From 2b583d571f98535dd44022421b56ba4d731a8e10 Mon Sep 17 00:00:00 2001
From: Holden Rohrer <hr@hrhr.dev>
Date: Thu, 16 Jan 2020 22:52:07 -0500
Subject: first math document of the semester :(

---
 tech-math/de/hw1.tex | 28 ++++++++++++++++++++++++++++
 1 file changed, 28 insertions(+)
 create mode 100644 tech-math/de/hw1.tex

diff --git a/tech-math/de/hw1.tex b/tech-math/de/hw1.tex
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+\def\pre#1{\leavevmode\llap{\hbox to \parindent{\hfil #1 \hfil}}}
+\baselineskip=14pt
+\parskip=5pt
+\nopagenumbers
+
+\noindent {\bf Q1)} %1.1:5
+
+\pre{(a)} The general solution is $p = 900 + ce^{t/2}$. If $p(0) = 850$, $850 = 900 + ce^{0/2} = 900 + c \to c = -50$ and $0 = 900 - 50e^{T/2} \to 18 = e^{T/2} \to T = 2\ln(18)$
+
+\pre{(b)} $p = 900 + ce^{t/2}$, so $p_0 = 900 + ce^{0/2} \to c = 900 - p_0$. $0 = 900 + (900-p_0)e^{T/2} \to 2\ln({900 \over 900-p_0} = T$.
+
+\pre{(c)} Using the equation derived in part b, $e^{T/2} = {900 \over 900-p_0} \to 900-p_0 = {900\over e^{T/2}} \to p_0 = 900-{900\over e^{T/2}} = 900-{900\over e^6}$.
+
+\noindent {\bf Q2)} %1.2:10
+\vskip3in
+
+\noindent {\bf Q3)} %1.3:26
+
+Substituting $y=t^r$, $t^2(t^r)'' - 4t(t^r)' + 4(t^r) = 0 = (r)(r-1)t^r - 4rt^r + 4t^r = t^r(r^2 - 5r + 4) = t^r(r-4)(r-1)$. This is true only when one or more of the terms is constantly 0, so $r = 1, 4$.
+
+\noindent {\bf Q4)} %2.1:20
+
+\rightskip=2.5in
+\pre{(a)} $\int e^{3y} dy = \int x^2 dx \to e^{3y}/3 = x^3/3 + C \to 3y = \ln(x^3 + C) \to y = {\ln(x^3+C)\over 3}$. Substituting the initial values, $0 = {\ln(2^3+C)\over3} \to C=-7$. This gives $y = {\ln(x^3-7)\over3}$.
+
+\pre{(c)} $y(x)$ is defined on all $x>\root 3\of{7} $.
+
+\bye
-- 
cgit