MATH383: A first course in differential equationsJanuary 30, 2020

Due date: Feb 6, 2020, 11:55PM.

Bibliography: Trench Chap. 4. The first exercise in each problem set is solved for you to use as a model.

1. Exercises 1-5, p. 138

   Ex 1

   Half life $\tau =3200\mathrm{years}$, initial amount $Q_0=20$g. Radioactive decay model $Q^{\text{'}}=-kQ$, $k=\left(\ln 2\right)/\tau$ has solution

   ${\displaystyle Q\left(t\right)=e^{-kt}{Q}_0=\exp [-\frac{t}{\tau }\ln 2]Q_0=20\times 2^{-t/\tau }\text{g}.}$

   Ex 2

   Ex 3

   Ex 4

   Ex 5

2. Exercises 15-19, pp. 138-9

   Ex 15

   Gold creation rate $r=1$ oz/hr, theft rate $s=\frac{1}{20}W\left(t\right)$ oz/hr. Model equation $W^{\text{'}}=r-s=r-\frac{1}{20}W$, and solution to homogeneous problem $W^{\text{'}}+\frac{1}{20}W=0$ is $W_h=e^{-t/20}$. Variation of parameters, $W=u{W}_h$

   ${\displaystyle u^{\text{'}}{W}_h=r\Rightarrow u^{\text{'}}=r{e}^{t/20}\Rightarrow u=20r{e}^{t/20}+c\Rightarrow W\left(t\right)=20r+c{e}^{-t/20}.}$

   Initial condition $W\left(0\right)=20r+c=1\Rightarrow c=1-20r\Rightarrow W\left(t\right)=20r(1-e^{-t/20})+e^{-t/20}$, ${lim}_{t\to \infty }W\left(t\right)=20r$.

   Check in Maxima

   (%i3)

   ode2('diff(W,t)=r-W/20,W,t);

   ${\displaystyle \text{(\%o3)}W={\mathrm{e}}^{-\frac{t}{20}}(20r{\mathrm{e}}^{\frac{t}{20}}+\%c)}$

   (%i4)

   ic1(%,t=0,W=1);

   ${\displaystyle \text{(\%o4)}W={\mathrm{e}}^{-\frac{t}{20}}(20r{\mathrm{e}}^{\frac{t}{20}}-20r+1)}$

   (%i5)

   Ex 16

   Ex 17

   Ex 18

   Ex 19

3. Exercises 1-5, p.148

   Ex 1

   Room temperature $T_0={70}^{\circ }\text{F}$, freezer temperature $T_1={12}^{\circ }\text{F}$. Model equation for thermometer temperature $T\left(t\right)$ is $T^{\text{'}}=-k(T-T_1),T\left(0\right)=T_0$. Solution is $T=T_1+(T_0-T_1)e^{-kt}$. From $T(1/2)=40$ obtain

   ${\displaystyle 12+58e^{-k/2}=40\Rightarrow k=-2\ln \frac{28}{58}=\ln {\left(\frac{29}{14}\right)}^2.}$

   Evaluate

   ${\displaystyle T\left(2\right)=12+58\exp [-2\ln {\left(\frac{29}{14}\right)}^2]=12+58{\left(\frac{14}{29}\right)}^4=15.{15}^{\circ }\text{F}.}$

   (%i13)

   float(12+58*(14/29)^4)

   ${\displaystyle \text{(\%o13)}15.15027266390586}$

   (%i14)

   Ex 2

   Ex 3

   Ex 4

   Ex 5

4. Exercises 1-5, p. 160

   Ex 1

   Weight $m=192\mathrm{lb}=87.3\mathrm{kg}$, model equation is $m{v}^{\text{'}}=mg-kv$, with $g=9.8m/s^2$, $k=2.5\mathrm{lbf}s/\mathrm{ft}=36.54\mathrm{kg}/s$, with solution $v\left(t\right)=mg/k[\exp (-kt/m)-1]$, and terminal velocity ${lim}_{t\to \infty }v\left(t\right)=mg/k=23.4m/s$.

   Ex 2

   Ex 3

   Ex 4

   Ex 5