dy/dt = -0.3y, y(0) = 90, when is y = 10 Step by Step

Step by step

The question

Question: find the time t when y = 10, if dy/dt = −0.3y and y(0) = 90.

dydt = −0.3y

y(0) = 90

1 Read the equation

The rate of change is −0.3 times the amount y.

The more there is, the faster it shrinks.

k < 0: decay.

the rate = −0.3 × y: the more y, the faster it changes

2 Separate and integrate

This is a separable equation.

Divide by y, times dt.

dyy = −0.3 dt

Integrate both sides, as in the separable lesson.

ln |y| = −0.3t + C

Use e on both sides.

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

y = Ae^(−0.3t)

3 Use the starting value

Put in t = 0 and y = 90.

e⁰ = 1.

90 = Ae^(0)

So A = 90: the amount at the start.

A = 90

The formula: y = 90e^(−0.3t).

y = 90e^(−0.3t)

Careful!

This is not a steady rate.

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

Careful: not y = 90 + −0.3 × 90t ✗

4 Answer the question

Put in y = 10.

Now t is the unknown.

10 = 90e^(−0.3t)

Divide both sides by 90.

19 = e^(−0.3t)

Take ln of both sides.

ln(19) = −0.3t

Divide.

So t ≈ 7.32.

t = ln(19)−0.3 ≈ 7.32

5 Check

Let's check.

At t = 0: 90, the start.

Right!

y(0) = 90 × 1 = 90

Put t ≈ 7.32 back in: about 10.01.

Right!

y(7.32) ≈ 10.01

Let's look back.

First: read the equation.

Then: separate and integrate.

Then: use the starting value.

Then: answer the question.

Last: check.

Answer