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

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SUMMARY

The discussion revolves around solving the differential equation Dp/dt = 4p - 3p^2 - p^3. Participants identify errors in factorization and integration steps, specifically noting that 4 - 3p - p^2 factors as (4 + p)(1 - p) rather than (4 - p)(1 + p). Additionally, they highlight an integration mistake in the third line, suggesting the correct form should be (1/4)ln(p) - (1/20)ln(4 + p) - (1/5)ln(1 - p). The conversation emphasizes the importance of checking work by differentiating results to ensure accuracy.

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  • Familiarity with integration techniques, including U-substitution.
  • Knowledge of logarithmic properties and their applications in algebra.
  • Ability to differentiate functions to verify integration results.
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Students and educators in mathematics, particularly those studying differential equations, integration techniques, and algebraic manipulation. This discussion is also beneficial for anyone looking to improve their problem-solving skills in calculus.

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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. :)
 
MaxwellsDemon said:
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...
 
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.
 
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?
 
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)
 
MaxwellsDemon said:
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.
 
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) ?
 
  • #10
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.
 
  • #11
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.
 
  • #12
sounds good. Thanks, you have been a lot of help :)
 
  • #13
I tried anyway. Sorry I couldn't have been more helpful...I'm just so lazy. :D
 

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