D.E.Q. Find the equilibrium and general solutions

[killersanta]

Homework Statement

Find the equilibrium and general solutions


dp/dt = 4P - 3P^2 - P^3

The Attempt at a Solution

Equilibrium:
dp/dt = -P(P^2 - 3P - 4)
dp/dt = -P(P+4)(P-1)
Equilibrium solutions: P = 0, P = 1, P= -4

General solutions:

1/(4P-3P^2 - P^3) dp = 1dt

This is where I'm having trouble. The right side is easy enough to intergrat, but the left side I'm lost. What is the first step to intergrating that?

 

[rock.freak667]

You can rewrite the cubic as -P(P+4)(P-1). So split it into partial fractions.

 

[killersanta]

How did I now see that, thanks...

1/-P(P+4)(P-1) = A/-P + B/(P+4) + C/(P-1)
1 = A(P+4)(P-1) - BP(P-1) - CP(P-4)
A little magic
1= (A-B-C)P^2 + (3A+B+4C)P - 4A
A = -1/4, A little more magic, B = -9/4, C = 1
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S(1/4P - 9/(4P+16) + 1/(P-1)) dp = t dt

(1/4)ln|4P| - (9/4)ln|4P+16| + ln|P-1| + C = t + C

-(8/4)ln |4P/(4P+16)| + ln|P-1| = t + C

-(8/4)ln |(4P/4P+16) * (p-1)| = t + C

ln |(4P^2 - 4P)/ (4P+16)| = -4t/8 + C

(4P^2 - 4P)/(4P+ 16) = Ce^(-4/8t)

This is were I'm stuck now, I think it is right up until here...

I did some stuff and got this:

P = (-64Ce^(-4/8t) - (16Ce^(-4/8t) * 4Ce^(-4/8t)) / (1 - 64Ce^(-4/8t)

But I can't help to think it's wrong, or at least there is a easier way...

 

[rock.freak667]

Your value for A should work out as 1/4.

 

[killersanta]

Wouldn't A = -1/4 ?

1/-P(P+4)(P-1) = A/-P + B/(P+4) + C/(P-1)

1 = A(P+4)(P-1) - BP(P-1) - CP(P-4)

1 = A(P^2 +3P -4) - BP^2 + BP - CP^2 + 4CP


1 = AP^2 + A3P -A4 - BP^2 + BP - CP^2 + 4CP

1= (A-B-C)P^2 + (3A+B+4C)P - 4A

1= (A-B-C)P^2 and 1= (3A+B+4C)P and 1= -4A

1= -4A ----> A = 1/-4??