Question

In many population growth problems, there is an upper limit beyond which the population cannot grow. Let us suppose that the earth will not support a population of more than 16 billion and that there were 2 billion people in 1925 and 4 billion people in 1975. Then, if y is the population t years after 1925, an appropriate model is the differential equation
dy/dt = ky(16-y).
(a) Solve this differential equation.

Answer 1

I got dy/(16ky-ky^2)=dt, then I took the integral of both sides, but I am not sure how to integrate the first side. I was thinking integration by parts, but it doesn't work out b/c there is no y in the numerator. U substitution doesn't seem to work out either

Answer 2

Do you know how to do partial fractions ?

Answer 3

see http://www.karlscalculus.org/calc11_5.html

Answer 4

Oops I meant to say that I tried partial fraction decomposition, not integration by parts.

Answer 5

I get A/y + B/(y-16)
which mean that AY-16A +BY = 1 how do I solve this since I have AY + BY = ?

Answer 6

\[ \frac{1}{ky(16-y)} = \frac{A}{ky} + \frac{B}{16-y} \]
\[ 16A - Ay + Bky = 1\\ (Bk-A)y + 16A = 0y+1
\]
matche corresponding coefficients

Answer 7

you get 2 equations
16A=1
and
Bk - A = 0

Answer 8

Thx

Answer 9

Do I need the k in the bottom. I pulled it out of the integral b/c it was a constant

Answer 10

you can pull it out, just don't lose it

Answer 11

I would write it as
\[ \frac{dy}{y(16-y)} = k\ dt \]

Answer 12

So i get that A = -1/16 and B=17/16 is that right?

Answer 13

How did you get A= -1/16 ?
Did you get this far:
\[ 16A - Ay + By = 1\\ (B-A)y + 16A = 0y+1 \]

Answer 14

why did you change (16-y) to (y-16) ?

Answer 15

I pulled -k out of the integral. which gave me y(y-16). So I got A/Y +B/Y-16

Answer 16

ok, then A= -1/16
now solve for A+B = 0
to get B= 1/16

Answer 17

I got -1/16k(-ln(y)+ln(y-16) +C)

Answer 18

and t on the other side
\[ \ln(y-16) - \ln(y) = -16kt + C \]
\[ \ln\left(\frac{y-16}{y}\right) = -16k t +C \]
\[ \frac{y-16}{y} = e^C e^{-16k t} \\\frac{y-16}{y} = B e^{-16k t} \]

Answer 19

-1/16k(-ln(y)+ln(y-16) +C) = t+C
If I multiply both sides by 16k, does 16k*C = C b/c 16k*C is still a constant?

Answer 20

You only need the constant on one side or the other.

Answer 21

because you probably want to solve for y, I would put the constant on the side with t

Answer 22

Did you change your e^c to B just to represent another constant?

Answer 23

yes, just to be clearer about how we get the arbitrary constant

Answer 24

Using the initial conditions that at t=0 y = 2 I get that c=ln(-7),
which doesn't work because ln doesn't work for negatives

Answer 25

\[ \frac{y-16}{y} = B e^{-16k t} \]
\[ 1 - \frac{16}{y} = B e^{-16k t} \\ \frac{16}{y}= 1 - B e^{-16k t} \\
\frac{y}{16}= \frac{1}{1 - B e^{-16k t} } \\
y = \frac{16}{1 - B e^{-16k t} }
\]

Answer 26

I got that, then tried to solve for c by using t=0 and y = 2 and got that c= ln(-7) which doesn't make sense

Answer 27

I am thinking that when you factored out -1/k rather than 1/k you changed the problem?

Answer 28

if you multiply both sides by -1
\[ \frac{16-y}{y} = -B e^{-16k t} \\ \frac{16}{y}-1 = -B e^{-16k t} \\
\frac{16}{y}=1 -B e^{-16k t} \\
y= \frac{16}{1 -B e^{-16k t}}
\]
you get
\[ 1 = \frac{8}{1-B} \\ 1-B=8 \\B=-7
\]
and
\[ y= \frac{16}{1 +7 e^{-16k t}} \]

Answer 29

now solve for k, using y=4 when t=50

Answer 30

Looking over all of that, it must mean that if the unknown constant B in
1- B exp( -16kt) is negative, then that is the same as the unknown constant C being positive in 1 + C exp(-16kt)

solving for C should give us +7 which is legit.

Answer 31

but this problem does not arise if you had kept it (16-y) at the very beginning.

Answer 32

I got
\[y=\frac{ 16 }{ \frac{ e ^{t/50} }{ 3 }+1 }\] and it is still not right. Could someone please look and see where my error is?

Answer 33

once you get
\[ y= \frac{16}{1 +7 e^{-16k t}} \]
evaluate this for t=50 (1975), y=4
\[ 4 = \frac{16}{1 +7 e^{-800 k}} \\
1 = \frac{4}{1 +7 e^{-800 k}} \\
1 +7 e^{-800 k}=4\\
e^{-800 t}= \frac{3}{7} \\
-800 k = \ln(3/7) \\
k= \frac{\ln(3/7)}{-800} \approx 0.00159122
\]

if we multiply k by -16, the equation becomes (approximately)
\[ y= \frac{16}{1 +7 e^{-0.016946 t}} \]

See the attached for a plot of this function

Answer 34

From the beginning
\[ \frac{dy}{dt} = ky(16-y) \\ \frac{dy}{y(16-y)}= k \ dt \\
\frac{dy}{y} + \frac{dy}{16-y} = 16 k \ dt \\
\int \frac{dy}{y} + \frac{dy}{16-y} = \int 16 k \ dt \\
\ln(y) - \ln(16-y) = 16 k t +C\\
\ln\left( \frac{y}{16-y}\right) = 16 k t+C\\
\frac{y}{16-y}= De^{16 k t}\\
\frac{16-y}{y}= E e^{-16kt} \\
\frac{16}{y}= 1+E e^{-16kt} \\
y= \frac{16}{1+E e^{-16kt}}
\]

Answer 35

Thank you, I got that y = 16/(8e^(16kt)). I'll look through your solution and see if I can find my error

Answer 36

to solve for the constant, use (0,2)
\[ \frac{16}{1+E}= 2 \\ 8= 1+E\\ E= 7\]
\[ y= \frac{16}{1+7 e^{-16kt}} \]

finally, use (50,4) to solve for k

Answer 37

why did you do this step? 3rd to last
(16−y)/y=Ee^(−16kt)

Answer 38

because if you invert y/(16-y) to (16-y)/y you can divide the top by y to get
16/y - 1 and you only have 1 y to solve for.

Answer 39

I tried dividing the bottom by the top, to give me 16/y -1. That isn't allowed is it?

Answer 40

can you use the equation editor to show your steps ?

Answer 41

whatever you did does not sound kosher.

Answer 42

I just did -y+16 divided by y
which gave me a 1 remainder 16 which meant -1 + 16/y

Answer 43

did you get to
\[ \frac{y}{16-y}= De^{16 k t} \]
?
to do what you did , you have to invert both sides:
\[ \frac{16-y}{y}= Ee^{-16 k t} \]
where E= 1/D and we negate the exponent.
now we simplify the left side to
\[ \frac{16}{y}-1 \]