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Quiz 3 Form B \hfil \break
Sept 25, 2017

1.  Sketch the direction field for the autonomous equation $y' = y^2 - 2y - 8$. Find the equilibrium solutions, and classify them as stable
or unstable.  Sketch the solution with initial value $y(0) = 1$. 

 $y^2 - 2y - 8 = (y +4)(y-2) = 0$ implies $y = 4$ and $y = -2$ are equil solns.


[2]~ Equilibrium solution:  $\underline{~~y = 4~~}$. ~~ Stability of this equilibrium solution  $\underline{~~unstable~~}$.

[2]~ Equilibrium solution:  $\underline{~~y = -2~~}$. ~~ Stability of this equilibrium solution  $\underline{~~stable~~}$.

Note the equilibrium solution is a constant solution, not a number.  Thus I took off 0.5pt per problem if you did not include $y =$ (i.e, 4 is incorrect, but the equation $y = 4$ is correct).

\hr


\textcolor{red}{I forgot to change the IVP in the following, so I accepted all of the answers below}

[2]~ If $y  = \phi(t)$ is the solution to the initial value problem $y' = y^2 - 5y - 6$, $y(0) = 3$, what happens to $y = \phi(t)$ as $t$ goes to $+\infty$?

 $y^2 - 5y - 6 = (y - 6)(y+1) = 0$ implies $y = 6$ and $y = -1$ are equil solns. % Note $y = 3 \in (-1, 6)$ %and $y = -1$ is a stable equil
$y \rightarrow -1$ as $t \rightarrow +\infty$ 


[2]~ If $y  = \phi(t)$ is the solution to the initial value problem $y' = y^2 - 2y - 8$, $y(0) = 5$, what happens to $y = \phi(t)$ as $t$ goes to $+\infty$?



 Note $y(0) = 5 > 4$.  Thus 
$y \rightarrow +\infty$ as $t \rightarrow +\infty$ 

[2]~ If $y  = \phi(t)$ is the solution to the initial value problem $y' = y^2 - 2y - 8$, $y(0) = 1$, what happens to $y = \phi(t)$ as $t$ goes to $+\infty$?



 Note $y(0) = 1 \in (-4, 2)$.  Thus 
$y \rightarrow -2$ as $t \rightarrow +\infty$ 

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[4]~  Sketch of Direction field and solution with initial value $y(0) = 3$:

Note one really needs to draw slopes at 9 different \hb
$y$-values (Two for $y > 4$, three for $-2 < y < 4$, \hb and two for $y < -2$ plus the 0-slopes at $y = 4$ and\hb
 $y = -2$) in order to see all possible solutions.

Note, you don't need to calculate the slopes by \hb plugging 
in numbers.  You just need to know where \hb
slope is positive vs negative, small vs large. 

\vskip -1.4in
\rightline{\includegraphics[height=1.34in]{quiz3DFieldFormB.pdf}}

\eject

Choose the best answer for the following problems:







[2]~ 2.)  If $u = y - x$, then  E.)  ${dy \over dx} = {du \over dx} +1$  


$y = u + x$ \& hence ${dy \over dx} = {d \over dx}(u+x) = {du \over dx} + {dx \over dx} = {du \over dx} +1$, since derivative is a linear fn.




\hr

[2]~ 3.)  The integrating factor used to solve the differential equation ${dy \over dx} - {y \over x} = x^2$ is



%\centerline

{D.)  ${1 \over x}$ }\hfill
since $u(x) = e^{\int{ -1 \over x}dx} = e^{-ln|x|} = e^{ln|x|^{-1}} = |x|^{-1}$

Check:  If $u(x) = {1 \over x} = x^{-1}$, then 

\rightline{$ u(x)[{dy \over dx} - {y \over x}] = x^{-1}[y' - x^{-1}{y}] =x^{-1}y' - x^{-2}{y} = (x^{-1}{y})' $}


%\eject
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[2]~ 4.)  The solution to the initial value problem $y' = {1 - x \over y}$, $y(0) = -\sqrt{3}$ is\hb  $y = - \sqrt{-x^2 + 2x + 3}$.~~
State the largest interval on which the solution is defined.


$-x^2 + 2x + 3 \geq 0$ and since our convention is to not allow points vertical tangent lines in our domain, $y \not= 0$.

Thus $-x^2 + 2x + 3 = (-x + 3)(x+1) > 0$.  Hence either $(-x + 3)$ and $(x+1)$
are both positive or both negative.  Thus

$-x + 3 > 0$ and $x+1 > 0$ and thus  $3 > x$ and $x > -1$.  I.e., $x \in (-1, 3)$ 
~~~~~OR
\vskip -8pt
$-x + 3 < 0$ and $x+1 < 0$ and thus  $3 < x$ and $x < -1$.  But $x$ can't be both greater than 3 and less than -1, so largest possible  domain is ~~~B.)~ (-1, 3).


\hr

[2]~ 5.)  A tank contains 100 liters of pure water. Saline solution with a variable
concentration $c(t) = e^{-t \over 100}$ grams of salt per liter is pumped into the
tank at rate 1 liter per minute. The tank is kept perfectly mixed and
also drains at a rate of 1 liter per minute, so the volume stays constant.
Write an initial value problem modeling $A(t)$, the amount of grams of
salt dissolved in the solution in the tank at time $t$ minutes.

Note since the water is pure, the tank starts off with 0 grams of water.  Thus $A(0) = 0$.  
\vskip -5pt
{Rate in = ${e^{-t \over 100} g \over 1 L}{1 L \over 1 minute}$
\hfill
Rate out = ${A(t) g \over 1 L}{1 L \over 1 minute}$
\hfill

Thus~~
A.)  ${dA \over dt} =  e^{-t \over 100} - {A \over 100}$, ~
$A(0) = 0$}


\hr

[2]~ 6.)  Note this is a 3.1 problem.  The general solution to $y'' - 2y' - 8y = 0$ is
\vskip -5pt
C.)  $y = c_1e^{-2t} + c_2e^{ 4t}$\hfill
~~~~~
 since $r^2 - 2r - 8 = (r -4)(r+2) = 0$ implies $r = 4, -2$.





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