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Quiz 5 Form A \hfil \break
Oct 30, 2017

[10]~~ 1.  Let $y_1$ and $y_2$ be solutions of
$ty''
+ 4y'
+
sin(t)
y = 0$; $t > 0$.
Let $W(t)$ be the Wronskian of $y_1(t)$ and $y_2(t)$. Given that $W(1) = 3$, find
$W(t)$.
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%\centerline
{$W(t) = \underline{~~3t^{-4}~~}$}


$p(t) = {4 \over t}$.  Thus by Abel's thm $W(t) =ce^{-\int  {4 dt \over t}}= ce^{-4\int  {dt \over t}}= ce^{-4 ln|t|} =e^{ln|t|^{-4}} = ct^{-4}$ .

Thus  $W(t) =ct^{-4} $ for some constant $c$.  Since $W(1) = 3$,   ~~$3 =  c(1)^{-4}$.  Thus $c = 3$.

Hence $W(t) =3t^{-4} $

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[10]~~2.)  Write $y = \sqrt{3}cos(5t) - sin(5t)$ in the form $y = Rcos(\omega t - \delta)$.  Determine the period, phase, and amplitude.

$y $$=  \underline{2cos(5t +  {\pi \over 6})}$ 
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period$=  \underline{~~ {{2\pi \over 5}}~~}$, phase$=  \underline{~~ -{\pi 
\over 6}~~}$, and amplitude$=  \underline{~~ 2~~}$.

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amplitude = $R = \sqrt{c_1^2 + c_2^2} = \sqrt{3 + 1} = \sqrt{4} = 2$

phase = $\delta$ where $tan(\delta) = {c_2 \over c_1} = {-1 \over \sqrt{3}}$.

Note if you plot $(Rcos(\delta), Rsin(\delta) = (\sqrt{3}, -1)$, you can see the angle the vector $ (\sqrt{3}, -1)$ forms with the positive $x$ axis is $-{\pi \over 6}$.  Thus $\delta = -{\pi \over 6}$

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Derivation:
$Rcos(\omega t - \delta) = Rcos(\delta) R(\omega t) + Rsin(\delta)Rsin(\omega t) =  \sqrt{3}cos(5t) - sin(5t)$

Thus $\omega = 5$, period = ${2\pi \over 5}$, and $ Rcos(\delta) =  
\sqrt{3}$, $Rsin(\delta) =  - 1$

Thus $tan(\delta) = {R sin(\delta) \over Rcos(\delta) } = {-1 \over \sqrt{3}}$.

Since  $cos(\delta) > 0$ and $sin(\delta) < 0$, $\delta$ is in quadrant 4 ( i.e, $-{\pi \over 2} < \delta < 0)$.  Thus $\delta = -{\pi \over 6}$  

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