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Quiz 6 Math 2560 Sect 81\hfill  Show your work \break
Dec 2, 2016

[3]~~ 1.)  Circle T for True and F for false:


1a.)  If $\phi$ is a solution to a first order {\bf linear homogeneous} differential equation, then $c\phi$ is 
also a 
solution to this equation.
 \hfill  T~~~~~~~
\v 
1b.)   If $\phi$ is a solution to a first order linear differential equation, then $c\phi$ is also a 
solution to this equation.
 \hfill  ~~~~~~~F
\v
 
 [4]~~ 2a.)  Solve:  ${\bf x}' = \left[\matrix{ 1 & 5 \cr 2 & 3}\right] {\bf x}$


Find eigenvalues:  

  $\left|\matrix{ 1-r & 5 \cr 2 & 3-r}\right| = 
(1-r)(3-r) - 10 = r^2 - 4r + 3 - 10 = r^2 - 4r - 7 = 0$
  
  $r = {4 \pm \sqrt{16 - 4(-7)} \over 2} = {4 \pm \sqrt{4(4 +7)} \over 2}
= {4 \pm 2\sqrt{11} \over 2} = 2 \pm \sqrt{11}$.

$2 + \sqrt{11} > 0$ and $2 - \sqrt{11} < 0$.  
Thus the critical point ${\bf x} = {\bf 0}$  is an unstable saddle point.

Find eigenvectors:

 $\left[\matrix{ 1-(2 \pm \sqrt{11}) & 5 \cr 2 & 3-(2 \pm \sqrt{11})}\right]$ =  $\left[\matrix{ -1 \mp \sqrt{11} & 5 \cr 2 & 1 \mp \sqrt{11}}\right]$
 
 $\left[\matrix{ -1 \mp \sqrt{11} & 5 \cr 2 & 1 \mp \sqrt{11}}\right]$
 $\left[\matrix{ 5 \cr  1 \pm \sqrt{11}}\right]$ = 
 $\left[\matrix{ 5(-1 \mp \sqrt{11}) +5(  1 \pm \sqrt{11}) \cr
 10+(1 \mp \sqrt{11})(  1 \pm \sqrt{11}) }\right]$= $\left[\matrix{ 0 \cr 0}\right]$

Thus  $\left[\matrix{ 5 \cr  1 \pm \sqrt{11}}\right]$ is an eigenvector 
with eigenvalue  $2 \pm \sqrt{11}$.
\vfill\vfill\vfill\vfill

\centerline{Answer:  ~~
${\bf x} 
= c_1 \left[\matrix{ 5 \cr  1 + \sqrt{11}}\right] 
e^{(2 + \sqrt{11})t} +  c_2 \left[\matrix{ 5 \cr  1 - \sqrt{11}}\right] 
e^{(2 - \sqrt{11})t}$ ~~}

[2]~~ 2b.)  Determine if the critical point is stable, asymptotically stable, or unstable.

\centerline{unstable}
\vskip 10pt

[1]~~ 2c.)  Draw the direction field and sketch a few trajectories.  

Graph $x_2 = {1 + \sqrt{11} \over 5}x_1$ and 
$x_2 = {1 - \sqrt{11} \over 5}x_1$  for $x_1 > 0$ and $x_1 < 0$ along with some other trajectories.


\centerline{\includegraphics[width=100ex]{quiz6em4fig.pdf}}













\end

\vfill

\eject



Quiz 5 Math 3600 \hfill  Show your work \break
Dec 2, 2016

[5]~~ 1.)  Circle T for True and F for false:

 The power series for $f(x) = -(2 + x)$ converges for all $x$.
 \hfill  T~~~~~~~
 
 Note the power series for the finite polynomial  $f(x) = -(2 + x)$  is itself,  $f(x) = -(2 + x)$ 
 
 
 
 \v\v
[6]~~ 2. Suppose
$A   \left[\matrix{5 \cr 6}\right] = \left[\matrix{5 \cr 13}\right]$,
$A   \left[\matrix{3 \cr 5}\right] = \left[\matrix{9 \cr 15}\right]$,
$A   \left[\matrix{-1 \cr 3}\right] = \left[\matrix{17 \cr
19}\right]$,
$A   \left[\matrix{2 \cr 1}\right] = \left[\matrix{-4 \cr -2}\right]$
\v
State the 2 eigenvalues of $A$:  \underbar{3, -2}
\v\v
State 5 eigenvectors of $A$:   {$\left[\matrix{3 \cr 5}\right], \left[\matrix{-3 \cr -5}\right], \left[\matrix{6 \cr 10}\right], \left[\matrix{2 \cr 1}\right], \left[\matrix{-2 \cr -1}\right]$}
\v
3.)  Given the non-linear system of equations ${dx \over dt} = x - 2 $ and ${dy \over dt} = x(y - 1)^2 $,

${dx \over dt} = x - 2 = 0$.  Thus $x = 2$

  and ${dy \over dt} = x(y - 1)^2 = 2(y-1)^2 =  0 $, implies $y = 1$



Jacobian:  $\left[\matrix{1 & 0 \cr (y-1)^2 & 2x(y-1)}\right]$

\vfill
[7]~~ 3a.)  The critical point of this system is \underbar{~~(2, 1)~~}.
\v
[8]~~ 3b.)  The corresponding linear system near this critical point is \underbar{\hskip 2in}.
\v
[7]~~ 3c.)  The eigenvalues of this linear system are \underbar{\hskip 2in}.  
\v
[7]~~ 3d.)  What conclusions can you then draw about the non-linear system:

\end
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 \end

[3]~~ 1.)  Circle T for True and F for false:

If $b^2 - 4ac < 0$ then the solution to the initial value problem
$ay'' +
by' + cy = 0$, $y(0) = -1$, $y'(0) = -3$ is real valued.

