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Math 3600 Differential Equations Exam \#1
\vskip -10pt
                               March 2, 2016 \hfill  SHOW ALL
WORK
~~~

\vskip -5pt
[12]~ 1.)  Find the largest eigenvalue and its corresponding eigenvectors for  $\left[\matrix{5 & 2 \cr 3 & 0}\right]  $

 $\left|\matrix{5 - r & 2 \cr 3 & 0 - r}\right| = (5-r)(-r) - 6 = r^2 - 5r - 6 = (r + 1)(r - 6)$.  Thus $r = -1, 6$
 
 
 For $r = 6$:  $\left[\matrix{5 - r & 2 \cr 3 & 0 - r}\right] = \left[\matrix{5 - 6 & 2 \cr 3 & 0 - 6}\right] = \left[\matrix{-1 & ~~2 \cr ~~3 &  -6}\right] $

$\left[\matrix{-1 & ~~2 \cr ~~3 &  -6}\right] \left[\matrix{2 \cr 1}\right] = 
\left[\matrix{0 \cr 0}\right] $

\v
\centerline{Answer: The largest eigenvalue of the above matrix is $\underline{~~6~~}$\hskip 1.5in}
\vskip 10pt
 \centerline{and its eigenvectors are all non-zero multiples of the vector  $\underline{~~\left[\matrix{2 \cr 1}\right]~~}$}
\vskip 10pt
[8]~ 2.)  Find all the singular points of the following differential equation and determine whether each one is regular or irregular.

\vskip 5pt
\centerline{ ~~~~$x^3(x - 3)y'' - 6xy' + 9xy = 0$}

%$1y'' - {6x \over x^3(x - 3)} y' + {9x \over x^3(x - 3)}y = 0$

$1y'' - {6 \over x^2(x - 3)} y' + {9 \over x^2(x - 3)}y = 0$.  Thus $x = 0, 3$ are singular points.

Euler equation:   $x^2y'' + \alpha xy' + \beta y = 0$.

Multiply by $x^2$: ~~ $x^2y'' - \left({6 \over x(x - 3)}\right) xy' + {9 \over (x - 3)}y = 0$.  Thus $x = 0$ is an irregular singular point.

Multiply by $(x-3)^2$:~~  $(x-3)^2y'' - \left({6 \over x^2}\right) (x-3)y' + {9 (x-3) \over x^2}y = 0$.  Thus $x = 3$ is an regular singular point.


Alternately:  $lim_{x \rightarrow 0}~x( {6x \over x^2(x - 3)} )$ is not finite.  Thus $x = 0$ is an irregular singular point.

 $lim_{x \rightarrow 3}~(x-3)( {6 \over x^2(x - 3)} )$ and   $lim_{x \rightarrow 3}~(x-3)^2(  {9 \over x^2(x - 3)} )$ are finite.   Thus $x = 3$ is an regular singular point.



%Multiply by $x^2$:  $x^2y'' - \left({6 \over x(x - 3)})\right) xy' + {9 \over (x - 3)}y = 0$.



[20]~ 3.)  Solve 
\vskip -35pt
$$y'' - 6y' + 9y = {e^{3t} \over t}$$

Solve homogeneous equation:  $y'' - 6y' + 9y = 0 $.

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

$r^2 - 6r + 9= (r-3)^2 = 0$.  Thus $r = 3$

Thus general homogeneous solution is $y = c_1e^{3t} + c_2te^{3t}$

Solve non-homogeneous equation:  $y'' - 6y' + 9y = {e^{3t} \over t}$

