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Math 3600 Differential Equations Exam \#1
\vskip -10pt
                               Sept 26, 2018 \hfill  SHOW ALL
WORK
~~~

[10] ~1a.)  Draw the direction field for the following differential equation: 
%\v
%\centerline
{$y' = y(2-y )$}
\vskip 10pt
%\centerline{\includegraphics[width=41ex]{exam1figure.pdf}}
{\includegraphics[width=30ex]{grapht-eps-converted-to.pdf}}\hfill
{\includegraphics[width=30ex]{grapht-eps-converted-to.pdf}}

[4] ~1b.)  On the direction field above, draw the solution  to the above differential equation  that satisfies the initial condition $y(0) = 1$.
\vskip 20pt

[6]~ 1c.)  Does the  differential equation whose direction field is given above have any equilibrium solutions?  If so, state whether they are stable, semi-stable or unstable.
\vskip 20pt\vskip 20pt\vskip 20pt

[5]~  2.)  Give an example of an initial value problem that does not have a unique solution.
\vfill
3.)   Circle T for true and F for false.  

[5]~ 2c.)  Suppose $y = \phi_1(t)$ and $y = \phi_2(t)$ are solutions to 
 $ay'' + by' + cy = 0$.
Then $y = c_1\phi_1(t) + c_2\phi_2(t)$ is also a solution to this linear homogeneous differential equation.
%\rightline

\hfill{T~~~~~~~~~~~F}

[5]~ 2d.)   Suppose $y = \phi_1(t)$ and $y = \phi_2(t)$ are linearly independent solutions to 
\break  $ay'' + by' + cy = 0$.
If $y = h(t)$ is   also a solution to  $ay'' + by' + cy = 0$, then there exists constants $c_1$ and $c_2$ such that $h(t) = c_1\phi_1(t) + c_2\phi_2(t)$.

\hfill{T~~~~~~~~~~~F}


\eject

[20]~ 4.) Find the general solution to  $ty' - 2y= t^3 e^t - 8$.  Also find the solution that passes thru the point (1, 3). 
%and the to  initial value problem solution for the  differential equation, $ty' - 2y= t^3 e^t - 8$, ~~$y(1) = 3$.  
How does the solution passing thru (1, 3)
behave as $t \rightarrow \infty$?



\vfill
\centerline{General solution: $\underline{~~\hskip 4in~~}$}
\vskip 20pt
\centerline{IVP solution: $\underline{~~\hskip 4in~~}$}
\vskip 20pt
\centerline{ $t \rightarrow \infty$,  $y \rightarrow \underline{~~\hskip 4in~~}$}




\eject


5.) Solve the following 2nd order differential equations

[15]~   5a.) $2y'' - y' + 10y= 0$ 


\vfill\vfill


\centerline{General solution: $\underline{~~\hskip 4in~~}$}

\vskip10pt
[15]~  5b.) $(x)(x'') =  (x')^2 $  ~~~~Hint: Let $x' = {dx \over dt} = v$, then $v' =$

%Let  $x' = {dx \over dt} =v$
%Let  $x'' = v' = {dv \over dt} = {dv \over dx} {dx \over dt}  = {dv \over dx}v$


%$x{dv \over dx}v = v^2$

%${dv \over v} = {dx \over x}$

%$ln|v| = ln|x| + C$

%$v = Cx$

%$ {dx \over dt} = Cx$

%$ {dx \over x} = Cdt$ 

%$ln(x) = Ct + k$

%$x = ke^{Ct}$

%Check: $ ( ke^{Ct})( kC^2e^{Ct}) = Cke^{Ct} Cke^{Ct}$


\vfill\vfill\vfill\vfill\vfill\vfill\vfill
\vfill

\centerline{General solution: $\underline{~~\hskip 4in~~}$}

\eject



[15]~ 6.) Show by induction that for Picard's iteration method, $\phi_n(t) = \displaystyle\sum_{k=1}^{n} {t^{2k} \over k!}$ \break
approximates the solution to the initial value problem, 
 $y' = 2t(1 + y)$, $y(0) = 0$ where $\phi_1(t) = {t^2}$.  You may use the proof outline below or write it from scratch.

%$\phi_1 = \int 2sds$


\vskip 10pt

Proof by induction on $n$.

