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Quiz 3 ~~~SHOW ALL WORK\hfill Oct 12, 2018




1.)  The solution to $y'' + 16y =  36cos(2t) $ ~is~ $y = c_1cos(4t) +  c_2sin(4t) + 3cos(2t)$
\hfil \break
Use this fact to answer the following two questions.



[5]~ 1a.)  Guess a possible non-homog soln for the following differential equation 
(do
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 not solve):  ~~ $y'' + 16y = 3sin(4t) -  e^{4t} $

Guess $y = t[Asin(4t) + Bcos(4t)] + Ce^{4t}$

For explanations and examples, \hfill\break
see http://homepage.divms.uiowa.edu/$\sim$idarcy/COURSES/100/3\_5listSans.pdf
\hfill\break
and http://homepage.divms.uiowa.edu/$\sim$idarcy/COURSES/100/3\_5exA.pdf




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[3]~ 1b.)  The general solution to  ~$y'' + 16y = 36cos(2t) +  32$ ~is

 $y = c_1cos(4t) +  c_2sin(4t) + 3cos(2t) + 2$


See http://homepage.divms.uiowa.edu/$\sim$idarcy/COURSES/100/FALL18/18\_10\_15.pdf
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 2.)  Circle T for true and F for false.  





[2]~ 2a.)
 $L(f) = af'' + b f' + cf$ is a linear function on the space of 
all twice differentiable functions.
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\rightline{T~~~~~~~~~~~}  
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[2]~  2b.) $L(f) = af'' + bf' + cf^2$ is a linear function on the space of 
all twice differentiable functions.
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\rightline{~~~~~~~~~~~F}  
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[2]~ 2c.)  Suppose $y = \phi_1(t)$ and $y = \phi_2(t)$ are solutions to $ay'' + by' + cy = 0$, \hfil \break $y = \psi_1(t)$ is a solution to $ay'' + by' + cy = g_1(t)$, and \hfil \break
 $y = \psi_2(t)$ is a solution to $ay'' + by' + cy = g_2(t)$, then the {\bf general} solution to \hfil \break
$ay'' + by' + cy = g_1(t) + g_2(t)$ ~is~  $y = c_1\phi_1(t) + c_2\phi_2(t) + \psi_1(t) + \psi_2(t)$.

\rightline{~~~~~~~~~~~F}  
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[2]~ 2c.)  Suppose $y = \phi_1(t)$ and $y = \phi_2(t)$ are solutions to $ay'' + by' + cy = 0$, \hfil \break $y = \psi_1(t)$ is a solution to $ay'' + by' + cy = g_1(t)$, and \hfil \break
 $y = \psi_2(t)$ is a solution to $ay'' + by' + cy = g_2(t)$, then $y = c_1\phi_1(t) + c_2\phi_2(t) + \psi_1(t) + \psi_2(t)$ is also a solution to \hfil \break
$ay'' + by' + cy = g_1(t) + g_2(t)$.

\rightline{T~~~~~~~~~~~}  
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[2]~ 2c.)  Suppose $y = \phi_1(t)$ and $y = \phi_2(t)$ are linearly {\bf independent solutions} to $ay'' + by' + cy = 0$, \hfil \break $y = \psi_1(t)$ is a solution to $ay'' + by' + cy = g_1(t)$, and \hfil \break
 $y = \psi_2(t)$ is a solution to $ay'' + by' + cy = g_2(t)$, then the general solution to \hfil \break
$ay'' + by' + cy = g_1(t) + g_2(t)$ ~is~  $y = c_1\phi_1(t) + c_2\phi_2(t) + \psi_1(t) + \psi_2(t)$.

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

Note for 2c, I forgot to include linearly independent, so if lost 2 points because I graded your problem incorrectly, please let me know.  
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[2]~ 2d.) $\Sigma_{n = 2}^\infty n(n-1)a_nx^{n-2} = \Sigma_{j = 0}^\infty (j+2)(j+1)a_{j+2}x^{j} =\Sigma_{n = 0}^\infty (n+2)(n+1)a_{n+2}x^{n}$ 
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\rightline{T~~~~~~~~~~~}  
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[2]~ 2e.) Suppose $f(x) = \Sigma a_n (x - 3)^n$ has a radius of convergence = $r$ about the point $x_0 = 3$. Then we can define the domain of $f$ to be $(3- r, 3 + r)$.
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\rightline{T~~~~~~~~~~~}  
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%2f.) Suppose $f(x) = \Sigma a_n (x + 1)^n$ has a radius of convergence = $4$ about the point $x_0 = -1$. Then we can define the domain of $f$ to be $(3, 5)$.

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

[2]~  2f.) Suppose $f(x) = \Sigma a_n (x + 1)^n$ has a radius of convergence = $4$ about the point $x_0 = -1$. Then we can define the domain of $f$ to be $(-5, 3)$.
\u
\rightline{T~~~~~~~~~~~}  
           


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[10]~ 2.)  Solve $y' = {1 \over x(2y+3)}$, ~$y(1) =-2 $


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Answer:$\underline{~~\hskip 3in~~}$

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${dy \over dx} = {1 \over x(2y+3)}$

Separate variables:  $(2y+3)dy = {dx \over x}$

$y^2 + 3y = ln|x| + C$.

$y^2 + 3y - ln|x| - C = 0$.


$y = {-3 \pm \sqrt{9 - 4 (- ln|x| - C)} \over 2}$

$y = {-3 \pm \sqrt{9 + 4 ln|x| + C} \over 2}$

$y = {-3 \pm \sqrt{9  + C)} \over 2}$

$y = {-3 \pm \sqrt{9 -8)} \over 2}$

$y = {-3 - 1 \over 2}$

$y = {-3 \pm \sqrt{9 + 4 ln|x| -8} \over 2}$

$y = {-3 - \sqrt{1 + 4 ln|x| } \over 2}$

2.2       14




Solve $sin(t) y' + y cos(t) = 5$ 
 \hfil \break(hint:  this is a short problem if you are observant).

$sin(t) y' + y cos(t) = 5$

 $(sin(t) y)'  = 5$  
 
  $\int (sin(t) y)' dt  = \int 5 dt$  

 $\sin(t) y = 5t + C$  

 $ y  = {5t + C \over sin(t)}$  




Answer:$\underline{~~ y  = {5t + C \over sin(t)}~~}$
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