General non-homogeneous solution is $y = c_1e^{3t} + c_2te^{3t} + ( u_1(t)e^{3t} + u_2(t) (te^{3t}))$ where we determine $u_i$ as follows:

$W(e^{3t}, te^{3t}) = 
\left|\matrix{e^{3t} & te^{3t} \cr 3e^{3t} & e^{3t} + 3te^{3t}}\right| = e^{6t} + 3te^{6t} - 3te^{6t} =  e^{6t}$ 

 $u_1(t) = \int  {\left|\matrix{0 & te^{3t} \cr 1 & e^{3t} + 3te^{3t}}\right| \over e^{6t}}({e^{3t} \over t})dt $
 $=  \int  {-te^{3t} \over e^{6t}}({e^{3t} \over t})dt$
 $=  \int  {-te^{6t} \over te^{6t}}dt$
  $=  -\int  dt$
  $= -t + C_1$
 
 
  $u_2(t) = \int  {\left|\matrix{e^{3t} & 0 \cr 3e^{3t} & 1}\right| \over e^{6t}}({e^{3t} \over t})dt  = \int  {e^{3t}  \over e^{6t}}({e^{3t} \over t})dt 
   = \int { dt \over t} = ln|t| + C_2$
 
 
NOTE:  need only 1 solution for each $u_i$ and thus don't need constants $C_1, C_2$.

Alternate method to find $u_i$:  $y =  u_1e^{3t} + u_2 te^{3t}$

  Then $y' =  u_1'e^{3t} +  3u_1e^{3t} + u_2' te^{3t} + u_2 [e^{3t} + 3te^{3t}]$
      
  Choose 2nd equation for solving for 2 unknowns $u_1, u_2$:    Let $u_1'e^{3t} +  + u_2' te^{3t} = 0$ 
      
      Then $y' =    3u_1e^{3t} + u_2 [e^{3t} + 3te^{3t}]$
      
            and $y'' =    3u_1'e^{3t} +    9u_1e^{3t} + u_2' [e^{3t} + 3te^{3t}]+ u_2 [3e^{3t} + 3e^{3t} + 9te^{3t}]$
      
$y'' =    3u_1'e^{3t} +   u_2' [e^{3t} + 3te^{3t}]  +  9u_1e^{3t} + u_2 [6e^{3t}  + 9te^{3t}]$

$y'' =    (3u_1'e^{3t} +   u_2' [3te^{3t}])  +   u_2' [e^{3t}]+  9u_1e^{3t} + u_2 [6e^{3t} + 9te^{3t}]$


$y'' =  3(0) +  u_2' e^{3t}+  9u_1e^{3t} + u_2 [6e^{3t} + 9te^{3t}]
=    u_2' e^{3t}+  9u_1e^{3t} + u_2 (6e^{3t} + 9te^{3t})$

Plug into DE:

$u_2' e^{3t} +  9u_1e^{3t} + u_2 (6e^{3t} + 9te^{3t}) - 6[ 3u_1e^{3t} + u_2 (e^{3t} + 3te^{3t})] + 9[u_1e^{3t} + u_2 te^{3t}] = {e^{3t} \over t}$

$u_2' e^{3t} +  (9u_1e^{3t}  - 18u_1e^{3t} +  9u_1e^{3t})  +  [6u_2e^{3t} + 9u_2te^{3t} - 6 u_2e^{3t} -18u_2te^{3t} + 9u_2 te^{3t}] = 0$

$u_2' e^{3t}    = {e^{3t} \over t}$.  Thus $u_2' = {1 \over t} = t^{-1}$ and $u_2(t) = ln|t|$


$u_1'e^{3t} + u_2' te^{3t} = 0$ implies $u_1'e^{3t} +   {1 \over t} te^{3t} = 0$ 

Thus $u_1'  = -1$ and $u_1(t) = -t$ 

$y = c_1e^{3t} + c_2te^{3t} + ( -te^{3t} + ln|t| (te^{3t})) = c_1e^{3t} + (c_2 - 1)te^{3t}  + ln|t| (te^{3t})$
\vfill
\centerline{Answer: $\underline{~~y = c_1e^{3t} + c_2te^{3t}  + ln|t| (te^{3t})~~}$}

\eject




[20]~ 4.)  Solve ~~ $x^2y'' + 8xy' -8y = 0$.

Guess $y = x^r$.  Then $y' = rx^{r-1}$ and $y'' = r(r-1)x^{r-2}$

Plug $y = x^r$ into $x^2y'' + 8xy' -8y = 0$.

