More on Modeling1.2: Separation of

Variables

January 18, 2007

•HW change: 1.2 #38 is not due this week.•Bring your CD-ROM to class Tuesday.

A mystery

Here’s a new population model…

• Find the equilibrium solutions.• For what values of P is the population increasing?

decreasing?• Sketch some solutions to the differential equation.• Can you think of a situation in which this model

makes sense? Be creative…

€

dP

dt= 0.5 1−

P

50

⎛

⎝ ⎜

⎞

⎠ ⎟P

3−1

⎛

⎝ ⎜

⎞

⎠ ⎟P

From last time…

Spread of a rumor(group work)

Quantities: (identify as indep var, dep var, or parameter)• P = population of city• N = people who have heard the rumor• t = time • k = proportionality constantAnswers:1. dN/dt2. P - N3. dN/dt = k(P - N)4. dN/dt = k(350 - N)

(N is in thousands, t could be days, weeks, etc. Your choice of units for t affects the value of k.)

What should solutions look like? equilibrium solutions?

mmmm… Chocolate!

Quantities:• T = temp (degrees F) of hot chocolate at time t• t = time in minutes, hours, etc. (Why doesn’t it matter?)

• k = proportionality constant

Equation:

(Should k be positive or negative?)

€

dT

dt= k(T − 25)

HUH?????

I asked you to look at the statement on p. 22: “So we should never be wrong.” What does that mean?

Check this out:

Paul says “y1(t) = 1 + t is a solution.”

Glen says “y2(t) = 1 + 2t is a solution.”

Bob says “y3(t) = 1 is a solution.”

Who is right? How can we tell?

€

dy

dt=y2 −1

t2 + 2t

Separable Diffy-Q’s

Example:

€

1.dT

dt= k(T − 25)

€

2.1

T − 25dT = k dt

€

3.1

T − 25∫ dT = k dt∫

€

4. lnT − 25 = kt + c

€

5. T − 25 = ekt+c = Aekt

€

6.T = Aekt + 25

T = 25 − Aekt ⎧ ⎨ ⎩

Suppose the chocolatestarted out at 150o and

was 100o 15 minutes later.How would you solve this

initial value problem?