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

  • #1
killersanta
63
0

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
6,223
31
You can rewrite the cubic as -P(P+4)(P-1). So split it into partial fractions.
 
  • #3
killersanta
63
0
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
6,223
31
Your value for A should work out as 1/4.
 
  • #5
killersanta
63
0
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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