$x^2 r(r-1)x^{r-2} + 8x rx^{r-1} -8x^r = 0$.

$(r^2 -r)x^{r} + 8rx^{r} -8x^r = 0$.

$(r^2 -r + 8r -8)x^r = 0$.

$(r^2 + 7r -8)x^r = 0$ implies $(r-1)(r + 8) = 0$. Thus $r = 1, -8$

Thus $y = x$ and $y = x^{-8}$ are two linearly independent solutions to this 2nd order linear homogeneous DE.
\vskip 20pt
\centerline{Answer: $\underline{~~y = c_1x + c_2x^{-8}~~}$}

\eject

[20]~ 5.)  Seek the power series solution for the given first order differential equation about the point $x_0 = 0$:


\centerline{ ~~~~$(x + 4)y' - 2y  = 0$}
\vskip -4pt
Note:  A solution for this problem is a finite polynomial and thus all but a finite number of terms will be 0, but you must plug in the infinite series.





\def\So{\sum_{n=0}^\infty}
\def\S{\sum_{n=1}^\infty}
\def\St{\sum_{n=2}^\infty}

Let $y = \So a_n x^{n }$.
~~Then $y' = \So n a_n x^{n-1}$.

 $ (x + 4)y' - 2y = (x + 4) \So a_n nx^{n -1} - 2 \So a_n x^{n }$ 
 
 $\hskip 1.2in =  x \So a_n nx^{n -1} + 4\So a_n nx^{n -1} - 2 \So a_n x^{n }$ 
 
 $\hskip 1.2in =   \So a_n nx^{n} + \So 4a_n nx^{n -1} -  \So 2a_n x^{n }$ 

 $\hskip 1.2in =   \So a_n nx^{n} + \S 4a_n nx^{n -1} -  \So 2a_n x^{n }$ 
 
 $\hskip 1.2in =   \So a_n nx^{n} + \So 4a_{n+1} (n+1)x^{n} -  \So 2a_n x^{n }$ 

 $\hskip 1.2in =   \So (a_n n + 4a_{n+1} (n+1) - 2a_n) x^{n }$ 


 $\hskip 1.2in =   \So (a_n( n - 2) + 4a_{n+1} (n+1) ) x^{n }$  = 0

Thus $a_n( n - 2) + 4a_{n+1} (n+1) = 0$ for all $n \geq 0$


$a_{n+1} = {2-n \over 4(n+1)} a_n$

$n = 0$:  $a_1 =  {2 \over 4} a_0 = {1 \over 2} a_0 $

$n = 1$:  $a_2 =  {2-1 \over 4(2)} a_1 = {1 \over 8} a_1= {1 \over 8} ( {1 \over 2} a_0 ) =  {1 \over 16} a_0 $

$n = 2$:  $a_3 =  {2 - 2 \over 4(3)} a_2 = 0 $

$a_{n+1} = {2-n \over 4(n+1)} (0) = 0$ for $n > 2$

Thus $y = a_0 + a_1x + a_2x^2 + a_3x^3 + ... = a_0 + {1 \over 2}a_0x +  {1 \over 16} a_0x^2 = a_0( 1 + {1 \over 2}x +  {1 \over 16} x^2)$

Check:  $y' = a_0({1 \over 2} +  {1 \over 8} x)$

$(x + 4)y' - 2y  = (x + 4)( a_0)({1 \over 2} +  {1 \over 8} x)) - 2a_0( 1 + {1 \over 2}x +  {1 \over 16} x^2) $

$ =
 a_0({1 \over 2}x +  {1 \over 8} x^2 + 2 + {1 \over 2}x - 2 - x -  {1 \over 8} x^2)  = 0$

\vskip 20pt
\centerline{Answer: $\underline{~~y = a_0( 1 + {1 \over 2}x +  {1 \over 16} x^2)~~}$}
\eject


[20]~  6.)  Given the recursive relation $a_{n+2} = 6a_{n+1} - 9a_{n}$ where $a_0 = -1$ and $a_1 = 3$, 
\vskip -5pt
prove that $a_n = 3^n(2n - 1)$.  You may use induction.


Proof by induction:  First we prove that $a_n = 3^n(2n - 1)$ for $n = 0, 1$:

