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Quiz 1 \hfil \break
Feb 5, 2016

[10]~ 1.)  Draw the direction field for $y' = y^2 - 1$

\includegraphics[width=40ex]{graphquiz1.pdf}

[10]~ 2.)  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 