For $n = 1$, ~~ $ \displaystyle\sum_{k=1}^{1}  {t^{2k} \over k!} = $ % $ {3(-1)^{1+1}~  t^{1+1} \over (1+1)!} =  {3~  t^{2} \over 2!}$

\vskip 20pt
Suppose for $n = j$,~~  $\phi_{j-1}(t) = \displaystyle\sum_{k=1}^{j-1}  {t^{2k} \over k!}$ 

\vskip 20pt
Claim:  $\phi_j =$  
\vskip 20pt

Proof of claim:  By Picard's iteration method, $\phi_{j} = $

%$\int_0^t f(s, \phi_{j-1}(s))ds = \int_0^t 2s(1 + \displaystyle\sum_{k=1}^{j-1}  {s^{2k} \over k!})ds$
%$= \int_0^t (2s + \displaystyle\sum_{k=1}^{j-1}  {2s^{2k+1} \over k!})ds$
%$= \int_0^t ( \displaystyle\sum_{k=0}^{j-1}  {2s^{2k+1} \over k!})ds$
%$=  \displaystyle\sum_{k=0}^{j-1}  {2t^{2k+2} \over (2k + 2)k!})ds$
%$=  \displaystyle\sum_{k=0}^{j-1}  {2t^{2k+2} \over 2(k + 1)k!})ds$
%$=  \displaystyle\sum_{k=0}^{j-1}  {t^{2k+2} \over (k + 1)!})ds$
%$=  \displaystyle\sum_{k=1}^{j}  {t^{2k} \over (k)!})ds$
\end



 ${dy \over dx}   = -{3x \over y - 4}$

$\int(y-4)dy = \int -3xdx$

${y^2 \over 2} - 4y = -{3 \over 2}x^2$

$y^2 - 8y = -3x^2 + C$

$y^2 - 8y + 3x^2 + C = 0$

$y = {8 \pm  \sqrt{64 - 4(3x^2 + C)} \over 2} = 4 \pm \sqrt{16 - 3x^2 + C} = y$


 $y(1) = -2$:  $-2 =  4 \pm \sqrt{16 - 3(1)^2 + C}$ implies $-6 =   -\sqrt{16 - 3 + C}$

Note initial value determines sign of $\pm$.   In this case, IVP only has a solution when we choose the negative sign.  The the IVP in this case means $y = 4 - \sqrt{16 - 3x^2 + C}$ where we determine $C$ below:

$36 =  {13 + C}$.  Thus $C = 36 - 13 = 23$ and $y =  4 - \sqrt{16 - 3x^2 + 23} =  4 - \sqrt{39 - 3x^2}$


\vfill
\centerline{Answer: $\underline{~~y =  4 - \sqrt{39 - 3x^2}~~}$}
\eject

3.)  Suppose $y' = y - t + 1$, 	~$y(0) = 0$.  

Let $\phi_0(t) = 0$ and define $\{\phi_n(t)\}$ by the method of successive approximation (i.e, Picards iteration method).  Determine the following:
\vfill
\def\i{\int_0^t}
\def\e{|_0^t}

$y' = f(t, y)$

$\phi_1(t) =   \i f(s, \phi_0(s)) ds =  \i f(s, 0) ds =  \i (0 - s + 1)ds =$

$(- {s^2 \over 2} + s)\e = - {t^2 \over 2} + t - 0$ 
\vfill
[3] ~3a) $\phi_1(t) = \underline{~~- {t^2 \over 2} + t~~}$
\vfill
$\phi_2(t) =   \i f(s, \phi_1(s)) ds =  \i f(s, - {s^2 \over 2} + s ) ds =  \i (- {s^2 \over 2} + s - s + 1)ds $

$=  \i (- {s^2 \over 2}  + 1)ds=
(- {s^3 \over 6} + s)\e = - {t^3 \over 6} + t - 0$ 
\vfill
[3] ~3b) $\phi_2(t) = \underline{~~ - {t^3 \over 6} + t~~}$
\vfill
$\phi_3(t) =   \i f(s, \phi_2(s)) ds =  \i f(s, - {s^3 \over 6} + s ) ds =  \i (- {s^3 \over 6} + s - s + 1)ds $

$=  \i (- {s^3 \over 6}  + 1)ds=
(- {s^4 \over 24} + s)\e = - {t^3 \over 24} + t - 0$ 

\vfill
[3] ~3c) $\phi_3(t) = \underline{~~- {t^4 \over 24} + t ~~}$
\vfill

\vfill
[4] ~3d) $\phi_n(t) = \underline{~~- {t^{n+1} \over (n+1)!} + t ~~}$
\vfill

\vfill
[3] ~3e) $lim_{n\rightarrow \infty} \phi_n(t) = \underline{~~t~~}$

