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Quiz 4 Form A ~~~~
Oct 16, 2017
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1.  A mass weighing 5 lbs stretches a spring 8 in. The mass is acted on by an external force of 9sin(2t) lbs.  The mass is pulled down 1 foot and then set in motion with an upward velocity of 3ft/s. Assume that there is no damping. Note $g = 32ft/s^2$.  State the initial value problem that describes the motion of this mass.



IVP:  $\underline{~ {5 \over 32}u''  + {15 \over 2}u = 9sin(2t), ~u(0) = +2, ~u'(0) = -8 ~}$


$mg = 5$.  Thus $m = {5 \over 32}$.
\hfill
$kL = mg$.  Thus $k(8/12) = k(2/3) =5$.   Thus $k = {15 \over 2}$.
\vskip 5pt
\centerline{$mu'' + 0u' + ku = 9sin(2t) $~~~~~~~~~~~~~~~~~~~~~~}




2.)  Given that the solution to $y'' + y = 0$ is $y = c_1cos(t) + c_2sin(t)$, what would be a good guess for a non-homogeneous solution to
$y'' + y = cos(2t)$?  Note you do not need to solve this differential equation.  You also don't need to determine the undetermined coefficients.

\centerline{Acceptable guess:  $\underline{~y = Acos(2t) + Bsin(2t) ~}$}
 \vskip 10pt
\centerline{Best guess:  $\underline{~y = Bcos(2t) ~}$}
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Since no $y'$ term, don't need $sin$ term.  However, both guesses will give you the correct non-homogeneous solution, so both answers are correct since I didn't ask for the best guess.


3.)  Suppose that $y_1(t) = t$ and $y_2(t) = t^2$ are solutions to the differential equation, $y'' + p(t)y' + q(t)y = 0$.  Find the general solution to   $y'' + p(t)y' + q(t)y = {1 \over t}$


General solution:  $\underline{~y = c_1t + c_2t^2 -tln|t| ~}$


$W(t, t^2) = \left|\matrix{t  & t^2 \cr 1 & 2t} \right| = 2t^2 - t^2 = t^2$.  %~~~~~Check by Abel's theorem:  $W(t, t^2) = e^{-\int p_1(t) dt}
%= e^{-int $


${1 \over t}\left|\matrix{0 & t^2 \cr 1 & 2t} \right|  = \left|\matrix{0 & t^2 \cr {1 \over t} & 2t} \right| =-t$ \hfill 
$u_1(t) =  \int {g(t) \over a}{W_1 \over W} = \int {-t \over t^2} dt = \int {-1 \over t} dt = - ln|t|$

$\left|\matrix{t  & 0 \cr 1 & {1 \over t}} \right| =1$
\hfill 
$u_2(t) =  \int {g(t) \over a}{W_2 \over W} = \int {1 \over t^2} dt = \int t^{-2} dt = -t^{-1}$

Non-homog:$ -tln|t| - t^{-1}t^2 = -tln|t| - t$

General solution: $y = k_1t + c_2t^2 -tln|t| - t = (k_1- 1)t + c_2t^2 -tln|t|  = c_1t + c_2t^2 -tln|t|$

\eject

FYI:  


 $y'' + p(t)y' + q(t)y = 0$.

$y_1(t) = t$, $y' = 1$, $y'' = 0$

and $y_2(t) = t^2$, $y' = 2t$, $y'' = 2$

 $0 + p(t) + q(t)t = 0$.

 $2 + 2p(t)t + q(t)t^2 = 0$.

 $0 + p(t)t + q(t)t^2 = 0$.


$2 + tp(t) = 0$.  Thus $p(t) = -{2 \over t}$

$0 +  -{2 \over t} + q(t)t = 0$.  Thus $q(t) = {2 \over t^2}$


 $y'' -{2 \over t}y' +  {2 \over t^2}y = 0$.

By Abel's thm, $W(t, t^2) = e^{\int{2 \over t} dt} = e^{2ln|t|} = e^{ln|t|^2} = t^2$ 

\end

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 $y(0) = 1$. 


[2]~ Equilibrium solution:  $\underline{\hskip 0.6in}$. ~~ Stability of this equilibrium solution  $\underline{\hskip 0.6in}$.

[2]~ Equilibrium solution:  $\underline{\hskip 0.6in}$. ~~ Stability of this equilibrium solution  $\underline{\hskip 0.6in}$.

[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 $+\infty$?

[4]~  Sketch of Direction field and solution with initial value $y(0) = 1$:

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\hr

Choose the best answer for the following problems:







[2]~ 2.)  If $u = y - x$, then


A.)  ${dy \over dx} = {du \over dx}$  \hfill
B.)  ${dy \over dx} = {dx \over du}$  \hfill
C.)  ${dy \over dx} = {dx \over dy}$  \hfill

D.)  ${dy \over dx} = {du \over dy}$  \hfill
E.)  ${dy \over dx} = {du \over dx} + 1$  \hfill
F.)  ${dy \over dx} = {du \over dx}+ {dx \over du}$  \hfill

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


A.)  $e^{x}$ \hfill
B.)  $e^{-x}$ \hfill
C.)  ${x}$ \hfill
D.)  ${1 \over x}$ \hfill



%\eject
\hr

[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.

A.) ~ $(-1, \infty)$ \hfill
B.)  ~$(-1, 3)$ \hfill
C.)  ~$(3, \infty)$ \hfill
D.)  ~$(-\infty, 1)$ \hfill
E.)  ~$(-\infty, 3)$ \hfill

F.) ~ $[-1, \infty)$ \hfill
G.)  ~$[-1, 3]$ \hfill
H.)  ~$[3, \infty)$ \hfill
I.)  ~$(-\infty, 1]$ \hfill
J.)  ~$(-\infty, 3]$ \hfill

\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.


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

C.)  ${dA \over dt} =  {Q \over 100} + e^{-t \over 100}$, ~
$A(0) = 0$ \hfill
D.)  ${dA \over dt} =  e^{-t \over 100} - {100Q}$, ~
$A(0) = 0$ \hfill

E.)  ${dA \over dt} =  {100Q } - e^{-t \over 100}$, ~
$A(0) = 0$ \hfill
F.)  ${dA \over dt} =  {100Q } + e^{-t \over 100}$, ~
$A(0) = 0$ \hfill

\hr

[2]~ 6.)  Note this is a 3.1 problem.  The general solution to $y'' + 2y' -8y = 0$ is

A.)  $y = c_1e^{2t} + c_2e^{4t}$ \hfill
B.)  $y = c_1e^{2t} + c_2e^{-4t}$ \hfill
C.)  $y = c_1e^{-2t} + c_2e^{4t}$ \hfill

D.)  $y = c_1e^{-2t} + c_2e^{-4t}$ \hfill
E.)  $y = c_1e^{2t} + c_2te^{2t}$ \hfill
F.)  $y = c_1cos(2t) + c_2sin(2t)$\hfill




\end