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

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

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[10]~ 2.)  Solve $y' = {1 \over x(2y+3)}$, ~$y(1) =-2 $


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${dy \over dx} = {1 \over x(2y+3)}$

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

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

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


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

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

$y = {-3 \pm \sqrt{9  + C)} \over 2}$

$y = {-3 \pm \sqrt{9 -8)} \over 2}$

$y = {-3 - 1 \over 2}$

$y = {-3 \pm \sqrt{9 + 4 ln|x| -8} \over 2}$

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


Answer:$\underline{~~\hskip 3in~~}$

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