~~~~$n = 0$:   $3^0(2(0) - 1) = -1 = a_0 $

~~~~$n = 1$:   $ 3^1(2(1) - 1) = 3 = a_1$

Induction hypothesis:  Suppose $a_k = 3^k(2k - 1)$ for $k = n, n+1$.

~~~~Then $a_n = 3^n(2n - 1)$ and $a_{n+1} = 3^{n+1}(2(n+1) - 1)$

Claim:  $a_{n+2} = 3^{n+2}(2(n+2) - 1)$

$a_{n+2} = 6a_{n+1} - 9a_{n} $

~~~~~~~$= 6[ 3^{n+1}(2(n+1) - 1)] - 9[ 3^n(2n - 1)]$

~~~~~~~$ =
2[ 3^{n+2}(2n+2 - 1)] -  3^{n+2}(2n - 1)$

~~~~~~~$ = 2[ 3^{n+2}(2n+1)] -  3^{n+2}(2n - 1)$

~~~~~~~$= 3^{n+2}(4n+2) -  3^{n+2}(2n - 1)$

~~~~~~~$= 3^{n+2}[4n+2 -  2n + 1]$

~~~~~~~$= 3^{n+2}[2n+ 3]$

~~~~~~~$= 3^{n+2}[2(n+2) - 1]$
\vskip 10pt
\hrule

Alternative answer ({\bf not} covered in this class -- see     MATH:4050 Intro to Discrete Math):  


Guess $a_n = x^n$.  Then $a_{n+1} = x^{n+1}$ and $a_{n+2} = x^{n+2}$

Then $a_{n+2} = 6a_{n+1} - 9a_{n}$  implies $x^{n+2} - 6x^{n+1} + 9x^{n} = 0$.

 Hence  $x^{n}(x^2 - 6x + 9) = x^n(x-3)^2 = 0$.  Thus $x = 3$

Claim:  $a_n = c_1(3^{n}) + c_2(n3^{n})$ satisfies  $a_{n+2} - 6a_{n+1} + 9a_{n} = 0$ 

$ c_1(3^{n+2}) + c_2((n+2)3^{n+2}) - 6[ c_1(3^{n+1}) + c_2((n+1)3^{n+1})] + 9[ c_1(3^{n}) + c_2(n3^{n})] $ 

~~$=  c_1[3^{n+2} - 6(3^{n+1}) + 9 (3^{n}) ] + c_2[(n+2)3^{n+2} -6 ((n+1)3^{n+1}) + 9 (n3^{n})] $ 

~~$ =c_1[3^{n+2} - 2(3^{n+2}) + (3^{n+2}) ] + c_2\{n[(3^{n+2} -6(3^{n+1}) + 9 (3^{n})] + [(2)3^{n+2} -6 (3^{n+1}) ]\}$ 


~~$= c_1[0 ] + c_2\{n[(3^{n+2} -2(3^{n+2}) + (3^{n+2})] + [(2)3^{n+2} -2 (3^{n+2}) ]\} = 0$

Thus  $a_n = c_1(3^{n}) + c_2(n3^{n})$

IVP:  $a_0 = -1$ and $a_1 = 3$, 

$n = 0$:  $-1 = c_1(3^{0}) + c_2(0) = c_1$

$n = 1$: $3 = c_1(3) + c_2(3) = -3 + 3c_2$.  Thus $c_2 = 2$

Hence $a_n = -(3^{n}) + 2(n3^{n}) = 3^n(2n - 1)$



\end






























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

[4]~ 2a.)  The equation $lny' = {t \over t+1} -  y(sin t^2)$ 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 $y = \phi_1$ and $y = \phi_2$ are solutions to 
 $ay'' + by' + cy = 0$.
If $y = h$ is   also a solution to  $ay'' + by' + cy = 0$, then there exists constants $c_1$ and $c_2$ such that $h = c_1\phi_1 + c_2\phi_2$.
%\rightline
\hfill{T~~~~~~~~~~~F}

\vskip 20pt


[4]~ 2d.)   Suppose $y = \phi_1$ and $y = \phi_2$ are linearly independent solutions to 
 $ay'' + by' + cy = 0$.
If $y = h$ is   also a solution to  $ay'' + by' + cy = 0$, then there exists constants $c_1$ and $c_2$ such that $h = c_1\phi_1 + c_2\phi_2$.
%\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  $\phi = \displaystyle\sum_{k=2}^{\infty}{3(-1)^{k}~  t^{k} \over k!}$ converges for all $t$
and  show that  $\phi = \displaystyle\sum_{k=2}^{\infty}{3(-1)^{k}~  t^{k} \over k!}$  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, $\phi_n = \displaystyle\sum_{k=1}^{n} {3(-1)^{k+1}~  t^{k+1} \over (k+1)!}$ approximates the solution to the initial value problem,  $y' = 3t - y$, $y(0) = 0$ where $\phi_1 = {3t^2 \over 2}$.  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 $n = j$,~~  $\phi_{j-1} = \displaystyle\sum_{k=1}^{j-1} {3(-1)^{k+1}~  t^{k+1} \over (k+1)!}$ 

\vskip 20pt

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



\end

\vfill

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

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

\end



6A.)  Suppose $y = \phi_1$ and $y = \phi_2$ are solutions to 
the ODE $ay'' + by' + cy = 0$.
Prove that $y = c_1\phi_1 + c_2\phi_2$ 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, 
$y' + py = g$, 
then 
$2\phi$ must also be a solution to
$y' + py = g$.

\rightline{~~~~~~~~~~~F}
\w
[4]~ 2c.)  Suppose 
$\phi$
is a solution to the equation, 
$y' + py^2 = 0$, 
then 
$2\phi$ must also be a solution to
$y' + py^2 = 0$.

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

[4]~ 2d.)  Suppose 
$\phi$
is a solution to the equation, 
$y' + py = 0$, 
then 
$2\phi$ must also be a solution to
$y' + py = 0$.

\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
