\magnification 1300
\vsize 9.6truein
%%\voffset 0.3truein
\hsize 6.9 truein
\nopagenumbers
\parskip=10pt
\parindent= 0pt
\def\u{\vskip 3.3in}
\def\v{\vskip 2.2in}
\def\w{\vskip 29pt}

Math 100 Differential Equations Exam \#1
\vskip -10pt
                               February 27, 2013 \hfill  SHOW ALL
WORK
~~~

[10]~ 1.) By giving a specific example, prove that $f: R \rightarrow R, ~f(x) = e^x$ is not onto.

Answer 1:  We know that $e^x > 0$ for all $x$.  Thus there is no $x$ such that $f(x) = e^x = -1$.  Hence -1 is not in the image of $f$.

Answer 2:  Suppose $f(x) = -1$.  Then $e^x = -1$.  But then $x = ln(-1)$.  But $ln(-1)$ is not defined.  Thus -1 is not in the image of $f$.



\vfill
2.)  Circle T for true and F for false.  Note that the answer to 2a is true.

[3]~ 2a.)   In more advanced math classes, you may be required to provide many more details when proving a function is onto.

\rightline{T~~~~~~~~~~~}

\w

[4]~ 2b.)  Suppose 
$\phi$
is a solution to the equation, 
$y' + p(t)y = g(t)$, 
then 
$2\phi$ must also be a solution to
$y' + p(t)y = g(t)$.

\rightline{~~~~~~~~~~~F}
\w
[4]~ 2c.)  Suppose 
$\phi$
is a solution to the equation, 
$y' + p(t)y^2 = 0$, 
then 
$2\phi$ must also be a solution to
$y' + p(t)y^2 = 0$.

\rightline{~~~~~~~~~~~F}
\w

[4]~ 2d.)  Suppose 
$\phi$
is a solution to the equation, 
$y' + p(t)y = 0$, 
then 
$2\phi$ must also be a solution to
$y' + p(t)y = 0$.

\rightline{T~~~~~~~~~~~}


\eject



[15]~ 3.)  Draw the direction field for $y' = {1 \over 2}y + 1$.  Determine if there are any equilibrium solutions.  If so,  determine if the equilibrium solution(s) are stable, unstable or semi-stable.

Equilibrium solution = constant solution.  Thus $y' = 0$.  

$ {1 \over 2}y + 1 = 0$ implies $ {1 \over 2}y = -1$.  Thus $y = -2$ is the equilibrium solution.  

\eject

[15]~  4.) Solve the following initial value problem:  $y'y = t + 3ty^2$, $y(0) = -2$


$y'y = t + 3ty^2$

${dy \over dt} y = t(1 + 3y^2)$

${y dy \over 1 + 3y^2} = tdt$

$\int {y dy \over 1 + 3y^2} = \int tdt$

${1 \over 6} \int {6y dy \over 1 + 3y^2} = \int tdt$

${1 \over 6} ln|1 + 3y^2| = {1 \over 2} t^2 + C$

$ ln|1 + 3y^2| = 3 t^2 + C$

$ e^{ln|1 + 3y^2|} = e^{3 t^2 + C}$

$|1 + 3y^2| = e^{3 t^2}e^C$

$1 + 3y^2 = \pm e^C e^{3 t^2}$

$1 + 3y^2 = Ce^{3 t^2}$

$ 3y^2 = Ce^{3 t^2} - 1$

$ y^2 = {Ce^{3 t^2} - 1 \over 3}$

$ y  = \pm \sqrt{{Ce^{3 t^2} - 1 \over 3}}$

 $y(0) = -2$: ~~   $ -2  = - \sqrt{{Ce^{0} - 1 \over 3}} = - \sqrt{{C - 1 \over 3}}$

 $ 4  = {{C - 1 \over 3}}$ implies $12 = C - 1$ implies $C = 13$.


\vfill
\centerline{Answer 4.) $\underline{~~ y  = - \sqrt{{13e^{3 t^2} - 1 \over 3}}~~}$}

\eject



\eject
5.)   Find the general solutions for the following three  differential equations.

[15]~  5A.)  $2y''  - 3y' + 5y = 0$

$y = e^{rt}$.  Then $y' = re^{rt}$ and $y'' = r^2e^{rt}$.

$2r^2e^{rt}  - 3re^{rt} + 5e^{rt} = 0$ implies $2r^2 - 3r + 5 = 0$

$2r^2  - 3r + 5 = 0$ implies $r = {3 \pm \sqrt{9 - 4(2)(5)} \over 4} = 
{3 \pm \sqrt{9 - 40} \over 4} = {3 \pm \sqrt{-31} \over 4}
 = {3 \pm \sqrt{-31} \over 4} = {3\over 4} \pm i{\sqrt{31} \over 4}$

\vfill \vfill

\centerline{Answer 5A.) $\underline{~~
y = c_1 e^{3t \over 4}cos({\sqrt{31} \over 4}~t) + c_2 e^{3t \over 4}sin({\sqrt{31} \over 4}~t)
~~}$}
\vfill \vfill

[15]~  5B.)  $y'' + 6y' + 9y = 0$

$r^2 + 6r + 9 = 0$

$(r + 3)^2 = 0$ implies $r = -3$

\vfill \vfill
\centerline{Answer 5B) $\underline{~~y = c_1 e^{-3t} + c_2 te^{-3t}~~}$}
\vfill \vfill
[15]~  5C.)  $3y'' (y')^2 = 1$

Let $v  y'$.  Then $v' = y''$.

$3v' (v)^2 = 1$ implies $3{dv \over dt} (v)^2 = 1$

 $3{dv} (v)^2 = dt$
 
 $\int3 (v)^2dv = \int dt$

$v^3 = t + C_1$.  Thus $v = (t + C_1)^{1 \over 3}$

${dy \over dt} =  (t + C_1)^{1 \over 3}$

${dy} =  (t + C_1)^{1 \over 3}dt$

$\int {dy} = \int (t + C_1)^{1 \over 3}dt$

$y = {3 \over 4}(t + C_1)^{4 \over 3} + C_2$

\vfill \vfill 
\centerline{Answer 5C.) $\underline{~~y = {3 \over 4}(t + C_1)^{4 \over 3} + C_2~~}$}
\eject

 \end