Question

If given y'=.05y-.04y-1000
how does one arrive at y=100,000-90,000e^(.01t)?

Answer 1

Here's the context for the entire problem, I'll post it incase it'll help.

Answer 2

Well first you have like terms on the right hand side.

Answer 3

\[y'=.01y-1000 \\ \frac{dy}{dx}=.01y-1000 \\ \frac{dy}{.01y-1000}=dx\]

Answer 4

integrate both sides

Answer 5

How would I integrate that?

Answer 6

you could use a substitution

Answer 7

but it involves knowing \[\int\limits_{}^{}\frac{du}{u} \text{ equals } \ln|u|+C\]

Answer 8

But how would that given me y=100,000-90,000e^(.01t)?

Answer 9

How about we go through the process and see.

Answer 10

I think you know the y=e^x and y=ln(x) are inverse functions
so it is very likely we will use that

Answer 11

Well, I got the left side to be ln(.01y-1000)C=a (some random constant?)

Answer 12

\[\frac{1}{.01}\ln|0.01y-1000|=x+C\]

Answer 13

Now solve for y.

Answer 14

I still don't get it. 100[(e^(.01y)-e^(1000)]=x+c

Answer 15

Does that step even make sense? :(

Answer 16

\[\ln|0.01y-1000|=\frac{1}{100}(x+c) \\ e^{\ln|0.01y-1000|}=e^{\frac{1}{100}(x+c)} \\ 0.01y-1000=e^{\frac{x}{100}}e^{\frac{c}{100}} \\ 0.01y-1000=e^{\frac{x}{100}}K\]

Answer 17

Y=100,000+Ke^(.01)?

Answer 18

yep it looks fine.

Answer 19

but it doesn't match the answer the teacher gave :/

Answer 20

You didn't give an initial condition.

Answer 21

That will allow us to find K

Answer 22

Oh, yeah, I was meaning to type out the entire problem.
I'll do it right now.

Answer 23

A certain lake contains 10,000 fish. They produce at the rate of 5% per year. They die of natural causes at 4% per year. Fishermen catch 1000 fish per year.
a) draw a box diagram for the rate of change in the number of fish (I did that and it matched the answer given).
b) Write down the corresponding differential equation for the number of fish in the lake, y(t), and solve it. (I got the proper diff eq. but I couldn't solve it).
c) are the fish doomed to extinction? How many fish will there be in the lake after a long time (t=H)?
d) At what rate can the fish be caught to keep the population stable?

Answer 24

\[y=100000+k e^\frac{t}{100}\]
and base on that above we do we have as the int ital condition?

Answer 25

Look at the first sentence

Answer 26

at t=0, y=10,000

Answer 27

use that to solve for K

Answer 28

\[10000=100000+ke^{0}\]

Answer 29

Oh! I get it now! :) Thank you!

Answer 30

So you understood the integration part and the solving for y part?

Answer 31

I get the solving for y part, I'm not sure if I entirely got the integration part. Like, I get the steps you used, but I'm not sure if I could get to that point by myself tomorrow during the exam. :/ Our professor only spent 5 minutes covering this topic, but it's worth about 20% of tomorrow's test.

Answer 32

\[\int\limits_{}^{}\frac{dy}{0.01y-1000}=\int\limits_{}^{}dx\]
You are good up to this point?

Answer 33

Because the first thing I did was just separate variables.

Answer 34

yes. I got that part.

Answer 35

I just don't understand where the 1/.01 came in while we were integrating.

Answer 36

So the integral on left hand side we can put it in the form as du/u
by letting u=0.01y-1000
du=0.01 dy
so
du/0.01=dy
100du=dy

Answer 37

\[\int\limits_{}^{}\frac{100 du}{u}=\int\limits_{}^{}1 dx \]

Answer 38

There was a constant multiple in front of y that wasn't 1

Answer 39

\[\int\limits_{}^{}\frac{1}{cx+k} dx=\frac{1}{c} \ln|cx+k|+C \]
This is because if we let u=cx+k then du= c dx so du/c=dx

Answer 40

\[\int\limits_{}^{}\frac{1}{cx+k} dx=\int\limits_{}^{}\frac{1}{u} \frac{du}{c}=\frac{1}{c}\int\limits_{}^{}\frac{du}{u}=\frac{1}{c}\ln|u|+K=\frac{1}{c}\ln|cx+k|+K\]

Answer 41

our c here of course being the 0.01 and the k being the -1000

Answer 42

I'm getting so lost... :/

Answer 43

Ok how about with some numbers

Answer 44

\[\int\limits_{}^{} \frac{1}{5x-3} dx \]
do you know how to integrate this?

Answer 45

Maybe the decimals throw you off. I'm not sure yet. But if you can integrate this then that is likely.

Answer 46

(1/5)ln(5x+3)?

Answer 47

(1/5)ln|5x-3|

Answer 48

So you integrated that without doing a substitution?

Answer 49

Oh! I see. It's the same concept.

Answer 50

and yes it is

Answer 51

yeah. It was just like a formula I remembered from last year.

Answer 52

ok you were getting mixed up because of the decimal or the variable or both?

Answer 53

I think the variables, having dy over it was more confusing.

Answer 54

\[\int\limits_{}^{}\frac{1}{0.01y-1000} dy \text{ less confusing to write this way then ? }\]

Answer 55

Yep!

Answer 56

I also noticed you had trouble with right hand side

Answer 57

Yeah, I was confused and then I realized it was the same as 1 dx

Answer 58

maybe because it wasn't that you could see the 1
dx=1 dx

Answer 59

Yep! :)

Answer 60

It is the same thing that happened to the left hand side

Answer 61

I wrote dy but dy can also be written as 1 dy

Answer 62

Anyways I'm glad you understand the integration part better now.
And you said you were okay with the solving for y part?

Answer 63

Yes. I got how that works.

Answer 64

And finding K.

Answer 65

Wait...

Answer 66

Oh never midn e^(c/100) had to be a constant of some sort and you arbitrarily made that K.
:) Got it.

Answer 67

Thank you so much for your patience in explaining.

Answer 68

No problem.
Just make sure you ask your questions no matter how dumb they seem to you. No question is dumb.

Answer 69

Later and good luck