dN/dt = 0.02N, N(0) = 300, find N(25) Step by Step

Step by step

The question

Question: find N(25), if dN/dt = 0.02N and N(0) = 300.

dNdt = 0.02N

N(0) = 300

1 Read the equation

The rate of change is 0.02 times the amount N.

The more there is, the faster it grows.

k > 0: growth.

the rate = 0.02 × N: the more N, the faster it changes

2 Separate and integrate

This is a separable equation.

Divide by N, times dt.

dNN = 0.02 dt

Integrate both sides, as in the separable lesson.

ln |N| = 0.02t + C

Use e on both sides.

e^C is just a number: call it A.

N = Ae^(0.02t)

3 Use the starting value

Put in t = 0 and N = 300.

e⁰ = 1.

300 = Ae^(0)

So A = 300: the amount at the start.

A = 300

The formula: N = 300e^(0.02t).

N = 300e^(0.02t)

Careful!

This is not a steady rate.

The rate grows with N, so the formula has e, not a plus.

Careful: not N = 300 + 0.02 × 300t ✗

4 Answer the question

Put in t = 25.

N(25) = 300e^(0.02 × 25)

The power: 0.5.

= 300e^(0.5)

So N(25) ≈ 494.62.

≈ 494.62

5 Check

Let's check.

At t = 0: 300, the start.

Right!

N(0) = 300 × 1 = 300

And the derivative of the formula is 0.02 times N: the equation again.

Right!

N′ = 300 × 0.02e^(0.02t) = 0.02 × N

Let's look back.

First: read the equation.

Then: separate and integrate.

Then: use the starting value.

Then: answer the question.

Last: check.

Answer