Calculus of a Single Variable, Chapter 6, 6.2, Section 6.2, Problem 19

The rate of change of N is the derivative of N with respect to t, or (dN)/(dt) . If the rate of change of N is proportional to N, then
(dN)/(dt) = kN , where k is the proportionality constant. This is the differential equation we need to solve.
To solve it, separate the variables:
(dN)/N = kdt
Integrating both sides results in
lnN = kt + C , where C is another constant. This can be rewritten in exponential form as
N = e^(kt + C) = N_0e^(kt) . Here, N_0 = e^C and it equals N(t) when t = 0.
When t = 0, N = 250, so
N(0) = N_0 = 250 and N(t) = 250e^(kt) is the solution of the differential equation above with the initial condition N(0) = 250.
To find k, we can use that when t = 1, N = 400:
N(1) = 250e^(k*1) = 400
e^k = 400/250 = 8/5 = 1.6
k = ln(1.6)
Plugging this back into N(t), we get
N(t) = 250e^(t*ln(1.6)) = 250*1.6^t .
Then, for t = 4, N(4) = 250*1.6^4 =1638.4
So, the solution of the equation modeling the given verbal statement is
N(t) = 250*1.6^t and for t = 4, N = 1638.4.