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

  • #1

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?
 

Answers and Replies

  • #2
rock.freak667
Homework Helper
6,230
31
You can rewrite the cubic as -P(P+4)(P-1). So split it into partial fractions.
 
  • #3
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
______________________________________________

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....
 
  • #4
rock.freak667
Homework Helper
6,230
31
Your value for A should work out as 1/4.
 
  • #5
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??
 

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