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

[10] ~1a.)  Draw the direction field for the following differential equation: 
%\v
%\centerline
{$y' = y(2-y )$}
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\centerline{\includegraphics[width=30ex]{grapht-eps-converted-to.pdf}}
%\includegraphics[width=41ex]{slopefd.pdf}}
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[4] ~1b.)  On the direction field above, draw the solution  to the above differential equation  that satisfies the initial condition $y(0) = 1$.
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~~~~~Note the solution satisfying the initial condition $y(0) = 1$ must pass thru the initial value (0, 1).
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[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.
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~~~~~An equilibrium solution is a constant solution, $y = C$.  
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~~~~~The graph of a constant solution, $y= C$ is a horizontal line.
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~~~~~$y = 2$ is stable and
$y = 0$ is unstable

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[5]~  2.)  Give an example of an initial value problem that does not have a unique solution.

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~~~~~The classic example is $y^{1 \over 3}, y(0) = 0$, which as an infinite number of solutions.

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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.
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\hfill{T~~~~~~~~~~~}

[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~~~~~~~~~~~}


\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 $t \rightarrow \infty$?


 $ty' - 2y= t^3 e^t - 8$ implies  $1y' + (-{2 \over t})y= t^2 e^t - 8 t^{-1}$. 


Integrating factor:  $u = e^{\int p(t)dt} = e^{\int -{2 \over t}dt } = e^{-2ln|t|} = e^{ln(|t|^{-2})} = t^{-2}$

Let $u(t) = t^{-2}$

 $t^{-2}y'  -{2  t^{-3}}y=  e^t - {8 t^{-3}}$

 $(t^{-2}y)'  =  e^t - {8 t^{-3}}$ \hfill {\bf Check: $(t^{-2}y)'  = t^{-2}y'  -{2  t^{-3}}y $ }

 $\int (t^{-2}y)'dt  = \int( e^t - {8 t^{-3}})dt$

 $t^{-2}y  =  e^t + {4 t^{-2}}+C$

 $y  =  t^2e^t + {4}+Ct^2$


$y(1) = 3$:  ~~ $3  =  (1)^2e^1 + {4}+C(1)^2 = e + 4 + C$ 

$3  = e + 4 + C$.  Thus $C = -1-e$

IVP soln:   
 $y  =  t^2e^t + {4}+(-1-e)t^2$

 $y  =  t^2e^t + {4}-t^2-et^2$

 $y  =  t^2(e^t - e - 1) + {4}$

 $t \rightarrow +\infty$,  $t^2 \rightarrow +\infty$ and  $e^t\rightarrow +\infty$ and hence 
  $e^t - e - 1\rightarrow +\infty$ .

Thus  $y  =  t^2(e^t - e - 1) + {4}\rightarrow +\infty$ 

\vfill
\centerline{General solution: $\underline{~~y  =  t^2e^t + {4}+Ct^2~~}$}
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\centerline{IVP solution: $\underline{~~ y  =  t^2(e^t - e - 1) + {4}~~}$}
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\centerline{ $t \rightarrow \infty$,  $y \rightarrow \underline{~~+\infty~~}$}

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\eject


5.) Solve the following 2nd order differential equations

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

Guess $y = e^{rt}$

$2r^2 - r + 10 = 0$.  Thus $r = {1 \pm \sqrt{(-1)^2 - 4(2)(10)} \over 2(2)} = {1 \pm \sqrt{(1- 80)} \over 4} = {1 \pm \sqrt{- 79} \over 4}
= {1  \over 4} \pm  i{\sqrt{ 79} \over 4}$

\vfill

\centerline{General solution: $\underline{~~y = c_1e^{t \over 4}cos({\sqrt{79} \over 4}t) +  c_2e^{t \over 4}sin({\sqrt{79} \over 4}t) ~~}$}

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

Also   $x' = {dx \over dt} =v$ implies $x'' = v' = {dv \over dt} =  {dx \over dt}{dv \over dx}  = v{dv \over dx}$

$x{dv \over dt} = v^2$, but this has 3 variables, so replace ${dv \over dt} =  {dx \over dt}{dv \over dx}  = v{dv \over dx}$

$xv{dv \over dx} = v^2$

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

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

$v = Cx$

$ {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

\centerline{General solution: $\underline{~~x = ke^{Ct}~~}$}
\vfill
\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.

%$\phi_1 = \int 2sds$


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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!}=\phi_1(t) $

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

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Claim:  $\phi_j =  \displaystyle\sum_{k=1}^{j}  {t^{2k} \over k!}$  
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Proof of claim:  By Picard's iteration method, $\phi_{j} = $
$\int_0^t f(s, \phi_{j-1}(s))ds $

\hskip 3in $= \int_0^t 2s(1 + \displaystyle\sum_{k=1}^{j-1}  {s^{2k} \over k!})ds$

\hskip 3in $= \int_0^t (2s + \displaystyle\sum_{k=1}^{j-1}  {2s^{2k+1} \over k!})ds$

\hskip 3in $= \int_0^t ( \displaystyle\sum_{k=0}^{j-1}  {2s^{2k+1} \over k!})ds$

\hskip 3in $=  \displaystyle\sum_{k=0}^{j-1}  {2t^{2k+2} \over (2k + 2)k!}$

\hskip 3in $=  \displaystyle\sum_{k=0}^{j-1}  {2t^{2k+2} \over 2(k + 1)k!}$

\hskip 3in $=  \displaystyle\sum_{k=0}^{j-1}  {t^{2k+2} \over (k + 1)!}$

\hskip 3in $=  \displaystyle\sum_{k=1}^{j}  {t^{2k} \over k!}$


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