From f103b8b637a7b868b864c5e7c49e3feada1291ff Mon Sep 17 00:00:00 2001 From: Holden Rohrer Date: Wed, 29 Jan 2020 12:07:30 -0500 Subject: update: mostly literature --- tech-math/de/hw2.tex | 35 +++++++++++++++++++++++++++++++++++ 1 file changed, 35 insertions(+) create mode 100644 tech-math/de/hw2.tex (limited to 'tech-math') diff --git a/tech-math/de/hw2.tex b/tech-math/de/hw2.tex new file mode 100644 index 0000000..03563ff --- /dev/null +++ b/tech-math/de/hw2.tex @@ -0,0 +1,35 @@ +\def\pre#1{\leavevmode\llap{\hbox to \parindent{\hfil #1 \hfil}}} +\baselineskip=14pt +\parskip=5pt +\nopagenumbers + +\noindent {\bf Q1)} %2.2:32 + +The general solution of $2y' + ty = 2$ is, $y(t) = e^{-t^2/4} \int^t_0 e^{s^2/4}ds + ce^{-t^2/4}$. +The second term approaches 0 as $t\to\infty$, so it reduces to $$e^{-t^2/4} \int^t_0 e^{s^2/4}ds = {\int^t_0 e^{s^2/4}ds \over e^{t^2/4}}.$$ +Applying L'hopital's Rule (with respect to $t$) because both functions trend towards infinity gives ${e^{t^2/4} \over {t\over2}e^{t^2/4}} = {2 \over t}$. +This shows that the solution trends asymptotically towards 0. + +\noindent {\bf Q2)} %2.3:4 + +Let $Q$ be the concentration (in pounds per gallon) of the tank and $t$ be elapsed time. +Where $w$ is water amount, $${dw\over dt} = 1\to w = t+C = t+200.$$ +$${dQ\over dt} = {3 - 2Q\over w} = {3 - 2Q\over t+200} \to {Q'\over 3-2Q} = {1\over 200+t}$$ +$${d\over dt}-{1\over2}ln(3-2Q) = {Q'\over 3-2Q} = {1\over 200+t} \to -{1\over2}ln(3-2Q) = ln(200+t) + C$$$$ +\to ln(3-2Q) = -2ln(200+t) + C \to Q = {3-e^{-2ln(200+t) + C}\over2}$$ +Plugging in the initial value $Q(0) = {1\over2} = {3-e^{-2ln(200) + C}\over2} \to 2 = e^{-2ln(200) + C} = {C\over40000} \to C = 80000$. + +At 500gal or 300min, $Q(300) = {3-e^{-2ln(200+300 + 80000)}\over2} \approx 1$. +The asymptotic concentration is the concentration of the water flowing in, $1{lb \over gal}$. +This means that, relatively quickly, the tank becomes as saline as the incoming stream. + +\noindent {\bf Q3)} %2.4:4 + +This is discontinuous where $(4-t^2)=0 \to t \in {-2,2}$. +The initial value is $-3$, so the solution is guaranteed to exist on $t\in(-\infty, -2)$. + +\noindent {\bf Q4)} %2.5:12 + + + +\bye -- cgit