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

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

Then

$$dN=(t-2e^t)dt$$

Integrate both sides

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

$$N(t)=\dfrac{1}{2}t^2-2e^t+c$$

Given $N(0)=400$

$$400=\dfrac{1}{2}(0)^2-2e^0+c=>c=402$$

$$N(t)=\dfrac{1}{2}t^2-2e^t+402$$

$$N(5)=\dfrac{1}{2}(5)^2-2e^5+402=118$$