Question

So, I am just starting a diff eq course. Trying to finish homework for the first day, and confused.

Can someone please help me with this?
"The field mouse population satisfies the differential equation:
dp/dt = 0.5p - 450

a) find the time at which the population becomes extinct if p(0) = 850
b) Find the time of extinction if p(0) = p_0, where 0 < p_0 < 900
c) Find the initial population p_0 if the population is to become extinct in 1 year.

I will post what I have done so far in the comments.
Thanks in advance.

Answer 1

a) So I am guessing I want to do this?
p'/0.5p = -450 and integrate?
That would give me ln(0.5p) = -450t
=> \[e ^{-450t}-.5p\]
That is just a guess? Am I even on the right track?

Answer 2

This is done by separation of "variables":\[\frac{ dp }{ dt }=0.5p-450 \Rightarrow \frac{ dp }{ p-900 }=\frac{ 1 }{ 2 } dt\]You now integrate both sides.
Can you do this?

Answer 3

TYPO, Iactually meant "separation of variables" ;)

Answer 4

I am somewhat aware of how to do separation of variables from independent study and Khan Academy, but this is some strange method he is doing based off of Newton's Law of Cooling and models. So I am a bit perturbed to use a more advanced technique. The final solution that the book is giving is:
t = 2 ln 18, does that make any sense for a)?

Answer 5

Oh, actually that is exactly what he is doing. He is taking the dp an dividing, so for some reason this IS separation just before the chapter that covers this theory...Will try this again.

Answer 6

Do you need more help at this time?

Answer 7

I am retrying. Just give me a second. Thank you for this. :)

Answer 8

So first part is algebra separating p from the other stuff, and moving dt to the side with -450, yes?

Answer 9

So the next step is to get:
ln (0.5p-450) = t?
By integrating both sides, then get:
0.5p - 450 = e^{t}

Answer 10

Well, yes, I only left out the factor 1/2 to get a nice p-900 on the left side instead

Answer 11

ok, so i have this:
0.5p - 450 = e^{t+c}
What now?

Answer 12

If I both sides * 2, I get p - 900 = 2e^{t+c}

Answer 13

I am guessing to take the ln of both sides now so I isolate t, and solve?

Answer 14

I would do this a little more cautious. Next step would be:\[\ln|p-900|=\frac{ 1 }{ 2 }t +C\]where C is a real constant.
Now \[|p-900|=e^{\frac{ 1 }{ 2 }t+C}=C_1e^{\frac{ 1 }{ 2 }t}\]where C1 is a positive constant.
So:\[p-900=\pm C_1e^{0.5t}=C_2e^{0.5t}\]wehere C2 is not equal to 0.
C2=0 gives the constant solution, so the general solution is:\[p(t)=900+Ce^{0.5t}\](C is real)

Answer 15

ok. Thanks. I will spend some time trying to rework this til it sinks in. It has been like 2 years since calculus, so trying to bring back to memory the old work. :)

Answer 16

Now we have to find C. This is done with p(0)=850:\[p(0)=900+Ce^{0}=900+C=850 \Leftrightarrow C=-50\]

Answer 17

yw!