
\input graphicx
\nopagenumbers
\magnification 1600
\vsize 10.4truein
\voffset -.4truein
\parskip 9pt
\parindent 0pt
\hsize 7.2truein
\voffset -.4truein
\hsize 10.2truein
\voffset -.7truein
\def\R{{\bf R}}
\def\v{{\vskip 20pt}}

Quiz 1 ~~~SHOW ALL WORK\hfil \break
Sept 7, 2018

[10]~ 1.)  Draw the direction field for $y' = (y + 1)(y-3)$

\hskip 2in
{\includegraphics[width=20ex]{grapht-eps-converted-to.pdf}}

\vskip -.8in
Click on 9/10 date for solution:
\vskip 0.5in
[10]~ 2.)  Solve $y' = {1 \over x(2y+3)}$, ~$y(1) =-2 $

${dy \over dx} = {1 \over x(2y+3)}$ \hskip 2in Compare to 2.2:  14

Separate variables:  $(2y+3)dy = {dx \over x}$

$y^2 + 3y = ln|x| + C$.

$y^2 + 3y - ln|x| - C = 0$.


$y = {-3 \pm \sqrt{9 - 4 (- ln|x| - C)} \over 2}$
$ = {-3 \pm \sqrt{9 + 4 ln|x| + C} \over 2}$
$ = {-3 \pm \sqrt{4 ln|x| + C} \over 2}$

General solution:  $ y= {-3 \pm 2\sqrt{ ln|x| + C} \over 2}$

IVP:  ~$y(1) =-2 $

$-2 =  {-3 \pm 2\sqrt{ ln|1| + C} \over 2}=  {-3 \pm 2\sqrt{ C} \over 2}$


$-4 = {-3 \pm 2\sqrt{ C} }$.  Thus $-1 = { \pm 2\sqrt{ C} }$.  Hence   $-1 = { - 2\sqrt{ C} }$.

Thus $\sqrt{ C} = {1 \over 2}$ and $C = {1 \over 4}$

Hence IVP solution:     $ y= {-3 - 2\sqrt{ ln|x| + {1 \over 4}} \over 2}
y= {-3 - 2\sqrt{ {4ln|x| + 1 \over 4}} \over 2}$
$ = {-3 - \sqrt{4 ln|x|  + 1} \over 2}$


Answer:$\underline{~~y= {-3 - \sqrt{1 + 4 ln|x| } \over 2}~~}$

\end

2.2       14




Solve $sin(t) y' + y cos(t) = 5$ 
 \hfil \break(hint:  this is a short problem if you are observant).

$sin(t) y' + y cos(t) = 5$

 $(sin(t) y)'  = 5$  
 
  $\int (sin(t) y)' dt  = \int 5 dt$  

 $\sin(t) y = 5t + C$  

 $ y  = {5t + C \over sin(t)}$  




Answer:$\underline{~~ y  = {5t + C \over sin(t)}~~}$
\end 
