Can someone help me with a differential equation problem?

Question

Suppose that an initial population of 10,000 bacteria grows exponentially at a rate of 2% per hour and that y=y(t) is the number of bacteria present t hours later.

a) Find an initial-value problem whose solution is y(t)

b) Find a formula for y(t)

c) How long does it take for the initial population of bacteria to double

Answer 1

Using the equation,  y(t) = y_oe^kt,  where y(t) is the value of y at time t, y_o is the value of y at time  t = 0, and k is the rate at which y increases as a function of time.

 

Therefore, y_o = 10,000 and k = .02.

 

a)  y(0) = 10,000e^0.02*0 = 10,000(1) = 10,000.

b)  y(t) = 10,000e^0.02t.

c)  y(t) = 2y_o,  so 2 = e^0.02t.

 

Take the natural log of each side:  ln2 = 0.02t,   divide by 0.02, 

 

t = (ln2)/(0.02)    so t = 34.657 hours.