update: mostly literature

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+\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