[2] ~3f)  Is $\phi(t) = lim_{n\rightarrow \infty} \phi_n(t)$ a solution to  
$y' = y - t + 1$, 	~$y(0) = 0$?  $\underline{~~yes~~}$
\vfill

\vfill
[2] ~3g)  Is $\phi(t) = lim_{n\rightarrow \infty} \phi_n(t)$ the unique solution to  
$y' = y - t + 1$, 	~$y(0) = 0$?  $\underline{~~yes~~}$

\vfill

\vfill
\eject

[15]~ 4a.)  Solve $y'' - 8y' + 16y = 0$
\vfill

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

Plugging in the guess into our equation:  

 $ r^2e^{rt} - 8 re^{rt} + 16e^{rt} = 0$

Since  $e^{rt} > 0$, we can divide both sides of the above equation by $ re^{rt}$ without loosing any solutions:

 $ r^2 - 8 r + 16 = 0$ implies $(r - 4)^2 = 0$ and thus $r = 4$.
 
 Thus $y = e^{4t}$ is a solution.  We can check by plugging in (as we did in class for a different example) that $y = te^{4t}$ is also a solution.

Sidenote:  $\{e^{4t}, te^{4t}\}$ is a linear independent set and thus a basis for our solution.  We can check linear independence by calculating the Wronskian.







\vfill
\centerline{Answer: $\underline{~~y = c_1e^{4t} + c_2te^{4t}~~}$}
\vfill
[15]~ 4b.)   Solve $y'' - y' + 3y = 0$
\vfill
Educated guess:  $y = e^{rt}$.  Then $y' = re^{rt}$ and  $y'' = r^2e^{rt}$

Plugging in the guess into our equation: 

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

Since  $e^{rt} > 0$, we can divide both sides of the above equation by $ re^{rt}$ without loosing any solutions:

 $ r^2 -  r + 3 = 0$ implies  $r = {1 \pm \sqrt{1 - 4(3)} \over 2}
  = {1 \pm \sqrt{11} \over 2} =  {1 \pm i\sqrt{11} \over 2}
 $.

\vfill
\centerline{Answer: $\underline{~~y = c_1e^{t \over 2} cos({\sqrt{11} \over 2}t) + c_2e^{t \over 2} sin({\sqrt{11} \over 2}t)~~}$}
\vfill
\eject
[15]~  5.)  Let $y = y_1(t)$ be a solution of $y' + p(t)y = 0$ and let $y = y_2(t)$ be a solution of $y' + p(t)y = g(t)$.  Show that  $y = y_1(t) + y_2(t)$ is a solution of $y' + p(t)y = g(t)$.


Proof: Since $y = y_1(t)$ is a solution of $y' + p(t)y = 0$, we know that  
$y_1' + p(t)y_1 = 0$.

Since $y = y_2(t)$ is a solution of $y' + p(t)y = g(t)$, ~~ 
$y_2' + p(t)y_2 = g(t)$




Claim:  $y = y_1(t) + y_2(t)$ is a solution of $y' + p(t)y = g(t)$.

 We will plug $y = y_1(t) + y_2(t)$ into the LHS to determine that the LHS = RHS:
 
 $(y_1(t) + y_2(t))' + p(t)(y_1(t) + y_2(t)) =$
$ y_1'(t) + y_2'(t) + p(t)y_1(t) + p(t)y_2(t) $
\vskip 10pt
\rightline{ $= [y_1'(t) + p(t)y_1(t)] + [y_2'(t) + p(t)y_2(t)] = 0 + g(t) = g(t)$}
 
 Hence $y = y_1(t) + y_2(t)$ is a solution of $y' + p(t)y = g(t)$.
 \vskip 10pt
 \hrule
 
 Alternate proof:  
  Since $y = y_1(t)$ is a solution of $y' + p(t)y = 0$, we know that  
 \vskip 10pt
\centerline{$y_1' + p(t)y_1 = 0 ~~~(1)$.}

Since $y = y_2(t)$ is a solution of $y' + p(t)y = g(t)$.  
