end{array}
    \right]
    \longrightarrow
    \left[
    \begin{array}{rrr|r}
      1 & 0 & 2 & 0 \\
      0 & 1 & -3 & 0 \\
      0 & 0 & 0 & 0      
    \end{array}
    \right]
    \]
  \item  For the matrix $A$ below, what is a basis for the solution space for $A$?
      (notice the similarity with the linear system).
    \[
    \left[
    \begin{array}{rrr}
      1 & 1 & -1 \\
      1 & 0 & 2 \\
      2 & -3 & 13
    \end{array}
    \right]
    \]
    \sol: 
    We can read this off from the reduced echelon matrix.
    A basis consists of the vector:
    \[
    v = [ -2, 3, 1]
    \]
  \item  What is the dimension of the solution space?
    \sol: Dimension one
  \end{enumerate}

\item (5 points)
  Determine if the set of vectors is linearly dependent or linearly independent
  (justify your answer!)
  \[
  S = \{(1,2,3,4), (1,2,-3,12), (-2,4,5,-17), (0,1,2,4), (0,0,2,-5) \}
  \]
  \sol:
  You can do it by hand (ouch!) or notice that we are given five vectors 
  in a space of dimension four - so therefore the set must be linearly dependent.


\item  (5 points)
  Consider the set $S$ of vectors below.
  (Justify your answers!)
  \[
  S = \{ (4,0,1), (1,2,1), (1,-6,-2) \}
  \]
  Is the vector $(1,-2,3)$ in $\mathrm{Span}(S)$?
  \sol:  We row reducing the matrix:
    \[
    \left[
      \begin{array}{rrr|r}
	4 & 1 & 1 & 1 \\
	0 & 2 & -6 & -2 \\
	1 & 1 & -2 & 3 
      \end{array}
      \right]
    \longrightarrow
    \left[
      \begin{array}{rrr|r}
	1 & 0 & 1 & 0 \\
	0 & 1 & -3 & 0 \\
	0 & 0 & 0 & 1 
      \end{array}
      \right]
    \]
    which means that we can't solve the equation
    \[
    c_1
    \left[
      \begin{array}{r}
	4 \\ 0 \\ 1
      \end{array}
      \right]
    + c_2
    \left[
      \begin{array}{r}
	1 \\ 2 \\ 1 
      \end{array}
      \right]
    + c_3
    \left[
      \begin{array}{r}
	1 \\ -6 \\ -2
      \end{array}
      \right]
    =
    \left[
      \begin{array}{r}
	1 \\ -2 \\ 3
      \end{array}
      \right]
    \]

\end{enumerate}

\end{document}