# Write and solve the differential equation that models the verbal statement. Evaluate the solution at the specified value of the independent variable. The rate of change of N is proportional to N. When t = 0, N = 250 and when t = 1, N = 400. What is the value of N when t = 4?

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...

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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.

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