 \vskip 10pt
\centerline{$y_2' + p(t)y_2 = g(t) ~~~(2)$.}

If we add equations (1) and (2), we obtain:

$$[y_1'(t) + p(t)y_1(t)] + [y_2'(t) + p(t)y_2(t)]  = 0 + g(t)$$

Thus $ y_1'(t) + y_2'(t) + p(t)y_1(t) + p(t)y_2(t)  = g(t)$

and  $(y_1(t) + y_2(t))' + p(t)(y_1(t) + y_2(t)) =g(t)$

Hence $y = y_1(t) + y_2(t)$ is a solution of $y' + p(t)y = g(t)$.
 \vskip 10pt
 %\vskip -30pt
 \hrule
 
 Alternate proof: 
  \vskip -10pt
 Claim:  $L(f) = f' + pf $ is a linear function where $f$ and $p$ are functions of $t$.
 
 Proof of claim:  Let $a, b$ be constants and $f, g$ be functions of $t$.
 
 $L(af + bg) = (af+ bg)' + p(af + bg)
 = af' + bg' + paf + pbg = af' + paf + bg' + pbg = 
 [a(f' + pf)] + [b(g' + pg)] = 
 L(f) + L(g)$
 
 
 We will now show that  $y = y_1(t) + y_2(t)$ is a solution of $y' + p(t)y = g(t)$:
 
 Since $y = y_1(t)$ is a solution of $y' + p(t)y = 0$, ~~$L(y_1) = 0$.
 
  Since $y = y_2(t)$ is a solution of $y' + p(t)y = g(t)$,~~$L(y_2) = g(t)$
  
  $L(y_1 + y_2) = L(y_1) + L(y_2) = 0 + g(t) = g(t)$.  Thus 
 $y = y_1(t) + y_2(t)$ is a solution of $y' + p(t)y = g(t)$.
 \vskip 10pt
 \hrule
 
 Note similar proofs would show that  $y = cy_1(t) + y_2(t)$ is a solution of $y' + p(t)y = g(t)$ for any constant $c$.
 
\end
















[20]~ 1.)  Solve $y'' - 6y' + 9 = 0$, $y(0) = 2$, $y'(0) = 4$.



\vfill
\centerline{Answer: $\underline{~~\hskip 4in~~}$}

\eject

 2.)  Circle T for true and F for false.  

[4]~ 2a.)  The equation $ln(t)y' = {t \over t+1} -  y(sin t^2)$ is a linear differential equation.

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

\vskip 20pt

[4]~ 2b.)  The equation $y' + y = y^2$ is a linear differential equation.
%\rightline
\hfill{T~~~~~~~~~~~F}

\vskip 20pt


[4]~ 2c.)  Suppose $y = \phi_1(t)$ and $y = \phi_2(t)$ are solutions to 
 $ay'' + by' + cy = 0$.
If $y = h(t)$ is   also a solution to  $ay'' + by' + cy = 0$, then there exists constants $c_1$ and $c_2$ such that $h(t) = c_1\phi_1(t) + c_2\phi_2(t)$.
%\rightline
\hfill{T~~~~~~~~~~~F}

\vskip 20pt


[4]~ 2d.)   Suppose $y = \phi_1(t)$ and $y = \phi_2(t)$ are linearly independent solutions to 
 $ay'' + by' + cy = 0$.
If $y = h(t)$ is   also a solution to  $ay'' + by' + cy = 0$, then there exists constants $c_1$ and $c_2$ such that $h(t) = c_1\phi_1(t) + c_2\phi_2(t)$.
%\rightline
\hfill{T~~~~~~~~~~~F}

\vskip 40pt



[4]~  3.)  By giving a specific counter-example, prove that $y = ln(x)$ is
 not a linear function.

\eject

[20]~ 4.)  

\centerline{\includegraphics[width=48ex]{exam1fig.pdf}}

4a.)  Circle the differential equation whose direction field is given above.

i.)  $y' = y^2$
\hfil
ii.) $y' = (y+2)^2$
\hfil 
iii.)  $y' = y(y+2)^2$

iv.) $y' = y(y+2)$
\hfil 
v.) 
$y'  = {y \over y+2}$~~~~~~~~~~~~~
\hfil 
vi.) 
