Answer to Question #213420 in Calculus for DANNY

Question

The population N(t) of a species of micro-organism in a laboratory setting at any time t is established to vary under the influence of a certain chemical at a rate given by dN/dt=t-2e^t show that N(t)=t-2e^t+c.hence if N(0)=400 determine the population when t=5

Answer 1

$\text{Since the rate is given by $\frac{dN}{dt}= t-2e^t$ }\\\text{The N(t) is given by the integral of $\int\frac{dN}{dt}dt$}\\\text{Therefore N(t) =$\int (t-2e^{t})dt=\int tdt-2\int e^tdt$}\\=\frac{t^2}{2}-2e^t+c\\\text{Given N(0) = 400, we substitute the value of t for 0, therefore c= 402}\\\text{N(t) =$\frac{t^2}{2}-2e^t+402$ }\\\text{Thus N(5) = $\frac{5^2}{2}-2e^5+402=117.67$}$