Expert: Paul Klarreich - 10/9/2007

Question
The demand and supply of a good are given by:
qD = (10-2p)/3 and qS = 2p-6

where p denotes the price of the good.
The rate of change of price depends on the excess of demand over supply. More specifically, if x = qD –qS, the rate of change of price is given by dp/dt = 3x^3

Given that p = 3 at t = 0, show that p tends towards its equilibrium value p = 7/2.

Answer
Questioner:   camilla
Category:  Advanced Math
Private:  No
 
Subject:  demand and supply
Question:  The demand and supply of a good are given by:

D = (10-2p)/3 and S = 2p-6

where p denotes the price of the good.
The rate of change of price depends on the excess of demand over supply. More specifically, if x = D - S, the rate of change of price is given by dp/dt = 3x^3

Given that p = 3 at t = 0, show that p tends towards its equilibrium value p = 7/2.
.................................................

If  x = D - S, then
   10 - 2p
x = -------- - (2p - 6)
      3

   10 - 2p    6p - 18
x = -------- - ------
      3          3

   28 - 8p
x = --------
      3

So your (differential) equation says:

dp     (28 - 8p)^3
-- = 3 -----------
dt        27

dp     (28 - 8p)^3
-- =   -----------
dt         9

Separate the variables:

  dp           dt
----------- =  ----
(28 - 8p)^3     9

Integrate:


{
| (28-8p)^-3 dp
}


Let  u =  28 - 8p;  du = - 8 dp;  dp = - du/8

{ u^-3  
| ----- du
}  -8  

u^-2
----- =
-2(-8)

   1
------------
16(28-8p)^2

that's the left side.  Now the right side is just  t/9 + C, so:

   1            t
------------ =  --- + C
16(28-8p)^2      9

   1            t
------------ =  --- + C
256(7-2p)^2      9

Use your conditions:  p = 3  at  t = 0


   1           0
----------- =  --- + C
256(7-1)^2      9

     1
C = ----
    256


   1            t     1
------------ =  --- + ---
256(7-2p)^2      9    256

Now I am not sure about your jargon at the end.  (I hate problems that use special business jargon.)

You wrote:  show that p tends towards its equilibrium value p = 7/2.

From this equation, you can conclude that when  t -> infinity, the left side must also approach infinity.  That means the denominator of the fraction must approach zero.

7 - 2p = 0  means

7 = 2p

p = 7/2

Does that do it for you?