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

Given -

$\dfrac{dN}{dt}=$ $t-2e^{t}$

Then , $dN=(t-2e^{t})dt$

integrate both sides we get ,

$\int dN=\int(t-2e^{t})dt$

$N(t)=\dfrac{t^{2}}{2}-2e^{t}+c$

given -

$N(0)=400$

$=400=\dfrac{0^{2}}{2}-2e^{0}+c$ $\implies$ $c=402$

$N(t)=$ $\dfrac{t^{2}}{2}-2e^{t}+402$

now putting t=5 we get ,

$N(5)=$$\dfrac{5^{2}}{2}-2e^{5}+402=118$