dy/dt = -0.2y, y(0) = 80, when is y = 20 Step by Step

Step by step

The question

Question: find the time t when y = 20, if dy/dt = −0.2y and y(0) = 80.

dydt = −0.2y

y(0) = 80

1 Read the equation

The rate of change is −0.2 times the amount y.

The more there is, the faster it shrinks.

k < 0: decay.

the rate = −0.2 × y: the more y, the faster it changes

2 Separate and integrate

This is a separable equation.

Divide by y, times dt.

dyy = −0.2 dt

Integrate both sides, as in the separable lesson.

ln |y| = −0.2t + C

Use e on both sides.

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

y = Ae^(−0.2t)

3 Use the starting value

Put in t = 0 and y = 80.

e⁰ = 1.

80 = Ae^(0)

So A = 80: the amount at the start.

A = 80

The formula: y = 80e^(−0.2t).

y = 80e^(−0.2t)

Careful!

This is not a steady rate.

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

Careful: not y = 80 + −0.2 × 80t ✗

4 Answer the question

Put in y = 20.

Now t is the unknown.

20 = 80e^(−0.2t)

Divide both sides by 80.

14 = e^(−0.2t)

Take ln of both sides.

ln(14) = −0.2t

Divide.

So t ≈ 6.93.

t = ln(14)−0.2 ≈ 6.93

5 Check

Let's check.

At t = 0: 80, the start.

Right!

y(0) = 80 × 1 = 80

Put t ≈ 6.93 back in: about 20.01.

Right!

y(6.93) ≈ 20.01

Let's look back.

First: read the equation.

Then: separate and integrate.

Then: use the starting value.

Then: answer the question.

Last: check.

Answer