Dp/dt = 4p-3p^2-p^3Diff. Equation.

[killersanta]

Can someone tell me were I went wrong. This just doesn't seem right. If some how it is, I need help with the next part. Not sure how.
Talking about problem #2.

http://img513.imageshack.us/img513/8535/part1mc.jpg

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http://img826.imageshack.us/img826/9122/part2m.jpg

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[MaxwellsDemon]

I see a mistake in the second line which probably creates problems further down... 4-3p-p^2 factors as (4+p)(1-p) not (4-p)(1+p)...gotta watch those negative signs, they'll getcha every time. :)

 

[killersanta]

I see a mistake in the second line which probably creates problems further down... 4-3p-p^2 factors as (4+p)(1-p) not (4-p)(1+p)...gotta watch those negative signs, they'll getcha every time. :)

haha, thanks... Though, it sucks I made the mistake so early...

 

[killersanta]

i need help again, how do I handled this?

http://img638.imageshack.us/img638/772/im000714j.jpg

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[MaxwellsDemon]

I think there's an intigration mistake in line three...should be (1/4)ln(p) - (1/20)ln(4+p) - (1/5)ln(1-p)...I think. That should help with the simplification.

 

[killersanta]

How do you interat it?

I broke it down, and use a U-sub on all of them.
Ex: INT 1/(80+20P) dp
U = 80 + 20P
du = 20 dp
1/20du = dp
1/20 INT 1/U du
1/20 ln | U | + C
Resub: 1/20 ln |80+20P| + C

Can you pull the 1/20 out in front?

 

[MaxwellsDemon]

You can always double check by differentiating your integration result to see if it is the same as what you started with. But yeah, remember you can pull a constant out front when integrating...if you had something like INT(1/6x) that's the same as (1/6)INT(1/x)

 

[killersanta]

You can always double check by differentiating your integration result to see if it is the same as what you started with. But yeah, remember you can pull a constant out front when integrating...if you had something like INT(1/6x) that's the same as (1/6)INT(1/x)

How do you handled after this point though. The other I knew how to do, just made a few little errors.

I know, LN A - LN B = LN A/B

I also know, 1/A LN |B| = LN Aroot of B

Once I get them all to LN Some-Root B, I can get them all to the same LN. and will I take the e of both sides the LN is gone...

What do I do that at that point I have A-Root-# * B Root-# / C root - #

I don't know how to handle all the roots.

 

[MaxwellsDemon]

Do you know what the answer is supposed to be? I've been playing around with the math a bit and I get to (1/20) ln [(p^5) / (4+p)((1-p)^4)] After that I just kind of gave up because I'm lazy. :) I guess I just assumed that simplified algebraically...is it supposed to be a nice clean answer like p = f(t) ?

 

[killersanta]

I don't know the answer, but it does need to be P (t) = F(t). Because after we are done, we need to use a few initial conditions... P(0)= 3, P(0)= -1/2, P(0)= -2.

 

[MaxwellsDemon]

Hmm, I'm not sure and its my bed time. I'll give it a little thought and get back to you tomorrow. Hopefully someone else on the forum can give you some advice in the mean time.

 

[killersanta]

sounds good. Thanks, you have been a lot of help :)

 

[MaxwellsDemon]

I tried anyway. Sorry I couldn't have been more helpful...I'm just so lazy. :D