Part I: Multiple Choice

$1$ Using Laplace Transform Table, $L\{e^{2t}cos(3t)\}$ is

$($a$)\frac{s-2}{s^2+9}$

$($c$)\frac{s-2}{s^2-4s+13}$

$($e$)$ None of these

$($b$)\frac{s}{(s-2{)}^2+9}$

$($d$)\frac{s}{s^2+9}$

$2$ Find the inverse Laplace $L^{-1}\{\frac{s^2+4s+4}{(s-1)(s+2)(s+5)(s-2)}\}.$

$($Hint: The work is easier if you first factor the numerator & simplify.$)$

$($a$)\frac{4}{7}(\frac{1}{s-2})-\frac{1}{14}(\frac{1}{s+5})-\frac{1}{2}(\frac{1}{s-1})$

$($b$)\frac{4}{7}{e}^{2t}-\frac{1}{14}{e}^{-5t}-\frac{1}{2}{e}^t+\frac{2}{3}{e}^{-2t}$

$($c$)\frac{4}{7}{e}^{2t}-\frac{1}{14}{e}^{-5t}-\frac{1}{2}{e}^t$

$($d$)\frac{4}{7}{e}^{-5t}-\frac{1}{14}{e}^{2t}-\frac{1}{2}{e}^t$

$($e$)$ None of these

$3$ A frozen chicken at $0$ is taken out of the freezer and placed on a table at $20.$ One hour later the temperature of the chicken is $18.$ The mathematical model for the temperature $T(t)$ as a function of time $t,$ assuming Newton's law of warming, is:

$($ a$)\frac{dT}{dt}=kT,T(0)=0,T(1)=18$

$($b$)\frac{dT}{dt}=T,T(0)=0,T(1)=18$

$($c$)\frac{dT}{dt}=k(T-20),T(0)=0,T(1)=18$

$($d$)\frac{dT}{dt}=(T-18),T(0)=0,T(1)=18$

$($e$)$ None of these

$($This in and differential equations class, please solve all questions and provide steps$)$