Solve the problem.
-The rate of change in the number of bacteria (in thousands) in a culture after the introduction of a bactericide is $\mathrm{dy} / \mathrm{dt}=100-\mathrm{y}$, where $\mathrm{y}$ is the number of bacteria (in thousands) at time $t$. How many bacteria are present at $t=12$ if the initial population was $1,000,000$ thousand? (Round results to the nearest thousand bacteria.)
A) $162,739 \times 10^{6}$ thousand bacteria
B) 94 thousand bacteria
C) 106 thousand bacteria
D) 6 thousand bacteria

Question

Quizplus · Accepted Answer

The differential equation $\mathrm{dy} / \mathrm{dt}=100-\mathrm{y}$ can be solved using separation of variables. The solution is $y(t) = 100 + Ce^{-t}$. Given the initial condition $y(0) = 1000$, we find $C = 900$. Thus, $y(t) = 100 + 900e^{-t}$. At $t=12$, $y(12) \approx 106$ thousand bacteria.

Solve the Problem dy/dt=100−y\mathrm{dy} / \mathrm{dt}=100-\mathrm{y}dy/dt=100−y , Where y\mathrm{y}y Is the Number of Bacteria (In Thousands) at Time

Solve the Problem $\mathrm{dy} / \mathrm{dt}=100-\mathrm{y}$ , Where $\mathrm{y}$ Is the Number of Bacteria (In Thousands) at Time