Question

Two kinds of bacteria are found in a sample of tainted food. It is found that the population size of type 1, N1 and of type 2, N2 satisfy the equation

dN/dt=-0.2/N1

dN/dt=0.6/N2

N1 is equal to 2000 at time equal to zero, while N2 is equal to 20 at time equal to zero.

Then the population sizes are equal (N1 = N2) at what time? (4 decimal places)

Answer 1

Let us solve the equation $\frac{dN_1}{dt}=-\frac{0.2}{N_1},\ \ N_1(0)=2000.$

$N_1dN_1=-0.2dt$

$\int N_1dN_1=-\int 0.2dt$

$\frac{(N_1)^2}{2}=-0.2t+C_1$

If $N_1(0)=2000$, then $\frac{2000^2}{2}=C_1$, and thus $C_1=2,000,000$.

So, $(N_1)^2=-0.4t+4,000,000.$

Let us solve the equation $\frac{dN_2}{dt}=\frac{0.6}{N_2},\ \ N_2(0)=20.$

$N_2dN_1=0.6dt$

$\int N_2dN_2=\int 0.6dt$

$\frac{(N_2)^2}{2}=0.6t+C_2$

If $N_2(0)=20$, then $\frac{20^2}{2}=C_2$, and thus $C_2=200$.

So, $(N_2)^2=1.2t+400$

If $N_1=N_2$, then $(N_1)^2=(N_2)^2$ , and thus $-0.4t+4,000,000=1.2t+400$. It follows that $1.6t=3,999,600$, and therefore, $t=2,499,750.$

Answer: $t=2,499,750.$