$y'  = {y +2 \over y}$~~~~~~~~~~~~~

vii.) $y'  = {y^{2} \over t}$
\hfil 
viii.) $y'  = t\sqrt{y} $~~~~~~~
%$y'  = {\sqrt{y} \over t}$
\hfil 
ix.) 
$y'  = {t  y^{2}}$~~~~~~~~~


\vskip 20pt

4b.)  Draw the solution to the differential equation  whose direction field is given above that satisfies the initial condition $y(1) = -3$.
\vskip 20pt

4c.)  Does the  differential equation whose direction field is given above have any equilibrium solutions?  If so, state whether they are stable, semi-stable or unstable.



\eject

[20]~ 5.)  Solve:  
$y' =  e^{4t} - {y \over t}$



\vfill
\centerline{Answer: $\underline{~~\hskip 4in~~}$}


\eject

[20]~ 6.)  Choose one of the following 2 problems.
 If you do not choose your best problem, I will substitute the
other problem, but with a 2 point penalty (if it
improves your grade).
Circle the letter corresponding to your chosen problem: A ~~ B

6A.)  Show that  $\phi(t) = \displaystyle\sum_{k=2}^{\infty}{3(-1)^{k}~  t^{k} \over k!}$ converges for all $t$
and  show that  $\phi(t) = \displaystyle\sum_{k=2}^{\infty}{3(-1)^{k}~  t^{k} \over k!}$  is a solution to $y' = 3t - y$.

%${{3(-1)^{k+1}~  t^{k+1} \over (k+1)!} \over {3(-1)^{k}~  t^{k} \over k!}} ={t \over k+1} $


%$\phi' = \displaystyle\sum_{k=2}^{\infty}{3(-1)^{k}~ (k) t^{k-1} \over k!} = \displaystyle\sum_{k=2}^{\infty}{3(-1)^{k}~  t^{k-1} \over (k-1)!}$


%$3t - y = 3t -  \displaystyle\sum_{k=2}^{\infty}{3(-1)^{k}~  t^{k} \over k!} = 3t -  \displaystyle\sum_{k=3}^{\infty}{3(-1)^{k-1}~  t^{k-1} \over (k-1)!}$

%$=   \displaystyle\sum_{k=2}^{2}{3(-1)^{k}~  t^{k-1} \over (k-1)!}  +  \displaystyle\sum_{k=3}^{\infty}{3(-1)^{k}~  t^{k-1} \over (k-1)!}  $

\eject

6B.)  Show by induction that for Picard's iteration method, $\phi_n(t) = \displaystyle\sum_{k=1}^{n} {3(-1)^{k+1}~  t^{k+1} \over (k+1)!}$ approximates the solution to the initial value problem,  $y' = 3t - y$, $y(0) = 0$ where $\phi_1(t) = {3t^2 \over 2}$.  You may use the proof outline below or write it from scratch.

\vskip 20pt

Proof by induction on $n$.

For $n = 1$, ~~ $ \displaystyle\sum_{k=1}^{1} {3(-1)^{k+1}~  t^{k+1} \over (k+1)!} = $ % $ {3(-1)^{1+1}~  t^{1+1} \over (1+1)!} =  {3~  t^{2} \over 2!}$

\vskip 20pt
Suppose for $n = j$,~~  $\phi_{j-1}(t) = \displaystyle\sum_{k=1}^{j-1} {3(-1)^{k+1}~  t^{k+1} \over (k+1)!}$ 

\vskip 20pt

Then by Picard's iteration method, $\phi_{j} = $



\end

\vfill

$\phi_n(t) = \displaystyle\sum_{k=1}^{n} {3(-1)^{k+1}~  t^{k+1} \over (k+1)!}$

$\phi_n(t) = \displaystyle\sum_{k=2}^{n} {3(-1)^{k}~  t^{k} \over k!}$

\end



6A.)  Suppose $y = \phi_1(t)$ and $y = \phi_2(t)$ are solutions to 
the ODE $ay'' + by' + cy = 0$.
Prove that $y = c_1\phi_1(t) + c_2\phi_2(t)$ is also a solution to 
$ay'' + by' + cy = 0$.

\eject
6B.)  Show that %$\phi_n = {\Sigma\limits_{k = 1}?$

\end
















[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
