Homework Two

10

\begin{displaymath}y(x) = \pm \sqrt{\frac{2}{3}x^3 + 2 C}.\end{displaymath}

11

\begin{displaymath}y(x) = - \frac{1}{\cos x +C},\end{displaymath}

also by inspection

\begin{displaymath}y(x)=0.\end{displaymath}

12

\begin{displaymath}y(x) = \sin\left[ \log\vert x\vert + C \right], \end{displaymath}

also by inspection,

\begin{displaymath}y(x)=\pm 1.\end{displaymath}

13

\begin{displaymath}y(x) = - \sqrt{\frac{x^2 + 1}{2}}, \end{displaymath}

defined for

\begin{displaymath}-\infty<x<\infty.\end{displaymath}

14

\begin{displaymath}y(x) = - \sqrt{\frac{2}{3}\log\vert 1+x^3\vert+1}.\end{displaymath}

15

\begin{displaymath}\tau = \frac{1}{r}\log{2} = 20.0 {\rm days.}\end{displaymath}

16

(a)

\begin{displaymath}Q(t) = 35.36 + 64.64 e^{ - 0.02828 t}.\end{displaymath}

(b)

\begin{displaymath}\lim\limits_{t\to\infty} = 35.36 {\rm mg}.\end{displaymath}

(c)

\begin{displaymath}T = 171.9 {\rm days}\end{displaymath}

(d)

\begin{displaymath}k = r \cdot 100{\rm mg} = 2.828 {\rm mg}. \end{displaymath}

17

\begin{displaymath}Q(t) = 120 \gamma (1 - e^{-\frac{t}{60}}), \end{displaymath}

limiting amount is

\begin{displaymath}120 \gamma.\end{displaymath}

18

(a)


\begin{displaymath}x(t) = c ( 1 - e^{-\frac{\phi}{V} t}) = 0.04 (1-e^{-\frac{t}{13000}}).\end{displaymath}

(b)


\begin{displaymath}\tau = \frac{V}{\Phi}\log\frac{1}{1-\frac{x(\tau)}{c}} = 36 {\rm min.}\end{displaymath}

Dr Yuri V Lvov 2017-12-10