Section 3.2
Population Models

1. dp/dt = kP
2. dP/dt = aP - bP²

Mixing problems

(Drawing 1)

6L/min
5g/L
dx/dt = 6(5) - 6(x/1000)
x = x(t) = mass of the salt in the tank.

Mixing problems II
(Drawing 1 considered with following note)
Note--flow out is 4L/min
What is the density at time t?
x = x(t) = mass of salt at time t
dx/dt = 6(5) - 4(x/(1000 + 2t))

Exercise. Solve for x.
dx/dt = 30 - (2x/(t + 500))
dx/dt + 2x/(t + 500) = -30
m(t) = e^{ò (2/(t + 500))dt}
(d/dt) = m*x = m*(-30)
m = e^{2ln|t + 500| + C}
m = (t + 500)² + C₁
d/dt = ((t + 500)² + C₁)x
dx/x = ((t + 500)² + C₁)dt
ln|x| + C₂ = ò (t² + 1000t + 500² + C₁)dt
--
dp/dt = p - 2p^e
m(t) = (t + 500)²
dx/dt = 30 - 2(x)/(t + 500)
dx/dt + (2/(t + 500))x = 30
P(t) = 2/(t + 500), Q = 30
m(t) = ò (2/(t + 500))dt = 2ln|t + 500| = ln(t + 500)²
(t + 500)²x = ò 30(t + 500)²dt
= 30 ò (t + 500)²dt
= 30 * (1/3)(t + 500)3 + C
= 10(t + 500)³ + C
Therefore 10(t + 500) + C/(t + 500)²
x(0) = 0
= 5000 + C/500²
Therefore C = -5000(500²)
Density at t = 10
mass/volume (x(10) =)/(1000 + 2*10 = 1020)

1/(p-a)(p-b) = (A/(p - a)) + (B/(p-b))
A = (cp + d)/(p-b)|_p=a
B = (cp + d)/(p-a)|_p=b

Exercise
Find A and B such that
(3x + 4)/(x - 1)(x - 2) = A/(x - 1) + B/(x - 2)
((3x + 4)/(x - 2))/(x - 1) + ((3x + 4)/(x - 1)) /(x - 2) = (3x + 4)/(x - 1)(x - 2)
Answer is A = -7, B = 10

Test Review

1. Separation of variables.
2. Linear
dy/dx + P(x)y = Q(x)
m(x) = e^{ò P(x)dx}
(d/dx)(m*y) = m*Q(x)
3. Exact
Mdx + Ndy = 0
¶M/¶y = ¶N/¶x