 \hfill  T~~~~~~~F

[4]~~ 2a.)  Solve:  ${\bf x}' = \left[\matrix{ 1 & 2 \cr 2 & 1}\right] {\bf x}$


\vfill\vfill\vfill\vfill

\centerline{Answer:  \underbar{\hskip 2in}}

[2]~~ 2b.)  Determine if the critical point is stable, asymptotically stable, or unstable.
\vskip 10pt

[1]~~ 2b.)  Draw the direction field and sketch a few trajectories.  
\vfill

\end
2.)   ${\cal L}^{-1}({s  \over s^2 + 6s + 10}) = \underline{\hskip 3in}$ 

Note: ${\cal L}(e^{at}cos(bt)) = {s - a \over (s-a)^2 + b^2}$ and 
 ${\cal L}(e^{at}sin(bt)) = {b \over (s-a)^2 + b^2}$ 

\end

Quiz 4 Math 2560 Sect 81\hfill  Show your work \break
Oct. 27, 2016 

\centerline{\includegraphics[width = 2.7in]{grapht.eps}}

1.  Circle T for True and F for false:

[5]~  1a.)  $L(g)=g′′+p(t)g′+q(t)g$ is a linear function on the space of all twice differentiable functions.  \hfill  T~~~~~~~F


[5]~  1b.)   $ \left[\matrix{5 \cr 6}\right], \left[\matrix{4 \cr 13}\right]$ are linearly independent.
 \hfill  T~~~~~~~F


2.)  Draw the direction field for ${dx_2 \over dx_1} = {20x + 22y \over 2x + 15y}$

Hint  $ \left[\matrix{2 & 15\cr 20 & 22 }\right]   \left[\matrix{ 1 \cr 2 }\right]=   \left[\matrix{32 \cr 64 }\right]$ and $ \left[\matrix{2 & 15\cr 20 & 22 }\right]   \left[\matrix{~~3 \cr-2 }\right]=   \left[\matrix{ -24\cr 16}\right]$


\centerline{\includegraphics[width = 2.7in]{graph.pdf}}

\end

Quiz 4 Math 2560 Sect 81\hfill  Show your work \break
Oct. 27, 2016 

1.)  Solve

\centerline{$y''' - y'' -  y' + y = 2e^{-t} + 3$}

\vskip 5in
\vfill\vfill\vfill\vfill\vfill\vfill\vfill\vfill\vfill\vfill\vfill\vfill\vfill\vfill\vfill\vfill\vfill\vfill\vfill
\centerline{Answer:  \underbar{\hskip 2in}}

\end

1.)  Suppose $y = c_1 e^{3t} + c_2te^{3t} + 4t^2e^{3t}$
is a solution to $y'' - 6y' + 9y = 8e^{3t}$.
Find the solution to the initial value problem:
\vskip 10pt
\centerline{$y'' - 6y' + 9y = 8e^{3t} + 27t$,~~ $y(0) = 5$,~~ $y'(0) = 2$.}

Note:  Solving this IVP is a 4 part problem, but I have already done the first two parts for you.

\vskip 5in
\vfill\vfill\vfill\vfill\vfill\vfill\vfill\vfill\vfill\vfill\vfill\vfill\vfill\vfill\vfill\vfill\vfill\vfill\vfill
\centerline{Answer:  \underbar{\hskip 2in}}


\end
Quiz 2 Math 3600 Sect 2\hfill  Show your work \break
Sept. 23, 2016

[20]~ 1.)  Transform the given initial value problem into an equivalent problem with initial point at the origin:

\centerline{${dy \over dt} = t^2 + y^2$, ~~~~$y(1) = 2$}
\vskip 2.45in

\centerline{Answer:  \underbar{\hskip 2in}}
\vskip 10pt
Quiz 2 Math 3600 Sect 2\hfill  Show your work \break
Sept. 23, 2016

[20]~ 1.)  Transform the given initial value problem into an equivalent problem with initial point at the origin:

\centerline{${dy \over dt} = t^2 + y^2$, ~~~~$y(1) = 2$}
\vskip 2.45in

\centerline{Answer:  \underbar{\hskip 4in}}

\end

[10]~ 1.)  Given that $y(x) = x^{3 \over 2}$ and $y(x) = {1 \over x}$  are solutions to $2x^2y'' + xy' - 3y = 0$, state the general solution to this 2nd order homogeneous linear differential equation:

\vfill

%$2t^2(2t^{-3}) + t(-t^{-2}) - 3(t^{-1}) =4(t^{-1}) - t^{-1} - 3(t^{-1}) = 0$

%$2t^2({3 \over 4}t^{-1 \over 2}) + t({3 \over 2} t^{1 \over 2}) - 3( t^{3 \over 2}) =({3 \over 2}t^{3 \over 2}) + ({3 \over 2} t^{3 \over 2}) - 3( t^{3 \over 2})= 0$


%$t^2(6t) + t(3t^{2}) - 9(t^{3}) =4(t^{-1}) - t^{-1} - 3(t^{-1}) = 0$

%$t^2(2) + t(2t) - 4(t^{2}) =4(t^{-1}) - t^{-1} - 3(t^{-1}) = 0$


%Answer:$\underline{~~\hskip 2in~~}$


[10]~ 2.)  Solve: $y' = y~sin(x) + y$.

\vfill
\vfill
\vfill
\vfill


%Answer:$\underline{~~\hskip 2in~~}$
\end 
