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diff --git a/tech-math/de/hw1.tex b/tech-math/de/hw1.tex new file mode 100644 index 0000000..dc9946c --- /dev/null +++ b/tech-math/de/hw1.tex @@ -0,0 +1,28 @@ +\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 |